Class 12

MP Board Class 12th Special Hindi सहायक वाचन Solutions Chapter 5 नैनो टेक्नोलॉजी (वैज्ञानिक निबंध, संकलित)

नैनो टेक्नोलॉजी अभ्यास प्रश्न

प्रश्न 1.
नैनो टेक्नोलॉजी क्या है? इसके क्षेत्र-विस्तार के बारे में सम्भावनाएँ बतलाइए। (2011, 14)
उत्तर:
नैनो टेक्नोलॉजी एक अतिसूक्ष्म दुनिया है, जिसका दायरा एक मीटर के अरबवें हिस्से अथवा उससे भी छोटा है। वास्तव में नैनो’ शब्द की उत्पत्ति ग्रीक भाषा के शब्द नैनों से हुई है, जिसका शाब्दिक अर्थ है-बौना अथवा सूक्ष्म। यह नाम जापान के वैज्ञानिक नौरिया तानीगूगूची ने 1976 में दिया। नैनो टेक्नोलॉजी एक इकाई है जो एक मीटर के अरबवें हिस्से के बराबर होती है। आश्चर्यजनक बात यह है कि यह टेक्नोलॉजी जितनी अधिक सूक्ष्म है,उतनी ही विशाल सम्भावनाएँ यह अपने आप में समेटे हुए है। वर्तमान में नैनो टेक्नोलॉजी चिकित्सा के क्षेत्र से लेकर उद्योगों तक में अपनी उपयोगिता सिद्ध कर रही है। यद्यपि भारतीय बाजारों में नैनो उत्पादों की संख्या अत्यन्त कम है,मगर वह दिन दूर नहीं जब भारत में भी नैनो टेक्नोलॉजी जनित उत्पादों का बोलबाला होगा। वर्तमान में भारतीय विशेषज्ञ भी इस क्षेत्र में शोध करने में रत हैं। स्पष्ट है कि भविष्य में नैनो टेक्नोलॉजी के क्षेत्र विस्तार की सम्भावनाएँ असीम हैं।

प्रश्न 2.
‘नैनोडोमेन’ की निर्माण प्रक्रिया समझाते हुए उसके आश्चर्यजनक परिणाम लिखिए। (2009)
उत्तर:
नैनो टेक्नोलॉजी एक इकाई है जिसका मान 1 मीटर के अरबवें हिस्से के बराबर होता है। एक से लेकर सौ नैनोमीटर को ‘नैनोडोमेन’ कहा जाता है। नैनो टेक्नोलॉजी के तहत मेटेरियल का आकार छोटा करके उसे नैनोडोमेन’ बना लिया जाता है। ऐसा करने पर उस पदार्थ के विभिन्न गुण; जैसे-इलेक्ट्रिकल, मैकेनिकल,थर्मल व ऑप्टीकल इत्यादि हर स्तर पर बदलना शुरू हो जाते हैं। यह एक नवीन विज्ञान है जो बेहद आश्चर्यचकित परिणाम प्रदान करता है। जब अणु और परमाणु नैनो क्षेत्र को प्रभावित करते हैं तो इनमें नवीन परिवर्तन होना प्रारम्भ हो जाते हैं। ये परिवर्तन अद्भुत होते हैं, जिसमें वस्तु के मूल गुण तक बदल जाते हैं।

प्रश्न 3.
नैनो मेटेरियल किसे कहते हैं? नैनो मेटेरियल तैयार करने की दो पद्धतियाँ कौन-कौन सी हैं? (2015)
उत्तर:
नैनो टेक्नोलॉजी के तहत मेटेरियल के आकार को छोटा करके उसे नैनोडोमेन में बदल लिया जाता है। इस प्रक्रिया में मेटेरियल के विभिन्न गुण; जैसे-इलेक्ट्रिकल,मैकेनिकल, थर्मल व ऑप्टीकल इत्यादि हर स्तर पर बदलने लगते हैं। इन परिवर्तनों से अत्यन्त आश्चर्यजनक परिणाम प्राप्त होते हैं। जैसे-जैसे परमाणु का आकार छोटा होता जाता है, उसके अन्दर दूसरी धातुओं व पदार्थों से आपसी प्रतिक्रिया की क्षमता बढ़ती जाती है। इस विशेषता का उपयोग करके नैनो मेटेरियल से एकदम नया उत्पाद सरलता से तैयार किया जा सकता है।

नैनो मेटेरियल को तैयार करने के लिए सदैव दो पद्धतियों को उपयोग में लाया जाता है-प्रथम, बड़े से छोटा करने की पद्धति और द्वितीय, छोटे से बड़ा करने की पद्धति। इन पद्धतियों से एक आश्चर्यचकित कर देने वाला नैनो मेटेरियल तैयार किया जा सकता है।

प्रश्न 4.
चिकित्सा क्षेत्र में नैनो टेक्नोलॉजी की उपयोगिता बताइए। (2009)
उत्तर:
वर्तमान में नैनो टेक्नोलॉजी दुनिया भर में अपनी विशेषताओं और उपयोगिताओं के कारण अत्यन्त तेजी से लोकप्रिय हो रही है। वैसे तो नैनो टेक्नोलॉजी की उपयोगिताओं का क्षेत्र प्रसार अनन्त एवं असीम है, किन्तु वर्तमान में इसका सर्वाधिक प्रभाव चिकित्सा के क्षेत्र में ही देखने को मिल रहा है। वैज्ञानिक इस प्रौद्योगिकी का उपयोग कर ‘गोल्ड पार्टिकल बैक्टीरिया ट्यूमर सेल्स’ का निर्माण कर रहे हैं,जो कैंसर की संमूची प्रक्रिया को ही परिवर्तित करने में सक्षम होंगे। इससे ट्यूमर के खतरनाक तत्व को समाप्त कर दिया जाता है। इसके अतिरिक्त विशेषज्ञ कलाई में पहन सकने योग्य एक ऐसी कलाई घड़ी के रूप में नैनो टेक्नोलॉजी आधारित युक्ति का विकास करने में प्रयत्नशील हैं, जिसके माध्यम से व्यक्ति अपने शरीर की कई बीमारियों का समय रहते पता लगा पायेंगे।

प्रश्न 5.
“सुपर कम्प्यूटर नैनो टेक्नोलॉजी का ही परिणाम है।” समझाइए।
उत्तर:
वर्तमान युग प्रौद्योगिकी एवं कम्प्यूटर का युग है। आज कम्प्यूटर हमारे दिन-प्रतिदिन के जीवन के मध्य इतना घुल-मिल गये हैं कि इनके बिना अब जीवन की कल्पना करना भी आसान नहीं है। कम्प्यूटर की चर्चा चलते ही एक बड़े से घनाभाकार डिब्बे की आकृति मानस-पटल पर अंकित हो उठती है। एक समय था जब कम्प्यूटर अपनी शैशवावस्था में था। उसका आकार काफी बड़ा और उसकी क्षमता सीमित थी। किन्तु नैनो टेक्नोलॉजी के आगमन ने कम्प्यूटर की तो मानो काया ही पलट कर रख दी है। नैनो टेक्नोलॉजी के उपयोग से जहाँ कम्प्यूटर के आकार को छोटा किया जाना सम्भव हो सका है, वहीं उसकी तमाम क्षमताओं में आश्चर्यजनक बढ़ोत्तरी की जा सकती है। कम्प्यूटर के क्षेत्र में नैनो टेक्नोलॉजी के प्रभावी दखल का सबसे ज्वलंत उदाहरण है-‘सुपर कम्प्यूटर’ वास्तव में,सुपर कम्प्यूटर इस नैनो टेक्नोलॉजी का ही परिणाम है। पलभर में अरबों गणनाएँ त्रुटिरहित सम्पन्न कर देना, मैमोरी इतनी विशाल कि असंख्य आँकड़े समा जायें इत्यादि विशेषताएँ सुपर कम्प्यूटर में नैनो टेक्नोलॉजी के उपयोग से ही सम्भव हो सकी हैं।

प्रश्न 6.
निम्नलिखित क्षेत्रों में नैनो टेक्नोलॉजी की उपयोगिता बतलाइए
(1) पेंट
(2) कपड़ा
(3) जल-शोधन,
(4) टी. वी डिस्प्ले।
उत्तर:
जीवन से जुड़े विभिन्न क्षेत्रों में नैनो टेक्नोलॉजी की उपयोगिता अब सर्वविदित है। कुछ विशिष्ट क्षेत्रों में इस नवीन क्रान्तिकारी प्रौद्योगिकी के उपयोग की चर्चा निम्नवत् की जा सकती है-
(1) पेंट उद्योग में :
पेंट उद्योग के क्षेत्र में नैनो टेक्नोलॉजी अपनी पैठ बनाती जा रही है, जैसे-टिटेनियम डाइ-ऑक्साइड पेंट। यदि इस पेंट को नैनो मेटेरियल बनाकर अन्य पेंट में मिला दिया जाये तो उसकी चमक और अन्य गुण बढ़ जाते हैं। इस प्रकार निर्मित पेंट का जीवन अन्य सामान्य पेंट की तुलना में काफी अधिक होता है।

(2) कपड़ा उद्योग में :
कपड़ा उद्योग में भी नैनो टेक्नोलॉजी अपना प्रभाव एवं उपयोगिता सिद्ध कर रही है। इस चमत्कारी प्रौद्योगिकी का उपयोग करके ‘नैनोबेस्ड क्लॉथ’ बनाए जा रहे हैं,जो व्यक्ति के पसीने को सरलता से सोख लेते हैं। साथ ही, इस तकनीक से बना कपड़ा उपलब्ध अन्य कपड़ों की तुलना में अधिक टिकाऊ होता है।

(3) जल-शोधन में :
नैनो टेक्नोलॉजी रूपी वरदान का उपयोग जल-शोधन के क्षेत्र में भी किया जा सकता है। भारत जैसे विकासशील देश में जहाँ तीन-चौथाई जनसंख्या को शुद्ध पेयजल तक उपलब्ध नहीं है, इस प्रौद्योगिकी का उपयोग किसी ‘देव-वरदान’ से कम सिद्ध नहीं होगा। इस नवीन प्रौद्योगिकी के माध्यम से जल में मौजूद एवं स्वास्थ्य के लिए अत्यन्त हानिकारक ऑर्सेनिक तत्त्व को समाप्त कर दिया जाता है अथवा एकत्र कर जल से पृथक् कर दिया जाता है। इस प्रक्रिया में नैनो मिनरल, नैनो गोल्ड, नैनो सिल्वर व नाइट्रेट इत्यादि नैनो उत्पादों का प्रयोग किया जाता है।

(4) टी. वी. डिस्ले में :
नैनो टेक्नोलॉजी के आम जनजीवन में उपयोग का सबसे सुन्दर उदाहरण है टी.वी. डिस्प्ले का क्षेत्र। टेलीविजन पर दिखाई देने वाली तस्वीर की ‘ब्राइटनेस’ व ‘कंट्रास्ट’ को बेहतर बनाने के लिए इस प्रौद्योगिकी का उपयोग किया जाता है। इस प्रक्रिया में नैनो फास्टर मेटेरियल का उपयोग किया जाता है, जिससे टी.वी. की ‘पिक्चर क्वालिटी काफी उन्नत हो जाती है। वास्तव में, नैनो टेक्नोलॉजी एक ऐसी प्रौद्योगिकी है जिसे एक नये युग का सूत्रपात माना जा सकता है।

प्रश्न 7.
भारत में नैनो टेक्नोलॉजी के शिक्षण-प्रशिक्षण की उपलब्धता पर प्रकाश डालिए।
उत्तर:
यदि वर्तमान में भारत में नैनो टेक्नोलॉजी के प्रचार-प्रसार और उपयोग की बात की जाये तो नैनो टेक्नोलॉजी के शिक्षण-प्रशिक्षण में अभी बहुत कम निजी व सरकारी संस्थान आगे आए हैं। वर्तमान में, इस प्रौद्योगिकी से सम्बन्धित मात्र एम. टेक. नैनो टेक्नोलॉजी पाठ्यक्रम ही देश में उपलब्ध है, जिसके लिए निर्धारित योग्यता इंजीनियरिंग की डिग्री अथवा भौतिक विज्ञान, रसायन विज्ञान या जैव प्रौद्योगिकी के साथ स्नातकोत्तर डिग्री अनिवार्य है। एम. टेक.के इस कोर्स की कुल अवधि दो वर्ष है, जिसमें प्रशिक्षुओं को छ: माह का प्रशिक्षण उपलब्ध कराया जाता है।

आशा है कि जिस प्रकार यह प्रौद्योगिकी दनिया-भर में तेजी से अपने पैर पसार रही है. भारत में भी जल्दी ही विश्वविद्यालयों एवं संस्थानों में इससे सम्बन्धित अनेक पाठ्यक्रम पढ़ाए जाने लगेंगे।

नैनो टेक्नोलॉजी अति लघु उत्तरीय प्रश्न

प्रश्न 1.
किस वैज्ञानिक युक्ति की सहायता से वैज्ञानिक पहली बार अणु व परमाणु को देख पाए?
उत्तर:
जर्मन वैज्ञानिक गार्ड विनिग व स्विट्जरलैंड के हैनरिच रारेर के संयुक्त प्रयास से तैयार ‘स्कैनिंग टानलिंग माइक्रोस्कोप’ की सहायता से वैज्ञानिक पहली बार अणु व परमाणु को देख पाए।

प्रश्न 2.
एक नैनो मीटर मानव बाल के कौन-से हिस्से के बराबर होता है?
उत्तर:
एक नैनो मीटर का आकार ‘मानव बाल के 50 हजारवें’ हिस्से के बराबर होता है।

नैनो टेक्नोलॉजी पाठ का सारांश

संकलित वैज्ञानिक निबन्ध, ‘नैनो टेक्नोलॉजी’ में वर्तमान में चल रहे वैज्ञानिक युग में एक नये एवं क्रान्तिकारी युग के सूत्रपात की अवधारणा और सम्भावना की बात कही गई है।

‘नैनो’ शब्द की उत्पत्ति ग्रीक भाषा के शब्द ‘नैनों’ से हुई है, जिसका अर्थ है-‘बौना’ या ‘सूक्ष्म’। यह नाम जापान के वैज्ञानिक नौरिया तानीगूगूची ने वर्ष 1976 में दिया था । वास्तव में, नैनो टेक्नोलॉजी एक इकाई है जो एक मीटर के अरब हिस्से के बराबर होती है। एक से लेकर सौ नैनोमीटर को ‘नैनोडोमेन’ कहा जाता है।

प्रौद्योगिकी के क्षेत्र में नैनो टेक्नोलॉजी को एक नए युग के सूत्रपात के रूप में देखा जा रहा है। नैनो टेक्नोलॉजी की दुनिया अति सूक्ष्म है। अद्भुत बात यह है कि जितनी अधिक यह सूक्ष्म है उतनी ही अधिक सम्भावनाएँ इसमें निहित हैं। नैनो टेक्नोलॉजी के तहत मेटेरियल का आकार छोटा करके उसे ‘नैनोडोमेन’ बना लिया जाता है। ऐसा करने पर उस पदार्थ के विभिन्न गुण; जैसे-इलेक्ट्रिकल, मैकेनिकल, थर्मल व ऑप्टीकल, हर स्तर पर बदल जाते हैं। नैनो उत्पाद बेहद हल्के,छोटे,मजबूत,पारदर्शी एवं अपने मूल मेटेरियल से पूरी तरह से भिन्न होते हैं।

वर्तमान प्रौद्योगिकी के इस युग में नैनो टेक्नोलॉजी का क्षेत्र असीम है और इसके विस्तार की सम्भावनाएँ भी अनन्त हैं। आज नैनो टेक्नोलॉजी चिकित्सा क्षेत्र से लेकर उद्योगों तक कहीं-न-कहीं अपनी महत्वपूर्ण भूमिका अदा कर रही है। भारत में भी इसका उपयोग और विस्तार अत्यन्त तेजी से हो रहा है।

MP Board Class 12th Hindi Solutions

MP Board Class 12th Maths Important Questions Chapter 5A सांतत्य तथा अवकलनीयता

सांतत्य तथा अवकलनीयता Important Questions

_{सांतत्य तथा अवकलनीयता वस्तुनिष्ठ प्रश्न}

प्रश्न 1.
सही विकल्प चुनकर लिखिए –

प्रश्न 1.
यदि x = at², y = 2at है, तो \(\frac{dy}{dx}\) होगा –
(a) t
(b) t²
(c) \(\frac{1}{t}\)
(d) \(\frac { 1 }{ t^{ 2 } } \)
उत्तर:
(c) \(\frac{1}{t}\)

प्रश्न 2.
यदि y = 2 \(\sqrt { cot(x^{ 2 }) } \) हो, तो \(\frac{dy}{dx}\) होगा –
(a) \(\frac { -2\sqrt { 2x } }{ sinx^{ 2 }\sqrt { sin2x^{ 2 } } } \)
(b) \(\frac { 2\sqrt { 2x } }{ sinx^{ 2 }\sqrt { sin2x^{ 2 } } } \)
(c) \(\frac { 2\sqrt { 2x } }{ sinx^{ 2 }\sqrt { sin2x^{ 2 } } } \)
(d) \(\frac { 2\sqrt { x } }{ sinx^{ 2 }\sqrt { sin2x^{ 2 } } } \)
उत्तर:
(a) \(\frac { -2\sqrt { 2x } }{ sinx^{ 2 }\sqrt { sin2x^{ 2 } } } \)

प्रश्न 3.
\(\frac{dy}{dx}\) (x³ + sinx²) का मान है –
(a) 3x² + cos x²
(b) 3x² + x sin²
(c) 3x² + 2x cos x²
(d) 3x² + x cos x²
उत्तर:
(c) 3x² + 2x cos x²

प्रश्न 4.
\(\frac{d}{dx}\) a^x का मान है –
(a) a^x
(b) a^xlog_ae
(c) a^xlog_ea
(d) \(\frac { a^{ x } }{ log_{ e }a } \)
उत्तर:
(c) a^xlog_ea

प्रश्न 5.
यदि y = 500e^7x + 6000e^-7x हो, तो, \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) का मान होगा –
(a) 45 y
(b) 47 y
(c) 49 y
(d) 50 y
(d) 50y.
उत्तर:
(c) 49 y

प्रश्न 2.
(a) रिक्त स्थानों की पूर्ति कीजिए –

1. cosx⁰ का x के सापेक्ष अवकल गुणांक ………………….. है।
2. e^{loge a}का x के सापेक्ष अवकल गुणांक ………………………..
3. log_e a का a के सापेक्ष अवकल गुणांक ……………………….
4. a^x का x के सापेक्ष अवकल गुणांक ……………………………… है।
5. sin 3x का 3x के सापेक्ष अवकल गुणांक …………………………… है।
6. यदि y = sin^-1(2x\(\sqrt { 1-x^{ 2 } } \) ) हो, तो \(\frac{dy}{dx}\) = ……………………… होगा।
7. sin x का cos.x के सापेक्ष अवकल गुणांक ………………… है।
8. \(\frac{d}{dx}\) (log tan x) का मान ………………………… है।
9. log(log sin x) का अवकलन गुणांक ……………………………….. होगा।

उत्तर:

1. – \(\frac { \pi }{ 180 } \) sin x⁰
2. 0
3. 0
4. log_ea.a^x
5. cos x
6. \(\frac { 2 }{ \sqrt { 1-x^{ 2 } } } \)
7. – cot x
8. cosec 2x
9. \(\frac { cotx }{ logsinx } \)

प्रश्न 2.
(b) रिक्त स्थानों की पूर्ति कीजिए –

1. यदि x = \(\sqrt { 1-y^{ 2 } } \) हो, तो \(\frac{dy}{dx}\) = ………………………. होगा।
2. sin x का n वाँ अवकलज ………………………………. होगा।
3. यदि y = \(\sqrt { x+\sqrt { x+………..\infty } } \) हो, तो \(\frac{dy}{dx}\) = ………………………. होगा।
4. यदि x = r cos θ, y = r sin θ हो, तो \(\frac{dy}{dx}\) = ………………………. होगा।
5. e^x का \(\sqrt{x}\) के सापेक्ष अवकलज ………………………………. होगा।

उत्तर:

1. \(\frac { \sqrt { 1-y^{ 2 } } }{ 1-2y^{ 2 } } \)
2. sin ( \(\frac { n\pi }{ 2 } \) + x )
3. \(\frac { 1 }{ 2y-1 } \)
4. -cot θ
5. 2\(\sqrt{x}\) e^x

प्रश्न 3.
निम्न कथनों में सत्य/असत्य बताइए –

1. e^{loge x} का अवकल गुणांक \(\frac{1}{x}\) है।
2. यदि f (x) = \(\sqrt{x}\); x > 0 तो f'(2) का मान \(\frac { 1 }{ 2\sqrt { 2 } } \) है।
3. कोई फलन f (x) किसी बिन्दु x = a पर अवकलनीय कहलाता है, जब Lf'(a) # Rf (a).
4. sec^-1a का x के सापेक्ष अवकल गुणांक 0 होता है।
5. यदि y = ae^mx + Be^-mx, तो = -m²y है।
6. यदि y = sin^-1 ( \(\frac { x-1 }{ x+1 } \) ) + cos^-1 ( \(\frac { x-1 }{ x+1 } \) ) हो, तो \(\frac{dy}{dx}\) = 0 होगा।
7. प्रत्येक अवकलनीय फलन सतत् होता है।
8. a^2x का अवकल गुणांक a^2xlog a.

उत्तर:

1. असत्य
2. सत्य
3. असत्य
4. असत्य
5. असत्य
6. सत्य
7. सत्य
8. असत्य।

प्रश्न 4.
सही जोड़ी बनाइये –

उत्तर:

1. (d)
2. (e)
3. (a)
4. (f)
5. (h)
6. (g)
7. (c)
8. (b)

प्रश्न 5.
एक शब्द/वाक्य में उत्तर दीजिए –

1. \(\frac { 6^{ x } }{ x^{ 6 } } \) का x के सापेक्ष अवकल गुणांक ज्ञात कीजिए।
2. y = e^{loge tanx} का x के सापेक्ष अवकल गुणांक ज्ञात कीजिए।
3. a^x का n वाँ अवकलज ज्ञात कीजिए।
4. y = sin(ax + b) हो, तो \(\frac{dy}{dx}\) का मान ज्ञात कीजिए।
5. यदि x² + y² = sin xy, तो \(\frac{dy}{dx}\) का मान ज्ञात कीजिए।
6. x के सापेक्ष log tan\(\frac{x}{2}\) का अवकल गुणांक ज्ञात कीजिए।
7. sin^-1 \(\frac { 2x }{ 1+x^{ 2 } } \) का x के सापेक्ष अवकल गुणांक ज्ञात कीजिए।
8. e -log_e^x का अवकल गुणांक ज्ञात कीजिए।

उत्तर:

1. \(\frac { 6^{ x } }{ x^{ 6 } } \) [ [log 6 – \(\frac{6}{x}\) ]
2. sec² x
3. a^x(log a)ⁿ
4. -a² y
5. \(\frac { ycosxy-2x }{ 2y-xcosxy } \)
6. cosec x
7. \(\frac { 2 }{ 1+x^{ 2 } } \)
8. \(\frac { -1 }{ x^{ 2 } } \)

सांतत्य तथा अवकलनीयता लघु उत्तरीय प्रश्न

प्रश्न 1.
फलन के सभी असांतत्य के बिन्दुओं को ज्ञात कीजिए, जबकि निम्नलिखित प्रकार से परिभाषित है –

हल:
x < 2 के लिए f (x) = 2x + 3 बहुपदी फलन है।
अत: x < 2 के लिए f (x) संतत फलन है। x > 2 के लिए f (x) = 2x – 3 बहुपदी फलन है।
अतः x > 2 के लिए f (x) संतत है।
अब हम केवल x = 2 पर f (x) की संततता का परीक्षण करेंगे।
x = 2 + h रखने पर
जब x → 2 तब h → 0
\(\underset { x\rightarrow 2^{ + } }{ lim } \) f(x) = \(\underset { h\rightarrow 0 }{ lim } \) 2(2 + h) – 3
= 2(2 + 0) – 3 = 4 – 3 = 1
x = 2 – h रखने पर
जब x → 2 तब h → 0
\(\underset { x\rightarrow 2^{ – } }{ lim } \) f(x) = \(\underset { h\rightarrow 0 }{ lim } \) 2(2 – h) + 3
= 2(2 – 0) + 3 = 7
f(2) = 2(2) + 3 = 7
\(\underset { x\rightarrow 2^{ – } }{ lim } \) f(x) = f(2) ≠ \(\underset { x\rightarrow 2^{ + } }{ lim } \) f(x)

प्रश्न 2.
फलन के सभी असातत्य बिन्दुओं को ज्ञात कीजिए, जबकि f निम्नलिखित प्रकार से परिभाषित है –

हल:
x ≠ 0 के लिए f (x) = \(\frac { |x| }{ x } \) संतत फलन है।
अत:
हम केवल x = 0 पर f (x) की संततता का परीक्षण करेंगे।
x = 0 + h रखने पर
जब x → 0 तब h → 0.

x = 0 – h रखने पर
जब x → 0 तब h → 0

दिया है:
f (0) = 0
\(\underset { x\rightarrow 0^{ + } }{ lim } \) f (x) ≠ \(\underset { x\rightarrow 0^{ – } }{ lim } \) f (x) ≠ f (0)
अत: दिया गया फलन f(x), x = 0 पर असंतत है।

प्रश्न 3.
बिन्दु x = 0 पर निम्नलिखित फलन f (x) के सातत्य की जाँच कीजिए –

हल:
f (x) = \(\frac { 1-cosx }{ x^{ 2 } } \), जब x ≠ 0
x = 0 + h रखने पर, x → 0 तो h → 0

पुनः x = 0 – h रखने पर, x → 0 तो h → 0

दिया है कि f (x) = \(\frac{1}{2}\) जब x = 0
या f (0) = \(\frac{1}{2}\)
इस प्रकार,
\(\underset { x\rightarrow 0^{ + } }{ lim } \) f(x) = \(\underset { x\rightarrow 0^{ + } }{ lim } \) f(x) = f(0)
अत: x = 0 पर f (x) संतत है।

प्रश्न 4.
फलन f निम्न प्रकार से परिभाषित है –

दर्शाइये कि बिन्दु x = 4 के अतिरिक्त प्रत्येक बिन्दु पर संतत है।
हल:

स्पष्ट है कि x = 4 के लिये फलन f (x) संतत नहीं है क्योंकि
\(\underset { x\rightarrow 4^{ – } }{ lim } \) f (x) ≠ \(\underset { x\rightarrow 4^{ + } }{ lim } \) तथा f (4) = 0
जब x < 4 तब f (x) = – 1 जो कि एक अचर फलन है। अत: यह संतत फलन है। जब x > 4 तब f (x) = 1 जो कि एक अचर फलन है। अत: यह संतत फलन है।
अतः f (x) बिन्दु x = 4 के अतिरिक्त सभी बिन्दुओं पर संतत है। यही सिद्ध करना था।

प्रश्न 5.
k का मान ज्ञात कीजिए यदि फलन –

बिन्दु x = \(\frac { \pi }{ 2 } \) + h रखने पर, जब x → \(\frac { \pi }{ 2 } \), तब h → 0

x = \(\frac { \pi }{ 2 } \) + h रखने पर, जब x → \(\frac { \pi }{ 2 } \) तब h → 0

x = \(\frac { \pi }{ 2 } \) – h रखने पर, जब x → \(\frac { \pi }{ 2 } \) तब h → 0

दिया है
f ( \(\frac { \pi }{ 2 } \) ) = 3
तथा दिया गया फलन संतत है।

⇒ \(\frac{k}{2}\) = \(\frac{k}{2}\) = 3
⇒ k = 6

प्रश्न 6.
k का मान ज्ञात कीजिए यदि फलन –

बिन्दु x = 2 पर संतत है।
हल: x = π + h रखने पर,
जब x → 7 तब h → 0

x = π – h रखने पर
जब x → π तब h → 0

दिया गया फलन x = π पर संतत है।

प्रश्न 7.
निम्न फलन बिन्दु x = 0 पर संतत है –

k का मान ज्ञात कीजिये।
हल:
दिया है

दिया है –
f (0) = \(\frac{1}{2}\)
चूँकि फलन f (x) बिन्दु x = 0 पर संतत है –

⇒ \(\frac { k^{ 2 } }{ 2 } \) = \(\frac { k^{ 2 } }{ 2 } \) = \(\frac{1}{2}\)
⇒ k² = 1 या k = ± 1

प्रश्न 8.
a और b के बीच संबंध स्थापित कीजिए जिनके लिए –

हल:
दिया है:

x = 3 + h रखने पर,
जब x → 3 तब h → 0

x = 3 – h रखने पर,
जब x → 3 तब h → 0

= a ( 3 – 0) + 1
= 3a + 1
f (3) = 3a + 1
∵ दिया गया पालन x = 3 पर संतत है।

⇒ 3a + 1 = 3b + 3
⇒ 3a = 3b + 2
⇒ a = b + \(\frac{2}{3}\)

प्रश्न 9.
सिद्ध कीजिए कि फलन f (x) = |x – 1|, x ∈ R, x = 1 पर अवकलनीय नहीं है। (NCERT)
हल:
दिया है:

x = 1 – h रखने पर, जब x → 1 तब h → 0

x = 1 + h रखने पर, जब x → 1 तब → 0

अतः दिया गया फलन x = 1 पर अवकलनीय नहीं है।

प्रश्न 10.
दर्शाइए कि फलन

पर अवकलनीय नहीं है।
हल:
हम जानते हैं,

प्रश्न 11.
सिद्ध कीजिए कि फलन

यही सिद्ध करना था।
x = 0 पर संतत है तथा अवकलनीय भी।
हल: यहाँ f (0) = 0


चूँकि f (0 + 0) = f (0 – 0) = f (0), अतएव x = 0 पर दिया हुआ फलन संतत है।
अब


स्पष्टतः Rf’ (0) = Lf’ (0)
अतः x = 0 पर दिया हुआ फलन अवकलनीय है।
अतः दिया हुआ फलन x = 0 पर संतत है तथा अवकलनीय भी है। यही सिद्ध करना था।

MP Board Class 12 Maths Important Questions

MP Board Class 12th Biology Important Questions Chapter 3 Human Reproduction

Human Reproduction Important Questions

Human Reproduction Objective Type Questions

Question 1.
Choose the correct answers:

Question 1.
The number of chromosome in human cell will be :
(a) 23
(b) 46
(c) 92
(d) 22
Answer:
(b) 46

Question 2.
The hormone estrogen is secreted by :
(a) Corpus luteum
(b) Graafian follicle
(c) Uterus
(d) Vagina.
Answer:
(b) Graafian follicle

Question 3.
The copulatory organ in female human being is :
(a) Fallopian tube
(b) Uterus
(c) Clitoris
(d) Vagina.
Answer:
(c) Clitoris

Question 4.
The gestation period in rabbit is about :
(a) 27 days
(b) 22 – 30 days
(c) 45 days
(d) 28 days.
Answer:
(b) 22 – 30 days

Question 5.
Hormone secreted during delivery is:
(a) Progesterone
(b) Thyroxine
(c) Relaxin
(d) Glucocorticoid.
Answer:
(c) Relaxin

Question 6.
Blastopore is formed in :
(a) Gastrula
(b) Blastula
(c) Morula
(d) Nurmula
Answer:
(a) Gastrula

Question 7.
Number of parents involved in asexual reproduction is :
(a) Many
(b) Three
(c) Two
(d) One
Answer:
(d) One

Question 8.
Sertoli cells are found in :
(a) Kidney
(b) Ovary
(c) Liver
(d) Testis.
Answer:
(d) Testis.

Question 9.
Which two are found in human semen in more quantity :
(a) Ribos and potassium
(b) Fructose and calcium
(c) Glucose and calcium
(d) DNA and testosteron.
Answer:
(b) Fructose and calcium

Question 10.
Which part of the fallopian tube is nearest to ovary :
(a) Ampula
(b) Isthmus
(c) Infundibulum
(d) DNA and testosteron.
Answer:
(c) Infundibulum

Question 11.
In ovarian cycle of human ovulation is starts from :
(a) 1st day
(b) 5th day
(c) 14th day
(d) 28th day.
Answer:
(c) 14th day

Question 12.
Which hormone controls the sperm formation :
(a) ADH
(b) FSH
(c) LH
(d) STH.
Answer:
(b) FSH

Question 13.
Which is responsible for movement of sperm :
(a) Cilia
(b) Flagella
(c) Basal body
(d) Nucleosome.
Answer:
(b) Flagella

Question 14.
Sperms are mature in :
(a) In oviduct
(b) In epididymis
(c) In vagina
(d) All of the above
Answer:
(b) In epididymis

Question 15.
Seminiferous tubules are found in :
(a) In testis
(b) In ovary
(c) In kidney
(d) In lungs.
Answer:
(a) In testis

Question 2.
Fill in the blanks :

1. The nature of semen is ………………
2. The male sex hormone testosterone is produced by ……………… in testis.
3. The structure between placenta and foetus is called ………………
4. Human being are ……………… breeders.
5. The site of fertilization and implantation in human female is ……………… and ………………
6. After ……………… year growth does not occur in girls.
7. ……………… hormone is secreted during child birth.
8. Segments of testis is consist of ……………… tubules.
9. Testis starts the secretion of ……………… hormone at puberty.
10. The process of fertilization is occurs in ……………… tube.
11. ……………… provides the nutrition to embryo.
12. ……………… hormones are secreted by graafian follicle.

Answer:

1. Alkaline
2. Leyding cells
3. Umblical cord
4. Continuous
5. Oviducts, uterus
6. 14
7. Relaxin
8. Seminiferous tubules
9. Testosteron
10. Follopean tube
11. Placenta
12. Estrogen and Progesteron.

Question 3.
Match the followings:
I.

Answer:

1. (e)
2. (d)
3. (a)
4. (c)
5. (b).

II.

Answer:

1. (b)
2. (c)
3. (d)
4. (e)
5. (a).

Question 4.
Write the answer in one word/sentences:

1. How many sperms will be produced from 24 spermatocytes during spermatogenesis?
2. Where does fertilization takes place in mammals?
3. How many polar bodies are formed during the formation of one gamete?
4. How many polar bodies are formed during the formation of ovum during oogenesis?
5. Write the name of three layers of gastrula.
6. Spermatogenesis is the example of.
7. How many germ layers are found in gastrula stage?
8. Write the duration of gastation period in human cases.
9. How many autosomes are found in human sperm?
10. Name the process in which formation and maturation of ova.
11. Name the main store house of sperms.
12. Skeletal muscle is originated by these layer.
13. Name the process in which zygote starts dividing by specific mitotic divisions.
14. In which month body hair developed in embryo.

Answer:

1. 1.96
2. Fallopian tube
3. 2
4. 3
5. Ectoderm, Mesoderm and Endoderm
6. Male gamete formation
7. Three
8. 280 days
9. 22
10. Oogenesis
11. Epididymis
12. Ectoderm
13. Cleavage
14. Fourth month.

Human Reproduction Very Short Answer Type Questions

Question 1.
After fertilization, how many days are needed for the birth of human child?
Answer:
280 days (9 months and 10 days).

Question 2.
To produce the new organisms of their own kind is called by a term, name it.
Answer:
Reproduction.

Question 3.
By which name, the reproduction happens by means of fusion of two different gametes?
Answer:
Sexual reproduction.

Question 4.
Which gametogenesis goes on till life span in human beings after attaining puberty?
Answer:
Spermatogenesis.

Question 5.
The stoppage of menstrual cycle in a 50 yrs. old female is known as.
Answer:
Menopause.

Question 6.
What is pregnancy?
Answer: The time period between fertilization to the birth of child is called pregnancy.

Question 7.
Where is carpora cavernosa found?
Answer:
Carpora cavernosa is found in penis.

Question 8.
What is the gestation period in elephant, dog and cat?
Answer:
Elephant 641 days, dog 58 – 68 days, cat 63 days.

Question 9.
Name the hormone that relaxes pubic symphysis during parturition.
Answer:
Relaxin hormone.

Human Reproduction Short Answer Type Questions

Question 1.
Why testes are found outside the abdominal cavity in males?
Answer:
In male, one pair of testes are found in the scrotum or scrotal sac, which are situated outside the abdomen. Temperature of this scrotum is always 2°C less than the body temperature, which is suitable for formation and growth of sperm, otherwise spermatogenesis will not occur. Therefore, testes are situated outside the abdomen in man.

Question 2.
What will happen if leydig cells of testis in males are destroyed?
Answer:
The endocrine cells of the testis are the leydig cells, they are situated in between the seminiferous tubules. Leydig cells secrete the male sex hormones androgens. Testes secrete four types of androgens:

1. Testosterone
2. Androsterone
3. Epiandosterone and
4. Dihydroepiandrosterone.

Out of these male sex hormone is testosterone.

Functions of Testosterone:

1. Testosterone stimulates testes to descend into the scrotum in embryos.
2. It helps in the development of secondary sexual characters.
3. Stimulates spermatogenesis.

If leydig cells are removed from the body of males, then the above mentioned functions will stop.

Question 3.
Describe the structure of human egg.
Or
Draw neat well – labelled diagram of human ovum which is ready for fertilization.
Answer:
The egg of human being is rounded and non – motile. It lacks yolk and contains a large Cytoplasm Zona pellucida. amount of cytoplasm. A nucleus (haploid) is present in the centre of egg. Whole egg is covered by a translucent and non – cellular membrane known as zona pellucida. The zona pellucida is irregularly covered by follicle cells. This layer is called as corona radiata. The cells of corona radiata are disintegrated before fertilization.

Question 4.
What is menstrual cycle ? Which hormones regulate menstrual cycle?
Answer:
The reproductive cycle in the female primates is called menstrual cycle. The uterus linings becomes thick and spongy to receive fertilised egg. If the egg is not fertilised, this lining is not needed any longer so, it slowly breaks and comes out through vagina along with blood and mucus. This is called menstruation. It is repeated at an average interval of about 28/29 days.

Following hormones regulate this cycle:

1. Gonadotropin
2. Estrogen
3. Luteinizing hormone
4. Follicular stimulating hormone
5. Progesterone

Question 5.
What is parturition? Which hormones are involved in induction of parturition?
Answer:
The process of delivery of the foetus (child birth) at the end of the pregnancy is called parturition. The signals for parturition originate from the fully developed foetus and the placenta which trigger the release of oxytocin from the maternal pitutary. Oxytocin acts on the uterine muscles and induces stronger uterine contractions leading to expulsion of the baby. Relaxin hormone released by the ovary widens the vagina to facilitate birth.

Following hormones are involved in induction of parturition:

1. Cortisol
2. Estrogen
3. Oxytocin.

Question 6.
In our society the women are often blamed for giving birth to daughters. Can you explain why this is not correct?
Answer:
Women are blamed for giving birth to daughters. This is wrong because sex of the baby is determined by the sperm that can have either X or Y – chromosome. Women have only one type of chromosome (X) in all the ova.

1. If the sperm having X-chromosome fertillises the ovum (X), the resulting zygote (XX) will become a female.
2. If the sperm having Y-chromosome fertillises the ovum (X), the resulting zygote (XY) will become a male.

Question 7.
How many eggs are released by a human ovary in a month? How many eggs do you think would have been released if the mother gave birth to identical twins? Would your answer change if the twins born were fraternal?
Answer:
Only one egg is released by a human (female) ovary in a month. Only one egg is released if the mother gave birth to identical twins. Yes, two or more eggs are released in case fraternal twins are born.

Question 8.
Name the hormones involved in regulation of spermatogenesis.
Answer:
The hormones involved in regulation of spermatogenesis are:

1. Gonadotropin releasing hormone
2. Luteinizing hormone (LH)
3. Follicle stimulating hormone and
4. Testosterone.

Question 9.
Define spermiogenesis and spermiation.
Answer:
Spermiogenesis:
The process involving transformation of spermatid into spermatozoa is called spermiogenesis.

Spermiation:
After spermiogenesis, sperm heads became embedded in the sertoli cells and are finally released from the seminiferous tubules by the process called spermiation.

Question 10.
Give composition of seminal plasma.
Answer:
Composition of seminal plasma : Fructose, calcium ion, some enzymes prostaglandins.

Question 11.
Draw a labelled diagram of male reproductive system.
Answer:

Question 12.
Draw a labelled diagram of female reproductive system.
Answer:

Question 13.
Write two major functions each of testis and ovary.
Answer:
Functions of Testis:

1. Production of sperms by seminiferous tubules.
2. Production of male sex hormone, testosterone by leydig cells.

Functions of Ovary:

1. Production of ova (eggs).
2. Production of female sex hormones, estrogen and progesterone.

Question 14.
Explain menarchy or menopause.
Answer:
Menopause:
In every human female, puberty period starts from 12 – 13 yrs of age to 45 – 50 yrs of the age. During this period except pregnancy at every interval of a month during 26th day to 28th day. If pregnancy does not occur then the internal wall of the uterus secretes out mucilaginous liquid along with blood. Secretion continues for 3 – 4 days called menses. As it comes at definite period so, it is called menstruation cycle. At the age of 40 – 50 yrs. menses stop and females reach to a stage called menopause. Ability of pregnancy also stops after attaining menopause.

Question 15.
Draw a well labelled diagram of the T.S. of human testis.
Answer:

Question 16.
What is the position of fallopian tubule in female reproductive organs? What are their significance?
Answer:
Fallopian tubules are a pair of small muscular tubes, one each on either side of the uterus. These tubules extend from the vicinity of the ovary to the ovary. Each tubule is about 10 cm in length. The free end of each tube lies near the ovary of its side. This end is funnel shaped and fimbriated. It is called ostium and infundibulum. Infundibulum opens in the abdominal cavity by means of abdominal ostium. The fallopian tubule is kept in position by a mesentery which is attached to the uterus.

Significance or Functions of Fallopian Tubes:

1. By their lashing movement of the cilia present in the lining of infundibulum and nearby area help in pulling the released ovum into fallopian tube.
2. Passage of ovum into uterus is aided by muscular movement of fallopian tube as well as beating of cilia present in the lining layer of tube.
3. Fertilization of ovum mostly takes place in the ampulla part of fallopian tube.

Question 17.
Draw a labelled diagram of a section through ovary.
Answer:

Question 18.
Draw a labelled diagram of a graafian follicle.
Answer:

Question 19.
Write the functions of the following :

1. Corpus luteum
2. Endometrium
3. Acrosome
4. Sperm tail
5. Fimbriae

Answer:
The functions of the following:

1. Corpus luteum secretes large amount of progesterone which is essential for the maintenance of endometrium of the uterus.

2. Endometrium is necessary for the implantation of the fertillised ovum, for contributing towards making of placenta and other events of pregnancy.

3. Acrosome is filled with enzymes that help in dissolving the outer cover of the ovum and entry of sperm nucleus.

4. Sperm tail facilitates motility of the sperm essential for reaching the ovum to fertilise it.

5. Fimbriae are finger – like projections at the mouth of fallopian tubules that help in collection of the ovum after ovulation.

Human Reproduction Long Answer Type Question

Question 1.
Describe the structure of a seminiferous tubules.
Answer:
Seminiferous tubules are highly coiled tubes, which are lined on the inside by:
1. Male germ cells called spematogonia that undergo meiotic division to form sperm cells.

2. Sertoli cells provide nutrition and molecular signals to the germ cells.

Question 2.
What is spermatogenesis? Briefly describe the process of spermatogenesis.
Answer:
Spermatogenesis:
Formation of sperms in the testis is called as spermatogenesis. It involves in the following steps:

1. Multiplication phase:
In this phase, sperm cells are formed in testes. The inner layer of seminiferous tubules of testes is formed of germinal epithelium. Some of these cells called primary germ cells divide mitotically into spermatogonia which become separated in the germinal layer. Other cells of this layer serve as nutrition for the dividing cells.

2. Growth phase:
In this phase, spermatogonia starts growing, absorbing nutrient substances. These large cells are called primary spermatocytes.

3. Maturation phase:
It is a very important phase. Primary spermatocytes divide twice. The first division is meiotic due to which the number of chromosomes is reduced to half. In this process, primary spermatocyte divides into two halves which are known as secondary spermatocytes. The second division is mitotic and no change takes place in the number of chromosomes. Thus, from two secondary spermatocytes four spermatids are formed. In this manner from one primary spermatocyte four spermatids are formed. These spermatids change into sperm cells of spermatozoa by a process called metamorphosis.

Question 3.
Draw a labelled diagram of human sperm and explain its defferent parts.
Answer:
Structure of sperm:
A mature sperm is a delicate microscopic, motile structure. A typi¬cal mammalian sperm consists of the following three parts:

1. Head
2. Middle piece and
3. Tail.

1. Head:
Head is knob like terminal part of the sperm. It is composed of a large nucleus and an acrosome. At the time of entry of the sperm into egg acrosome secretes spermlysin which dissolve the egg membrane and thus facilitates entry of sperm into the egg or ovum.

2. Middle piece:
It is short and lies between head and tail. It contains two granules called the proximal and distal centrioles in front side and towards posterior side cylindrical middle part of sperm. It is considered as the power house of sperm as it contains compact mass of mitochondria, which provides energy for metabolism and movement of sperm.

3. Tail:
It is situated on posterior part of sperm. It moves with the help of axial filament. The posterior part of the tail is called as end piece and it is not covered by membrane.

Question 4.
What are the major functions of male accessory ducts and glands?
Answer:
The main functions of male accessory ducts and glands are as follows:

1. Rete testis : They transport sperms from seminiferous tubule to vas efferentia.
2. Vas efferentia : Transports sperms to epdidymis.

Question 5.
Describe oogenesis with suitable diagram.
Or
Explain the different phases of oogenesis with ray diagram.
Or
What is oogenesis? Describe its various phases.
Answer:
The process of formation of ova from oogonia in the ovaries is called oogenesis. It consists of three phases:

1. Multiplication phase
2. Growth phase and
3. Maturation phase.

1. Multiplication phase:
The primordial germ cells divide by mitosis to produce oogonia. These oogonia divide by repeated mitotic divisions forming clusters of oogonia called ovigerous cords. These lie close to germinal epithelium. When oogonia stop dividing they are called oocytes. In each cluster of oocytes only one enters to the growth phase and is known as primary oocyte, while the remaining oocytes form follicle cells and provide nourishment to the developing ovum, the primary oocyte.

2. Growth phase:
Growth phase of primary oocyte is very long. It varies from a few days to many years. It takes about 6 – 14 days in hen after ovulation, but three years in frog and in women all the oocytes are present at the time of birth but no one grows till the attainment of puberty, i.e., 12 – 14 years. However afterwards they grow one by one. During growth phase, following changes occur:

(i) Increase in size:
The primary oocyte increases many folds. For example, primary oocyte of the frog in the beginning has a diameter of about 50μ and on maturation it is about 1000μ to 2000μ. In man also on maturation oocyte increases in size about 7 times. The size increases due to the accumulation of reserve food like proteins and fats in the form of yolk. Due to heavy weight it is usually concentrated towards and lower portion of the egg forming vegetative pole. The portion of the cytoplasm with egg pronucleus remains often separated from the yolk and occurs towards the upper portion of the egg forming animal pole.

(ii) Number of mitochondria increases and in certain cases (birds, amphibia) they are concentrated in the same places to form mitochondrial clouds.

(iii) Increase the amount of ER and activity of golgi complex.

(iv) Synthesis of yolk or Vitellogenesis:
Chemically, yolk is a lipoprotein composed of protein, phospholipids and neutral fats along with little quantity of glycogen. The yolk is synthesized in the liver of female in soluble form and is transported through circulation to the follicle cells surrounding maturing ovum. From follicle cells it is absorbed by the ovum and is deposited in the form of yolk granules and platelets in the ooplasm. The mitochondria and golgi complex are said to be responsible for the conversion of soluble yolk into insoluble granules or platelets.

(v) Formation of thin vitelline membrane around the oocyte.

(vi) Increase the size of nucleus:
Due to the increase in the amount of DNA, nuclear sap and number of nucleoli, nucleus increases in size.

(vii) Gene amplification:
In growth phase the nucleolar genes which code for ribosomal RNA and are located in the nucleolar organizer region multiply to facilitate rapid synthesis of ribosomal RNA. This multiplication of genes without mitosis is known as gene amplification or redundancy.

3. Maturation phase:
In this phase, the nucleus of oocyte undergoes two maturation divisions. The first division is meiotic, as a result two haploid (n) cells are produced. In this division, cytokinesis is unequal, the large daughter cell with almost entire cytoplasm and yolk forms the secondary oocyte. While the smaller one with haploid nucleus (n) and almost without cytoplasm forms the first polar body which is given off from the surface of oocyte at the animal pole. The secondary oocyte with haploid number of chromosomes undergoes second maturation division or second meiotic division (it is meiotic division).

This division is also unequal the large one containing yolk is called ovum and small cell is the second polar body. The first polar body may also divide thus, producing total three bodies which degenerate soon. So, as a result of oogenesis only one functional ovum is formed from a primary oocyte. In most of the vertebrates, the first meiotic division occurs with the commencement of the growth phase, and second maturation division occurs when egg is activated by the entry of sperm.

Question 6.
Describe the structure of ovary.
Answer:
These are a pair of female gonads or primary sex organs lying one on each side of uterus. Ovary is attached to abdominal wall as well as utems by means of ligaments. It is surrounded by a fold of peritoneum named mesovarium. Ovary is internally differentiated into four parts : Germinal epithelium, tunica albuginea, cortex and medulla.

Germinal epithelium is the outermost layer of cuboidal to flattened cells. Germinal epithelium is followed by a sheath of condensed connective tissues which is termed as tunica albuginea. It is followed b.y cortex. The central part of ovary contains medulla. A large number of groups of specialized cells are present in the cortex which are termed as ovarian follicles. These follicles are found in four developmental stages.

1. Incipient follicle:
The central part of these follicles contains a large cell which is covered by many smaller cells.

2. Primary follicle:
These follicles are developed from incipient follicles.

3. Vascular follicle:
It is formed from primary follicles. The oocyte of these follicle is covered by a many layered thick wall.

4. Graafian follicle:
Mature ovarian follicle is termed as graafian follicle. It is covered by two sheaths derived from cortex. The follicle contains a single oocyte. A group of follicular cells surrounds the oocyte or ovum. It is called cumulus ovaricus. Another group produces membrane granulosa. Oocyte has two noncellular membranes, inner vitelline membrane and outer zona pellucida.
Note:

Question 7.
Describe the process of fertilization and give its significance.
Answer:
Fertilization:
Fertilization is the fusion of male and female gametes to produce a single diploid cell, called zygote. Fertilization in human female takes place in fallopian tube. In a sexual mating or coitus the male ejaculates semen into the vaginal passage of the female using the copulatory organ, the penis. In a single coitus as many as 200 million sperms are introduced into the female genital tract but out of this huge number of sperms only one is destined to fuse with the ovum, provided the fallopian tube lodges of fully developed secondary oocyte.

Sperms travel through the vaginal passage and enter the uterus through the cervix. They travel further through the uterus and finally enter the fallopian tube.The vaginal passage is highly acidic to prevent any bacterial infection but this acidic medium is not suitable for the survival of sperms. Many sperms die on the way. The liquid medium of the semen is alkaline which can neutralize the acidity of vagina to some extend and keep the sperms alive and active. The sperms move with the speed of 1-5 to 3 mm per minute.

The ovum gets surrounded by a large number of sperms but usually only one fuses with the ovum. The sperm penetrates the ovum using certain chemical substances of enzymatic nature.These chemicals are called spermlysins which are present in the acrosome. Certain receptors on the cell surface of the ovum enable the sperms to penetrate the wall of ovum. The ovum is surrounded by follicle cells.

These cells are joined together by a glue like substance known as mucopolysaccharide, an acid, called hyaluronic acid. The sperm produces spermlysin, known as hyaluronidase. The over all changes in a sperm before the fertilization is called capacitation. The hyaluronidase enzyme facilitates the sperm to pen-etrate through the corona radiata (follicle cells), zona pellucida and the plasma membrane of ovum. The nucleus and the cytoplasmic components get inside the ovum, leaving the tail outside. The entry of sperm stimulates the ovum and the signal is transmitted to the egg surface incapacitating all the other cells surrounding the ovum.

Nucleus of sperm move towards the nucleus of ovum and they fuses with each other to form zygote. It takes 2 – 2\(\frac { 1 }{ 2 }\) hours to complete fertilization process. Once the ovum has been fertilized a barrier forms around it that normally prevents other sperms from entering. Now fertilized egg reaches to the uterus and within seven days of fertilization it is transplanted to the wall of the uterus.

Significance of Fertilization:

1. Egg becomes active after entry of sperm and completes its second maturation division.
2. Formation of fertilization membrane prevent entry of other sperms in the ovum. In human this membrane is not formed.
3. It restores the diploid number (2n) of chromosomes in the zygote.
4. It combines characters of male and female resulting in the introduction of varia-tions. These variations make the offspring better equipped for struggle against environmental conditions to ensure the existence.
5. The ovum is stimulated to cleavage.
6. After fertilization ovum rotate inside the membrane.
7. Ovum do not contain centriole and obtain it from sperm during this process thus zygote continuously divides.
8. It is necessary for the egg to attain maturity.

Question 8.
Describe development of embryo up to the formation of three germ layer. Give the names of organs formed by three germ layer.
Or
Define cleavage. Describe the changes that occur in embryo till gastrulation.
Answer:
The term cleavage refers to a series of rapid mitotic divisions of the zygote following fertilization forming a many celled blastula. Following are the various steps of embryonic development in human up to the formation of three germ layer:

1. Formation of morula:
The fertilized zygote divides. It undergoes successive quick mitotic cell divisions called cleavage. First cleavage is holoblastic, unequal and meridional. It divides the oyum completely into two unequal blastomeres. The plane of cleavage passes
through animal vegetal axis, i.e., it is meridional. Large blastomere divides little earlier than the small one giving a transitional three cell stage.

As a result of further cleavages, a solid mass of mulberry shaped embryo is formed called morula. Morula is of the size of zygote but consists of 32 cells. These cells are of two types , The outer layer of smaller cells around, inner larger cells. Within 72 hours of fertilization, morula reaches the uterus.

2. Formation of blastula:
Transformation of morula into a blastula starts by the rearrangement of blastomeres. This leads to the formation of a central cavity inside the morula. The outer cells of morula absorb the nutritive fluid secreted by the uterine mu¬cous membrane and transform into trophoblast. The fluid absorbed by these cells collects in the central cavity called biastocoel. This cavity separates the trophoblast from the inner mass of larger cells except on one side and is termed blastocyst. The inner cell mass is pushed to one pole as a small knob. This knob gives rise to the embryo and is termed as embryonal knob.

3. Formation of gastrula:
In this embryonic stage development of three germ layer occurs. In this stage morphogenic movement of cells of embryo occurs, as a result of this three germ layers are formed. A cavity develops at the centre called as archenteron which opens outside through blastopore.

4. Formation of three germ layers:

Formation of endoderm:
The enlargement of blastodermic vesicle is followed by the separation of few cells from the inner cell mass. These cells push out from the blastocoel to become the initial cells of the innermost layer of gastrula, the pattern of a tube within a tube. The inner tube is called primitive gut. It is differentiated into gut tract which is within the body and a yolk sac that communicates with the gut of the embryo. The remaining cells of the inner cell mass are organized to form the embryonic disc.

Formation of mesoderm:
After the formation of endoderm an increased rate of cell proliferation takes place at the caudal end of the embryonic disc. This results in the localised increase in the thickness of the disc. These cells subsequently get detached from the embryonic disc and get organized to a well demarcated mesodermal layer.

Formation of ectoderm:
After the formation of endoderm and mesoderm, the remaining cells of the embryonic disc arrange themselves in a layer to form ectoderm.

Question 9.
Write differences between spermatogenesis and oogenesis.
Answer:
Differences between Spermatogenesis and Oogenesis:

Spermatogenesis:

• Sperms are produced by this process.
• In this process primary spermatocytes are formed by maturation of germinal epithelium cells.
• Primary spermatocyte divides to form four spermatids.
• There is equal division.
• Large number of sperms are formed by this process.

Oogenesis:

• Ovums are produced by this process.
• In this process primary oocytes are formed by maturation of germinal epithelium.
• Primary oocytes divides to form one ovum and three polar bodies.
• There is unequal division.
• Less number of ovums are formed by this process.

MP Board Class 12th Biology Important Questions

MP Board Class 12th Economics Important Questions Unit 1 Introduction

Micro Economics Introduction Important Questions

Micro Economics Introduction Objective Type Questions

Question 1.
Choose the correct answers:

Question 1.
Indian economy is:
(a) Centrally planned economy
(b) Market economy
(c) Mixed economy
(d) None of these.
Answer:
(c) Mixed economy

Question 2.
Who used the word ‘micro’ for the first time:
(a) Marshall
(b)Boulding
(c) Keynes
(d) Ragnar Frisch
Answer:
(d) Ragnar Frisch

Question 3.
Who said economics is the ‘Science of wealth:
(a) Prof. Robbins
(b) Prof. J.K. Mehta
(c) Prof. Marshall
(d) Prof. Adam Smith
Answer:
(d) Prof. Adam Smith

Question 4.
What is the shape of production possibility curve:
(a) Concave to the origin
(b) Concave
(c) Straight line
(d) None of the above.
Answer:
(a) Concave to the origin

Question 5.
The reason for downward shape of production possibility curve is:
(a) Increasing opportunity cost
(b) Decreasing opportunity cost
(c) Same opportunity cost
(d) Negative opportunity cost
Answer:
(b) Decreasing opportunity cost

Question 6.
The point of optimum utilization of resources lies on which side of PPC curve:
(a) Towards left
(b) Towards right
(c) Inside
(d) Upwards
Answer:
(d) Upwards

Question 2.
Fill in the blanks:

1. Every person has ………………… quantity of goods.
2. ……………………economics is the study of individual economic units.
3. Opportunity cost ………………….. in production list.
4. Micro and Macro-economics are ……………………… to each other.
5. Economic growth is related to ………………………… economics.
6. Economy is a group of ………………….. units.
7. Mixed economy is composed of ………………………. and …………………………
8. The goods and services which help in the production of other goods and services are ………………………. goods.

Answer:

1. Few
2. Micro-economics
3. Changes
4. Complement
5. Macro
6. Production
7. Socialism, Capitalism
8. Intermediate

Question 3
State true or false:

1. Production possibility curve is convex towards main point.
2. Central problem in the capitalist economy is solved by price mechanism.
3. An economy can be capitalist, socialist or opportunist.
4. Today all economics of the world are almost mixed economics.
5. Macro-economics is the study of the problems of unemployment, price inflation etc.
6. In socialism, the feeling of personal profit is prominent.

Answer:

1. False
2. True
3. False
4. True
5. True
6. False

Question 4.
Match the following:

Answer:

1. (c)
2. (d)
3. (a)
4. (e)
5. (b)

Question 5.
Answer the following in one word/ sentence:

1. Which economy is adopted in India ?
2. What is the production possibility curve towards its origin ?
3. The governmentalization of exploitation is the fault of which economy ?
4. The struggle class is the specialty of which economy ?
5. Which economy has limited the scope of private sector ?
6. What is the heart of all the institutions of capitalism ?
7. What are central problems of an economy known as ?
8. Based on priorities, who determines the production area and quantity of production ?

Answer:

1. Mixed economy
2. Concave
3. Socialist
4. Capitalism
5. Socialism
6. Profit
7. Basic work
8. Central Planning Authority

Micro Economics Introduction Very Short Answer Type Questions

Question 1.
What do you mean by macro-economics ?
Answer:
Macro-economics aims at dealing with the aggregates and averages. It is not interested in the individual items but as total savings, total consumption, total employment, aggregate demand, etc.

Question 2.
What do you mean by micro-economics ?
Answer:
Micro-economics studies the economic actions of individuals i.e., particular firm, individual households, wages, interest, profit etc. In other words it is infact a microscopic study of the economy.

Question 3.
What is the central point of macro-economics ?
Answer:
The central point of marco-economics is the analysis of national income.

Question 4.
What do you mean by goods ?
Answer:
All physical, tangible objects which satisfy human wants and need i.e., objects which possess utility, are called goods.
For example : Tables, cars, etc.

Question 5.
What do you mean by economic activity ?
Answer:
Economic activities are those activities which increases the material, welfare of man and which are used for increasing wealth and welfare.

Question 6.
How are services ?
Answer:
Services are non – material, intangible goods which can not be seen, touched or stored and have power to satisfy human wants and needs. For example, services provided by doctor.

Question 7.
Write the meaning of ‘a person’.
Answer:
By person, we mean the decision taking units.

Question 8.
How many branches are there of Economics ?
Answer:
Economics has two branches:

1. Micro-economics
2. Macro-economics

Question 9.
What to you mean by resources ?
Answer:
Those goods and services which are used to produce other goods are known as resources. They are traditionally known as factors of production.

Question 10.
What do you mean by economy ?
Answer:
It refers to the sum total of economic activities in an area, which may be a village, a city, a district, or a country as a whole. They provide source of livelihood.

Micro Economics Introduction Short Answer Type Questions

Question 1.
What do you mean by macro-economics ? Write its two characteristics.
Answer:
Macro-economics:
Macroeconomics is the study of aggregate factors such as employment, inflation and gross domestic product and evaluating how they influence the economy as a whole.

The characteristics of macro-economics are as follows:
1. Broader perspective:
The concept of macro-economics is broader. In it small units are not given importance but with the help of it National and International economic problems.

2. Broad analysis:
In macro-economics broad analysis is given importance. For example, under the subject matter of macro-economics we study the monetary and fiscal policies of the government. It studies the general problems of national level like the effects on – monetary policy, fiscal policy, general price level, general employment etc. If the effect of public finance and public expenditure is good on society then we can conclude that the . effect of it is also good on each person of the society.

Question 2.
Explain the main types of Macro-economics.
Answer:
Following are the types of Macro-economics:

1. Macro static:
Macro static method explains the certain aggregative  relations in 1 stationary state. It does not reveal the process by which the national economy reaches the equilibrium. It deals with the final equilibrium of the economy at a particular point of time. It does not analyse the path by which the economy reaches equilibrium. It presents a “still” picture of the economy as a whole at a particular point of time. Pro. Keynes has explained by this equation:

Y = C +1.
Here, Y = Total income, C = Total consumption expenditure, I = Total investment expenditure.

2. Comparative macro statics:
The various macro variables in the economy are subject to charge with the passage of time. Total consumption, total investment and total income etc. go on changing and so the economy reaches different levels of equilibrium.

3. Macro dynamics:
This is a realistic and new method of economic analysis. Under this method, we study how the equilibrium in the economy is reached as a result of changes in the macro variables and aggregates. This method enables us to see the movie picture of the entire economy as progressive whole. This method is a complex method. It is recently developed by Frish, Robertson, Hicks, Tinburzon, Samulson etc.

Question 3.
What is economic problem ? Why do economic problem arises ?
Answer:
An economic problem is basically the problem of choice which arises because of scarcity of resources. Human wants are unlimited but means to satisfy them are limited. Therefore, all human wants cannot be satisfied with limited means.

An economic problem arises because of the following causes:
1. Human wants are unlimited :
With the satisfaction of one want, another want arises. These is a difference in the intensity of wants also. Thus, there is an unending circle of wants, when they arise, are satisfied and arise again.

2. Limited resources:
Means are limited for the satisfaction of wants. Scarcity of resources is a relative term, for satisfying a particular want, resources can be in abundance, but for the satisfaction of all the wants, resources are scarce. Thus, scarcity of resources gives rise to economic problems.

Question 4.
“Scarcity is the mother of all economic problems”. Explain.
Answer:
The statement is true, no matter, how well a particular economy is endowed with resource, these resources will be relatively scarce to fulfill its unlimited wants. Moreover, these scarce resources have alternative uses and can be allocated to the production of different goods and services. Thus, it is due to the scarce availability of resources to fulfill the different and competing unlimited wants that an economy faces the economic problem of choice. Thus, it is rightly said that Scarcity of resources is the mother of all economic problems.

Question 5.
What do you understand by positive economic analysis ?
Answer:
Positive economic analysis is confined to cause and effect relationship. In other words, it states “What is it relates to, what the facts are, were or will be about various economic phenomena in the economics, e.g., it deals with the analysis of questions like what are the causes of unemployment.

Question 6.
What do you understand by normative economic analysis ?
Answer:
Normative economic analysis is concerned with ‘What ought to be’. It examines the real economic events from moral and ethical angles and judge whether certain economic events are desirable or undesirable, e.g., it deals with the analysis of questions like what should be the prices of food grains, unemployment is better than poverty, rich persons should be faxed more.

Question 7.
Write main assumptions of production possibility curve.
Answer:
Following are the main assumptions of production possibility curve:

1.  Economy produces only two goods X and Y in different proportions.
2.  Amount of resources available in an economy are given and fixed.
3.  Resources are not specific i.e., they can be shifted from the production of one goods to the other goods.
4.  Resources are fully employed, i.e., there is no wastage of resources.
5.  State of technology in an economy is given and remains unchanged.
6.  Resources are efficiently employed.

Question 8.
What is production possibility curve ? Explain with example.
Answer:
Production possibility curve shows graphical presentation of various combination of two goods that can be produced with available technologies and given resources assuming that the resources are fully and efficiently employed. The production possibility curve is also known as transformation curve. PPC can be explained with the help of example and imaginary schedule.

Question 9.
What do you mean by opportunity cost ?
Answer:
The opportunity cost of a good is the value of the next best alternative good is forgone for it. In other words opportunity cost of any commodity is the amount of other good which has been given up in order to produce that commodity.

Question 10.
A school teacher takes two years of leave for study to get the degree of doctorate of philosophy from a university. His monthly income is Rs 35,200 and the fees of research degree is Rs 40,000. What will be his opportunity cost for doctorate degree ?
Solution:
Opportunity cost = (35,200 × 12 x 2) + 40,000
= (35,200 × 24) + 40,000
=  Rs 8,84,800.

Question 11.
Why is production possibility curve downward sloping ?
Answer:
Production possibility is a curve which shows various production possibilities with the help of given limited resources and technology. It is also known as production possibility frontier and transformation curve. It i&downward sloping from left to right because in a situation of fuller utilization of the given resources, production of both the goods cannot be increased together. More of goods X can be produced only with less of goods Y as resources are scarce. Because of this un use relationship between production of both the goods, PPC is downward sloping.

Question 12.
Why is production possibility curve concave to the origin ?
Answer:
Production possibility curve is concave to the origin because to produce each additional unit of goods X, more and more unit of goods Y is to be sacrificed. Opportunity cost of producing every additional unit of goods ‘A’ tends to increase in terms of the loss of production of goods Y. It is so because factors of production are not perfect substitute of each other. As we move down along the PPC, the opportunity cost increases. And this causes the concave share of PPC.

For example, if in any state, production of mango is more and it may not be same in other state, the other state may lead in the production of sugar cane. So, it is not Goods (A) necessary that a person who is efficient in the production of one goods will also lead in the production other good. This shows increase in marginal opportunity cost reason of marginal opportunity cost that PP curve is concave to the origin.

Question 13.
What is studied in Economics ?
Answer:
It is an important question that what should be studied under subject matter of economics. Wants are unlimited but resources are limited. It is decided in economics that row resources should be economically and more efficiently used. Study of subject-matter of economics includes the rational use of resources. The ultimate air of any economy is to have maximum economic welfare.

In any economy, what economic system we use is based on macro and micro-economics. In micro-economics, we study principle of consumers, price determination, producers behavior, monetary policy, inflation, Government budget, exchange rate etc. is studied.

Micro Economics Introduction Long Answer Type Questions

Question 1.
Distinguish between Micro-economics and Macro-economics.
Answer:
Differences between Micro-economics and Macro-economics;

Question 2.
Define micro-economics and explain its characteristics.
Answer:
The word ‘micro’ is originated from the Greek word micros; which means small. Micro-economics studies the economic actions of individuals. Under this branch the whole economy is divided into small individual parts i.e., a particular firm, individual household, wages, interest, profit etc. According to Chamberlain, “The micro model is built solely on the individual and deals with interpersonal relations only.”

Characteristics:
Followings are the characteristics of micro-economics:
1. Study of individual units :
The first characteristics of micro-economics is that it studies individual units. It helps to explain personal income (individual income), individual production, individual consumption etc. Micro-economics studies the individual problems and helps in analyzing the whole economic system.

2. Study of small variable:
Micro-economics gives importance to the study of small variable. These variables have such little influence that they do not affect the whole economy.
For e.g., a single consumer by his consumption or a single producer by his production cannot have influence over the demand and supply of whole economic system.

3. Determination of individual price :
Micro-economics is also called the price theory. It determines the individual price of different products by analyzing the demand and supply i. e., behavior of buyers and sellers.

4. Based on the concept of full-employment:
While studying micro-economics the concept of full employment is taken into consideration.

Question 3.
Write any five importance of Micro-economics.
Answer:
The importance of micro-economics can be studied under the following points:
1. Essential for the knowledge of whole economy:
The sum of individual demand, individual production etc., make the aggregated demand and combined supply. Hence, micro-economics analysis is essential for knowing the position of whole economy.

2. Helpful in solving economic problems :
The problems of pricing of product and pricing of factors of production are main economic problems. Each factor of production demands more remuneration for it, each seller wants to get maximum price for his product. What price should be paid, at what price the problems should be sold all these problems are studied in micro-economics,

3. Helpful in determining economic policies:
The economic policies of government are studied in micro-economics in order to know as to how they influence the functioning of individual units.

4. Investigation of condition of economic welfare:
The study of micro-economics also reveals the effects of public expenditure and public revenue along with the position of individual consumption individual living standard etc. The classical economists favored micro-economic analysis as a measuring rod of economic welfare.

5. Helpful in decision making for individual units:
Micro-economics analysis proves helpful in taking rational decision about the problems of individual units profit of individual firm, individual income etc. Each consumers wants to get maximum satisfaction from his limited means, each firms want to get maximum profit.

Question 4.
What are the limitations of macro-economics ?
Or
Write defects of macro-economics.
Answer:
The limitations of macro-economics are as follows :

1. Ignorance of individual units :
Macro-economics gives importance to aggregate rather than individual units. The sum of small individual economic units make the wall of whole economy. The whole economy depend on the economy of individual unit. Thus, the small individual units play the role of foundation which are ignored in the macro-economic analysis.

2. Conclusions are not practical :
The conclusions drawn from macro-economic analysis are after misleading. For example, if the general price index number is unchanged, it cannot be concluded that the price level is unchanged. The increase in prices of certain commodities and decrease in prices of other commodities may make the general price level in – charged but the fluctuations are found in the prices of individual commodities.

3. Measurement of aggregates is difficult:
The measurement of aggregates itself presents serious problems in certain cases. Despite several improvements in statistical techniques in recent years, it has not been possible to obtain reliable measures of aggregates.

4. Macro-economics does not depict correct picture of Individual units :
Macro-economics does not depict the true picture of individual units. For example, if the aggregate demand increases, the production will rise but there may be certain firms whose cost and output may increase or decrease.

Similarly if the level of income increases the consumption of some goods may increase and that of some may decrease also. The out of those goods will be reduced whose consumption tends to decrease and the output of those goods will be reduced whose consumption tends to decrease and the output of those goods will be in-creased whose consumption tends to increase.

5. It may not influence all the sectors of economy:
The principles or analysis of Macro-economics may not influence all the sectors of economy. For example, a general rise in prices may not affect all the sections of the community in a similar manner. Some sections may be affected more adversely than others. Some industries may get more benefit than others.

6. No consideration of composition of the aggregate:
Macro-economic analysis takes into consideration the size and types of the aggregate but does not take into consideration the composition of the aggregate. Unless and until the composition and all the components of the aggregate are not analysed, the forecasts of Macro-economic analysis are baseless the suggestions given on their basis will be of no use.

Question 5.
Explain any five points of importance of macro-economics.
Or
Write any five advantages of macro-economics.
Answer:
The importance of macro-economics is clear from the following facts:

1. Helpful in making and executing governmental economic policies:
The study of macro-economics is essential for formulation of the economic policies of the government. Government has to make policies of general level of production, general price level. Macro-economists help.in formulation of such policies.

2. Helpful for Micro-economics:
The study of macro-economics is helpful for studying micro-economics. No micro-economic law can be formulated without studying macro-economics.

3. Measurement of economic development:
Macro-economics deals with the aggregates (output, consumption, income, capital formation) which help in measuring the economic development of the country.

4. Study of inflation and deflation:
The study of aggregate demand and aggregate supply of currency enables us to analyse the rate of inflation and deflation in the country. By this we can get the idea of economic conditions and standard of living of people.

5. Helpful in the study of fluctuation in economy:
Macro-economics proves helpful in the study of fluctuations in the economy through the analysis of national aggregates like; income, output saving, investment etc.

Question 6.
Write any five limitations of Micro-economics.
Answer:
Following are the limitations of micro-economics:

1. Conclusions drawn are not accurate:
What is true in the case of individual units may not be true in the case of aggregates. For example, individual saving is good but if the entire community starts saving more, effective demand will be reduced.

2. Based on unrealistic assumptions:
Micro-economist assumes other things being equal, and is based on the assumption of full employment in society. This is unrealistic assumption.

3. Concentration on small parts:
Instead of studying the total economy, micro-economics studies only small parts of it. It fails to enlighten us the collective functioning of the national economy.

4. Unable to analyse certain problems:
There are certain economic problems which cannot be analysed with the aid of Micro-economics. For example, important problems relating to public finance, monetary and fiscal policy etc., are beyond purview of microeconomics.

5. Infeasible:
Micro-economics fails to provide us with a description of the real world as it is. It is not in a position to take into accountable the entire economic data of the real world.

Question 7.
What is opportunity cost ? Explain with suitable example.
Answer:
Opportunity cost is defined as the value of a factor in its next best alternative use or it is the cost of foregone alternatives, or it can be defined as the value of the benefit that is sacrificed by choosing an alternative. We can also say that opportunity cost of any commodity is the amount of other goods which has been given up in order to produce that commodity. It is also known as opportunity lost or transfer earning of a factor e.g., a person is working in college ‘A’ at the salary of ? 70,000/- per month, he has two more options to work:

1. To work in college “C” at Rs 65,000 per month.
2. To work in college “D” at Rs 62,000 per month.

In this case opportunity cost of working in college “A” is Rs 65,000 i.e., salary he would get in college “C” (cost of foregone alternative) because it is the value of next best alternative between college “C” and college “D”. Thus, opportunity cost is very important concept in economics. It is because our re-sources are limited, we are always making choices from the available alternatives. Thus, the opportunity cost of using a resource is defined as the value of the next best use to which that resource could be put.

Question 8.
What is production possibility curve ? Explain with example.
Answer:
Production possibility curve:
Production possibility curve shows graphical presentation of various combination of two goods that can be produced with available technologies and given resources assuming that the resources are fully and efficiently employed. The production possibility curve is also known as transformation curve. PPC can be explained with the help of example and imaginary schedule.
Example:

We can show the basic problem of choice or all forms of economic problems with the help of this production possibility curve. This is shown in the table and diagram. Here, for the sake of simplicity we have presumed that only two commodities i.e., wheat and machines are being produced in an economy with limited capital.

From the table, it is clear that when the production of machine is zero then production of rice is 500 tone. If, the production of machines increase from 1 to 2 then the production of rice will be decreased from 500 to 450 ton. If all resources are implied on production of machines then production of machines will become 4 and production of rice will be zero. So suitable combination is essential for production.

Question 9.
What do you mean by Marginal Opportunity Cost ? Explain with the help of an imaginary schedule.
Answer:
Marginal Opportunity Cost can be explained as if the resources are transferred from one use to the other, it is obvious that there would be loss in production of one goods in order to increase output of the other. This loss is marginal opportunity cost which is technically termed as marginal rate of transformation. It is also called as rate of sacrifice.

i.e., \(\frac { ΔY }{ ΔX }\)
Here,
ΔY = Increase in production of Goods.
ΔX = Increase in production in Goods of X.

From the above table it is clear that in order to produce one unit of commodity ‘A’ he has to sacrifice all the units of commodity ‘B’. After combination ‘E’ to ‘F’ marginal opportunity cost increases to 25.0. Thus, it is clear that production possibility curve is also known as transformation curve or rate of sacrifice.

Question 10.
“Micro-economics and Macro-economics are not competitive to each other but complementary to each other”. Explain.
Answer:
Interdependence of micro and macro can be studies as follows:

1. Macro – economics depends on micro-economics:
Macro-economics and microeconomics depend on each other. Both are interdependent, macro-economics contributes to the micro-economics. For example, the theory of investment belongs to micro-economics. It is derived from the behavior of the individual entrepreneur. The theory of aggregate investment function can also be derived from micro-economic theory of investment. Thus, we can say, that macro-economics depends on micro-economics.

2. Micro-economics depends on macro-economics:
Micro-economics depend upon macro-economics to a certain extent. For example, the rate of interest is a subject which belongs to micro-economics but it is influenced by macro-economics aggregates. Thus, micro-economics depends upon macro-economics.

Conclusion:
Macro and micro-economics both are interdependent on each other. Neither is complete without the other. We must study macro-economics because it deals with average variables, such as national income and national output. We must study microeconomics because national output and national income are eventually the result of decision of millions of business firms and individuals. Thus, we can conclude that both are complementary to each other.

Question 11.
Search for new entrepreneurs so that expansion of industries can be done and more use of capital intensive technology can be done in the country.
Answer:
When the economy is below its potential due to unemployment, the economy operates inside the PPC. When the government starts employment generation scheme, it enables the economy to utilise its existing resources in the optimum manner. The resources which were sitting idle, now get

job and the economy functions at its maximum capacity and moves from inside the PPC to points on the PPC. Thus, economy moves from point ‘a’ below PPC, to any point on PPC as shown in the figure.

Question. 12.
Explain the subject matter of macro-economics under five headings.
Answer:
We can study the subject matter of economics under the following heads :

1. Income and level of employment:
Income and level of employment is the main subject matter of Macro-economics. Level of employment and income depends on effective demand. Effective demand is determined by total expenditure.

2. Price theory :
It includes the study of inflation and deflation. Macro-economics also studies the normal level of prices of goods. It also studies the causes of inflation.

3. Theory of economic growth:
Macro-economics studies the economic growth by economic planning we can solve the problems of economic growth.

4. Theory of distribution :
During distribution of shares of national income many problems arise. How much should be paid as wages, what should be rent, how much interest should be paid all are the main elements of the subject matter of macro economics.

5. Theory of public finance:
Public revenue, public expenditure, public debts, taxation, fiscal policies are the subject matter of macro-economics.

Question 13.
Explain the five factors determining demand.
Answer:
1. Price of commodity:
Price is a dominating factor. The demand for a commodity is increased when the price of it decreases and the demand for it decreases when the price increases. For example, if the price of a commodity is Rs 50 then its demand is 100 units. If the price increases to Rs 100, its demand will be 50 units. In other words, there is an inverse relation between price and demand.

2. Price of substitutes:
If the price of substitutes of a commodity rises, the price of the commodity will also rise. If the price of substitute of a commodity goes down, the price of the commodity will also go down. For example, tea and coffee are substitutes to each other. If the price of tea rises, the demand for coffee will increase. In other words, the increase or decrease in the price of the substitute of a commodity results in the increase or decrease in the price of the commodity.

3. Price of complementary goods:
If the price of complementary goods increases, the price of the commodity will decrease. For example, the increase in the price of petroleum results in the decrease of demand for motor-cars and scooters etc. If the price of petroleum decreases, the demand for motor-car and scooter etc. will increase.

4. Income of consumer :
There is a direct correlation between the income of consumer and demand. If the income of a consumer increases, the demand for commodities will also increase. If the income of a consumer decreases, the demand for commodities will also decrease. The main reason for this tendency is that the purchasing power of a consumer increases or decreases as a result of increase or decrease in income.

5. Tastes and preferences of consumers:
The consumers, often, consume those commodities which suit to their taste and preference. If a commodity does not suit to their taste and preference, there will no remarkable change its demand as a result of increase or decrease in its price. If the price of a commodity of consumer’s taste and preference goes down, the demand for it will certainly increase and vice-versa.

Question 14.
Distinguish between Market Economy and Centrally Planned Economy.
Answer:
Differences between Centrally planned economy and Market economy:

MP Board Class 12th Economics Important Questions

MP Board Class 12th Maths Important Questions Chapter 3 Matrices

Matrices Important Questions

Matrices Objective Type Questions:

Question 1.
If A = \(\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}\) and A + A’ = I, then the value of α is:
(a) \(\frac { \pi }{ 6 } \)
(b) \(\frac { \pi }{ 3 } \)
(c) π
(d) \(\frac { 3\pi }{ 2 } \)
Answer:
(b) \(\frac { \pi }{ 3 } \)

Question 2.
If A = \(\left[\begin{array}{lll}
{2} & {0} & {0} \\
{0} & {2} & {0} \\
{0} & {0} & {2}
\end{array}\right]\), then A⁵ is equal to:
(a) 5 A
(b) 10 A
(c) 16 A
(d) 32 A
Answer:
(c) 16 A

Question 3.
If a matrix is both symmetric and skew – symmetric, then:
(a) A is a diagonal matrix
(b) A is zero matrix
(c) A is a square matrix
(d) None of these
Answer:
(b) A is zero matrix

Question 4.
If A = \(\begin{bmatrix} \alpha & \beta \\ \lambda & -\alpha \end{bmatrix}\) is such that A² = I, then:
(a) 1 + α² + βλ = 0
(b) 1 – α² + βλ = 0
(c) 1 – α² – βλ = 0
(d) 1 + α² – βλ = 0
Answer:
(c) 1 – α² – βλ = 0

Question 5.
If A = \(\begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix}\), then Aⁿ = ………………………:
(a) A = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\), if n is even natural number
(b) A = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\), if n is odd natural number
(c) A = \(\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}\), if n ∈ N
(d) None of these
Answer:
(a) A = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\), if n is even natural number

Question 2.
Fill in the blanks:

1. If A = \(\begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}\) and B = \(\begin{bmatrix} 1 & 3 \\ 2 & 5 \end{bmatrix}\), then AB = …………………………….
2. If A = diag [1, -1, 2] and B = diag [2, 3, -1], then value of 3A + 4B will be ……………………
3. A matrix A is said to be idempotent if ………………………
4. A matrix A is said to be orthogonal if ……………………..
5. If [x, 1] \(\begin{bmatrix} 1 & 0 \\ -2 & 0 \end{bmatrix}\) = 0, then find the value of x will be …………………………….

Answer:

1. \(\begin{bmatrix} 10 & 26 \\ 7 & 19 \end{bmatrix}\)
2. diag [11, 9, 2]
3. A² = A
4. AA’ = A’A = I
5. x = 2

Question 3.
Write True/False:

1. Multiplication of matrix is always commutative?
2. Two matrix are said to be comparable if they have same number of rows and columns?
3. If A is a square matrix, then A. adj A = |A| I?
4. A square matrix A is said to be symmetric if A = – A^T?
5. Matrix A and B are inverse of each other if AB = BA?

Answer:

1. False
2. True
3. True
4. False
5. False

Question 4.
Match the Column:

Answer:

1. (d)
2. (e)
3. (a)
4. (b)
5. (c)

Question 5.
Write the answer in one word/sentence:

1. If A and B are two square matrix of same order, then what is the value of Adj (AB)?
2. A square matrix A is said to be Involountary matrix if?
3. If A = \(\begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}\), then find the value of A²?
4. If A = [1, 2, 3], then find the value of AA^T?
5. If X + Y = \(\begin{bmatrix} 1 & -2 \\ 3 & 4 \end{bmatrix}\) and X – Y = \(\begin{bmatrix} 3 & 2 \\ -1 & 0 \end{bmatrix}\), then find the value of X?

Answer:

1. Adj.(AB) = (Adj B). (Adj A)
2. A² = I
3. -I
4. [1, 4]
5. \(\begin{bmatrix} 2 & 0 \\ 1 & 2 \end{bmatrix}\)

Matrices Short Answer Type Questions

Question 1.
If A = \(\begin{bmatrix} a^{ 2 }+b^{ 2 } & b^{ 2 }+c^{ 2 } \\ a^{ 2 }+c^{ 2 } & a^{ 2 }+b^{ 2 } \end{bmatrix}\) and B = \(\begin{bmatrix} 2ab & 2bc \\ -2ac & -2ab \end{bmatrix}\), then find the value of A + B? (NCERT)
Solution:
A + B =

Question 2.
If A = \(\begin{bmatrix} cos^{ 2 }x & sin^{ 2 }x \\ sin^{ 2 }x & cos^{ 2 }x \end{bmatrix}\) and B = \(\begin{bmatrix} sin^{ 2 }x & cos^{ 2 }x \\ cos^{ 2 }x & sin^{ 2 }x \end{bmatrix}\), then find A + B? (NCERT)
Solution:
A + B = \(\begin{bmatrix} cos^{ 2 }x & sin^{ 2 }x \\ sin^{ 2 }x & cos^{ 2 }x \end{bmatrix}\) + \(\begin{bmatrix} sin^{ 2 }x & cos^{ 2 }x \\ cos^{ 2 }x & sin^{ 2 }x \end{bmatrix}\)

⇒ A + B = \(\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\)

Question 3.
If A = and B = then find 3A – 5B? (NCERT)
Solution:

Question 4.
Simplify: cos θ \(\begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}\) + sinθ \(\begin{bmatrix} sin\theta & -cos\theta \\ cos\theta & sin\theta \end{bmatrix}\)
Solution:
cos θ \(\begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}\) + sinθ \(\begin{bmatrix} sin\theta & -cos\theta \\ cos\theta & sin\theta \end{bmatrix}\)

Question 5.
From the following equation find the value of x and y?
2 \(\begin{bmatrix} x & 5 \\ 7 & y-3 \end{bmatrix}\) + \(\begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix}\) = \(\begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix}\)? (NCERT)
Solution:

From defnition of matrix,
2x + 3 = 7
⇒ 2x = 4 ⇒ x = 2
⇒ 2y – 4 = 14
⇒ 2y = 18 ⇒ y = 9
∴x = 2, y = 9.

Question 6.
Find the value of X and Y if X + Y = \(\begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix}\) and X – Y = \(\begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix}\)? (NCERT)
Solution:
Given X + Y = \(\begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix}\) ……………….. (1)
and X – Y = \(\begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix}\) ………………….. (2)
adding eqns. (1) and (2),
2X = \(\begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix}\) + \(\begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix}\)
⇒ 2X = \(\begin{bmatrix} 5+3 & 2+6 \\ 0+0 & 9-1 \end{bmatrix}\)
⇒ 2X = \(\begin{bmatrix} 8 & 8 \\ 0 & 8 \end{bmatrix}\)
⇒ X = \(\frac{1}{2}\) \(\begin{bmatrix} 8 & 8 \\ 0 & 8 \end{bmatrix}\) = \(\begin{bmatrix} 4 & 4 \\ 0 & 4 \end{bmatrix}\)
Substracting eqn. (2) from eqn. (1),
2Y = \(\begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix}\) – \(\begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix}\)
⇒ 2Y = \(\begin{bmatrix} 5-3 & 2-6 \\ 0-0 & 9+1 \end{bmatrix}\)
⇒ Y = \(\frac{1}{2}\) \(\begin{bmatrix} 2 & -4 \\ 0 & 10 \end{bmatrix}\) = \(\begin{bmatrix} 1 & -2 \\ 0 & 5 \end{bmatrix}\)

Question 7.
Find the value of x and y
2 \(\begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix}\) + \(\begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix}\) = \(\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}\) (NCERT)
Solution:
Given:
2 \(\begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix}\) + \(\begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix}\) = \(\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}\)
⇒ \(\begin{bmatrix} 2 & 6 \\ 0 & 2x \end{bmatrix}\) + \(\begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix}\) = \(\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}\)
⇒ \(\begin{bmatrix} 2+y & 6+0 \\ 0+1 & 2x+2 \end{bmatrix}\) = \(\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}\)
⇒ \(\begin{bmatrix} 2+y & 6 \\ 1 & 2x+2 \end{bmatrix}\) = \(\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}\)
By defnition of matrix,
2 + y = 5 ⇒ y = 3
2x + 2 = 8
⇒ x + 1 = 4
⇒ x = 3
∴ x = 3, y = 3.

Question 8.
If \(\left[\begin{array}{c}
{x+y+z} \\
{x+z} \\
{y+z}
\end{array}\right]\) = [ \(\begin{matrix} 9 \\ 5 \\ 7 \end{matrix}\) ] find the value of x, y, and z?
Solution: Given \(\left[\begin{array}{c}
{x+y+z} \\
{x+z} \\
{y+z}
\end{array}\right]\) = [ \(\begin{matrix} 9 \\ 5 \\ 7 \end{matrix}\) ]
By defnition of matrix,
x + y + z = 9
x + z = 5
y + z = 7
From eqns. (1) and (2),
x + y + z = 9
⇒ 5 + y = 9 ⇒ y = 4
From eqns. (1) and (3),
x + (y + z) = 9
⇒ x + 7 = 9
⇒ x = 2
Putting the value of x in eqn. (2),
2 + z = 5
⇒ z = 3
∴ x = 2, y = 4, z = 3.

Question 9.
If \(\begin{bmatrix} x+y & 2 \\ 5+z & xy \end{bmatrix}\) = \(\begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}\), then fund the value of x, y and z? (NCERT)
Solution:
Given:
\(\begin{bmatrix} x+y & 2 \\ 5+z & xy \end{bmatrix}\) = \(\begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}\)
By defnition of matrix,
x + y = 6 ……………….. (1)
xy = 8 ………………. (2)
5 + z = 5
⇒ z = 0
From eqn. (1), y = 6 – x
xy = 8
⇒ 6x – x² = 8
⇒ x² – 6x + 8 = 0
⇒ x² – 4x – 2x + 8 = 0
⇒ x (x – 4) – 2 ( x – 4) = 0
⇒ ( x – 2) (x – 4) = 0
⇒ x = 2, 4
When x = 2 then y = 6 – 2 = 4
When x = 4 then y = 6 – 4 = 2
So x = 2, y = 4, z = 0
x = 4, y = 2, z = 0.

Question 10.
If A = \(\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}\) then prove that A² – 4A + 5I = 0?
Solution:
Given:
A = \(\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}\)
A² = A.A = \(\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}\) \(\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}\)
= \(\begin{bmatrix} 1.1+(-1)2 & 1(-1)+(-1).3 \\ 2.1+3.2 & 2(-1)+3.3 \end{bmatrix}\)
= \(\begin{bmatrix} 1-2 & -1-3 \\ 2+6 & -2+9 \end{bmatrix}\)
= \(\begin{bmatrix} -1 & -4 \\ 8 & 7 \end{bmatrix}\)
4A = 4\(\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}\) = \(\begin{bmatrix} 4 & -4 \\ 8 & 12 \end{bmatrix}\)
5I = 5\(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) = \(\begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}\)
= \(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)
= 0. Proved.

Question 11.
If A = \(\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}\) then prove that A² – 6A + 17I = 0?
Solution:
A = \(\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}\)
A² = A.A = \(\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}\) \(\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}\)
= \(\begin{bmatrix} 4-9 & -6-12 \\ 6+12 & -9+16 \end{bmatrix}\) = \(\begin{bmatrix} -5 & -18 \\ 18 & 7 \end{bmatrix}\)
∴ L.H.S = A² – 6A + 17I
= \(\begin{bmatrix} -5 & -18 \\ 18 & 7 \end{bmatrix}\) – 6 \(\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}\) + 17 \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
= \(\begin{bmatrix} -5 & -18 \\ 18 & 7 \end{bmatrix}\) – \(\begin{bmatrix} 12 & -18 \\ 18 & 24 \end{bmatrix}\) + \(\begin{bmatrix} 17 & 0 \\ 0 & 17 \end{bmatrix}\)
= \(\begin{bmatrix} -5-12+17 & -18+18+0 \\ 18-18+0 & 7-24+17 \end{bmatrix}\)
= \(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\) = 0 = R.H.S. proved.

Question 12.
If A = \(\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\) then prove that A² – 5A + 7I = 0. (NCERT)
Solution:
A² = A.A = \(\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\) × \(\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\)
⇒ A² = \(\begin{bmatrix} 3\times 3-1\times 1 & 3\times 1+1\times 2 \\ -1\times 3+2\times -1 & -1\times 1+2\times 2 \end{bmatrix}\)
⇒ A² = \(\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}\)
5A = 5\(\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\)
⇒ 5A = \(\begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}\)
7I = 7\(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
⇒ 7I = \(\begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}\)
∴ A² – 5A + 7I = \(\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}\) – \(\begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}\) + \(\begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}\)
⇒ A² – 5A + 7I = \(\begin{bmatrix} 8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7 \end{bmatrix}\)
⇒ A² – 5A + 7I = \(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)
⇒ A² – 5A + 7I = 0.

Question 13.
If A = \(\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}\) and I = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) then find the value of k if A² = kA – 2I? (NCERT)
Solution:
Given:
A² = kA – 2I
⇒ kA = A² + 2I
⇒ k \(\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}\) = \(\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}\) × \(\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}\) + 2 \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
⇒ \(\begin{bmatrix} 3k & -2k \\ 4k & -2k \end{bmatrix}\) = \(\begin{bmatrix} 3\times 3-2\times 4 & 3\times -2+2\times 2 \\ 4\times 3-2\times 4 & 4\times -2+2\times 2 \end{bmatrix}\) + \(\begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}\)
= \(\begin{bmatrix} 3k & -2k \\ 4k & -2k \end{bmatrix}\) = \(\begin{bmatrix} 1 & -2 \\ 4k & -4 \end{bmatrix}\) + 2\(\begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}\)
⇒ \(\begin{bmatrix} 3k & -2k \\ 4k & -2k \end{bmatrix}\) = \(\begin{bmatrix} 1+2 & 2+0 \\ 4+0 & -4+2 \end{bmatrix}\)
⇒ \(\begin{bmatrix} 3k & -2k \\ 4k & -2k \end{bmatrix}\) = \(\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}\)
⇒ 2k = 2
⇒ k = 1.

Question 14.
If f(x) = x² – 2x – 3, then find the value of f(A) if A = \(\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}\)
Solution:
f(x) = x² – 2x – 3
∴ f(A) = A² – 2A – 3I
⇒ f(A) = \(\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}\) × \(\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}\) – 2 \(\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}\) – 3 \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)

⇒ f(A) = \(\begin{bmatrix} 5 & 4 \\ 4 & 5 \end{bmatrix}\) – \(\begin{bmatrix} 2 & 4 \\ 4 & 2 \end{bmatrix}\) – \(\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}\)
⇒ = \(\begin{bmatrix} 5 & 4 \\ 4 & 5 \end{bmatrix}\) – \(\begin{bmatrix} 5 & 4 \\ 4 & 5 \end{bmatrix}\) = \(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)
⇒ = 0.

Question 15.
If matrix A = \(\left[\begin{array}{ccc}
{0} & {a} & {-3} \\
{2} & {0} & {-1} \\
{b} & {1} & {0}
\end{array}\right]\) skew – symmetric matrix, then find the value of ‘a’ and ‘b’? (CBSE 2018)
Solution:
If A is skew symmmetric matrix then A’ = -A
Given:
A = \(\left[\begin{array}{ccc}
{0} & {a} & {-3} \\
{2} & {0} & {-1} \\
{b} & {1} & {0}
\end{array}\right]\)
A’ = \(\left[\begin{array}{ccc}
{0} & {2} & {b} \\
{a} & {0} & {1} \\
{-3} & {-1} & {0}
\end{array}\right]\)
– A = \(\left[\begin{array}{ccc}
{0} & {-a} & {3} \\
{-2} & {0} & {1} \\
{-b} & {-1} & {0}
\end{array}\right]\)
∵ A’ = – A
\(\left[\begin{array}{ccc}
{0} & {2} & {b} \\
{a} & {0} & {1} \\
{-3} & {-1} & {0}
\end{array}\right]\) = \(\left[\begin{array}{ccc}
{0} & {-a} & {3} \\
{-2} & {0} & {1} \\
{-b} & {-1} & {0}
\end{array}\right]\)
∴ 2 = -a or a = -2
-3 = -b or b = 3.

Question 16.
If A = \(\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}\), then prove that AA^-1 = I?
Solution:
Given:
A = \(\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}\)
A^-1 = \(\frac { adjA }{ |A| } \)
|A| = |\(\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}\)|
= cos²α – (-sin² α)
= cos²α + sin²α = 1 ………………………… (1)
∴|A| = 1
We know adj A = \(\begin{bmatrix} C_{ 11 } & C_{ 21 } \\ C_{ 12 } & C_{ 22 } \end{bmatrix}\)
Where C₁₁ = (-1)² cos α = cos α
C₁₂ = (-1)³ sin α = -sin α
C₂₁ = (-1)³ (-sin α ) = sin α
C₂₂ = (-1)⁴cos α = cos α
∴ adj A = \(\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}\) …………….. (2)
∴ A^-1 = \(\frac { adjA }{ |A| } \) = \(\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}\)
∴ A.A^-1 = \(\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}\) \(\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}\)

= \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) = I. Proved.

Question 17.
If A = \(\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}\) then prove that:
A. (Adj A) = |A| I?
Solution:
Given A = \(\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}\)
A₁₁ = cos α, A₁₂ = sin α, A₂₁ = – sin α, A₂₂ = cos α
Adj A = \(\begin{bmatrix} A_{ 11 } & A_{ 21 } \\ A_{ 12 } & A_{ 22 } \end{bmatrix}\) = \(\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}\)
A.(Adj A) = \(\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}\) \(\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}\)

∴ A.(Adj A) = I = |A| I. Proved.

Question 18.
Prove that the square matrix A = \(\begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}\) is orthogonal matrix?
Solution:
Given:
A = \(\begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}\)
∴ A’ = \(\begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}\)
Now A.A’ = \(\begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}\) \(\begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}\)

= \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) = I
Similarly A’.A = I
Then A.A’ = A’A = I
So A is orthogonal matrix.
Proved.

Question 19.
If A = \(\begin{bmatrix} 3 & 2 \\ 7 & 5 \end{bmatrix}\) and B = \(\begin{bmatrix} 6 & 7 \\ 8 & 9 \end{bmatrix}\), then find the value of (AB)^-1?
Solution:
Given:
A = \(\begin{bmatrix} 3 & 2 \\ 7 & 5 \end{bmatrix}\), B = \(\begin{bmatrix} 6 & 7 \\ 8 & 9 \end{bmatrix}\)
= \(\begin{bmatrix} 3.6+2.8 & 3.7+2.9 \\ 7.6+5.8 & 7.7+5.9 \end{bmatrix}\) = \(\begin{bmatrix} 34 & 39 \\ 82 & 94 \end{bmatrix}\)
|AB| = \(\begin{bmatrix} 34 & 39 \\ 82 & 94 \end{bmatrix}\)
= \(\begin{bmatrix} 34 & 39 \\ 82 & 94 \end{bmatrix}\) = 3196 – 3198 = -2 ≠ 0
AB₁₁ = 94, AB₁₂ = -82, AB₂₁ = -39, AB₂₂ = 34
adj AB = \(\begin{bmatrix} 94 & -39 \\ -82 & 34 \end{bmatrix}\)
(AB)^-1 = \(\frac { adjAB }{ |AB| } \) = \(\frac{1}{ -2}\) \(\begin{bmatrix} 94 & -39 \\ -82 & 34 \end{bmatrix}\)
= \(\begin{bmatrix} -47 & \frac { 39 }{ 2 } \\ 41 & -17 \end{bmatrix}\)

Question 20.
If A = \(\begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix}\) then prove that:
2A^-1 = 9I – A? (CBSE 2018)
Solution:
Given:
A = \(\begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix}\)
∴ A₁₁ = 7, A₁₂ = – (- 4), A₂₁ = – (- 3), A₂₂ = 2
∴ adj A = \(\begin{bmatrix} A_{ 11 } & A_{ 21 } \\ A_{ 12 } & A_{ 22 } \end{bmatrix}\) = \(\begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix}\)
Now |A| = \(\begin{vmatrix} 2 & -3 \\ -4 & 7 \end{vmatrix}\) = 14 – 12 = 2
∴ A^-1 = \(\frac { adjA }{ |A| } \) = \(\frac{1}{2}\) \(\begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix}\)
⇒ 2A^-1 = \(\begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix}\)
Again, 9I – A = 9 \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) – \(\begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix}\)
= \(\begin{bmatrix} 9 & 0 \\ 0 & 9 \end{bmatrix}\) – \(\begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix}\) = \(\begin{bmatrix} 9-2 & 0+3 \\ 0+4 & 9-7 \end{bmatrix}\)
∴ 9I – A = \(\begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix}\)
From eqns. (1) and (2),
2A^-1 = 9I – A. proved.

Question 21.
For matrix A and B prove that (AB)’ = B’A’ where A = [ \(\begin{matrix} 1 \\ -4 \\ 3 \end{matrix}\) ], B = [-1 2 1]? (NCERT)
Solution:
AB = [ \(\begin{matrix} 1 \\ -4 \\ 3 \end{matrix}\) ]_3×1 [-1 2 1]_1×3
AB = \(\left[\begin{array}{rrr}
{-1 \times 1} & {1 \times 2} & {1 \times 1} \\
{-4 \times-1} & {-4 \times 2} & {-4 \times 1} \\
{-3 \times 1} & {3 \times 2} & {3 \times 1}
\end{array}\right]\)
⇒ AB = [ \(\begin{matrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{matrix}\) ]
(AB)’ = [ \(\begin{matrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{matrix}\) ] …………………… (1)
B = [ -1 2 1]’
B’ = [ \(\begin{matrix} -1 \\ 2 \\ 1 \end{matrix}\) ]
A’ = [ \(\begin{matrix} -1 \\ -4 \\ 3 \end{matrix}\) ]
A’ = [1 -4 3]_{1 ×3}
B’A’ = [ \(\begin{matrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{matrix}\) ]
From eqns. (1) and (2),
(AB)’ = B’A’. proved.

Matrices Long Answer Type Questions – II

Question 1.
If A = \(\begin{bmatrix} 1 & 4 \\ 3 & 5 \end{bmatrix}\) then prove that:
A.adj A = (adj A). A = |A| I?
Solution:
Given:
A = \(\begin{bmatrix} 1 & 4 \\ 3 & 5 \end{bmatrix}\)
Then, |A| = \(\begin{bmatrix} 1 & 4 \\ 3 & 5 \end{bmatrix}\) = 5 – 12 = -7
A₁₁ = 5, A₁₂ = -3, A₂₁ = -4, A₂₂ = 1
∴ adj A = \(\begin{bmatrix} A_{ 11 } & A_{ 21 } \\ A_{ 12 } & A_{ 22 } \end{bmatrix}\) = \(\begin{bmatrix} 5 & -4 \\ -3 & 1 \end{bmatrix}\)
⇒ A. adj A = \(\begin{bmatrix} 1 & 4 \\ 3 & 5 \end{bmatrix}\) × \(\begin{bmatrix} 5 & -4 \\ -3 & 1 \end{bmatrix}\)

⇒ A.adj A = \(\begin{bmatrix} -7 & 0 \\ 0 & -7 \end{bmatrix}\) = -7\(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
∴ A.adjA = |A| I
and (adj A) . A = \(\begin{bmatrix} 5 & -4 \\ -3 & 1 \end{bmatrix}\) × \(\begin{bmatrix} 1 & 4 \\ 3 & 5 \end{bmatrix}\)

⇒ (adj A).A = \(\begin{bmatrix} -7 & 0 \\ 0 & -7 \end{bmatrix}\)
⇒ (adj A).A = – 7 \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
⇒ (adj A).A = |A| I
From eqns. (1) and (2),
A.adj A = (adj A) .A = |A| I. Proved.

Question 2.
If A = \(\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\) then prove that:
A.(adj A) = (adj A). A = |A| I?
Solution:
Solve like Q.No. 1.

Question 3.
If matrix A = [ \(\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix}\) ] then prove that: A^-1 = A?
Solution:
Given:
A = [ \(\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix}\) ]
By formula A^-1 = \(\frac { adjA }{ |A| } \)
Then, |A| = [ \(\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix}\) ] = 1.(0-1) = -1
A₁₁ = (-1)² (0 – 0) = 0
A₁₂ = (-1)³ (0 – 0) = 0
A₁₃ = (-1)⁴ ( 0 – 1) = -1, A₂₁ = (-1)³ (0 – 0) = 0
A₂₂ = (-1)⁴ ( 0 – 1) = -1, A₂₃ = (-1)⁵ (0 – 0) = 0
A₃₁ = (-1)⁴ ( 0 – 1) = -1, A₃₂ = (-1)⁵ (0 – 0) = 0
A₃₃ = (-1)⁶ ( 0 – 0) = 0
adj A = \(\left[\begin{array}{ccc}
{A_{11}} & {A_{21}} & {A_{31}} \\
{A_{12}} & {A_{22}} & {A_{32}} \\
{A_{13}} & {A_{23}} & {A_{33}}
\end{array}\right]\) = [ \(\begin{matrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{matrix}\) ]
= [- \(\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix}\) ]
∴ image 14 = [ \(\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix}\) ]
∴ A^-1 = A.

Question 4.
Matrix A = [ \(\begin{matrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{matrix}\) ], find inverse of matrix A?
Solution:
Given A = [ \(\begin{matrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{matrix}\) ]
∴ |A| = 2 (4 × 2 – 7 × 1) – 3 (3 × 2 – 3 × 1) + 1 (3 × 7 – 3 × 4)
= 2 ( 8 – 7) – 3 ( 6 – 3) + 1 (21 – 12)
= 2(1) – 3(3) + 1(9)
= 2 – 9 + 9 = 2
If |A| ≠ 0 then A^-1 will exist.
Now, A₁₁ = + (8 – 7) = 1, A₁₂ = – (6 – 3) = -3
A₁₃ = + (21 – 12), A₂₁ = – (6 – 7) = 1
A₂₂ = +(4 – 3) = 1, A₂₃ = – (14 – 9) = -5
A₃₁ = + (3 – 4) = -1, A₃₂ = – (2 – 3) = 1
A₃₃ = + (8 – 9) = -1
∴ adj A = \(\left[\begin{array}{ccc}
{A_{11}} & {A_{21}} & {A_{31}} \\
{A_{12}} & {A_{22}} & {A_{32}} \\
{A_{13}} & {A_{23}} & {A_{33}}
\end{array}\right]\) = [ \(\begin{matrix} 1 & 1 & -1 \\ -3 & 1 & 1 \\ 9 & -5 & -1 \end{matrix}\) ]
∵ A^-1 = \(\frac { adjA }{ |A| } \)
∴A^-1 = \(\frac{1}{2}\) [ \(\begin{matrix} 1 & 1 & -1 \\ -3 & 1 & 1 \\ 9 & -5 & -1 \end{matrix}\) ]
⇒ A^-1 = \(\left[\begin{array}{ccc}
{\frac{1}{2}} & {\frac{1}{2}} & {\frac{-1}{2}} \\
{\frac{-3}{2}} & {\frac{1}{2}} & {\frac{1}{2}} \\
{\frac{9}{2}} & {\frac{-5}{2}} & {\frac{-1}{2}}
\end{array}\right]\)

Question 5.
If A = [ \(\begin{matrix} 1 & 2 & 3 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix}\) ], then find the value of A^-1?
Solution:
Solve like Q.No.4.
Answer:
[ \(\begin{matrix} 1 & -3 & 2 \\ -3 & 3 & -1 \\ 2 & -1 & 0 \end{matrix}\) ]

Question 6.
If A = [ \(\begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix}\) ], then find the value of A^-1?
Solution:
Given:
A = [ \(\begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix}\) ]
|A| = [ \(\begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix}\) ]
⇒ |A| = 1 (1 – 4) + 2 ( 4 – 2) + 2 ( 4 – 2)
= – 3 + 4 + 4 = 5
If |A| ≠ 0, then A^-1 exists
A₁₁ = \(\begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix}\) = 1 – 4 = -3
A₁₂ = – \(\begin{vmatrix} 2 & 2 \\ 2 & 1 \end{vmatrix}\) = – ( 2 – 4) = 2
A₁₃ = \(\begin{vmatrix} 2 & 1 \\ 2 & 2 \end{vmatrix}\) = 4 – 2 = 2
A₂₁ = – \(\begin{vmatrix} 2 & 2 \\ 2 & 1 \end{vmatrix}\) = – ( 2 – 4) = 2
A₂₂ = \(\begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix}\) = 1 – 4 = – 3
A₂₃ = – \(\begin{vmatrix} 1 & 2 \\ 2 & 2 \end{vmatrix}\) = – ( 2 – 4) = 2
A₃₁ = \(\begin{vmatrix} 2 & 2 \\ 1 & 2 \end{vmatrix}\) = 4 – 2 = 2
A₃₂ = – \(\begin{vmatrix} 1 & 2 \\ 2 & 2 \end{vmatrix}\) = – (2 – 4) = 2
A₃₃ = \(\begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix}\) = 1 – 4 = -3
∴adj A = \(\left[\begin{array}{ccc}
{A_{11}} & {A_{21}} & {A_{31}} \\
{A_{12}} & {A_{22}} & {A_{32}} \\
{A_{13}} & {A_{23}} & {A_{33}}
\end{array}\right]\)
= [ \(\begin{matrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{matrix}\) ]
∴A^-1 = \(\frac { adjA }{ |A| } \) = \(\frac{1}{5}\) [ \(\begin{matrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{matrix}\) ]

Question 7.
If A = \(\begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix}\) then prove that: A² – 2A + 3I = 0?
Solution:
Given:
A = \(\begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix}\)
A² = A.A = \(\begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix}\) \(\begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix}\)
= \(\begin{bmatrix} 2.2+3(-1) & 2.3+3.0 \\ -1.2+0(-1) & -1.3+0.0 \end{bmatrix}\)
= \(\begin{bmatrix} 1 & 6 \\ -2 & -3 \end{bmatrix}\)
2A = \(\begin{bmatrix} 4 & 6 \\ -2 & 0 \end{bmatrix}\)
3I = \(\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}\)
∴ A² – 2A + 3I = \(\begin{bmatrix} 1 & 6 \\ -2 & -3 \end{bmatrix}\) – \(\begin{bmatrix} 4 & 6 \\ -2 & 0 \end{bmatrix}\) + \(\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}\)
= \(\begin{bmatrix} 4 & 6 \\ -2 & 0 \end{bmatrix}\) – \(\begin{bmatrix} 4 & 6 \\ -2 & 0 \end{bmatrix}\) = \(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)
= 0. Proved.

Question 8.
If A = \(\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}\) then prove A² – 6A + 17I = 0 and find A^-1?
Solution:
Given:
A = \(\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}\)
A² = A.A = \(\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}\) \(\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}\)
= \(\begin{bmatrix} 4-9 & -6-12 \\ 6+12 & -9+16 \end{bmatrix}\) = \(\begin{bmatrix} -5 & -18 \\ 18 & 7 \end{bmatrix}\)
∴L.H.S = A² – 6A + 17I
= \(\begin{bmatrix} -5 & -18 \\ 18 & 7 \end{bmatrix}\) – 6 \(\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}\) + 17\(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
= \(\begin{bmatrix} -5 & -18 \\ 18 & 7 \end{bmatrix}\) – \(\begin{bmatrix} 12 & -18 \\ 18 & 24 \end{bmatrix}\) + \(\begin{bmatrix} 17 & 0 \\ 0 & 17 \end{bmatrix}\)
= \(\begin{bmatrix} -5-12+17 & -18+18+0 \\ 18-18+0 & 7-24+17 \end{bmatrix}\) = \(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)
= 0 = R.H.S. Proved.
Then |A| = \(\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}\) = 8 + 9 = 17
∴ A₁₁ = (-1)¹⁺¹ 4 = 4
A₁₂ = (-1)¹⁺² (3) = -3
A₂₁ = (-1)²⁺¹ (-3) = 3
A₂₂ = (-1)²⁺² (2) = 2.
and adj A = \(\begin{bmatrix} A_{ 11 } & A_{ 12 } \\ A_{ 12 } & A_{ 22 } \end{bmatrix}\) = \(\begin{bmatrix} 4 & 3 \\ -3 & 2 \end{bmatrix}\)
∴A^-1 = \(\frac { adjA }{ |A| } \)
= \(\frac{1}{17}\) \(\begin{bmatrix} 4 & 3 \\ -3 & 2 \end{bmatrix}\)

Question 9.
If A = \(\begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix}\), then prove A² + 4A – 42 I = 0 and find A^-1?
Solution:
Solve like Q.No. 8.
Answer:
A^-1 = \(\frac{1}{42}\) \(\begin{bmatrix} -4 & 5 \\ 2 & 0 \end{bmatrix}\).

Question 10.
(A) Solve the following equations by matrix method:
x + y + z = 3
2x – y + z = 2
x – 2y + 3z = 2.
Solution:
If A = [ \(\begin{matrix} 1 & 1 & 1 \\ 2 & -1 & 1 \\ 1 & -2 & 3 \end{matrix}\) ], X = [ \(\begin{matrix} x \\ y \\ z \end{matrix}\) ] and B = [ \(\begin{matrix} 3 \\ 2 \\ 2 \end{matrix}\) ]
|A| = \(\left|\begin{array}{ccc}
{1} & {1} & {1} \\
{2} & {-1} & {1} \\
{1} & {-2} & {3}
\end{array}\right|\)
= 1( – 3 + 2) – 2 (3 + 2) + 1(1 + 1)
= -1 – 10 + 2 = -9
A₁₁ = \(\begin{vmatrix} -1 & 1 \\ -2 & 3 \end{vmatrix}\) = – 3 + 2 = – 1
A₁₂ = – \(\begin{vmatrix} 2 & 1 \\ 1 & 3 \end{vmatrix}\) = – ( 6 – 1) = – 5
A₁₃ = \(\begin{vmatrix} 2 & -1 \\ 1 & -2 \end{vmatrix}\) = – 4 + 1 = – 3
A₂₁ = – \(\begin{vmatrix} 1 & 1 \\ -2 & 3 \end{vmatrix}\) = – (3 + 2) = – 5
A₂₂ = \(\begin{vmatrix} 1 & 1 \\ 1 & 3 \end{vmatrix}\) = 3 -1 = 2
A₂₃ = – \(\begin{vmatrix} 1 & 1 \\ 1 & -2 \end{vmatrix}\) = – ( -2 -1) = 3
A₃₁ = \(\begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix}\) = 1 + 1 = 2
A₃₂ = – \(\begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}\) = – (1 – 2) = 1
A₃₃ = \(\begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix}\) = -1 -2 = -3
and adj A = [ \(\begin{matrix} -1 & -5 & 2 \\ 5 & 2 & 1 \\ -3 & 3 & -3 \end{matrix}\) ]
∴ A^-1 = \(\frac { adjA }{ |A| } \)
⇒ A^-1 = \(\frac{-1}{9}\) [ \(\begin{matrix} -1 & -5 & 2 \\ 5 & 2 & 1 \\ -3 & 3 & -3 \end{matrix}\) ]
⇒ A^-1 = image 14
X = A^-1B

∴ x = 1, y = 1, z = 1.

Question 10.
(B) Solve the following equations by matrix method:
x + y + z = 6
x + 2y = 3z = 14
x + 4y + 9z = 36
Solution:
x + y + z = 6
x + 2y + 3z = 14
x + 4y + 9z = 36.
Where
A = [ \(\begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 9 \end{matrix}\) ], B = [ \(\begin{matrix} 6 \\ 14 \\ 36 \end{matrix}\) ] , X = [ \(\begin{matrix} x \\ y \\ z \end{matrix}\) ]
= 6 – 6 + 2 = 2


X = A^-1 B
⇒ [ \(\begin{matrix} x \\ y \\ z \end{matrix}\) ] = [ \(\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}\) ]
∴ x = 1, y = 2, z = 3.

Question 11.
If A’ = \(\left[\begin{array}{rr}
{3} & {4} \\
{-1} & {2} \\
{0} & {1}
\end{array}\right]\) and B = \(\left[\begin{array}{rrr}
{-1} & {2} & {1} \\
{1} & {2} & {3}
\end{array}\right]\) then prove the following:
(i) (A + B)’ = A’ + B’
(ii) (A – B)’ = A’ – B’. (NCERT)
Solution:
(i) Given

Question 12.
If A = \([latex]\left[\begin{array}{rrr}
{-1} & {2} & {3} \\
{5} & {7} & {9} \\
{-2} & {1} & {1}
\end{array}\right]\)[/latex] and B = \(\left[\begin{array}{rrr}
{-4} & {1} & {-5} \\
{1} & {2} & {0} \\
{1} & {3} & {1}
\end{array}\right]\) then prove that:
(i) (A + B)’ = A’ + B’
(ii) (A – B)’ = A’ – B’. (NCERT)
Solution:
solve like Q.No.11.

Question 13.
Express matrix A = \(\begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}\) as sum of a symmetric and a skew symmetric matrix? (NCERT)
Solution:

Given:

Eqn. (1) and symmetric matrix and eqn. (2) is skew symmetric matrix.

Question 14.
(A) By using elementary operations, find the inverse of matrix A = \(\begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix}\)
Solution:
Using A = AI

Question 14.
(B) By using elementary operation, find the inverse of matrix A = \(\begin{bmatrix} 3 & 10 \\ 2 & 7 \end{bmatrix}\)?
Solution:
Solve like Q.No. 14 (A).
Answer:
\(\begin{bmatrix} 7 & -10 \\ -2 & 3 \end{bmatrix}\)

Question 15.
Solve the following system of equations by using matrix method: (NCERT, CBSE 2011)
\(\frac{2}{x}\) + \(\frac{3}{y}\) + \(\frac{10}{z}\) = 4
\(\frac{4}{x}\) – \(\frac{6}{y}\) + \(\frac{5}{z}\) = 1
\(\frac{6}{x}\) + \(\frac{9}{y}\) – \(\frac{20}{z}\) = 2, x, y, z, ≠ 0.
Solution:
Let \(\frac{1}{x}\) = u,
\(\frac{1}{y}\) = v
\(\frac{1}{z}\) = w, then
2u + 3v + 10w = 4
4u – 6v + 5w = 1
6u + 9v – 20 w = 2
Applying formula AX = B
where

⇒ |A| = 2 × (120 – 45) -3 (- 80 – 30) + 10 (36 + 36)
⇒ |A| = 150 + 330 + 720 = 1200
⇒ |A| ≠ 0, hence A^-1 exists.
Applying formula
X = A^-1B


As \(\frac{1}{x}\) = u, \(\frac{1}{y}\) = v, and \(\frac{1}{z}\) = \(\frac{1}{z}\) = w
∴\(\frac{1}{x}\) = \(\frac{1}{2}\), \(\frac{1}{y}\) = \(\frac{1}{3}\) and \(\frac{1}{z}\) = \(\frac{1}{5}\)
⇒ x = 2, y = 3 and z = 5. is required solution.

Question 16.
Find A^-1 where A = \(\left[\begin{array}{ccc}
{1} & {2} & {-3} \\
{2} & {3} & {2} \\
{3} & {-3} & {-4}
\end{array}\right]\), corresponding equation is:
x + 2y – 3z = -4
2x + 3y + 2z = 2
3x – 3y – 4z = 11. (CBSE 2008, 10, 12)
Solution:
Given:
A = \(\left[\begin{array}{ccc}
{1} & {2} & {-3} \\
{2} & {3} & {2} \\
{3} & {-3} & {-4}
\end{array}\right]\)
∴ |A| = \(\left[\begin{array}{ccc}
{1} & {2} & {-3} \\
{2} & {3} & {2} \\
{3} & {-3} & {-4}
\end{array}\right]\)
⇒ |A| = 1(- 12 + 6) – 2 (- 8 – 6) – 3 (- 6 – 9)
⇒ |A| = 1 (-12 + 6) -2 (-8 -6) -3 (-6 -9)
⇒ |A| ≠ 0
⇒ Hence A^-1 exists.
Hence



Equation of above matrix
x + 2y – 3z = – 4
2x + 3y + 2z = 2
3x – 3y – 4z = 11
Solution of above equations
AX = B
Where A = \(\left[\begin{array}{ccc}
{1} & {2} & {-3} \\
{2} & {3} & {2} \\
{3} & {-3} & {-4}
\end{array}\right]\), X = [ \(\begin{matrix} x \\ y \\ z \end{matrix}\) ] , B = [ \(\begin{matrix} -4 \\ 2 \\ 11 \end{matrix}\) ]
Applying formula
X = A^-1B

Question 17.
Find the product of matrix \(\left[\begin{array}{ccc}
{-4} & {4} & {4} \\
{-7} & {1} & {3} \\
{5} & {-3} & {-1}
\end{array}\right]\) \(\left[\begin{array}{ccc}
{1} & {-1} & {1} \\
{1} & {-2} & {-2} \\
{2} & {1} & {3}
\end{array}\right]\) and with the help of product of matrix solve the equations? (CBSE 2012)
x – y + z = 4
x – 2y – 2z = 9
2x + y + 3z = 1.
Solution:
Let

Writing above equation in matrix form
AX = C
Where


⇒ x = 3, y = -2, z = -1, (by equating of two matrix)

Question 18.
The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs. 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs. 90. The cost of 6 kg onion, 2 kg wheat and 3 kg rice is Rs. 70. Find the cost of each item per kg by matrix method?
Solution:
Let the cost of 1 kg onion = Rs. x
1 kg wheat = Rs. y
and 1 kg rice = Rs. z
According to equation
4x + 3y + 2z = 60
2x + 4y + 6z = 90
6x + 2y + 3z = 70
Matrix form will be
AX = B
Where

⇒ Hence A^-1 exist.
Applying formula
X = A^-1 B



∴ x = Rs. 5, y = Rs. 8, z = Rs. 8.

MP Board Class 12 Maths Important Questions

 

MP Board Class 12th Maths Important Questions Chapter 9 अवकल समीकरण

अवकल समीकरण Important Questions

अवकल समीकरण वस्तुनिष्ठ प्रश्न

प्रश्न 1.
सही विकल्प चुनकर लिखिए –

प्रश्न 1.
अवकल समीकरण \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) + x² \(\frac{dy}{dx}\) = e^x की घात है –
(a) 1
(b) 2
(c) 3
(d) अस्तित्व नहीं है।
उत्तर:
(c) 3

प्रश्न 2.
अवकल समीकरण (1 + x) y dx + (1 – y) x dy = 0 का हल होगा –
(a) log xy + x + y = c
(b) log y + x – y = c
(c) log xy – x – y = c
(d) log xy – x + y = c.
उत्तर:
(b) log y + x – y = c

प्रश्न 3.
उन सभी वृत्तों का अवकल समीकरण जो मूलबिन्दू से गुजरते हैं तथा जिनके केन्द्र X – अक्ष पर स्थित है –
(a) x² = y² + xy \(\frac{dy}{dx}\)
(b) x² = y² + 3xy \(\frac{dy}{dx}\)
(c) y² = x² + 2xy \(\frac{dy}{dx}\)
(d) y² = x² – 2xy \(\frac{dy}{dx}\)
उत्तर:
(c) y² = x² + 2xy \(\frac{dy}{dx}\)

प्रश्न 4.
अवकल समीकरण \(\frac{dy}{dx}\) + y = e^-x, y(0) = 0 का हल होगा –
(a) y = e^-x (x -1)
(b) y = xe^x
(c) y = xe^-x + 1
(d) y = xe^-x
उत्तर:
(d) y = xe^-x

प्रश्न 5.
सरल रेखा जो अवकल समीकरण \(\frac{dy}{dx}\) = m को संतुष्ट करती हो तथा Y – अक्ष पर धनात्मक दिशा में 3 अन्तःखण्ड काटती हो, है –
(a) y = mx + c
(b) = mx + 3
(c) y = mx – 3
(d) y = – mx + 3
उत्तर:
(b) = mx + 3

प्रश्न 2.
रिक्त स्थानों की पूर्ति कीजिये –

1. समीकरण x² + y² = a² के संगत अवकल समीकरण ……………………….. है।
2. वक्र y = e^cx से संबंधित अवकल समीकरण …………………………….. है, जहाँ c स्वेच्छ अचर है।
3. रेखीय अवकल समीकरण \(\frac{dy}{dx}\) + Py = Q में समाकलन गुणांक …………………………. है।
4. रेखीय अवकल समीकरण \(\frac{dy}{dx}\) + Py = Q में P और …………………………… हैं।
5. अवकल समीकरण (x + y + 1) dy = dx ………………………… रूप का है।
6. अवकल समीकरण e^-x+y \(\frac{dy}{dx}\) = 1 का हल हो ……………………… है।

उत्तर:

1. y \(\frac{dy}{dx}\) + x = 0
2. x\(\frac{dy}{dx}\) = y log y
3. e^∫pdx
4. अचर
5. रेखीय अवकल समीकरण

प्रश्न 3.
निम्न कथनों में सत्य/असत्य बताइए –

1. अवकल समीकरण y = x ( \(\frac{dy}{dx}\) )² + \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) की कोटि 2 है।
2. अवकल समीकरण ( \(\frac { d^{ 3 }y }{ dx^{ 3 } } \) )^4/5 – 2 ( \(\frac{dy}{dx}\) ) ( \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) )² = 0 की घात 5 है।
3. अवकल समीकरण x \(\frac{dy}{dx}\) – y = 2x² का समाकलन गुणांक e^-x है।
4. अवकल समीकरण dy = sin x dx का हल y + cos x – c = 0 है।
5. अवकल समीकरण ydx + (x – y³) dy = 0 का हल xy = \(\frac { y^{ 4 } }{ 4 } \) + c है।

उत्तर:

1. सत्य
2. असत्य
3. असत्य
4. सत्य
5. सत्य।

प्रश्न 4.
एक शब्द/वाक्य में उत्तर दीजिए –

1. अवकल समीकरण (1 + y²) + (2xy – cot y) \(\frac{dy}{dx}\) = 0 का समाकल गुणांक लिखिए।
2. अवकल समीकरण (1 + x²)dy = (1 + y²) dx का हल ज्ञात कीजिए।
3. अवकल समीकरण y = x( \(\frac{dy}{dx}\) )² + \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) की कोटि व घात का योग लिखिए।
4. अवकल समीकरण dy = sin x dx का हल y + cos x – c = 0 है।
5. अवकल समीकरण \(\frac{dy}{dx}\) + \(\frac{1}{x}\) = \(\frac { e^{ y } }{ x^{ 2 } } \) का हल ज्ञात कीजिए।

उत्तर:

1. 1 + y²
2. x – y = c (1 + xy)
3. 3
4. log x
5. 2xe^-y = cx² + 1.

अवकल समीकरण अति लघु उत्तरीय प्रश्न

प्रश्न 1.
अवकल समीकरण \(\frac{dy}{dx}\) + y = e^-x की कोटि तथा घात ज्ञात कीजिये।
उत्तर:
1, 1

प्रश्न 2.
अवकल समीकरण ( \(\frac{dy}{dx}\) )³ = \(\sqrt { 1+(\frac { dy }{ dx } )^{ 2 } } \) की कोटि तथा घात ज्ञात कीजिये।
उत्तर:
1, 6

प्रश्न 3.
अवकल समीकरण \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) + \(\sqrt { 1+(\frac { dy }{ dx } )^{ 3 } } \) = 0 की कोटि तथा घात ज्ञात कीजिये।
उत्तर:
2, 2

प्रश्न 4.
वृत्त के समीकरण x² + y² = a² के संगत अवकल समीकरण क्या होगा?
उत्तर:
y \(\frac{dy}{dx}\) + x = 0

प्रश्न 5.
सरल रेखा y = mx + c के लिये अवकल समीकरण बनाइये।
उत्तर:
\(\frac{dy}{dx}\) = m

प्रश्न 6.
अवकल समीकरण \(\frac{dy}{dx}\) = 4y को हल कीजिये।
उत्तर:
y = c.e^4x

प्रश्न 7.
x² \(\frac{dy}{dx}\) = 2 का व्यापक हल ज्ञात कीजिये।
उत्तर:
y = c – \(\frac{2}{x}\)

प्रश्न 8.
अवकल समीकरण dy = sin x dx का हल ज्ञात कीजिये।
उत्तर:
y + cos x = c

प्रश्न 9.
\(\frac{dy}{dx}\) + Px = Q रूप वाले अवकल समीकरण का व्यापक हल ज्ञात कीजिये।
उत्तर:
xe^∫pdy = ∫Q.e^∫pdy.dy + c

प्रश्न 10.
अवकल समीकरण (1 – y²) \(\frac{dy}{dx}\) + yx = ay का समाकल गुणांक ज्ञात कीजिये।
उत्तर:
\(\frac { 1 }{ \sqrt { 1-y^{ 2 } } } \)

अवकल समीकरण लघु उत्तरीय प्रश्न

प्रश्न 1.
अवकल समीकरण x log x dy – y dx = 0?
हल:
दिया है:
⇒ xlog x dy = y dx
⇒ \(\frac{1}{y}\) \(\frac{dy}{dx}\) = \(\frac{1}{xlogx}\) dx
⇒ ∫\(\frac{1}{y}\) \( dy = ∫[latex]\frac{1}{xlogx}\) dx
⇒ log y = ∫\(\frac{1}{t}\) dt (माना log x = t, \(\frac{1}{x}\) \( dx = dt)
⇒ log y = log t + log c
⇒ log y = log log x + log c

प्रश्न 2.
अवकल समीकरण dy/dx = e^x-y + x.e^-y को हल कीजिए।
हल:
दिया है:
[latex]\frac{dy}{dx}\) = e^x-y + x.e^-y
⇒ \(\frac{dy}{dx}\) = e^-y(e^x + x)
⇒ e^y dy = (e^x + x) dx
दोनों पक्षों का समाकलन करने पर,
∫e^y dy = ∫(e^x + x) dx
e^y = e^x + \(\frac { x^{ 2 } }{ 2 } \) + c

प्रश्न 3.
सिद्ध कीजिए कि y = 4 sin 3x अवकल समीकरण \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) + 9y = 0 का एक हल है।
हल:
दिया है:
y = 4 sin 3x ……………… (1)
x के सापेक्ष अवकलन करने पर,
∴ \(\frac{dy}{dx}\) = 12 cos 3x
पुनः x के सापेक्ष अवकलन करने पर,
\(\frac { d^{ 2 }y }{ dx^{ 2 } } \) = -36 sin 3x = -9 × 4 sin 3x
⇒ \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) = -9y, [समी. (1) से]
⇒ \(\frac { d^{ 2 }y }{ dx^{ 2 } } \)+ 9y = 0 यही सिद्ध करना था।

प्रश्न 4.
अवकल समीकरण \(\frac{dy}{dx}\) = sec x (sec x + tan x) का हल ज्ञात कीजिए।
हल:
दिया गया अवकल समीकरण है:
\(\frac{dy}{dx}\) = sec x (sec x + tan x)
⇒ dy = (sec²x + sec x tan x) dx
⇒ ∫dy = ∫sec²x dx + ∫sec x tan x dx
∴ y = tan x + sec x + c

प्रश्न 5.
अवकल समीकरण \(\frac{dy}{dx}\) = sec² x + 3x² को हल कीजिए।
हल:
\(\frac{dy}{dx}\) = sec² x + 3x²
⇒ dy = (sec² x + 3x²) dx
⇒ ∫dy = ∫sec² x dx + 3∫x² dx
⇒ y = tan x + \(\frac { 3x^{ 3 } }{ 3 } \) + c
⇒ y = tan x + x³ + c

प्रश्न 6.
अवकल समीकरण \(\frac{dy}{dx}\) = sec² x + 2x का हल ज्ञात कीजिए।
हल:
प्रश्न क्र. 5 की भाँति हल करें।

प्रश्न 7.
अवकल समीकरण \(\frac{dy}{dx}\) = (3x² + 2) को हल कीजिए।
हल:
दिया है:
\(\frac{dy}{dx}\) = (3x² + 2)
⇒ dy = (3x² + 2) dx
⇒ ∫dy = ∫(3x² + 2) dx
⇒ y = 3 × \(\frac { x^{ 3 } }{ 3 } \) + 2x + c = x³ + 2x + c

प्रश्न 8.
अवकल समीकरण x² \(\frac{dy}{dx}\) = 2 को हल कीजिए।
उत्तर:
x² \(\frac{dy}{dx}\) = 2
⇒ dy = 2.x^-2 dx
⇒ ∫dy = 2∫x^-2 dx
⇒ y = 2( \(\frac{-1}{x}\) ) + c

प्रश्न 9.
अवकल समीकर \(\frac{dy}{dx}\) = x³ + sin 4x का हल ज्ञात कीजिये।
दिया है:
\(\frac{dy}{dx}\) = x³ + sin 4x
⇒ dy = (x³ + sin 4x) dx
⇒ ∫dy = ∫x³ dx + ∫sin 4x dx
⇒ y = \(\frac { x^{ 4 } }{ 4 } \) + ( \(\frac { -cos4x }{ 4 } \) ) + c
⇒ y = \(\frac { x^{ 4 } }{ 4 } \) – \(\frac{cos4x}{4}\) + c

प्रश्न 10.
अवकल समीकरण \(\frac{dy}{dx}\) + 2x = e^3x का हल ज्ञात कीजिये।
हल:
दिया है:
\(\frac{dy}{dx}\) + 2x = e^3x
⇒ \(\frac{dy}{dx}\) = e^3x – 2x
⇒ dy = (e^3x – 2x) dx
⇒ ∫dy = ∫e^3x dx – 2∫x dx
⇒ y = e^3x. \(\frac{1}{3}\) – \(\frac { 2x^{ 2 } }{ 2 } \) + c = \(\frac{1}{3}\) e^3x – x² + c

प्रश्न 11.
अवकल समीकरण \(\frac{dy}{dx}\) = \(\frac { cos^{ 2 }y }{ sin^{ 2 }x } \) का हल ज्ञात कीजिये।
हल:
\(\frac{dy}{dx}\) = \(\frac { cos^{ 2 }y }{ sin^{ 2 }x } \)
⇒ \(\frac { 1 }{ cos^{ 2 }y } \) dy = \(\frac { 1 }{ sin^{ 2 }x } \) dx
⇒ sec² ydy = cosec² xdx
⇒ ∫sec² ydy = ∫cosec² xdx
⇒ tan y = – cot x + c

प्रश्न 12.
अवकल समीकरण (x² + 1) \(\frac{dy}{dx}\) = 1 को हल कीजिये।
हल:
दिया है:
(x² + 1) \(\frac{dy}{dx}\) = 1
⇒ \(\frac{dy}{dx}\) = \(\frac { 1 }{ (1+x^{ 2 }) } \)
⇒ dy = \(\frac { 1 }{ (1+x^{ 2 }) } \) dx
⇒ ∫dy = ∫\(\frac { 1 }{ (1+x^{ 2 }) } \) dx
⇒ y = tan^-1 x + c

प्रश्न 13.
समीकरण \(\frac{dy}{dx}\) = sin x sin y को हल कीजिये।
हल:
\(\frac{dy}{dx}\) = sin x sin y
⇒ cosec y dy = sin x dx
समाकलन करने पर,
-log_e(cosec y + cot y) = – cos x + c
⇒ cos x – log_e(cosec y + cot y) = c

प्रश्न 14.
अवकल समीकरण \(\frac{dy}{dx}\) = y sin x को हल कीजिये।
हल:
\(\frac{dy}{dx}\) = y sin x
⇒ \(\frac{1}{y}\) \(\frac{dy}{dx}\) = sin x
⇒ ∫\(\frac{1}{y}\) dy = ∫sin x dx
⇒ log y = – cos x + c

प्रश्न 15.
अवकल समीकरण \(\frac{dy}{dx}\) = x cos x को हल कीजिये।
हल:
दिया है:
\(\frac{dy}{dx}\) = x cos x
⇒ dy = x cos x dx
⇒ ∫dy = ∫x cos x dx
⇒ y = x sin x – ∫1. sin x dx + c
⇒ y = x sin x + cos x + c

प्रश्न 16.
अवकल समीकरण \(\frac{dy}{dx}\) = 1 – x + y – xy को हल कीजिए।
हल:
दिया गया अवकल समीकरण है –
\(\frac{dy}{dx}\) = 1 – x + y – xy
⇒ \(\frac{dy}{dx}\) = (1 – x) + y(1 – x)
⇒ \(\frac{dy}{dx}\) = (1 – x) (1 + y)
⇒ \(\frac{dy}{1+y}\) = (1 – x)dx
⇒ ∫\(\frac{dy}{1+y}\) = ∫(1 – x)dx
⇒ log_e(1 + y) = x – \(\frac { x^{ 2 } }{ 2 } \) + c

प्रश्न 17.
अवकल समीकरण \(\frac{dy}{dx}\) = (1 + x)(1 + y²) को हल कीजिए।
हल:
दिया गया अवकल समीकरण है:
\(\frac{dy}{dx}\) = (1 + x)(1 + y²)
⇒ \(\frac { 1 }{ 1+y^{ 2 } } \) dy = (1 + x) dx
समाकलन करने पर,
⇒ tan^-1 y = x + \(\frac { x^{ 2 } }{ 2 } \) + c

प्रश्न 18.
अवकल समीकरण हल कीजिए –
\(\frac{dy}{dx}\) = cot²x
⇒ dy = cot² x dx
⇒ ∫dy = ∫cot² x dx
⇒ y = ∫(cosec² x – 1) dx
⇒ y = – cot x – x + c

अवकल समीकरण दीर्घ उत्तरीय प्रश्न-I

प्रश्न 1.
(A) अवकल समीकरण \(\frac{dy}{dx}\) + y tan x = sec x को हल कीजिए।
हल:
दिया है:
\(\frac{dy}{dx}\) + y tan x = sec x ………… (1)
यह एक रैखिक अवकल समीकरण है, इसकी तुलना \(\frac{dy}{dx}\) + Py = Q से करने पर,
P = tan x, Q = sec x
∴ I.F. = e^{∫p dx} = e^{∫tan x dx} = e^{log secx}
⇒ I.F. = sec x
अतः अवकल समी. (1) का अभीष्ट हल है:
y × (I.F.) = ∫Q × (I.F). dx + c
⇒ y × (sec x) = ∫sec x × (sec x) dx + c
= ∫sec² xdx + c
⇒ y sec x = tan x + c

(B) अवकल समीकरण \(\frac{dy}{dx}\) + y tan x = sin x को हल कीजिए।
हल:
प्रश्न क्रमांक 1 (A) की भाँति हल करें।

प्रश्न 2.
अवकल समीकरण \(\frac { dy }{ dx } \) = \(\frac { \sqrt { 1-y^{ 2 } } }{ \sqrt { 1-x^{ 2 } } } \) को हल कीजिए।
हल:
\(\frac{dy}{dx}\) = \(\frac { \sqrt { 1-y^{ 2 } } }{ \sqrt { 1-x^{ 2 } } } \)

प्रश्न 3.
अवकल समीकरण 3x²dy = (3xy + y²)dx को हल कीजिए।
हल:
दिया गया अवकल समीकरण है:
3x² dy = (3xy + y²) dx
⇒ \(\frac{dy}{dx}\) = \(\frac { 3xy+y^{ 2 } }{ 3x^{ 2 } } \) ……….. (1)
माना y = vx
⇒ \(\frac{dy}{dx}\) = v + x \(\frac{dv}{dx}\)
समी. (1) में मान रखने पर,

प्रश्न 4.
अवकल समीकरण हल कीजिए –
(1 + x)² \(\frac{dy}{dx}\) + 2xy = 4x²
हल:
दिया गया अवकल समीकरण है:
(1 + x²) \(\frac{dy}{dx}\) + 2xy = 4x²
⇒ \(\frac{dy}{dx}\) + \(\frac { 2xy }{ 1+x^{ 2 } } \) = \(\frac { 4x^{ 2 } }{ 1+x^{ 2 } } \)
इसकी तुलना रैखिक अवकल समीकरण \(\frac{dy}{dx}\) + Py = Q से करने पर,

अतः अभीष्ट हल होगा:

प्रश्न 5.
अवकल समीकरण (1 + x²) \(\frac{dy}{dx}\) + 2xy = cos x को हल कीजिए।
हल:
दिया है:
(1 + x²) \(\frac{dy}{dx}\) + 2xy = cos x
⇒ \(\frac{dy}{dx}\) + \(\frac { 2x }{ (1+x^{ 2 }) } \). y = \(\frac { cosx }{ 1+x^{ 2 } } \) …………. (1)
P = \(\frac { 2x }{ (1+x^{ 2 }) } \), Q = \(\frac { cosx }{ 1+x^{ 2 } } \)

अतः अवकल समी. (1) का अभीष्ट हल है
y.(I.F.) = ∫Q.(I.F.) dx + c
⇒ y(1 + x²) = ∫\(\frac { cosx }{ 1+x^{ 2 } } \) dx + c
⇒ y(1 + x²) = ∫cos x dx + c
⇒ y(1 + x²) = sin x + c

प्रश्न 6.
किसी वस्तु के बनाने का सीमांत लागत मूल्य c'(x) = \(\frac{dc}{dx}\) = 2 + 0.15 x समीकरण से दिया जाता है। इस वस्तु के बनाने पर कुल लागत मूल्य c(x) ज्ञात कीजिए। (दिया है: c(0) = 100)
हल:
दिया है:
समीकरण c'(x) = \(\frac{dc}{dx}\) = 2 + 0.15 x
समाकलन करने पर,
∫c'(x) dx = ∫(2 + 0.15 x) dx
c(x) = 2x + 0.15 \(\frac { x^{ 2 } }{ 2 } \) + A ………. (1)
अब यदि x = 0 तो
c(0) = 2 × 0 + \(\frac{0.15}{2}\) × o² + A
⇒ c(0) = A [∵c(0) = 100]
∴ A = 100,
समी. (1) में मान रखने पर,
c(x) = 2x + 0.075 x² + 100

प्रश्न 7.
अवकल समीकरण x\(\sqrt { 1+y^{ 2 } } \) dx + y \(\sqrt { 1+x^{ 2 } } \) dy = 0 को हल कीजिए।
हल:
दिया गया अवकल समीकरण है:
x\(\sqrt { 1+y^{ 2 } } \) dx + y \(\sqrt { 1+x^{ 2 } } \) dy = 0
⇒ \(\frac { y }{ \sqrt { 1+y^{ 2 } } } \) dy = – \(\frac { x }{ \sqrt { 1+x^{ 2 } } } \)
दोनों पक्षों का समाकलन करने पर,
∫\(\frac { y }{ \sqrt { 1+y^{ 2 } } } \) dy = -∫\(\frac { x }{ \sqrt { 1+x^{ 2 } } } \) dx + c
⇒ \(\sqrt { 1+y^{ 2 } } \) = – \(\sqrt { 1+x^{ 2 } } \) + c
⇒ \(\sqrt { 1+x^{ 2 } } \) + \(\sqrt { 1+y^{ 2 } } \) = c

प्रश्न 8.
अवकल समीकरण हल कीजिए –
(x + y + 1) \(\frac{dy}{dx}\) = 1?
हल:
दिया गया अवकल समीकरण है:
(x + y + 1) \(\frac{dy}{dx}\) = 1
⇒ \(\frac{dx}{dy}\) = x + y + 1
⇒ \(\frac{dx}{dy}\) – x = y + 1
यह y के सापेक्ष x का अवकल समीकरण है।
इसकी तुलना रैखिक अवकल समीकरण \(\frac{dx}{dy}\) + Px = Q से करने पर,
P = – 1 तथा Q = y + 1

अत: अभीष्ट हल होगा:

प्रश्न 9.
अवकल समीकरण sec² x tan y dx + sec² y tan xdy = 0 को हल कीजिए।
हल:
दिया है:
sec² xtan ydx + sec² ytan xdy = 0
⇒ sec² tan x dy = -sec² tan ydx
⇒ \(\frac { sec^{ 2 }y }{ tany } \) dy = \(\frac { sec^{ 2 }x }{ tanx } \) dx
⇒ ∫\(\frac { sec^{ 2 }y }{ tany } \) dy = -∫\(\frac { sec^{ 2 }x }{ tanx } \) dx + c
⇒ log y = – log x + log c
⇒ log x + log y = log c
⇒ log xy = log c
⇒ xy = c

प्रश्न 10.
अवकल समीकरण हल कीजिए।
\(\frac{dy}{dx}\) = y tan x – 2 sin x?
हल:
दिया गया अवकल समीकरण
\(\frac{dy}{dx}\) – y tan x = -2 sin x
इसकी तुलना \(\frac{dy}{dx}\) + Py = Q से करने पर,
P = – tan x, Q = – 2 sin x
अभीष्ट हल y.(I.F.) = ∫Q.I.F.dx + c
⇒ y cos x = -2∫sin x cos x dx + c
⇒ y cos x= -∫sin 2x dx + c
⇒ y cos x = \(\frac{cos 2x}{2}\) + c

प्रश्न 11.
अवकल समीकरण \(\frac{dy}{dx}\) + 2y = 4x को हल कीजिए।
हल:
\(\frac{dy}{dx}\) + 2y = 4x
इसकी तुलना \(\frac{dy}{dx}\) + Py = Q से करने पर,
P = 2,Q = 4x

⇒ y = 2x – 1 + c.e^-2x

प्रश्न 12.
अवकल समीकरण cos²x \(\frac{dy}{dx}\) + y = 2 को हल कीजिए।
हल:
cos² x \(\frac{dy}{dx}\) + y = 2
⇒ \(\frac{dy}{dx}\) + sec² x.y = 2 sec² x
इसकी तुलना \(\frac{dy}{dx}\) + Py = Q से करने पर,
P = sec² x, Q = 2 sec² x

प्रश्न 13.
अवकल समीकरण cos x \(\frac { dy }{ dx } \) + y = sin x को हल कीजिये।
हल:
cos x \(\frac{dy}{dx}\) + y = sin x
⇒ \(\frac{dy}{dx}\) + secx. y = tan x
इसकी तुलना \(\frac{dy}{dx}\) + Py = Q से करने पर,
P = sec x, Q = tan x

प्रश्न 14.
अवकल समीकरण (1 + y²) dx = (tan^-1 y – x) dy को हल कीजिए। (CBSE 2015)
हल:
दिया गया अवकल समीकरण है –
(1 + y²) dx = (tan^-1y – x) dy
\(\frac{dx}{dy}\) + \(\frac { x }{ 1+y^{ 2 } } \) = \(\frac { tan^{ -1 }y }{ 1+y^{ 2 } } \)

प्रश्न 15.
अवकल समीकरण (1 + y²) + (x – e^{tan-1 y}) \(\frac{dy}{dx}\) = 0 को हल कीजिए। (CBSE 2016)
हल:
दिया गया अवकल समीकरण है –
(1 + y²) + (x – e^{tan-1 y}) \(\frac{dy}{dx}\) = 0

समी (1) की तुलना \(\frac{dx}{dy}\) + Px = Q से करने पर,

अतः समीकरण का हल होगा –
x.I.F = ∫I.F. × Qdy

tan^-1y = t रखने पर,
\(\frac{d}{dy}\) tan^-1 y = \(\frac{dt}{dy}\)
⇒ \(\frac { 1 }{ 1+y^{ 2 } } \) dy = dt
∴ x.e^{tan-1 y} = ∫e^t.e^t dt = ∫e^2t dt
⇒ x.e^{tan-1 y} = \(\frac{1}{2}\) e^2t + c
⇒ x.e^{tan-1 y} = \(\frac{1}{2}\) e^{2 tan-1 y} + c

प्रश्न 16.
अवकल समीकरण (1 + x²) \(\frac{dy}{dx}\) + 2xy = \(\frac { 1 }{ 1+x^{ 2 } } \) को हल कीजिए, जहाँ y = 0 तथा x = 1(NCERT)
हल:
दिया गया अवकल समीकरण है –
(1 + x²) \(\frac{dy}{dx}\) + 2xy = \(\frac { 1 }{ 1+x^{ 2 } } \)
⇒ \(\frac{dy}{dx}\) + \(\frac { 2xy }{ 1+x^{ 2 } } \) = \(\frac { 1 }{ 1+x^{ 2 } } \) )²
समी. (1) की तुलना \(\frac{dy}{dx}\) + Py = Q से करने पर,
यहाँ P = \(\frac { 2x }{ 1+x^{ 2 } } \), Q = \(\frac { 1 }{ 1+x^{ 2 } } \) )²

अतः समीकरण का हल होगा –
y.I.F. = ∫I.F. × Qdx

y = 0 तथा x = 1 रखने पर,
0(1+1)² = tan^-1 + c
⇒ 0 = \(\frac { \pi }{ 4 } \) + c
⇒ c = – \(\frac { \pi }{ 4 } \)
c का मान समी. (2) में रखने पर, अवकल समीकरण का हल होगा
y.(1 + x²) = tan^-1 x – \(\frac { \pi }{ 4 } \)

प्रश्न 17.
अवकल समीकरण \(\frac{dy}{dx}\) + cot x = 4x cosec x का एक विशिष्ट हल ज्ञात कीजिए। दिया गया है कि – y = 0 तथा x = \(\frac { \pi }{ 2 } \) (NCERT; CBSE 2012)
हल:
दिया गया अवकल समीकरण है –
\(\frac{dy}{dx}\) + y cot x = 4x cosec x
समी. (1) की तुलना \(\frac{dy}{dx}\) + Py = Q से करने पर,
P = cot x, Q = 4x cosecx

अतः समीकरण का हल होगा –
y.I.F. = ∫1.F. × Qdx
⇒ y.sin x = ∫sin x × 4x cosec x dx
⇒ y sin x = 4∫\(\frac { xsinx }{ sinx } \) dx
⇒ y sin x = 4∫xdx
⇒ y sin x = \(\frac { 4x^{ 2 } }{ 2 } \) + c
⇒ y sin x = 2x² + c
x = \(\frac { \pi }{ 2 } \) तथा y = 0 रखने पर,
0(sin \(\frac { \pi }{ 2 } \) ) = 2( \(\frac { \pi }{ 2 } \) )² + c
⇒ 0 = \(\frac { 2\pi ^{ 2 } }{ 4 } \) + c
∴ c = \(\frac { -\pi ^{ 2 } }{ 2 } \)
c का मान समी. (2) में रखने पर, अवकल समीकरण का अभीष्ट हल होगा –
y.sinx = 2x² – \(\frac { \pi ^{ 2 } }{ 2 } \)

MP Board Class 12 Maths Important Questions

MP Board Class 12th Physics Important Questions Chapter 15 Communication Systems

Communication Systems Important Questions 

Communication Systems Objective Type Questions

Question 1.
Choose the correct answer of the following:

Question 1.
The elements of communication system are :
(a) One transmitter
(b) Only receiver
(c) Only communication channel
(d) All of the above.
Answer:
(d) All of the above.

Question 2.
Sound wave are not directly transmitted after converting them into electric signals because:
(a) They propagates with speed of sound
(b) Their frequency does not remain constant
(c) For their transmission very high antenna is needed
(d) Their energy is very high.
Answer:
(c) For their transmission very high antenna is needed

Question 3.
The super imposing of audio waves with carrier wave is called :
(a) Transmission
(b) Reception
(c) Modulation
(d) Defection.
Answer:
(c) Modulation

Question 4.
The frequency range used for T.V. transmission is :
(a) 30 – 300 MHz
(b) 30 – 300 GHz
(c) 30 – 300 KHz
(d) 30 – 300 Hz.
Answer:
(a) 30 – 300 MHz

Question 5.
The periodic time of communication satellite is :
(a) 1 year
(b) 1 day
(c) 12 hours
(d) 12 minutes.
Answer:
(b) 1 day

Question 6.
T. V. signals are reflacted by :
(a) Mesosphere
(b) Ionosphere
(c) Troposphere
(d) None of these.
Answer:
(d) None of these.

Question 7.
The short wave band of radio waves are transmitted through :
(a) Sky wave propagation
(b) Ground wave propagation
(c) Artificial satellite
(d) Direct sending from transmitter to receiver.
Answer:
(a) Sky wave propagation

Question 8.
T.V. network uses :
(a) Microwaves
(b) Radio waves of very high frequency
(c) Gamma rays
(d) X – ray.
Answer:
(b) Radio waves of very high frequency

Question 9.
The waves used in telecommunication are :
(a) Infrared rays
(b) UV – rays
(c) Microwaves
(d) Cosmic rays.
Answer:
(c) Microwaves

Question 2.
Fill in the blanks :

1. …………….. is that signal which varies periodically with time.
2. Digital signal is represented by two binary number …………….. and……………..
3. The range of audio signals is ……………..
4. The frequency of carrier waves is of order of ……………..
5. The order of frequency of radio waves which can be transmitted by total internal reflection through ionosphere is ……………..
6. The band width of AM waves is ……………..
7. The T.V. transmission was invented by……………..
8. WWW means ……………..
9. …………….. is a device with the help of which the location of any place can be obtained.

Answer:

1. Analog signals
2. 0 and 1
3. 20 Hz to 20000 Hz
4. MHz
5. 30 – 300 MHz
6. Frequency of modulating wave
7. J. L. Baird
8. World Wide Web
9. GPS.

Question 3.
Match the columns:
I.

Answer:

1. (e)
2. (d)
3. (b)
4. (c)
5. (a)

II.

Answer:

1. (e)
2. (c)
3. (d)
4. (b)
5. (a)

Question 4.
Write the answer in one word/sentence:

1. Write any two light sources which are used in optical communication?
2. On what principle does the optical fibre work?
3. Who invented the T, V. transmission?
4. What is FAX?
5. What is meant by channel width? Write channel width of AM and FM radio station.
6. What is demodulation or detection?

Answer:

1. Light emitting diode (LED) and Diode LASER
2. Total internal reflection
3. J. L. Baird
4. The electronic reproduction of a document at a distant place is known as facsimile telegraphy or FAX
5. Channel width is that frequency range in which signals can be transmitted from a station. The channel width of AM is 10
6. kHz and of FM is 150 kHz
7. The process of extracting the audio signal from the modulated wave is known as demodulation or detection.

Communication Systems Very Short Answer Type Questions

Question 1.
What is principle of semiconductor LASER?
Answer:
In these LASER Gallium Arsenide (Ga – As) is used. This LASER the ability to produce LASER rays in the range 0.75 to 0.9 um.

Question 2.
Why is a parallel wire line not suitable for microwave transmission?
Answer:
In microwave transmission, half the operating wavelength approaches the separation between the two parallel wire lines. Therefore energy loss in the form of radiation becomes maximum.

Question 3.
What do you understand by channel and channel noise?
Answer:
The frequency range prescribed for a given transmission is called channel. The unwanted signals present by known or unknown reason in the transmitted signal is called noise.

Question 4.
What is importance of modulation index?
Answer:
The modulation index determines the strength and quality of the transmitted signal. If the modulation index is small the amount of variation in the carrier amplitude will be small.

Question 5.
Write full form of LED and LASER.
Answer:
Full form of LED is “Light Emitting Diode”. Full form of LASER is “Light Amplification by Stimulated Emission of Radiations”.

Question 6.
Which device is used to transmit the T.V. signals up to long distances?
Answer:
Communication satellites are used to transmit the T.V. signals up to long distances.

Question 7.
What is a communication system?
Answer:
Communication system is that system through which the signals are transmitted from one place and required at the other place.

Question 8.
What are message signals?
Answer:
A message signal is a single valued function of time that conveys the information. The electrical analog of information or basic message is called message signal. These are obtained by suitable transducers.

Question 9.
What are analog signals?
Answer:
An analog signal is continuous single value which at any instant lies within the range of a maximum and minimum value.

Question 10.
What is a digital signal?
Answer:
Digital signal is a discontinuous signal value which appears in steps in predetermined levels rather than having the continuous change.

Question 11.
What is noise?
Answer:
Noise refers to the unwanted signals that tends to disturb the transmission and processing of message signals in a communication system.

Question 12.
What is modulation?
Answer:
Modulation is a process in which the signal of low frequency (audio signals) are superimposed over a signal of high frequency (carrier signal).

Question 13.
What is FAX?
Answer:
Fax is a type of communication in which an electronic copy of a document is sent to a distance place.

Question 14.
Which type of signal is used in communication through computer?
Answer:
Digital Signal.

Question 15.
Write basic difference between the light emitted by LED and LASER.
Answer:
The light emitted by LASER is completely coherent and monochromatic while the light emitted by the LED is not coherent.

Question 16.
What is bandwidth? What is bandwidth of audio signals?
Answer:
The frequency range of a signal is called its bandwidth. The bandwidth audio signals is 20 Hz to 20 KHz.

Communication Systems Short Answer Type Questions

Question 1.
Write name of two system used for pulse modulation.
Answer:

1. PAM : Pulse Amplitude Modulation.
2. PCM : Pulse Code Modulation.

Question 2.
Write name of two systems which are used to convert the digital data into analog data.
Answer:

1. Amplitude Shift Keying (ASK)
2. Phase Shift Keying (PSK).

Question 3.
FM signals are less sensitive as compared to AM signals. Why?
Answer:
In FM transmission the information (message) in the carrier waves is in the form ofvariation (change) in frequency. In the AM the noise get amplitude modulated. Hence the amplitude carrier wave varies. That is why the FM signals are less sensitive as compared to AM signals.

Question 4.
Write limitations of frequency modulation.
Answer:

1. Area of reception for FM is much smaller.
2. About 10 times wider channel is required by FM.
3. FM receivers and transmitters are very complex and costly.

Question 5.
Write the meaning of LASER and two uses.
Answer:
Meaning of LASER:
Light Amplification by stimulated Emission of Radiations.

Uses:

1. Communication and
2. The medical field to determine vary long distances.

Question 6.
Write any two uses of optical communication.
Answer:

1. It is free from noise.
2. Its bandwidth is large, so number of channels can be transmitted simultaneously.

Question 7.
What is Light Emitting Diode?
Answer:
It is forward biased P – N junction which emits the light. It converts electrical energy into light energy.

Question 8.
What is optical fibre? On what principle does it work?
Answer:
An optical fibre is a device with the help of which the optical signals can be transmitted over through a zig-zag path without loss of energy. It is based on principle of total internal reflection.

Question 9.
What is meant by population inversion and optical pumping?
Answer:
Population inversion:
The process by which the number of atoms are increased . in the excited state as compared to ground state by stimulated absorption, is called population inversion.

Optical pumping:
The process by which population inversion is obtained, is called optical pumping.

Question 10.
Give two characteristics of LASER rays.
Answer:

1. LASER rays are monochromatic.
2. These are highly coherent.

Question 11.
What are the elements of communication system? Explain with the help of block diagram and also describe the different kinds of communication system.
Or
What is communication system? What are its main parts? Explain with block diagram.
Answer:
Communication system is that system through which the signals are transmitted from one place and received at the other place.

Elements of communication system : Following are the elements of a communication system:

1. Transmitter : Its function is to transmit the information after modifying it, to a form suitable for transmission.
Block diagram of transmitter :

2. Communication channel : The free space through which the electromagnetic waves sent by the transmitter, reach the receiver is called communication channel. It may be a coupled wire, coaxial cable or radio wave.

3. Receiver : Its function is to receive information.
Block diagram of receiver:

Question 12.
What is noise? Why digital communication is popular? What are A/D and D/A converter?
Answer:
Noise : Noise is unwanted disturbance or foreign element interfering with the desired information or signal. Noise can come up in any part of the communnication system but it has its worst effect when the signal is weakest.

The digital communication is popular nowaday because of the following characteristics :

1. Its quality is good.
2. Digital signals are in the form of pulses, they can produce easily by using logic gates.
3. These signals do not get distorted by noise.
4. These signals can be stored as digital data.
5. In this transmission many information can be transmitted using one channel.

A/D and D/A converter:
The electric circuit which can convert the analog signal to digital signal is called A/D converter. The electric circuit which can convert the digital signal into analog signal is called D/A converter.

Question 13.
Write difference between Analog and Digital signals.
Answer:
Difference between Analog and Digital signals :

Analog signal:

• In this signal value of current or voltage changes continuously with time.
• It is a continuous function of time.
• A simple analog signal is represented by a sine wave.
• The signal obtained by converting a speech, music or intensity of reflected light are analog signals.

Digital signal:

• In these signals the current or voltage has only two descrete levels 0 and 1.
• It is discontinuous function of time.
• Digital signal is represented in the form of pulses.
• The letters of a book, output of digital computers or the information obtained from a FAX are digital signals.

Communication Systems Long Answer Type Questions

Question 1.
Explain need of modulation and define the term modulation.
Or
What is modulation? Why it is needed for the transmission of signals?
Answer:
Modulation:
Modulation is a process in which the signal of low frequency (audio signals) superimposed over a signal of high frequency (carrier signal). Such that same properties of carrier wave like amplitude, frequency or phase varies in accordance with the instantaneous value of audio signals.

Need of Modulation:
Modulation is necessary for a low frequency signal when it is to be sent to a distant place so that the information may not die out in the way itself as well as for the proper identification of a signal and to keep the height of antenna small also.

For the transmission of electromagnetic waves the length of antenna should be of order of wavelength of the transmitted waves. Since, λ= \(\frac {c}{ ν}\) therefore the required length of antenna for the audible range should be equal to \(\frac { 3\times { 10 }^{ 8 } }{ 20,000 }\) = 1.5 x 10⁴m to \(\frac { 3\times { 10 }^{ 8 } }{ 20 }\) = 1.5 x 10⁷

This length of antenna is not possible in practical. Now if the waves of 300kHz or more than it are to be transmitted then the required length of antenna will be \(\frac { 3\times { 10 }^{ 8 } }{ 3,00,000 }\) = 100 m or less than it. The antenna of this size can be constructed easily. Hence to transmit the audio signals they are superimposed with the radio waves of frequency of order of Mega Hertz. These waves are called carrier waves and this process is called modulation.

Question 2.
How many types of modulation are there? Explain each types of modula-tion.
Answer:
The equation of carrier wave is given as :
e_c = E_c. cos (ω_ct + θ)
Where e_c is instantaneous value of carrier signal. E_c is amplitude of carrier wave, ω_c is angular frequency and θ is phase angle. Thus, there are three types of modulation corresponding to E_c, m_c and θ.

1. Amplitude modulation:
When an audio frequency (modulating) signal is superimposed that the amplitude of modulated wave is linear function of instantaneous value of modulating signal, then this type of modulation is called as amplitude modulation.

2. Frequency modulation:
Frequency modulation is that modulation in which the frequency of carrier wave varies in accordance with the instantaneous value of modulating signal. In this modulation the amplitude and phase of modulating signal are equal to that of carrier wave.

3. Phase Modulation:
In the phase modulation, the modulating waves are super imposed with carrier waves such that the phase of the modulated wave is the linear function of the amplitude of the modulating wave but the frequency and amplitude of the modulated signal remains same as the frequency and amplitude of the carrier wave.

Question 3.
What are limitations of amplitude modulation?
Answer:
The limitations (disadvantages) of amplitude modulation are as follows :

1. Efficiency of Amplitude modulation is smaller:
In AM modulation the message signals are contained in side bands, but not contained in the carrier wave. It is found that in amplitude modulation only one third power is contained inside bands, remaining power is contained in carrier wave. Hence efficiency decreases.

2. Amplitude modulation is more likely to suffer from noise.

3. In Amplitude modulation the fidelity of reception is less:
The range of audio signal is 20 Hz to 20 kHz. Hence the bandwidth must be 40 kHz. But the disturbance created by the nearby radio station should be taken into account and hence the bandwidth is kept only of about 20 kHz.

4. Its transmission range is low. Due to less power it is not possible to transmit the signals up to long distance. Inspite of these limitations the AM is mostly used for the transmission of audio signals.

Question 4.
Write advantages and disadvantages of frequency modulation.
Answer:
Advantages:

1. Frequency modulation is inherently and practically free from the effects of noise.
2. In frequency modulation, noise can further be decreased by increasing the deviation 8.
3. FM receiver can further be improved with the help of limiters to remove amplitude changes, if any which controls the noise level.
4. In FM it is possible to operate many independent transmitters on same frequency without interference.

Disadvantages:

1. A bout 10 times wider channel is required by FM.
2. Area of reception for FM is much smaller.
3. FM receivers and transmitters are very complicated and costly.

Question 5.
Explain the data transmission and retreival with the help of block diagram.
Answer:
The important use of data transmission and retrieval is in the microprocessor and computers. All the informations and signals which are to be transmitted are converted into digital signals by coding, which are then modulated with the carrier signals and are sent to far distant places. These signals are received by receivers. These received signals are amplified and demodulated in their original form. The block diagram of this process is shown in fig.

Question 6.
What is FAX machine? Draw its block diagram and explain its working.
Answer:

1. FAX : The electronic reproduction of a document at a distant place is known as facsimile telegraphy or Fax. To send the document through FAX following functions are performed.

• Optical scanning
• Conversion of data for transmission and reception.
• Printing a copy of images of the original document at the receiving end.

The block diagram of working of FAX is shown in fig.

Question 7.
Write notes on ‘MODEM’.
Or
What is MODEM? Write its working and uses.
Answer:
The block diagram of a modem is shown ahead : A modem is basically a modulator and a demodulator circuit.

Working:
At the transmitting end i.e., at section A, a modem change the digital data (obtained from source) into analogue form in audio frequency range.

So, it is easily modulated and transmitted by communication network through telephone lines and transmitter and sent to receiver for section B. At the receiving end i.e., at sermon B, the modem converts the analogue signal to digital signal.

Uses : It is used for short distance communication by connecting one computer to another through telephone lines.

Kinds of MODEM : MODEM are of two types.

• Internal MODEM : These are internally attached in the computer.
• External MODEM : These are connected externally with the computer.

Question 8.
Describe the following in short:

1. E – mail
2. Internet
3. World Wide Web
4. Celular phone
5. Pager.

Answer:
1. E – mail:
Its full form is electronic mail the message produced by using word processing, programs are transmitted over the world wide network called internet and they get stored in a computer called mail server. Any person connected to the internet can contact mail server. Any person connected to the internet can contact the mail server to check whether it is holding mail for him.

2. Internet:
Internet is a world wide network which connects the millions of computers which are linked together for data transmission. It is a global system of inter connected computer networks.

3. World Wide Web:
Its abreviation is WWW. It was invented by Tim Berners Lee in 1989 – 91 which is highly enlarged encylopedia which is accessible to every one for knowledge.

4. Celular Phone:
It is basically mini wireless phones for exchanging messages as well as for mobile telephonic conversation.

5. Pager:
It is a wireless device which records the messages in writing.

Question 9.
What is line communication? Explain.
Answer:
In a communication system the communication channel is a medium which provides the physicals path between transmitter and receiver. There are two types of communication medium:

1. Guided medium:
This medium connects transmitter and receiver in the case of point to point communication, double line cable, coaxial cable and optical fibre are the examples of guided medium.

2. Unguided medium:
This medium connects one transmitter and many receivers. Sky wave communication is an example of this medium. Its band frequency rays from few thousand kHz to few GHZ.

In the sky wave communication there is no point to point connection between transmitter and antenna. But in some communication applications the point to point connection is needed for example in telephone and telegraph the transmitter and receiver are connected by wire lines. It is called as line communication.

Question 10.
What is Global Positioning System (G.P.S.)? Write its applications.
Answer:
It is a space based satellite navigation system that provides users with accurate information on position time wherein the world even through the local streets and in all weather conditions.

To use GP.S. system of a satellite the user must have a GP.S. device fitted with transmitter and receiver for sending and receiving radio wave signal.

Applications of G.P.S

1. Navigation on land, water and air.
2. Keeping standard time all over the world.
3. Used in unmanned automatic vehicles movement and mobile telephony.
4. In desiging map of a location.

Communication Systems Numericals Questions

Question 1.
When there is 50% modulation the transmitter emits the 20 kW power calculate the power of carrier wave.
Solution:
Given : m_a = 50% = \(\frac {50}{100}\) = P_t = 20 kW = 20,000 W
Formula:

Question 2.
When a modulating of 500 Hz is given to a frequency modulation generator then it produces a frequency deviation of 2.25 kHz. Calculate modulation depth.
Solution:
Given : δ_max = 2.25 kHz = 2250 Hz, f_m = 500 Hz

Question 3.
The maximum and minimum amplitude of a amplitude modulated wave are 16 mV and 4 mV. What is the depth of modulation?
Solution:
Given : E_max = 16mV, E_min = 4mV
Formula:

Question 4.
A 20 kHz modulating signal is modulated with a carrier wave of 4 MHz. What will be the upper side band and lower side band? What will be the channel width?
Solution:
Given: f_m = 20 kHz, f_c = 4 MHz =4,000 kHz
f_USB = f_c + f_m = 4000 + 20 = 4020 kHz.
f_LSB = f_c – f_m= 4000 – 20 = 3980 kHz.
Channel width = f_USB – f_LSB
= 4020 – 3980 = 40 kHz.

Question 5.
Find out the modulation index of a frequency modulated wave whose modulating frequency is 2 kHz and maximum frequency deviation is 10 kHz.
Solution:
Given : f_m = 2 kHz and δ_max = 10 kHz
Formula : m_f = \(\frac { { \delta }_{ max } }{ { f }_{ m } }\)
m_f = \(\frac {10}{2}\) = 5

MP Board Class 12th Physics Important Questions

MP Board Class 12th Hindi Swati Solutions पद्य Chapter 8 जीवन दर्शन

जीवन दर्शन अभ्यास

जीवन दर्शन अति लघु उत्तरीय प्रश्न

प्रश्न 1.
‘जगती के मधुवन’ में पुराने साथी कौन-कौन हैं?
उत्तर:
जगती के मधुवन में पुराने साथी जीवन और पुष्प हैं।

प्रश्न 2.
‘फूल और मालिन’ दोनों कब मुरझा जायेंगे? (2009, 15)
उत्तर:
संसार में समय के साथ सभी चीजें नष्ट होती हैं। फूल और मालिन दोनों मधुवर्षण करके कल (भविष्य में) मुरझा जायेंगे।

प्रश्न 3.
‘विघ्न बाधाएँ’ कब मार्ग छोड़ती हैं? (2017)
उत्तर:
जब साहस के साथ युवक अपने मार्ग पर बढ़ता है, तो विघ्न बाधाएँ उसका मार्ग छोड़ देती हैं।

प्रश्न 4.
‘बन उन्हीं-सा’ पंक्ति के माध्यम से कवि किसके समान बनने की बात कह रहा
उत्तर:
जो लोग अपनी प्रतिकूल परिस्थिति को बदल देते हैं और निर्भय होकर जय गीत गाते हुए आगे बढ़ते हैं,कवि उन्हीं-सा बनने की बात कह रहा है।

प्रश्न 5.
कवि किस बात की शपथ लेने की याद दिलाता है?
उत्तर:
कवि नवयुवक का आह्वान करते हुए कहता है कि तू प्रबलतम आँधी का अपने साहस से सामना कर और तुझे शपथ है कि आगे पैर बढ़ाकर किसी भी प्रकार पीछे मत हटना।

जीवन दर्शन लघु उत्तरीय प्रश्न

प्रश्न 1.
“हृदय तुम्हारा-सा ही मेरा, इसको यों न मरोड़ो”। इस पंक्ति का आशय स्पष्ट कीजिए।
उत्तर:
फूल मालिन से कहता है कि मेरा हृदय भी तुम्हारे हृदय के समान ही कोमल है, इसको यों मरोड़ कर मत तोड़ो। भावना के समक्ष फूल और मालिन एक जैसी ही परिस्थिति को अनुभव करते हैं। संसार में जीवन और फूल दोनों ही परोपकारी हैं। जिस प्रकार जीवन अपना प्रेम दूसरों में बाँटता है उसी प्रकार पुष्प अपनी सुगन्ध दूसरों को लुटाता है। कोमल हृदय में कोमल परोपकारी भावना छिपी हुई है।

प्रश्न 2.
फूल मालिन से, न तोड़ने की प्रार्थना क्यों कर रहा है?
उत्तर:
फूल संसार में और (अधिक समय तक) जिन्दा रहकर अपना मधु और सुगन्ध संसार को अर्पण करना चाहता है। फूल हमेशा प्रसन्न रहता है और लोगों में प्रसन्नता बाँटता है। वह जानता है कि उसका जीवन कभी भी समाप्त हो सकता है। पर मालिन से यही अनुरोध करता है कि वह उसे न तोड़े,तो वह संसार का और अधिक उपकार कर सकेगा। जीवन जैसे एक फूल ही है,जो सौन्दर्य और सुगन्ध चारों ओर बिखेरता है।

प्रश्न 3.
‘पुरानी धरती’ से कवि का क्या आशय है? (2016)
उत्तर:
‘पुरानी धरती’ से कवि का तात्पर्य गौरवशाली भारतभूमि से है। भारत की भूमि प्राचीन काल से ही वीरों की जन्मभूमि रही है। स्वर्ग के समान सुखदायी भूमि भारत की ही है। यहाँ पर ब्रह्म ने भी साकार रूप में अवतार लिया है। प्राचीन भारतभूमि अपनी सभ्यता,संस्कृति, धार्मिकता और ज्ञानदायिनी के रूप में विश्व में प्रसिद्ध रही है। उसी प्राचीन भूमि को कवि ने ‘पुरानी धरती’ के नाम से सम्बोधित किया है।

प्रश्न 4.
‘त्यागमय इतिहास’ कहकर कवि क्या भाव व्यक्त करना चाहता है? (2012)
उत्तर:
कवि ने संकलित कविता में भारत के नवयुवकों का आह्वान किया है कि वे अपने शौर्यपूर्ण कार्यों से देश की उन्नति करें। उन्हें अपने देश के लिए अपने सुखों का त्याग करके देशहित में कार्य करना चाहिए। जब अनेक वीरों ने बलिदान दिया तभी भारतवर्ष ने स्वतन्त्रता प्राप्त की और प्रगति के पथ पर अग्रसर हुआ। इतिहास में भारत के वीरों के त्याग की गाथाएँ भरी पड़ी हैं। इसलिए कवि ने भारत के इतिहास को त्यागमय कहा है।

प्रश्न 5.
विजय सुनिश्चित कब होती है?
उत्तर:
जिन लोगों को विजय का विश्वास होता है उन्हीं लोगों को विजय प्राप्त होती है। विजय का यही विश्वास जीवन दर्शन बनकर बिगड़ी परिस्थितियों को बदलने की सामर्थ्य रखता है। आत्मबल के साथ लिए हुए संकल्प के साथ ही विजय का पथ आसान होता जाता है। जिन लोगों में साहस और अपने पथ पर आगे बढ़ने की शक्ति है, विजय उनके कदमों को चूमती है। मनुष्य का शौर्य ही उसकी विजय को निश्चित करता है।

जीवन दर्शन दीर्घ उत्तरीय प्रश्न

प्रश्न 1.
संकलित कविता के आधार पर सुमन जी के दार्शनिक विचार लिखिए।
उत्तर:
शिवमंगल सिंह ‘सुमन’ का काव्य जीवन के बहुविध क्षेत्रों को प्रकट करता है। उनकी कविताओं में प्रेम और संघर्ष दोनों ही रहते हैं। संकलित कविता में सुमन जी जीवन और फूल को एक समान स्वीकार करते हैं। जिस प्रकार फूल जीवन भर अपने मधु और सुगन्ध को दान करता है उसी प्रकार जीवन भी जन-जन के प्रेम के मधु को एकत्रित करके वितरित कर देता है। कवि की चेतना, प्रिया बनकर जीवन को विस्तारमय प्रसन्नता भी देती है और कभी जीवन में विपन्नता भी लाती है। इस कविता में कवि का स्वर आशावादी है इसलिए वह जीवन के आनन्द को शाश्वत बनाने की कामना करता है।

प्रश्न 2.
फूल और मालिन दोनों के सन्दर्भ में मधुवर्षण का अर्थ स्पष्ट कीजिये।
उत्तर:
फूल स्वयं में प्रसन्न रहकर अपना मधु और सुगन्ध औरों में वितरित करता है और मालिन रूपी कवि की चेतना जन-जन के प्रेम रूपी मधु को एकत्रित करती है। यही चेतना प्रिया बनकर जीवन को विस्तारमय प्रसन्नता भी देती है। दोनों के कार्य परोपकार को लिए हुए हैं। फूल का सम्पूर्ण जीवन मधु एकत्रित करके और सुगन्ध जुटा कर सर्वत्र वितरित करने में व्यतीत होता है। संसार में जीवन भी स्नेह रूपी मधु एकत्रित करके सर्वत्र वितरित कर परोपकार का कार्य करता है। जीवन जैसे एक फूल ही है जो अपने सौन्दर्य और सुगन्ध को बिखेरता रहता है।

प्रश्न 3.
विजय किनके चरणों में लोटती है और क्यों?
उत्तर:
संकलित कविता में कवि ने युवकों को विजय के आत्मबल से परिपूर्ण किया है। विजय उन्हें ही प्राप्त होती है जिन्हें विजय का विश्वास होता है। विजय का यही विश्वास जीवन दर्शन बनकर बिगड़ी परिस्थितियों को बदलने की सामर्थ्य रखता है। विजय उन्हीं के चरणों में लोटती है जो अपने पथ की बाधाओं पर विजय पाते हुए अपने धर्म-पथ पर आगे बढ़ते जाते हैं। वे अपने साहस और शक्ति से विश्व को भी झुका देते हैं। विजय के विश्वास को लेकर जो व्यक्ति अपने पथ पर आगे बढ़ता है विजय उसका स्वागत करती है।

प्रश्न 4.
विश्व कब झुकता है? उसे झुकाने की सामर्थ्य किसमें है? (2009)
उत्तर:
विश्व तभी झुकता है जब वीर अपनी शक्ति और सामर्थ्य के बल पर उसे झुका देता है। उसे झुकाने की सामर्थ्य उन वीर युवकों में है जो प्रतिकूल परिस्थिति को अपने अनुकूल बना लेते हैं और अभय होकर विजय के गीत गाते हुए अपने सद्मार्ग पर बढ़ते रहते हैं। उन कर्मवीरों के सामने ऐसा कोई कार्य नहीं जिसे वे नहीं कर सकते हों। वे मार्ग की आँधियों से लड़ते हुए अपना पथ सुगम बना लेते हैं और विजय का विश्वास लिये आगे बढ़ते हैं। विश्व उसी को मानता है जो शक्तिशाली है और सामर्थ्यवान है। वह अपने मार्ग से कभी हटता नहीं और विजय प्राप्त करके ही वापस लौटता है।

प्रश्न 5.
‘सृष्टि का चित्र नाश की पटभूमिका पर’ अंकित करने की बात कवि क्यों कह रहा
उत्तर:
नाश की भूमि पर नवनिर्माण की गाथा लिखने का साहस भी नवयुवक के बढ़ते हुए कदमों को है। यह ऐतिहासिक सत्य है कि नाश की पृष्ठभूमि पर ही सृजन का चित्र बनता है। जो लोग निर्भय होकर मृत्यु को अपनी हथेली पर रखकर आगे बढ़ते हैं वे ही विजयश्री का वरण करते हैं। त्याग और बलिदान के बाद ही सृजन और शान्ति स्थापित होती है। युद्ध के बाद शान्ति स्थापित होती है और नवनिर्माण होता है। इसीलिए कवि ने सृष्टि का चक्र नाश की पृष्ठभूमि पर अंकित करने की बात कही है।

प्रश्न 6.
निम्नलिखित पद्यांशों की सप्रसंग व्याख्या कीजिये
(1) हम तुम बहुत पुराने साथी …………. कण-कण कर जोड़ो।
(2) जब जग मुझे तोड़ने ………….. मुझे न तोड़ो।
(3) लौटती है विजय ………… तेरी सुनिश्चित।
(4) छोड़ देंगी मार्ग …………. तेरी सुनिश्चित।
उत्तर:
(1) शब्दार्थ :
जगती = संसार; मधुवन = ब्रजभूमि के एक वन का नाम।

सन्दर्भ :
प्रस्तुत पंक्तियाँ ‘देखो मालिन मुझे न तोड़ो’ नामक कविता से उद्धृत हैं। इसके रचयिता शिवमंगल सिंह ‘सुमन’ हैं।

प्रसंग :
प्रस्तुत पंक्तियों में कवि पुष्प से कहलवाता है कि हे मालिन ! तुम मुझे मत तोड़ो।

व्याख्या :
पुष्प मालिन से कहता है कि संसार रूपी मधुवन में हम पुराने साथी हैं। दूसरे अर्थ में जीवन और पुष्प पुराने समय से साथ-साथ हैं और अपने जीवन में परोपकार में रत हैं। हम दोनों तन-मन से कोमल हैं और अपने घर और वन में फल-फूल रहे हैं। पुष्प उपवन में मधु संचय करते हैं उसी प्रकार कवि की चेतना जन-जन के प्रेम के मधु को एकत्रित करती है। पुष्प मालिन से कहता है कि हे मालिन मुझे ! मत तोड़ो।

(2) सन्दर्भ :
पूर्ववत्।

प्रसंग:
इन पंक्तियों में फूल ने मालिन से कहा है कि जब कोई मुझे तोड़ने आता है, तो मैं सुधि-बुधि खो बैठता हूँ।

व्याख्या:
जब लोग मुझे तोड़ने आते हैं, तो मैं हँसता हुआ रो देता हूँ और हे मालिन ! जब तुम मुझे तोड़ने के लिए हाथ उठाती हो,तो मैं अपनी सुधि-बुधि खो देता हूँ। जीवन की तरह ही पुष्प का भी हृदय कोमल है,वह कामना करता है कि उसे मरोड़ कर नष्ट मत करो, हे मालिन ! मुझे न तोड़ो।

(3) सन्दर्भ :
प्रस्तुत पंक्तियाँ ‘बढ़ सिपाही’ नामक कविता से उद्धृत की हैं और इसके रचयिता विष्णुकान्त शास्त्री हैं।

प्रसंग :
यहाँ पर कवि नवयुवकों का आह्वान करते हुए कहता है कि वे अपने प्रगति-पथ पर आगे बढ़ते रहें, विजय उनकी सुनिश्चित है।

व्याख्या :
कवि कहता है कि हे सिपाही (नवयुवक) ! तू विजय पथ पर आगे बढ़ता जा, तेरी विजय सुनिश्चित है। जो लोग निर्भय होकर आगे बढ़ते हैं और अपने रास्ते की बाधाओं को नकारते हुए धर्मपथ पर डटे रहते हैं, उन्हीं के चरणों में विजय लोटती है अर्थात् उन्हीं साहसी व्यक्तियों को विजय प्राप्त होती है। विश्व उनके सामने ही झुकता है जो उसे झुकाने की शक्ति रखते हैं। वे अपनी प्रतिकूल परिस्थितियों को भी बदल देते हैं और अभय के गीत गाते हुए आगे बढ़ते हैं। तू आगे बढ़ और उन जैसा ही वीर बनकर अपनी इच्छानुसार फल प्राप्त कर। तेरी विजय निश्चित है।

(4) सन्दर्भ :
पूर्ववत्।

प्रसंग :
इन पंक्तियों में कवि कहता है कि तू विघ्न-बाधाओं को सहन करता हुआ चल, समय तेरा स्वागत करेगा।

व्याख्या :
कवि नवयुवक का आह्वान करते हुए कहता है कि तू मार्ग की बाधाओं को सहन कर लेगा तो वे तेरा मार्ग छोड़ देंगी और समय तेरे साहस को देखकर तेरा स्वागत करेगा। तेरी सफलता पर प्रसन्न होकर आकाश गर्व के साथ तेरी कहानी कहेगा। धरती तेरे पग चिह्नों को अपने ऊपर अंकित कर लेगी। तू अपने त्याग से गौरवपूर्ण इतिहास का निर्माण करेगा। विजय प्राप्त करना ही तेरी कहानी होगी अर्थात् तेरे भाग्य में विजय ही लिखी हुई है।

प्रश्न 7.
फूल और मालिन की किन समानताओं की ओर कवि ने संकेत किया है?
उत्तर:
फूल और मालिन की समानताओं की ओर संकेत करते हुए कवि ने कहा है कि ये दोनों पुराने साथी हैं,तन-मन से कोमल हैं, गृह और वन में फूल-फल रहे हैं, दोनों में यौवन और आकर्षण है। दोनों ही कल अपना मधुवर्षण करके मुरझा जायेंगे। फूल उपवन में मधु इकट्ठा करता है और मालिन रूपी चेतना जन-जन के मन से मधु एकत्रित करती है।

जीवन दर्शन काव्य सौन्दर्य

प्रश्न 1.
निम्नलिखित शब्दों में से तत्सम तथा तद्भव शब्दों की पृथक्-पृथक् सूची बनाइये।
सृष्टि, गर्व, तुच्छ, पट, उज्ज्वल, बादल, फूल, मधु, पुराना, पीछे।
उत्तर:

प्रश्न 2.
निम्नलिखित पंक्तियों में काव्य-गुण पहचान कर लिखिए
(i) “दोनों तन-मन से कोमल हैं, फूल रहे गृह, वन में।”
(ii) “प्राण में भर अटल साहस खेल ले, इनको खिला ले।”
उत्तर:
(i) मालिन और पुष्प की समानता बताई है। ‘फूलना फलना’ मुहावरे का प्रयोग सटीक है।
(ii) वीर रस की कविता है। अनुप्रास का प्रयोग दृष्टिगत होता है।

प्रश्न 3.
निम्नलिखित पंक्तियों में कौन-सा रस प्रधान है पहचान कर लिखिए
“जब जग मुझे तोड़ने आता, मैं हँस-हँस रो देता,
जब तुम मुझ पर हाथ उठाती, मैं सुधि बुधि खो देता,
हृदय तुम्हारा सा ही मेरा, इसको यों न मरोड़ो।”
उत्तर:
करुण रस।

प्रश्न 4.
निम्नलिखित शब्दों के विलोम शब्द लिखिए
निश्चित, आदर, आकर्षण, कोमल, साहस, यौवन, मुरझाना।
उत्तर:

प्रश्न 5.
पाठ में आए पुनरुक्तिप्रकाश के उदाहरण छाँटकर लिखिए।
उत्तर:
कण-कण,हँस-हँस।

प्रश्न 6.
निम्नलिखित पंक्तियों का भाव सौन्दर्य स्पष्ट कीजिए
(क) “गगन गायेगा गरज कर गर्व से तेरी कहानी”
(ख) “काल अभिनन्दन करेगा आज तेरा समय सादर।”
उत्तर:
(क) वीर रस की पंक्तियाँ हृदय में उत्साह जाग्रत करती हैं। अनुप्रास अलंकार की छटा दर्शनीय है।
(ख) वीर रस का प्रयोग हुआ है। मानवीकरण किया गया है।

देखो मालिन मुझे न तोड़ो भाव सारांश

प्रस्तुत कविता ‘देखो मालिन मुझे न तोड़ो’ सुप्रसिद्ध कवि ‘शिवमंगल सिंह सुमन’ द्वारा रचित है। इस कविता में कवि ने स्पष्ट किया है कि मानव जीवन भी पुष्प के समान सुगन्ध रूपी प्रेम लोगों में बाँटता है। कवि आशावादी है।

संकलित कविता में वे जीवन और फूल को एक समान अनुभव करते हैं। जीवन एक फूल के समान है जो सौन्दर्य और सुगन्ध को बिखेरता है। कवि मालिन को सम्बोधित करते हुए कहता है कि जैसे वह फूल-फूल से रस लेती है, वैसे ही कवि की चेतना जन-जन के प्रेम के मधु को एकत्रित करती है। जीवन और पुष्प दोनों में आकर्षण है और एक दिन यह आकर्षण समाप्त हो जायेगा और दोनों धूल में मिल जायेंगे। कवि की चेतना प्रिया बनकर जीवन को प्रसन्नता देती है और कभी जीवन में विपन्नता भी लाती है। कवि का स्वर आशावादी है, इसलिए वह जीवन के आनन्द को शाश्वत बनाने की कामना करता है।

देखो मालिन मुझे न तोड़ो संदर्भ-प्रसंग सहित व्याख्या

(1) हम तुम बहुत पुराने साश्री
जगती के मधुबन में
दोनों तन-मन से कोमल हैं।
फूल रहे गृह, बन में
हम उपवन का, तुम जन-मन का मधु, कण-कण कर जोड़ो,
देखो मालिन, मुझे न तोड़ो।

शब्दार्थ :
जगती = संसार; मधुवन = ब्रजभूमि के एक वन का नाम।

सन्दर्भ :
प्रस्तुत पंक्तियाँ ‘देखो मालिन मुझे न तोड़ो’ नामक कविता से उद्धृत हैं। इसके रचयिता शिवमंगल सिंह ‘सुमन’ हैं।

प्रसंग :
प्रस्तुत पंक्तियों में कवि पुष्प से कहलवाता है कि हे मालिन ! तुम मुझे मत तोड़ो।

व्याख्या :
पुष्प मालिन से कहता है कि संसार रूपी मधुवन में हम पुराने साथी हैं। दूसरे अर्थ में जीवन और पुष्प पुराने समय से साथ-साथ हैं और अपने जीवन में परोपकार में रत हैं। हम दोनों तन-मन से कोमल हैं और अपने घर और वन में फल-फूल रहे हैं। पुष्प उपवन में मधु संचय करते हैं उसी प्रकार कवि की चेतना जन-जन के प्रेम के मधु को एकत्रित करती है। पुष्प मालिन से कहता है कि हे मालिन मुझे ! मत तोड़ो।

काव्य सौन्दर्य :

1. भाषा में सरलता और प्रवाह पाठक की अनुभूति के अनुकूल है।
2. पुष्प का मानवीकरण किया गया है।
3. कवि की शैली में सरलता एवं स्वाभाविकता है।

2. हम-तुम दोनों में यौवन है
दोनों में आकर्षण
दोनों कल मुरझा जायेंगे
कर क्षण-भर मधुवर्षण
आओ, क्षण भर हँस खिल मिल लें, कल की कल पर छोड़ो,
देखो मालिन, मुझे न तोड़ो।

शब्दार्थ :
यौवन = जवानी; आकर्षण = खिंचाव; मधुवर्षण = शहद बरसाना।

सन्दर्भ :
पूर्ववत्।

प्रसंग:
इन पंक्तियों में कवि ने बताया है कि संसार में जीवन और पुष्प दोनों में जीवन है।

व्याख्या :
कवि कहता है कि जीवन और पुष्प दोनों में यौवन है और दोनों में ही आकर्षण है। एक दिन वे अपना प्रेम और मधु दान कर मुरझा जायेंगे। इसलिए आओ, क्षणभर हँस-खेल कर अपना जीवन सुखपूर्वक व्यतीत कर लें,कल की कल देखी जायेगी। अर्थात् भविष्य का तो पता नहीं है,क्या होगा। अतः वर्तमान को खुशी से जी लेना चाहिए।

काव्य सौन्दर्य :

1. जीवन और पुष्प का मानवीकरण किया गया है।
2. कवि ने वर्तमान को महत्त्वपूर्ण बताया है।
3. अनुप्रास अलंकार से कविता की शोभा बढ़ गई है।

(3) जब जग मुझे तोड़ने आता
मैं हँस-हँस रो देता
जब तुम मुझ पर हाथ उठाती
मैं सुधि-बुधि खो देता
हृदय तुम्हारा-सा ही मेरा इसको यों न मरोड़ो,
देखो मालिन मुझे न तोड़ो। (2011, 15)

शब्दार्थ :
सुधि-बुधि = होश-हवाश।

सन्दर्भ :
पूर्ववत्।

प्रसंग:
इन पंक्तियों में फूल ने मालिन से कहा है कि जब कोई मुझे तोड़ने आता है, तो मैं सुधि-बुधि खो बैठता हूँ।

व्याख्या:
जब लोग मुझे तोड़ने आते हैं, तो मैं हँसता हुआ रो देता हूँ और हे मालिन ! जब तुम मुझे तोड़ने के लिए हाथ उठाती हो,तो मैं अपनी सुधि-बुधि खो देता हूँ। जीवन की तरह ही पुष्प का भी हृदय कोमल है,वह कामना करता है कि उसे मरोड़ कर नष्ट मत करो, हे मालिन ! मुझे न तोड़ो।

काव्य सौन्दर्य:

1. भाषा में सहजता और सरलता है।
2. पुष्प का मानवीकरण किया गया है।
3. ‘हँस-हँस’ में पुनरुक्तिप्रकाश है।

बढ़ सिपाही भाव सारांश

प्रस्तुत कविता ‘बढ़ सिपाही’ राष्ट्रीय भावनाओं के द्योतक कवि ‘आचार्य विष्णकान्त शास्त्री’ द्वारा रचित है। इस कविता के माध्यम से कवि युवकों को विजय पथ पर अग्रसित करने हेतु प्रेरित करना चाहता है।

वह उनमें आत्मबल भरना चाहता है। उनका मूल मन्त्र है कि विजय उनको ही मिलती है जिन्हें अपने आत्मबल और विजय का विश्वास होता है। विजय का यही विश्वास जीवन दर्शन बनकर प्रतिकूल परिस्थितियों को बदलने की सामर्थ्य रखता है। नाश को भूमि पर नवनिर्माण करने का साहस भी नवयुवक के आगे बढ़ते हुए कदमों को है। नवयुवक कभी कदम आगे बढ़ाकर पीछे नहीं हटते और यदि वे ऐसा सोचते भी हैं, तो कवि उन्हें शपथ दिलाता है कि वे बढ़कर पीछे न हटें। वे आगे बढ़कर ही विजय प्राप्त कर सकते हैं। तेरे त्याग से ही गौरवशाली इतिहास का निर्माण होगा। गगन और धरती तेरी विजय से धन्य हो जायेगी।

बढ़ सिपाही संदर्भ-प्रसंग सहित व्याख्या

(1) विजय पथ पर बढ़ सिपाही
विजय है तेरी सुनिश्चित।
लोटती है विजय चरणों पर उन्हीं के, जो बढ़े हैं,
तुच्छ कर सब आपदाएँ, धर्मपथ पर जो अड़े हैं।
विश्व झुकता है उन्हीं के सामने जो हैं झुकाते,
बदलकर बिगड़ी परिस्थिति, हो अभय जय-गीत गाते।।
बढ़ सँभल कर, बन उन्हीं सा, प्राप्त कर तू फल अभीप्सित।
विजय है तेरी सुनिश्चित। (2014)

शब्दार्थ :
तुच्छ = छोटा; आपदाएँ = संकट; धर्मपक्ष = धर्म का रास्ता; अभीप्सित = वांछित।

सन्दर्भ :
प्रस्तुत पंक्तियाँ ‘बढ़ सिपाही’ नामक कविता से उद्धृत की हैं और इसके रचयिता विष्णुकान्त शास्त्री हैं।

प्रसंग :
यहाँ पर कवि नवयुवकों का आह्वान करते हुए कहता है कि वे अपने प्रगति-पथ पर आगे बढ़ते रहें, विजय उनकी सुनिश्चित है।

व्याख्या :
कवि कहता है कि हे सिपाही (नवयुवक) ! तू विजय पथ पर आगे बढ़ता जा, तेरी विजय सुनिश्चित है। जो लोग निर्भय होकर आगे बढ़ते हैं और अपने रास्ते की बाधाओं को नकारते हुए धर्मपथ पर डटे रहते हैं, उन्हीं के चरणों में विजय लोटती है अर्थात् उन्हीं साहसी व्यक्तियों को विजय प्राप्त होती है। विश्व उनके सामने ही झुकता है जो उसे झुकाने की शक्ति रखते हैं। वे अपनी प्रतिकूल परिस्थितियों को भी बदल देते हैं और अभय के गीत गाते हुए आगे बढ़ते हैं। तू आगे बढ़ और उन जैसा ही वीर बनकर अपनी इच्छानुसार फल प्राप्त कर। तेरी विजय निश्चित है।

काव्य सौन्दर्य :

1. वीर रस की कविता है।
2. कहीं कहीं अनुप्रास अलंकार की छटा दर्शनीय है।
3. यह युवकों के लिए प्रेरणास्पद कविता है।

(2) दे रहे आह्वान तुझको मत्त होकर मेघ काले,
उठ रही झंझा प्रबलतम जोर इनका आजमा ले।
शपथ तुझको जो हटाया एक पग भी आज पीछे,
प्राण में भर अटल साहस खेल लें, इनको खिला ले।
नाश की पटभूमिका पर, सृष्टि का कर चित्र अंकित।
विजय है तेरी सुनिश्चित। (2009)

शब्दार्थ :
आह्वान = बुलाना; मत्त = मस्त; झंझा = आँधी; प्रबलतम = जोर की; सृष्टि = निर्माण।

सन्दर्भ एवं प्रसंग :
पूर्ववत्।

व्याख्या :
कवि कहता है कि हे नवयुवक ! आकाश में घिरते हुए काले बादल तेरा आह्वान कर रहे हैं। बड़े बेग से आती हुई आँधी तेरे सामने है,तू इससे लड़कर अपने बल की परीक्षा कर ले। जो लोग अपने पथ में अग्रसर होकर पीछे लौटने की सोचते हैं उनको कवि शपथ देता है कि वे अपने पैर आगे ही बढ़ाएँ,पीछे कदापि न लौटें। अपने प्राणों में अटल साहस भर लो और साथ में अन्य लोगों को भी ले लो। नाश की पटभूमि पर तू निर्माण का चित्र बना। अर्थात् नाश की परवाह न करके निर्माण का मार्ग बनाएँ। तू निरन्तर आगे बढ़ता चल,तेरी विजय निश्चित है।

काव्य सौन्दर्य :

1. वीर रस की कविता है।
2. सरलता है और स्वाभाविकता है।
3. कविता ओज से दैदीप्यमान है।

3. छोड़ देंगी मार्ग तेरा विघ्न बाधाएँ सहम कर,
काल अभिनन्दन करेगा आज तेरा समय सादर।
मगन गायेगा गरज कर गर्व से तेरी कहानी,
वक्ष पर पदचिह्न लेगी धन्य हो धरती पुरानी॥
कर रहा तू गौरवोज्ज्वल त्यागमय इतिहास निर्मित।
विजय है तेरी सुनिश्चित।। (2017)

शब्दार्थ :
विन बाधाएँ = रुकावटें; अभिनन्दन = स्वागत; वक्ष = सीने पर; पदचिह्न = पैरों के निशान; गौरवोज्वल = गौरव से उज्ज्वल; त्यागमय = त्याग से भरा; निर्मित = बना हुआ।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
इन पंक्तियों में कवि कहता है कि तू विघ्न-बाधाओं को सहन करता हुआ चल, समय तेरा स्वागत करेगा।

व्याख्या :
कवि नवयुवक का आह्वान करते हुए कहता है कि तू मार्ग की बाधाओं को सहन कर लेगा तो वे तेरा मार्ग छोड़ देंगी और समय तेरे साहस को देखकर तेरा स्वागत करेगा। तेरी सफलता पर प्रसन्न होकर आकाश गर्व के साथ तेरी कहानी कहेगा। धरती तेरे पग चिह्नों को अपने ऊपर अंकित कर लेगी। तू अपने त्याग से गौरवपूर्ण इतिहास का निर्माण करेगा। विजय प्राप्त करना ही तेरी कहानी होगी अर्थात् तेरे भाग्य में विजय ही लिखी हुई है।

काव्य सौन्दर्य :

1. कविता में वीर रस की अभिव्यक्ति सुन्दर है।
2. अनुप्रास अलंकार की छटा अद्भुत है।
3. वीर पुरुष के सिद्धान्त को प्रतिपादित किया गया है।

MP Board Class 12th Hindi Solutions

MP Board Class 12th Maths Important Questions Chapter 13 प्रायिकता

प्रायिकता Important Questions

प्रायिकता वस्तुनिष्ठ प्रश्न

प्रश्न 1.
सही विकल्प चुनकर लिखिए –

प्रश्न 1.
एक थैले में 5 भूरे तथा 4 सफेद मोजे रखे हैं। एक पुरुष थैले में से दो मोजे निकालता है। दोनों का एक ही रंग होने की प्रायिकता है –
(a) \(\frac{5}{108}\)
(b) \(\frac{18}{108}\)
(c) \(\frac{30}{108}\)
(d) \(\frac{48}{108}\)

प्रश्न 2.
एक पासे को तीन बार फेंका गया। प्रत्येक बार पूर्व प्राप्त संख्या से बड़ी संख्या प्राप्त करने की प्रायिकता है –
(a) \(\frac{5}{72}\)
(b) \(\frac{5}{34}\)
(c) \(\frac{13}{216}\)
(d) \(\frac{1}{18}\)

प्रश्न 3.
यह दिया है कि घटनाएँ A तथा B ऐसी हैं कि P(A) = \(\frac{1}{4}\), P( \(\frac{A}{B}\) )  \(\frac{1}{2}\) तथा P( \(\frac{B}{A}\) ) =
\(\frac{2}{3}\) तो P(B) का मान है –
(a) \(\frac{1}{3}\)
(b) \(\frac{2}{3}\)
(c) \(\frac{1}{2}\)
(d) \(\frac{1}{6}\)

प्रश्न 4.
एक सिक्का 4 बार उछाला गया है। कम से कम एक शीर्ष आने की प्रायिकता है –
(a) \(\frac{1}{16}\)
(b) \(\frac{2}{16}\)
(c) \(\frac{14}{16}\)
(d) \(\frac{15}{16}\)

प्रश्न 5.
A के सत्य बोलने की प्रायिकता है \(\frac{4}{5}\) तथा B के सत्य बोलने की प्रायिकता \(\frac{3}{4}\) है। एक ही तथ्य के पूछने पर उनके एक-दूसरे के विरोधी उत्तर देने की प्रायिकता है –
उत्तर:
(a) \(\frac{3}{10}\)
(b) \(\frac{1}{4}\)
(c) \(\frac{7}{20}\)
(d) \(\frac{2}{5}\)

प्रश्न 2.
रिक्त स्थानों की पूर्ति कीजिये –

1. P \(\left(\frac{A \cup B}{C}\right)\)  = P \(\left(\frac{A}{C}\right)\)  + ………………………
2. P(A∩B) = …………….. था ………………………..
3. A और B दो स्वतंत्र घटनाएँ हों, तो P(A∩B) = ……………………..
4. A और B दो स्वतंत्र घटनाएँ हों, तो P\(\left(\frac{A}{C}\right)\) = …………………………….
5. यदि यादृच्छिक चर x के संगत प्रायिकता P(X) हो, तो घटना E का X का माध्य E(X) = ……………………….. होगा।
6. यदि यादृच्छिक चर X के संगत P(X) हो, तो X का प्रसरण Var (X) = ……………………….. होगा।

उत्तर:

1. P \(\left(\frac{B}{C}\right)\) – P \(\left(\frac{A \cup B}{C}\right)\)
2. P(B).P \(\left(\frac{A}{B}\right)\)  या P(A).P \(\left(\frac{A}{B}\right)\) = या P(A).P \(\left(\frac{B}{A}\right)\)
3. P(A).P(B)
4. P(A), P(B) ≠ 0
5. ΣX.P(X)
6. ΣX².P(X)- [ΣX.P(X)]².

प्रश्न 3.
निम्न कथनों में सत्य/असत्य बताइए –

1. यदि A और B कोई दो घटनाएँ हों तब P(A – B) = P(A) – P (A∩B)
2. यदि A और B कोई दो घटनाएँ हों तब P(A∪B) – P(A∩B) = P(A) + P(B).
3. P \(\left(\frac{A}{B}\right)\)  =  \(\frac{P(A \cap B)}{P(A)}\)
4. यदि A और B प्रतिदर्श समष्टि S की कोई दो घटनाएँ हैं और F एक अन्य घटना इस प्रकार है कि P(F) ≠ 0, तब P \(\left(\frac{A \cup B}{F}\right)\)  = P \(\left(\frac{A}{F}\right)\)  + P \(\left(\frac{B}{F}\right)\)  – P \(\left(\frac{A \cup B}{F}\right)\)
5. किन्ही दो घटनाओं A और B के लिए P(A∪B) = P(A∩B) + P( \(\bar { A } \)∩B) + P(A∩\(\bar { B } \))

उत्तर:

1. सत्य
2. असत्य
3. असत्य
4. सत्य
5. सत्य।

प्रश्न 4.
एक शब्द/वाक्य में उत्तर दीजिए –

1. एक लीप वर्ष में 53 शुक्रवार आने की प्रायिकता ज्ञात कीजिए।
2. एक सिक्का 4 बार उछाला गया है कम-से-कम एक शीर्ष आने की प्रायिकता ज्ञात कीजिए।
3. A और B दो ऐसी घटनाएँ हैं जहाँ, P(A) = \(\frac{1}{4}\), P \(\left(\frac{A}{B}\right)\)  = \(\frac{1}{2}\) तथा P \(\left(\frac{B}{A}\right)\)  = \(\frac{2}{3}\), तो P(B) का मान ज्ञात कीजिए।
4. 5 पत्र तथा 5 पते लिखे लिफाफे हैं। यदि यादृच्छया पत्रों को लिफाफे में रखा जाए, तो 3 पत्रों को सही लिफाफे में रखे जाने की प्रायिकता क्या होगी?
5. गणित की एक समस्या को तीन विद्यार्थियों द्वारा हल करने की संभावनाएँ क्रमशः \(\frac{1}{2}\), \(\frac{1}{3}\), \(\frac{1}{4}\) हैं। समस्या हल हो जाने की प्रायिकता क्या है?

उत्तर:

1. \(\frac{2}{7}\)
2. \(\frac{15}{16}\)
3. \(\frac{1}{3}\)
4. \(\frac{1}{12}\)
5. \(\frac{3}{4}\)

प्रायिकता लघु उत्तरीय प्रश्न

प्रश्न 1.
(A) दो पासे एक साथ एक बार उछाले जाते हैं। योगफल 8 आने की प्रायिकता ज्ञात कीजिए।
हल:
दिया है:
दो पासे एक साथ एक बार उछाले जाते हैं
∴ n(S) = 6² = 36
योगफल 8 आने की घटना
A = {(2,6); (6,2); (5,3); (3,5); (4,4)}
∴ n(A) = 5
∴ योगफल 8 आने की प्रायिकता
P(A) = \(\frac { n(A) }{ n(S) } \) = \(\frac{5}{36}\)

(B) दो पासे एक साथ उछाले जाते हैं। ऊपरी फलक पर योगफल 9 आने की प्रायिकता ज्ञात कीजिए।
हल:
प्रश्न क्र. 1 (A) की भाँति हल करें।

(C) दो पासे एक साथ उछाले जाते हैं। ऊपरी फलक पर योगफल 7 आने की प्रायिकता ज्ञात कीजिए।
हल:
प्रश्न क्र. 1 (A) की भाँति हल करें।

प्रश्न 2.
एक घटना का प्रतिकूल संयोगानुपात 3 : 4 है, तो उसके न घटने की प्रायिकता ज्ञात कीजिए।
हल:
दिया है:
घटना के प्रतिकूल संयोगानुपात = 3 : 4
∵ घटना के न घटने की प्रायिकता P( \(\bar { A } \) ) = \(\frac{b}{a+b}\)
⇒ P( \(\bar { A } \) ) = \(\frac{4}{3+4}\) = \(\frac{4}{7}\)

प्रश्न 3.
52 ताश के पत्तों की एक गड्डी में से एक पत्ता यदृच्छया निकाला जाता है, तो उसके चेहरे वाला पत्ता होने की प्रायिकता ज्ञात कीजिए।
हल:
यहाँ, प्रतिदर्श समष्टि n(S) = 52
∵ चेहरे वाले पत्ते = 4 गुलाम + 4 बेगम + 4 बादशाह
∴ n(A) = 4 + 4 + 4 = 12
अतः अभीष्ट प्रायिकता P(A) = \(\frac { n(A) }{ n(S) } \) = \(\frac{12}{52}\) = \(\frac{3}{13}\)

प्रश्न 4.
यदि किसी लीप वर्ष को यदृच्छया चुन लिया जाये तो उस वर्ष में 53 रविवार होने की प्रायिकता ज्ञात कीजिए।
हल:
लीप वर्ष में 366 दिन होते हैं अर्थात् 52 सप्ताह और 2 दिन। अब शेष दो दिन बचे जिनकी सम्भावना इस प्रकार है:
(रविवार, सोमवार), (सोमवार, मंगलवार), (मंगलवार, बुधवार); (बुधवार, गुरुवार), (गुरुवार, शुक्रवार), (शुक्रवार, शनिवार), (शनिवार, रविवार)
प्रतिदर्श समष्टि n(S) = 7
यदि रविवार की घटना E है, तो n(E) = 2
अतः 53 रविवार होने की प्रायिकता = \(\frac { n(E) }{ n(S) } \) = \(\frac{2}{7}\).

प्रश्न 5.
घोड़े A के किसी दौड़ में जीतने की प्रायिकता \(\frac{1}{4}\) तथा घोड़े B के उसी दौड़ में जीतने की प्रायिकता \(\frac{1}{8}\) है, तो इनमें से किसी एक के जीतने की प्रायिकता क्या है?
हल:
घोड़े A के जीतने की प्रायिकता P(A) = \(\frac{1}{4}\)
घोड़े B के जीतने की प्रायिकता P(B) = \(\frac{1}{8}\)
∵ A और B परस्पर अपवर्जी घटनाएँ हैं, अत: P(A∩B) = 0
∴ P(A∪B) = P(A) + P(B) = \(\frac{1}{4}\) + \(\frac{1}{8}\) = \(\frac{2+1}{8}\) = \(\frac{3}{8}\)

प्रश्न 6.
(A) 52 पत्तों की ताश की गड्डी से यदृच्छया एक पत्ता खींचने पर उसके बादशाह या हुकुम का पत्ता होने की प्रायिकता ज्ञात कीजिए।
हल:
मान लीजिए बादशाह खींचने की घटना A तथा हुकुम का पत्ता खींचने की घटना B है, तो
n(S) = 52, n(A)= 4, n(B) = 13, n(A∩B) =1
[क्योंकि ताश की गड्डी में चार बादशाह होते हैं; 13 हुकुम के पत्ते होते हैं तथा एक हुकुम का बादशाह होता है।]
∴P(A) = \(\frac { n(A) }{ n(S) } \) = \(\frac{4}{52}\), P(B) = \(\frac { n(B) }{ n(S) } \) = \(\frac{13}{52}\)
P(A∩B) = \(\frac{1}{52}\)
∴ P(A∪B) = P(A) + P(B) – P(A∩B)
= \(\frac{4}{52}\) + \(\frac{13}{52}\) – \(\frac{1}{52}\) = \(\frac{16}{52}\) = \(\frac{4}{13}\).

(B) ताश की गड्डी में से एक पत्ता खींचा जाता है, प्रायिकता ज्ञात कीजिये कि यह न तो इक्की है न ही बादशाह।
हल:
कुल बादशाह = 4, कुल इक्का = 4
8 पत्तों में से कोई एक पत्ता प्राप्त करने की प्रायिकता = ⁸C₁
52 पत्तों में से कोई एक पत्ता खींचने की प्रायिकता = ⁵²C₁
∴ P(A) = \(\frac { ^{ 8 }C_{ 1 } }{ ^{ 52 }C_{ 1 } } \) = \(\frac{8}{52}\) = \(\frac{2}{13}\)
इनमें से कोई पत्ता न होने की प्रायिकता = 1 – P(A) = 1 – \(\frac{2}{13}\) = \(\frac{11}{13}\).

प्रश्न 7.
(A) एक पासे को एक बार उछाला जाता है। प्रायिकता ज्ञात कीजिए कि सम अंक या 5 से कम अंक प्राप्त हो।
हल:
एक पासे को एक बार उछालने पर प्राप्त प्रतिदर्श समष्टि
S = {1, 2, 3, 4, 5, 6}
n(S) = 6
E₁ = {2, 4, 6} ∴ n(E₁) = 3
E₂ = {1, 2, 3, 4} ∴ n(E₂) = 4
E₁∩E₂ = {2, 4} ∴ n(E₁∩E₂) = 2
∴ अभीष्ट प्रायिकता
P(E₁∪E₂) = P(E₁) + P(E₂) – P(E₁∩E₂)
= \(\frac { n(E_{ 1 }) }{ n(S) } \) + \(\frac { n(E_{ 2 }) }{ n(S) } \) – \(\frac { n(E_{ 1 }∩E_{ 2 }) }{ n(S) } \)
= \(\frac{3}{6}\) + \(\frac{4}{6}\) – \(\frac{2}{6}\) = \(\frac{7-2}{6}\) = \(\frac{5}{6}\)

(B) पासे के एक युग्म को फेंकने पर योग 9 या 11 न आने की प्रायिकता ज्ञात कीजिये।
हल:
एक पासे के युग्म को एक साथ उछालने के तरीके = 6 × 6 = 36
अतः n(S) = 36
माना कि E योग 9 या 11 न आने की घटना है।
∴ E = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (4,1), (4,2), (4,3), (4,4), (4,6), (5,1), (5, 2), (5,3), (5,5), (6,1), (6,2), (6,4), (6,6)}
∴ n(E) = 30
अतः अभीष्ट प्रायिकता P(E) = \(\frac { n(E) }{ n(S) } \) = \(\frac{30}{36}\) = \(\frac{5}{6}\)

प्रश्न 8.
यदि तीन सिक्के एक साथ उछाले जायें, तो घटना में कम-से-कम एक शीर्ष प्राप्त करने की प्रायिकता क्या होगी?
हल:
तीन सिक्के एक साथ उछाले जाने पर प्रतिदर्श समष्टि की संख्या n(S) = 2³ = 8
माना एक की शीर्ष न होने की घटना \(\bar { A } \) हो, तो
\(\bar { A } \) = {(T, T, T)}
∴ n( \(\bar { A } \) ) = 1
अतः एक की शीर्ष न होने की प्रायिकता
P( \(\bar { A } \) ) = \(\frac { n(\bar { A } ) }{ n(S) } \) = \(\frac{1}{8}\)
∴ कम – से – कम एक शीर्ष प्राप्त करने की प्रायिकता P(A) = 1 – P( \(\bar { A } \) )
= 1 – \(\frac{1}{8}\) = \(\frac{7}{8}\)

प्रश्न 9.
एक सिक्का दो बार उछाला जाता है। दोनों बार शीर्ष आने की प्रायिकता ज्ञात कीजिए।
हल:
माना पहली बार उछालने में शीर्ष आने की घटना E₁ तथा दूसरी बार उछालने में शीर्ष आने की घटना E₂ है।
∴ P(E₁) = \(\frac{1}{2}\) तथा P(E₂) = \(\frac{1}{2}\)
ये दोनों घटनाएँ स्वतन्त्र हैं।
∴ P(E₁∩E₂) = P(E₁) × P(E₂)
= \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{4}\).

प्रश्न 10.
किसी दौड़ में A, B तथा C घोड़े के जीतने के अनुकूल संयोगानुपात क्रमश: 1 : 2, 1 : 3 तथा 1 : 4 हैं, तो किसी एक के जीतने की प्रायिकता ज्ञात कीजिए।
हल:
P (E₁) = प्रथम घोड़े के विजयी होने की प्रायिकता = \(\frac{1}{1+2}\) = \(\frac{1}{3}\)
P (E₂) = द्वितीय घोड़े के विजयी होने की प्रायिकता = \(\frac{1}{1+3}\) = \(\frac{1}{4}\)
P(E₃) = \(\frac{1}{1+4}\) = \(\frac{1}{5}\)
अतः किसी एक घोड़े के विजयी होने की प्रायिकता = P(E₁) + P(E₂) + P(E₃)
= \(\frac{1}{3}\) + \(\frac{1}{4}\) + \(\frac{1}{5}\)
= \(\frac{20+15+12}{60}\) = \(\frac{47}{60}\).

प्रश्न 11.
गणित का एक प्रश्न तीन छात्रों A, B और C को हल करने के लिए दिया जाता है, जिसके हल करने की प्रायिकताएँ क्रमश: 1/2, 1/3 और 1/4 हैं। प्रश्न के हल न होने की प्रायिकता ज्ञात कीजिए।
हल:
यदि तीन छात्रों A, B और C द्वारा प्रश्न के हल होने की प्रायिकता क्रमश: P(A), P(B) व P(C) हो, तो
P(A) = \(\frac{1}{2}\), P(B) = \(\frac{1}{3}\), P(C) = \(\frac{1}{4}\)
∴ P( \(\bar { A } \) ) = 1 – P(A) = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
P( \(\bar { B } \) ) = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
और P( \(\bar { C } \) ) = 1 – P(C) = 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\)
अतः प्रश्न के हल न होने की प्रायिकता = P( \(\bar { A } \) ) P( \(\bar { B } \) ) P( \(\bar { C } \) )
= \(\frac{1}{2}\) × \(\frac{2}{3}\) × \(\frac{3}{4}\) = \(\frac{1}{4}\)

प्रायिकता दीर्घ उत्तरीय प्रश्न – I

प्रश्न 1.
रायपुर में 20% व्यक्ति अंग्रेजी का अखबार पढ़ते हैं, 40% व्यक्ति हिन्दी तथा 5% व्यक्ति दोनों प्रकार के अखबार पढ़ते हैं। कितने प्रतिशत व्यक्ति कोई भी अखबार नहीं पढ़ते हैं?
हल:
P(A) = अंग्रेजी अखबार पढ़ने वाले व्यक्ति
= \(\frac{20}{100}\) = \(\frac{1}{5}\)
P(B) = हिन्दी अखबार पढ़ने वाले व्यक्ति
= \(\frac{40}{100}\) = \(\frac{2}{5}\)
P(A∩B) = दोनों भाषाओं के अखबार पढ़ने वाले व्यक्ति
= \(\frac{5}{100}\) = \(\frac{1}{20}\)
दोनों भाषाओं के अखबार नहीं पढ़ने वाले व्यक्ति = P( \(\bar { A } \)∩\(\bar { B } \) )
= 1 – P(A∪B)
= 1 – [P(A) + P(B) – P(A∩B)]
= 1 – [ \(\frac{1}{5}\) + \(\frac{2}{5}\) – \(\frac{1}{20}\) ] = 1 – \(\frac{3}{5}\) + \(\frac{1}{20}\)
= \(\frac{20-12+1}{20}\) = \(\frac{9}{20}\).

प्रश्न 2.
A, प्रकरणों में 75% सत्य बोलता है और B, 80% सत्य बोलता है। यदि दोनों में एक ही प्रकरण पर विरोधाभास हो, तो इसका क्या प्रतिशत होगा? अथवा मोहन 75% प्रकरणों में तथा सोहन 80% प्रकरणों में सच बोलता है। इस घटना की प्रायिकता ज्ञात कीजिये जबकि मोहन सच तथा सोहन झूठ बोलता है।
हल:
A के सत्य बोलने की प्रायिकता P(A) = \(\frac{75}{100}\) = \(\frac{3}{4}\)
B के सत्य बोलने की प्रायिकता P(B) = \(\frac{80}{100}\) = \(\frac{4}{5}\)
A के सत्य न बोलने की प्रायिकता P( \(\bar { A } \) ) = 1 – \(\frac{3}{4}\) = \(\frac{1}{4}\)
B के सत्य न बोलने की प्रायिकता P( \(\bar { B } \) ) = 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\)
P(A और B में विरोधाभास है) = P (A सत्य बोलता है और B असत्य बोलता है) या P(A असत्य बोलता है और B सत्य बोलता है)
= P(A)P( \(\bar { B } \) ) + P( \(\bar { A } \) )P(B)
= \(\frac{3}{4}\) × \(\frac{1}{5}\) + \(\frac{1}{4}\) × \(\frac{4}{5}\) = \(\frac{7}{20}\)
अतः विरोधाभास का प्रतिशत = \(\frac{7}{20}\) × 100 = 35%

प्रश्न 3.
सिद्ध क्रीजिए कि
P(A) + P(not A) = I.
हल:
माना प्रतिदर्श समुच्चय S है तथा घटना A, S का उपसमुच्चय है। तब,
S के सापेक्ष (A’) = A का पूरक
= not A
स्पष्ट है कि A और A’ परस्पर अपवर्जी घटनाएँ हैं, क्योंकि
A∩A’ = ϕ
समुच्चय सिद्धान्त से किन्हीं दो समुच्चयों A और B के लिये,
n(A∪B) = n(A) + n(B) – n(A∩B)
अब B के स्थान पर A’ लिखने पर,
n(A∪A’) = n(A) + n(A’) – n(A∩A’)
⇒ n(S) = n(A) + n(A’) – n(ϕ),
∴ n(S) = n(A) + n(A’) – 0
∴ 1 = \(\frac { n(\bar { A } ) }{ n(S) } \) + \(\frac { n(\bar { A’ } ) }{ n(S) } \)
था 1 = P(A) + P(A’)
अतः P(A) + P(not A) = 1. यही सिद्ध करना था।

प्रश्न 4.
यदि P(A) = \(\frac{1}{2}\), P(B) = \(\frac{1}{9}\) तथा P(A∩B) = \(\frac{1}{18}\) हो, तो निम्न के मान ज्ञात कीजिए –
हल:
दिया है:
P(A) = \(\frac{1}{2}\), P(B) = \(\frac{1}{9}\) तथा P(A∩B) = \(\frac{1}{18}\)

प्रश्न 5.
10 बच्चों के एक समूह में जिसमें 6 लड़के और 4 लड़कियाँ हैं, 3 बच्चे यदृच्छया चुने जाते है। प्रायिकता ज्ञात कीजिए –
(i) कोई लड़की नहीं रखता हो।
(ii) कम – से – कम 1 लड़की रखता हो।
हल: 10 बच्चों में से 3 बच्चे चुनने के तरीके ¹⁰C₃ = \(\frac { 10! }{ 7!3! } \) = \(\frac{10×9×8×7!}{7!×3×2}\) = 120
(i) कोई लड़की नहीं रखता हो –
6 लड़कों में से 3 लड़के चुनने के तरीके ⁶C₃ = 20
अतः अभीष्ट प्रायिकता = \(\frac{20}{120}\) = \(\frac{1}{6}\)
(ii) कम – से – कम एक लड़की रखता हो –
2 लड़का + 1 लड़की चुनने के तरीके = ⁶C₂, × ⁴C₁ = 15 × 4 = 60
1 लड़का + 2 लड़की चुनने के तरीके = ⁶C₁ × ⁴C₂
= 6 × 6 = 36
3 लड़की चुनने के तरीके = ⁴C₃ = 4
अतः अभीष्ट प्रायिकता = \(\frac{60+36+4}{120}\) = \(\frac{100}{120}\) = \(\frac{5}{6}\).

प्रश्न 6.
ताश की गड्डी फेंटते समय एक-एक करके 4 ताश गिर पड़ते हैं। प्रायिकता ज्ञात कीजिए कि एक ताश पान का, दूसरा ईंट का, तीसरा हुकुम का तथा चौथा चिड़ी का होगा।
हल:
ताश के कुल पत्तों की संख्या = 52
पहला पत्ता 52 प्रकार से गिर सकता है।
पहले पत्ते के पान का पत्ता होने के अनुकूल प्रकारों की संख्या = 13
∴ प्रायिकता = \(\frac{13}{52}\) = \(\frac{1}{4}\)
शेष पत्ते = 52 – 1 = 51
इनमें से एक पत्ता 51 प्रकार से गिर सकता है।
इस पत्ते के ईंट का पत्ता होने के अनुकूल प्रकारों की संख्या = 13
∴ प्रायिकता = \(\frac{13}{51}\)
शेष पत्ते = 52 – 1 = 50
इनमें से एक पत्ता 51 प्रकार से गिर सकता है।
इस पत्ते के हुकुम का पत्ता होने के अनुकूल प्रकारों की संख्या = 13
∴प्रायिकता = \(\frac{13}{50}\)
शेष पत्ते = 50 – 1 = 49
इनमें से एक पत्ता 49 प्रकार से गिर सकता है।
इस पत्ते के चिड़ी का पत्ता होने के अनुकूल प्रकारों की संख्या = 13
∴ प्रायिकता = \(\frac{13}{49}\)
अतः मिश्र प्रायिकता के सिद्धान्त से अभीष्ट प्रायिकता = \(\frac{1}{4}\) × \(\frac{13}{51}\) × \(\frac{13}{50}\) × \(\frac{13}{49}\).

प्रश्न 7.
गणित का एक प्रश्न तीन विद्यार्थियों को हल करने के लिए दिया गया। उसके हल करने की प्रायिकता \(\frac{1}{2}\), \(\frac{1}{3}\) और \(\frac{1}{4}\) हैं। यदि वे सभी हल करने का प्रयत्न करें तो किसी एक के प्रश्न हल किये जाने की प्रायिकता ज्ञात कीजिए।
हल:
गणित का प्रश्न तीन विद्यार्थियों द्वारा हल किये जाने की प्रायिकता माना क्रमशः P₁, P₂, और P₃, हैं।
अतः P₁ = \(\frac{1}{2}\), P₂ = \(\frac{1}{3}\), P₃ = \(\frac{1}{4}\)
प्रश्न को उन तीनों के द्वारा हल न किये जाने की प्रायिकता क्रमशः
q₁ = 1 – P₁ = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
q₂ = 1 – P₂ = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
q₃ = 1 – P₃ = 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\)
अतः तीनों के द्वारा साथ-साथ प्रश्न हल न किये जाने की प्रायिकता = q₁ .q₂.q₃
= \(\frac{1}{2}\) × \(\frac{2}{3}\) × \(\frac{3}{4}\) = \(\frac{1}{4}\)
अतः प्रश्न के हल किये जाने की प्रायिकता = कम-से-कम एक विद्यार्थी के प्रश्न को हल कर सकने की
प्रायिकता = 1 – q₁.q₂.q₃ = 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\).

प्रश्न 8.
एक थैले में 6 काली गेंदें, 5 सफेद गेंदें और 2 नीली गेंदें हैं। उनमें से दो गेंदें यदृच्छया बाहर निकाली जाती हैं। प्रायिकता ज्ञात कीजिए कि दोनों गेंदें सफेद हों।
हल:
दिया है:
काली गेंद = 6, सफेद गेंद = 5, नीली गेंद = 2
∴ कुल गेंद = (6 + 5 + 2) = 13
अतः n(S) = ¹³C₂
माना सफेद गेंद निकलने की घटना A है, तब
n(A) = ⁵C₂
∴ सूत्र, P(A) = \(\frac { n(A) }{ n(S) } \) = \(\frac { ^{ 5 }C_{ 2 } }{ ^{ 13 }C_{ 2 } } \)
P(A) = \(\frac { 5!/3!.2! }{ 13!/11!.2! } \) = \(\frac { \frac { 5\times 4 }{ 2\times 1 } }{ \frac { 13\times 12 }{ 2\times 1 } } \) = \(\frac { 5\times 4 }{ 13\times 12 } \) = \(\frac{5}{39}\).

प्रश्न 9.
एक थैले में 6 लाल, 4 सफेद और 5 नीली गेंदें हैं। यदि थैले में से एक-एक करके गेंद निकाली जाये तथा उन्हें वापस न रखा जाये तो पहले के लाल, दूसरे के सफेद और तीसरे के नीले होने की प्रायिकता ज्ञात कीजिए।
हल:
दिया है:
एक थैले में 6 लाल (R), 4 सफेद (W) और 5 नीली (B) गेंदें हैं।
इस प्रकार कुल गेंदों की संख्या = 6 + 4 + 5 = 15
∴ पहली गेंद के लाल होने की प्रायिकता P(R) = \(\frac{6}{15}\)
अब थैले में कुल 15 – 1 = 14 गेंदें हैं
∴ दूसरे गेंद के सफेद होने की प्रायिकता P(W) = 4.
इसके बाद थैले में कुल 14 – 1 = 13 गेंदें हैं।
∴ तीसरे गेंद के नीली होने की प्रायिकता P(B) = \(\frac{5}{13}\)
अत: अभीष्ट (मिश्र) प्रायिकता = \(\frac{6}{15}\) × \(\frac{4}{14}\) × \(\frac{5}{13}\) = \(\frac{2}{5}\) × \(\frac{2}{7}\) × \(\frac{5}{13}\) = \(\frac{4×1}{91}\) = \(\frac{4}{91}\).

प्रश्न 10.
दो घनाकार पासे एक साथ फेंके जाते हैं। पहले पासे पर सम संख्या अथवा दोनों का योगफल 9 आने की प्रायिकता ज्ञात कीजिए।
हल:
दो घनाकार पासों को साथ-साथ उछालने के तरीके = 6 × 6 = 36
∴ n(S) = 36
मानलो E₁ = पहले पासे पर सम संख्या आने की घटना
तथा E₂ = योग 9 प्राप्त करने की घटना
∴ E₁ = {(2, 1), (2, 2), (2, 3), (2, 4), (2,5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4,5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6,5), (6, 6)}
∴ n(E₁) = 18
E₂ = {(3, 6), (4, 5), (5, 4), (6, 3)}
∴ n(E₂) = 4
(E₁∩E₂) = {(4, 5), (6, 3)}
∴ n(E₁∩E₂) = 2
अत: P(E₁) = \(\frac { n(E_{ 1 }) }{ n(S) } \) = \(\frac{18}{36}\)
P(E₂) = \(\frac { n(E_{ 2 }) }{ n(S) } \) = \(\frac{4}{36}\)
तथा
P(E₁∩E₂) = \frac { n(E_{ 1 }∩E_{ 2 }) }{ n(S) } = \(\frac{2}{36}\)
∴ P(E₁∪E₂) = P(E₁) + P(E₂) – P(E₁∩E₂)
= \(\frac{18}{36}\) + \(\frac{4}{36}\) – \(\frac{2}{36}\) = \(\frac{20}{36}\)
⇒ P(E₁∪E₂) = \(\frac{5}{9}\).

प्रश्न 11.
दो थैलों में से एक में 3 काली और 4लाल गेंदें हैं और दूसरे में 8 काली और 10 लाल गेंदें हैं। यदि किसी एक थैले को चुनकर उसमें से एक गेंद निकाली जाये तो उसके लाल होने की प्रायिकता ज्ञात कीजिए।
हल: I थैला
3B + 4R = 7 कुल गेंद
दो थैले में से 1 थैला चुनने की प्रायिकता है = \(\frac{1}{2}\)
1 थैले से 1 लाल गेंद निकालने की प्रायिकता = \(\frac{4}{7}\)
अतः I थैला चुनना तथा इसी थैले से,
1 लाल गेंद निकालने की प्रायिकता P₁ = \(\frac{1}{2}\) × \(\frac{4}{7}\) = \(\frac{2}{7}\)
पुनः यदि II थैला चुना गया तो, इसकी प्रायिकता = \(\frac{1}{2}\)
II थैले से 1 लाल गेंद निकालने की प्रायिकता = \(\frac{10}{18}\) = \(\frac{5}{9}\)
अतः II थैला चुनना तथा इसी थैले से,
1 लाल गेंद निकालने की प्रायिकता P₂ = \(\frac{1}{2}\) × \(\frac{5}{9}\) = \(\frac{5}{18}\)
उपर्युक्त दोनों घटनाएँ परस्पर अपवर्जी हैं। इनमें से एक ही घटना घटेगी।
अतः अभीष्ट प्रायिकता P = P₁ + P₂ = \(\frac{2}{7}\) + \(\frac{5}{18}\) = \(\frac{36+35}{126}\) = \(\frac{71}{126}\)

प्रश्न 12.
दो थैलों में एक में 5 लाल और 7 सफेद गेंदें हैं और दूसरे में 3 लाल और 12 सफेद गेंदें हैं। दोनों थैलों में से किसी एक से एक गेंद निकाली जाती है। प्रायिकता ज्ञात कीजिये कि वह गेंद लाल है।
हलः
प्रश्न क्र. 11 की भाँति हल करें।
उत्तर:
\(\frac{37}{120}\)

प्रश्न 13.
एक थैले में 50 बोल्ट तथा 150 नट हैं। आधे बोल्ट और आधे नट जंग लगे हैं। यदि यदृच्छया एक नट थैले से निकाला जाये तो इसके जंग लगे हुये या बोल्ट होने की प्रायिकता ज्ञात कीजिए।
हल:
माना प्रतिदर्श समष्टि S हैं। तब,
n (S) = 200, (∵ 50 बोल्ट + 150 नट = 200)
प्रश्नानुसार,
आधे बोल्ट और आधे नट जंग लगे हैं
इनकी संख्या = 25 + 75 = 100
∴ E₁: जंग लगी वस्तु निकलने की घटना
तब P(E₁) = \(\frac { n(E_{ 1 }) }{ n(S) } \) = \(\frac{100}{200}\)
E₂: बोल्ट निकलने की घटना
तब P(E₂) = \(\frac { n(E_{ 2 }) }{ n(S) } \) = \(\frac{50}{200}\)
उक्त दोनों घटनाओं E₁ और E₂ में 25 जंग लगे बोल्ट उभयनिष्ठ है।
∴ n(E₁∩E₂) = 25
∴ P(E₁∪E₂) = P(E₁) + P(E₂) – P(E₁∩E₂)
= \(\frac{100}{200}\) + \(\frac{50}{200}\) – \(\frac{25}{200}\) = \(\frac{125}{200}\) = \(\frac{5}{8}\).

प्रश्न 14.
A किसी लक्ष्य को 5 बार में से 4 बार भेद सकता है। B, 4 में से 3 बार और C, 3 बार में से 2 बार। वे एक साथ निशाना लगाते हैं। कम-से-कम दो व्यक्तियों द्वारा निशाना लगाये जाने की प्रायिकता ज्ञात कीजिये।
हल:
A द्वारा निशाना लगाये जाने की प्रायिकता = \(\frac{4}{5}\)
B द्वारा निशाना लगाये जाने की प्रायिकता = \(\frac{3}{4}\)
C द्वारा निशाना लगाये जाने की प्रायिकता = \(\frac{2}{3}\)
कम-से-कम दो व्यक्तियों द्वारा निशाना लगाये जाने के निम्न प्रकार होंगे –
(i) जब A, B, C तीनों निशाना लगा लें जिसकी प्रायिकता = \(\frac{4}{5}\) × \(\frac{3}{4}\) × \(\frac{2}{3}\) = \(\frac{24}{60}\)
(ii) जब B, C के निशाने लग जाये पर A का निशाना न लगे, इसकी प्रायिकता
= (1 – \(\frac{4}{5}\) × \(\frac{3}{4}\) × \(\frac{2}{3}\) = \(\frac{1}{5}\) × \(\frac{3}{4}\) × \(\frac{2}{3}\) = \(\frac{6}{60}\)
(iii) जब C, A के निशाने लग जाये पर B का निशाना न लगे, इसकी प्रायिकता
= \(\frac{4}{5}\) × (1 – \(\frac{3}{4}\) ) × \(\frac{2}{3}\) = \(\frac{4}{5}\) × \(\frac{1}{4}\) × \(\frac{2}{3}\) = \(\frac{8}{60}\)
(iv) A, B के निशाने लग जाये पर C का निशाना न लगे, इसकी प्रायिकता
= \(\frac{4}{5}\) × \(\frac{3}{4}\) × (1 – \(\frac{2}{3}\) ) = \(\frac{4}{5}\) × \(\frac{3}{4}\) × \(\frac{1}{3}\) = \(\frac{12}{60}\)
सभी घटनायें परस्पर अपवर्जी हैं अतः अभीष्ट प्रायिकता
= \(\frac{24}{60}\) + \(\frac{6}{60}\) + \(\frac{8}{60}\) + \(\frac{12}{60}\) = \(\frac{50}{60}\) = \(\frac{5}{6}\)

प्रश्न 15.
एक कक्षा में 30% विद्यार्थी भौतिकी में, 25% गणित तथा 10% दोनों में फेल होते हैं। एक छात्र यदृच्छया चुना जाता है तो प्रायिकता ज्ञात कीजिये कि वह
(i) गणित में फेल होता है यदि भौतिकी में फेल है
(ii) भौतिकी में फेल होता है जबकि वह गणित में फेल है
हलः
भौतिकी में फेल होने की प्रायिकता P(A) = \(\frac{30}{100}\)
गणित में फेल होने की प्रायिकता P(B) = \(\frac{25}{100}\)
भौतिकी और गणित दोनों में फेल होने की प्रायिकता P(A∩B) = \(\frac{10}{100}\)
(i) P( \(\frac{B}{A}\) ) = \(\frac { P(B∩A) }{ P(A) } \) = \(\frac { \frac { 10 }{ 100 } }{ \frac { 30 }{ 100 } } \) = \(\frac{1}{3}\)
(ii) P( \(\frac{A}{B}\) ) = \(\frac { P(A∩B) }{ P(B) } \) = \(\frac { \frac { 10 }{ 100 } }{ \frac { 25 }{ 100 } } \) = \(\frac{10}{25}\) = \(\frac{2}{5}\).

प्रश्न 16.
दो थैले A और B में क्रमशः 8 हरी और 9 सफेद और 5 हरी और 4 सफेद गेंदें रखी हैं। किसी एक थैले में से यादृच्छया एक गेंद निकाली गई है जो कि हरे रंग की है। प्रायिकता ज्ञात कीजिए कि यह गेंद थैले B से निकाली गई है। (NCERT)
हल:
माना E₁: थैले A के चयन होने की घटना
∴ P(E₁) = \(\frac{1}{2}\) ………………… (1)
E₂: थैले B के चयन होने की घटना
P(E₂) = \(\frac{1}{2}\) ………………… (2)
पुनः माना C : थैले से एक हरे रंग की गेंद निकालने की घटना
∴ सप्रतिबंधी प्रायिकता की परिभाषा से,
P( \(\frac { C }{ E_{ 1 } } \) ) = P(एक हरे रंग की गेंद थैले A से निकालना)
P( \(\frac { C }{ E_{ 2 } } \) ) = P(एक हरे रंग की गेंद थैले B से निकालना)
⇒ P( \(\frac { C }{ E_{ 1 } } \) ) = \(\frac{8}{8+9}\) = \(\frac{8}{17}\) [∴ थैले A में 8 हरी और 9 सफेद]
⇒ P( \(\frac { C }{ E_{ 2 } } \) ) = \(\frac{5}{5+4}\) = \(\frac{5}{9}\)
अब अभीष्ट प्रायिकता अर्थात् थैले B से एक गेंद निकालने की प्रायिकता, जबकि यह ज्ञात है कि वह हरे रंग की है।

प्रश्न 17.
एक कंपनी दो फैक्टरी में साईकिल बनाती है। पहली फैक्टरी 60% और दूसरी फैक्टरी 40% साईकिल बनाती है। पहली फैक्टरी से 80% साईकिल उच्च स्तर की और 90% साईकिल दूसरी फैक्टरी उच्च स्तर की बनायी जाती है। एक साईकिल यादृच्छया उच्च स्तर की चुनी जाती है उसके दूसरी फैक्टरी से होने की प्रायिकता ज्ञात कीजिए। [CBSE 2003]
हल:
माना E₁: एक साईकिल पहली फैक्टरी से चुने जाने की घटना
∴ P(E₁) = \(\frac{60}{100}\)
माना E₂: एक साईकिल दूसरी फैक्टरी से चुने जाने की घटना
∴ P(E₂) = \(\frac{40}{100}\)
पुनः माना E उच्च स्तर की एक साईकिल चुने जाने की घटना है, तब
P( \(\frac { E }{ E_{ 1 } } \) ) = उच्च स्तर की एक साईकिल चुने जाने की प्रायिकता जो दूसरी फैक्टरी से बनाई गयी है।
⇒ P( \(\frac { E }{ E_{ 1 } } \) ) = \(\frac{80}{100}\), [दिया है 80%]
P( \(\frac { E }{ E_{ 2 } } \) ) = उच्च स्तर की एक साईकिल चुने जाने की प्रायिकता जो दूसरी फैक्टरी से बनाई गयी है।
⇒ p( \(\frac { E }{ E_{ 2 } } \) ) = \(\frac{90}{100}\), [दिया है 90%]
∴ अब अभीष्ट प्रायिकता (अर्थात् थैले B से एक गेंद निकालने की प्रायिकता, जबकि यह ज्ञात है कि वह हरे रंग की है।

प्रश्न 18.
एक टी. वी. (T. V.) बनाने के कारखाने में मशीनें ( यंत्र) A, B और C कुल उत्पादन का क्रमश: 30%, 20% और 50% टी. वी. (T. V.) बनाती हैं। इन मशीनों के उत्पादन का क्रमशः 7%, 5% और 2% टी. वी. खराब (त्रुटिपूर्ण ) पायी जाती हैं। टी. वी. के कुल उत्पादन में से एक टी. वी. यादृच्छया जाँच के लिए निकाला गया है और वह खराब (त्रुटिपूर्ण) पाया गया। इसकी प्रायिकता ज्ञात कीजिए कि वह मशीन A द्वारा बनाया गया है। (CBSE)
हल:
माना कि घटनाएँ T₁, T₂, और T₃, निम्नानुसार हैं –
घटना T₁ : टी. वी. मशीन A द्वारा बनाया गया है।
घटना T₂ : टी. वी. मशीन B द्वारा बनाया गया है।
घटना T₃ : टी. वी. मशीन C द्वारा बनाया गया है।
यहाँ ध्यान देने योग्य बात यह है कि घटनाएँ परस्पर अपवर्जी हैं।
पुनः माना E : टी. वी. (T. V.) खराब होने की घटना है।
दिया है: P(T₁) = 30% = \(\frac{30}{100}\)
P(T₂) = 20% = \(\frac{20}{100}\)
और P(T₃) = 50% = \(\frac{50}{100}\)
पुनः P( \(\frac { E }{ T_{ 1 } } \) ) = 7% = \(\frac{7}{100}\)
P( \(\frac { E }{ T_{ 2 } } \) ) = टी. वी. के खराब होने की प्रायिकता जबकि वह मशीन A द्वारा बनाया गया है।
⇒ P( \(\frac { E }{ T_{ 2 } } \) ) = 5% = \(\frac{5}{100}\)
इसी प्रकार
P( \(\frac { E }{ T_{ 3 } } \) ) = 2% = \(\frac{2}{100}\)
अब बेज़ प्रमेय (Baye’s Theorem) से,
P( \(\frac { T_{ 1 } }{ E } \) ) = मशीन A द्वारा बनाये गये टी. वी. के खराब होने की प्रायिकता

प्रश्न 19.
मोहन द्वारा झूठ बोलने की प्रायिकता = है। एक सिक्का उछालने पर मोहन द्वारा चित (Head) बताया जाता है; इसकी प्रायिकता ज्ञात कीजिए कि सिक्का उछालने पर वास्तव में चित आता (NCERT)
हल:
माना A: एक सिक्का उछालने पर चित आने की घटना
∴ P(A) = \(\frac{1}{2}\)
B: एक सिक्का चित नहीं आने की घटना है
∴ P(B) = 1 – P(A) = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
पुनः माना E: मोहन द्वारा एक सिक्का के एक फेंक में यह बताने कि एक सिक्का के एक फेंक में चित आने की घटना है।
अब P( \(\frac{E}{A}\) ) = 1 – \(\frac{1}{5}\) = \(\frac{4}{5}\)
अब बेज़ प्रमेय (Bayes’ theorem) से
P( \(\frac{A}{E}\) ) = मोहन द्वारा यह बताने पर कि सिक्का के एक फेंक में चित आया है।

प्रश्न 20.
बैग A में 3 सफेद और 4 लाल गेंदें और बैग B में 5 सफेद और 6 लाल गेंदें हैं। इन थैलों से एक गेंद यादृच्छया निकाली जाती है और वह लाल पायी गयी। बैग B से इस गेंद के निकलने की प्रायिकता ज्ञात कीजिए।
हल:
माना E₁ = बैग A के चुनाव की घटना
E₂ = बैग B के चुनाव की घटना
∴ P(E₁) = \(\frac{1}{2}\); P(E₂) = \(\frac{1}{2}\)
⇒ P(E₁) = P(E₂) = \(\frac{1}{2}\)
पुनः माना एक एक एक गंद लाल होने की घटना
अब प्रश्नानुसार,
P ( \(\frac { R }{ E_{ 1 } } \) ) = \(\frac{4}{3+4}\) = \(\frac{4}{7}\) चूँकि बैग A में 3 सफेद और 4 लाल गेंदें हैं।]
P ( \(\frac { R }{ E_{ 2 } } \) ) = \(\frac{6}{5+6}\) = \(\frac{6}{11}\) [चूँकि बैग B में 5 सफेद और 6 लाल गेंदें हैं।]
अब बेज़-प्रमेय (Bayes’ theorem) से,
P ( \(\frac { E_{ 2 } }{ R } \) ) = एक गेंद बैग B से चुने जाने की प्रायिकता जबकि वह लाल हो

प्रश्न 21.
तीन अभिन्न थैले A, B और C दिए गए हैं, प्रत्येक में दो-दो पुस्तकें हैं। थैले A में दोनों गणित की पुस्तकें हैं, थैले B में दोनों रसायन की पुस्तकें हैं और थैले C में एक गणित और एक रसायन की पुस्तक है। एक विद्यार्थी यादृच्छया एक थैला चुनता है और उसमें से यादृच्छया एक पुस्तक निकालता है। यदि पुस्तक गणित की है, तो दूसरी पुस्तक भी थैले A से गणित की निकलने की प्रायिकता ज्ञात कीजिए। (NCERT)
हल:
माना कि E₁, E₂ और E₃ क्रमशः थैले A, B और C के चयन को दर्शाता है तो उनके चयन की प्रायिकता
P(E₁) = P(E₂) = P(E₃) = \(\frac{1}{3}\) ……… (1)
D: गणित की पुस्तक निकलने की घटना तब सप्रतिबंधी प्रायिकता की परिभाषा से
P( \(\frac { D }{ E_{ 1 } } \) ) = P (एक गणित की पुस्तक थैले A से निकलना)
= \(\frac{2}{2}\) = 1 ……….. (2)
P( \(\frac { D }{ E_{ 2 } } \) ) = P (एक गणित की पुस्तक थैले B से निकलना)
= \(\frac{0}{2}\) = 0 ………. (3) [∵ प्रश्नानुसार थैले B में दोनों रसायन की पुस्तक हैं।]
P( \(\frac { D }{ E_{ 3 } } \) ) = P (एक गणित की पुस्तक थैले C से निकलना)
= \(\frac{1}{2}\) ………. (4)
अब अभीष्ट प्रायिकता (थैले A से गणित की पुस्तक निकलने की प्रायिकता)
P ( \(\frac { E_{ 1 } }{ D } \) )
अब बेज़-प्रमेय से,

प्रश्न 22.
एक विद्यार्थी के बारे में जाँच किया जाता है कि वह 5 में से 2 बार सत्य बोलता है। वह एक पासे को फेंकता है और रिपोर्ट करता है कि पासे के ऊपरी फलक पर संख्या 6 आयी है। इसकी प्रायिकता ज्ञात कीजिए कि पासे पर सही में संख्या 6 आयी है।
हल:
माना E: उस विद्यार्थी द्वारा पासे के फेंक में यह बताने की घटना कि पासे के ऊपरी फलक पर संख्या 6 आयी है।
A: पासे के ऊपरी फलक पर संख्या 6 आने की घटना
B: पासे के ऊपरी फलक पर संख्या 6 नहीं आने की घटना …………. (1)
∴ P(A) = \(\frac{1}{6}\)
P(B) = P(नहीं A) = P( \(\bar { A } \) )
⇒ P(B) = 1 – P(A) = 1 – \(\frac{1}{6}\) ………… (2)
⇒ P(B) = \(\frac{5}{6}\)
P( \(\frac{E}{A}\) ) = विद्यार्थी द्वारा यह बताने पर कि पासे के एक फेंक में उसके ऊपरी फलक पर संख्या 6 है जबकि पासे पर वास्तव में संख्या 6 नहीं आयी है।
P( \(\frac{E}{A}\) ) = विद्यार्थी द्वारा सत्य बोलने की प्रायिकता = \(\frac{2}{5}\) ………… (3)
अब
P ( \(\frac{E}{B}\) ) = विद्यार्थी द्वारा यह बताने की प्रायिकता कि पासे के एक फेंक में उसके ऊपरी फलक पर संख्या 6 आयी है, जबकि वास्तव में संख्या 6 आयी है।
P( \(\frac{E}{B}\) ) = विद्यार्थी द्वारा सत्य नहीं (झूठ) बोलने की प्रायिकता
P( \(\frac{E}{B}\) ) = 1 – \(\frac{2}{5}\) = \(\frac{3}{5}\)
अब बेज़-प्रमेय से,
P( \(\frac{A}{E}\) ) = विद्यार्थी द्वारा यह बताने की प्रायिकता कि पासे के एक फेंक में उसके ऊपरी फलक पर संख्या 6 आयी है, जबकि वास्तव में संख्या 6 आयी है

MP Board Class 12th Maths Important Questions

MP Board Class 12th Maths Important Questions Chapter 13 Probability

Probability Important Questions

Probability Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
A bag contains 5 brown and 4 white socks. A man drawn two socks from the bag. The probability that both are of the same colour is:
(a) \(\frac { 5 }{ 108 } \)
(b) \(\frac { 18 }{ 108 } \)
(c) \(\frac { 30 }{ 108 } \)
(d) \(\frac { 48 }{ 108 } \)
Answer:
(d) \(\frac { 48 }{ 108 } \)

Question 2.
A die is roiled three times. The probability of getting a larger number then the previous number each time is:
(a) \(\frac { 5 }{ 72 } \)
(b) \(\frac { 5 }{ 54 } \)
(c) \(\frac { 13 }{ 216 } \)
(d) \(\frac { 1 }{ 18 } \)
Answer:
(d) \(\frac { 1 }{ 18 } \)

Question 3.
It is given that the events A and B are such that P(A) = \(\frac{1}{4}\), P( \(\frac{A}{B}\) ) = \(\frac{1}{2}\) and P ( \(\frac{B}{A}\) ) = \(\frac{2}{3}\), then P(B) is:
(a) \(\frac{1}{3}\)
(b) \(\frac{2}{3}\)
(c) \(\frac{1}{2}\)
(d) \(\frac{1}{6}\)
Answer:
(a) \(\frac{1}{3}\)

Question 4.
A coin is tossed 4 times. Probability of getting at least one head is:
(a) \(\frac{1}{16}\)
(b) \(\frac{2}{16}\)
(c) \(\frac{14}{16}\)
(d) \(\frac{15}{16}\)
Answer:
(d) \(\frac{15}{16}\)

Question 5.
The probability of A speaking a truth is \(\frac{4}{5}\) and that of B speaking a truth is \(\frac{3}{4}\) The probability that they will contradict each other in answering a fact is:
(a) \(\frac{3}{10}\)
(b) \(\frac{1}{4}\)
(c) \(\frac{7}{20}\)
(d) \(\frac{2}{5}\)
Answer:
(c) \(\frac{7}{20}\)

Question 2.
Fill in the blanks:

1. P( \(\frac { A∪B }{ C } \) ) = P ( \(\frac{A}{C}\) ) + ……………………….
2. P(A∩B) = ………………….. or ………………………….
3. If A and B are two independent events then, P (A ∩ B) = ………………………………..
4. If A and B are two independent events then, P ( \(\frac{A}{B}\) ) = ……………………………….
5. If the probability of random variable X is P (X), then mean E (X) will be ……………………………..
6. If probability of random variable X is P(X), then variance Var (X) will be ……………………………

Answer:

1. P( \(\frac{B}{C}\) ) – P ( \(\frac { A∩B }{ C } \) )
2. P(B).P ( \(\frac{A}{B}\) ) or P(A) . P ( \(\frac{B}{A}\) )
3. P(A).P(B)
4. P(A), P(B) ≠ 0
5. ΣX.P(X)
6. ΣX².P(X) – [ΣX.P(X)]².

Question 3.
Write True/False:

1. If A and B are any two events, then P (A – B) = P(A) – P (A∩B).
2. If A and B are any two events, then P (A ∪ B) – P (A∩B) = P(A) + P(B).
3. P( \(\frac{A}{B}\) ) = \(\frac { P(A∩B) }{ P(A) } \).
4. If A and B are any two events of a sample space S and F is another event such that P(F) ≠ 0, then P ( \(\frac { P(A∪B) }{ P(F) } \) ) = P ( \(\frac{A}{F}\) ) + P ( \(\frac{B}{F}\) ) – P ( \(\frac { P(A∪B) }{ P(F) } \) )
5. For any two events A and B. P (A∪B) = P (A∩B ) + P ( \(\bar { A } \) ∩ B) + P (A ∩\(\bar { B } \) )

Answer:

1. True
2. False
3. False
4. True
5. True.

Question 4.
Write the answer in one word/sentence :

1. Find the probability that a leap year has 53 Friday.
2. A coin is tossed 4 times. Find the probability of getting at least one head.
3. It is given that the events A and B are such that P (A) = \(\frac{1}{4}\), P ( \(\frac{A}{B}\) ) = \(\frac{1}{2}\) and P ( \(\frac{B}{A}\) ) = \(\frac{2}{3}\), then find the value of P(B).
4. There are 5 letters and 5 addressed envelops. If the letters are placed at random what is the probability that exactly 3 letters are placed in right envelops?
5. A problem in mathematics is given to 3 students whose chance of solving it are \(\frac{1}{2}\), \(\frac{1}{3}\), \(\frac{1}{4}\). What is the probability that the problem is solved?

Answer:

1. \(\frac{2}{7}\)
2. \(\frac{15}{16}\)
3. \(\frac{1}{3}\)
4. \(\frac{1}{12}\)
5. \(\frac{3}{4}\)

Probability Short Answer Type Questions

Question 1.
(A) Two dice are thrown simultaneously. Find the probability of getting a sum?
Solution:
When two dice are thrown simultaneously.
n(S) = 6² = 36.
Sample space of getting a sum 8 is
A = {(2, 6); (6, 2); (5, 3); (3, 5); (4, 4)}
∴ n(A) = 5
∴ Required probability
P(A) = \(\frac { n(A) }{ n(S) } \) = \(\frac{5}{36}\).

(B) Two cubical dice are thrown simultaneously. Find the probability of getting a sum of 9?
Solution:
Solve as Q.No. 1 (A)

(C) Two cubical dice are thrown simultaneously. Find the probability of getting a sum of 7?
Solve as Q.No. 1 (A)

Question 2.
The odds against happening of an event is 3 : 4? Find the probability of its failing?
Solution:
The odds against the happening of an event = 3 : 4
∵ Probability of its failing P ( \(\bar { A } \) ) = \(\frac { b }{ a+b } \)
⇒ P ( \(\bar { A } \) ) = \(\frac { 4 }{ 3+4 } \) = \(\frac { 4 }{ 7 } \).

Question 3.
A card is drawn from a pack of 52 cards. Find the probability that its face?
Solution:
Here, n(S) = 52
∵Cards having face = 4 knave + 4 queens + 4 kings
∴ n(A) = 4 + 4 + 4 = 12
Required probability P(A) = \(\frac { 12 }{ 52 } \) = \(\frac { 3 }{ 13 } \)

Question 4.
What is the probability that a leap year selected at random will contain 53 Sundays?
Solution:
A leap year consists of 366 days. It has 52 complete weeks and 2 days extra. The equally likely cases for the occurrence of these extra days are:

1. Monday and Tuesday,
2. Tuesday and Wednesday,
3. Wednesday and Thursday,
4. Thursday and Friday,
5. Friday and Saturday,
6. Saturday and Sunday,
7. Sunday and Monday.

Out of these 7 exclusive cases, the last two cases are favourable.
∴ n(S) = 7 and n(E) = 2
∴ The required probability = \(\frac { n(E) }{ n(S) } \) = \(\frac{2}{7}\).

Question 5.
The probability of the horse A of winning a race is \(\frac{1}{4}\) and the probability of the horse B winning the same race is \(\frac{1}{8}\)? What is the probability that one of the horse will win the race?
Solution:
Given, P(A) = Probability of winning the horse A = \(\frac{1}{4}\)
and P(B) = Probability of winning the horse B = \(\frac{1}{8}\)
Since, the events of winning the race by A and B are mutually exclusive, therefore
P(A∪B)= P(A) + P(B)
\(\frac{1}{4}\) + \(\frac{1}{8}\) = \(\frac{2+1}{8}\) = \(\frac{3}{8}\).

Question 6.
(A) A card is drawn from an ordinary pack of cards, find the probability of getting a king or a spade?
Solution:
Let the event A denotes the event of drawing a king and B of drawing a spade. Then we
n(S) = 52, n(A) = 4, n(B) = 13, n(A∩B) = 1
[Since there are 4 kings in pack of cards the cards of spade are 13 includes its king]
∴ P(A) = \(\frac { n(A) }{ n(S) } \) = \(\frac{4}{52}\), P(B) = \(\frac { n(B) }{ n(S) } \) = \(\frac{13}{52}\)
P(A∩B) = \(\frac{1}{52}\)
∴ P(A∪B)= P(A) + P(B) = P(A∩B )
= \(\frac{4}{52}\) + \(\frac{13}{52}\) – \(\frac{1}{52}\) = \(\frac{16}{52}\) = \(\frac{4}{13}\).

(B) A card is drawn from a pack of cards, find the probability that neither it is an ace nor a King?
Solution:
Total No. of King = 4, No. of ace = 4
And drawing one card from 8 cards = ⁸_C1
And drawing one card from 52 cards = ⁵²_C1
∴ P(A) = \(\frac { ^{ 8 }{ C_{ 1 } } }{ ^{ 52 }{ C_{ 1 } } } \)
∴ Probability of neither ace nor king = P( \(\bar { A } \) ) = 1 – P(A)
= 1 – \(\frac{2}{13}\) = \(\frac{11}{13}\)

Question 7.
(A) A dice is thrown once. Find the probability of getting even No. “or ” No. less than 5?
Solution:
When a dice is thrown once the sample space is
S = {1, 2, 3, 4, 5, 6}
n(S) = 6
E₁ = {2,4,6}, [∴ n(E₁) = 3]
E₁ = {1, 2, 3, 4}, [∴ n(E₂) = 4]
E₁∩E₂= {2, 4}, [∴ n(E₁ ∩E₂)
= \(\frac { n(E_{ 1 }) }{ n(S) } \) + \(\frac { n(E_{ 2 }) }{ n(S) } \) – \(\frac { n(E_{ 1 }\cap E_{ 2 }) }{ n(S) } \)
= \(\frac{3}{6}\) + \(\frac{4}{6}\) – \(\frac{2}{6}\) = \(\frac{7-2}{6}\) = \(\frac{5}{6}\).

(B) A pair of dice is thrown. Find the probability that the sum is not 9 or 11?
Solution:
Total number of ways in which 2 dice can be thrown = 6 × 6 = 36
n(S) = 36
Event of getting sum 9 = {(3, 6), (4, 5), (5, 4), (6, 3)}
Total No. = 4.
∴ Probability of getting sum 9 = \(\frac{4}{36}\) = \(\frac{1}{9}\)
Now event of getting sum 11 = {(5, 6), (6, 5)}
∴ Probability of getting sum 11 = \(\frac{2}{36}\) = \(\frac{1}{18}\)
Now probability of getting sum 9 or 11
P(A) = \(\frac{1}{9}\) + \(\frac{1}{18}\) = \(\frac{2+1}{18}\) = \(\frac{3}{18}\) = \(\frac{1}{6}\)
∴ probability of not getting sum 9 or 11
P( \(\bar { A } \) ) = 1 – P(A)
= 1 – \(\frac{1}{6}\) = \(\frac{6 – 1}{6}\) = \(\frac{5}{6}\).

Question 8.
If 3 coins are tossed simultaneously, then find the probability of getting at least one head?
Solution:
Here, n(S) = 2³ = 8
If \(\bar { A } \) A denotes not getting a head then A = {{T, T, T)}
∴ n( \(\bar { A } \) ) = 1
Probability of not getting a head P( \(\bar { A } \) ) = \(\frac { n(\bar { A) } }{ n(S) } \) = \(\frac{1}{8}\)
∴ Probability of getting at least one head P(A) = 1 – P( \(\bar { A } \) )
= 1 – \(\frac{1}{8}\) = \(\frac{7}{8}\)

Question 9.
A coin is tossed twice. Find the probability of getting head, both the times?
Solution:
Let the event of getting head in first throw be E₂ and the second throw be E₂.
∴ P(E₁) = \(\frac{1}{2}\) = P(E₂) = \(\frac{1}{2}\)
Both the events are independent
∴ P(E₁ ∩ E₂) = P(E₁) × P(E₂)
= \(\frac{1}{2}\) × \(\frac{1}{2}\)
∴ Probability of getting head both the times = P(E₁ ∩ E₂)
= \(\frac{1}{4}\)

Question 10.
In a given race, the odds in favour of three horses A, B and C are 1 : 2, 1 : 3 and 1 : 4. Find the chance that one of them will win the race?
Solution:
The probability of winning the horse, A = \(\frac { 1 }{ 1+2 } \) = \(\frac { 1 }{ 3 } \)
P(B) = The probability of winning the horse, B = \(\frac { 1 }{ 1+3 } \) = \(\frac { 1 }{ 4 } \)
P(C) = The probability of winning the horse, C = \(\frac { 1 }{ 1+4 } \) = \(\frac { 1 }{ 5 } \)
These events are mutually exclusive, therefore the probabilities of winning any one horses
= P(A) + P(B) + P(C)
= \(\frac{1}{3}\) + \(\frac{1}{4}\) + \(\frac{1}{5}\)
= \(\frac{20+15+12}{60}\) = \(\frac{47}{60}\)

Question 11.
A problem in mathematics is given to three students A, B and C whose chances of solving it are \(\frac{1}{2}\), \(\frac{1}{3}\), \(\frac{1}{4}\) respectively. Find the probability that the problem will not be solved?
Solution:
The chances of A, B, C solving the problem are respectively
\(\frac{1}{2}\), \(\frac{1}{3}\), \(\frac{1}{4}\)
∴ The chances of A, B, C not solving the problem are respectively
1 – \(\frac{1}{2}\), 1 – \(\frac{1}{3}\), 1 – \(\frac{1}{4}\), i.e; \(\frac{1}{2}\), \(\frac{2}{3}\), \(\frac{3}{4}\)
∴ The probability that none of the students A, B, C is able to solve the probelm
= \(\frac{1}{2}\) × \(\frac{2}{3}\) × \(\frac{3}{4}\) = \(\frac{1}{4}\).

Probability Long Answer Type Questions – I

Question 1.
In Raipur 20% persons read English newspaper 40% persons read Hindi newspaper and 5% person read both. What percentage of persons are not read any newspaper?
Solution:
P(A) = The probability of reading English newspaper
= \(\frac{20}{100}\) = \(\frac{1}{5}\)
P(B) = The probability of reading Hindi newspaper
= \(\frac{40}{100}\) = \(\frac{2}{5}\)
P(A∩B) = The probability of reading both English and Hindi newpaper
= \(\frac{5}{100}\) = \(\frac{1}{20}\)
No. of persons not reading any newspaper

Question 2.
A speaks truth in 75% of the cases and B in 80% of the cases. In what percentage of cases are they likely to contradict each other in stating the same fact?
Or
Mohan speaks truth in 75% of cases. Sohan speaks truth in 80% of the cases. In what percentage of cases did the Mohan speaks truth and Sohan speaks lie (or when they contradict each other)?
Solution:
P(A) = Probability that A speaks the truth = \(\frac{75}{100}\) = \(\frac{3}{4}\)
and P(B) = Probability that B speaks the truth = \(\frac{80}{100}\) = \(\frac{4}{5}\)
∴ P( \(\bar { A } \) ) = Probability that A does not speak the truth = 1 – \(\frac{3}{4}\) = \(\frac{1}{4}\)
and P( \(\bar { A } \) ) = Probability that B does not speak the truth = 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\)
It is clear that A and B will contradict each other if one of them speaks the truth and the other does not.
∴ P(A and B contradict) = P(A) P( \(\bar { B } \) ) + P( \(\bar { A } \) ) P(B)
= \(\frac{3}{4}\) × \(\frac{1}{5}\) + \(\frac{1}{4}\) × \(\frac{4}{5}\) = \(\frac{7}{20}\) = \(\frac{35}{100}\)
Hence, in 35% cases, A and B will contradict each other.

Question 3.
Prove that P(A) + P( \(\bar { A } \) ) = 1?
Solution:
In total a + b trails, if an events can happen in a ways and fails in b ways, when all of these ways being equally likely to occur, then the probability of happening of event A,
P(A) = \(\frac { a }{ a+b } \)
and probability of failing of event A,
P( \(\bar { A } \) ) = \(\frac { b }{ a+b } \)
Adding eqns. (1) and (2), we get
P(A) + P( \(\bar { A } \) ) = \(\frac { a }{ a+b } \) + \(\frac { b }{ a+b } \)
= \(\frac { a+b }{ a+b } \) = 1.
∴ P(A) + P( \(\bar { A } \) ) = 1. Proved.

Question 4.
If P(A) = \(\frac{1}{2}\), P(B) = \(\frac{1}{9}\) and P(A∩B) = \(\frac{1}{18}\), then find the value of following:

(i) P( \(\frac{A}{B}\) ),

(ii) P( \(\frac{B}{A}\) )

(iii) P(A∪B)

Solution:
Given: P(A) = \(\frac{1}{2}\), P(B) = \(\frac{1}{9}\) and P(A∩B) = \(\frac{1}{18}\)

Question 5.
A group of 10 children contains 6 boys and 4 girls. Three children are chosen at random from this group. Find the probability that this group chosen:

1. Does not contain any girl.
2. Contains at least one girl.

Solution:
Total number of children = 6B + 4G = 10
3 children out of total 10 may be chosen in ¹⁰_{C3 ways}

1. 3 boys out of 6 boys may be chosen in ⁶_C3 ways
∴ Required probability = \(\frac { ^{ 6 }C_{ 3 } }{ ^{ 10 }C_{ 3 } } \) = \(\frac{1}{6}\)

2. At least 1 girl may be taken as follows
2 boys + 1 girl, number of ways = ⁶_C2 × ⁴_C1
or 3 girls, number of ways = ⁴_{C3
Hence, the required probability}

Question 6.
4 cards are fallen down by one during the shuffling of the cards. Find the probability that one card is heart, the other is diamond, the third is spade and the fourth is a club?
Solution:
Number of cards = 52
The first can fall in 52 ways.
The favourable ways for the first card to be heart = 13
Probability = \(\frac{13}{52}\) = \(\frac{1}{4}\)
The number of remaining cards = 52 – 1 = 51
Another card can be fall in 51 ways.
The favourable ways for this card to be a spade = 13
Probability = \(\frac{13}{50}\)
Number of remaining cards = 50 – 1 = 49
∴ Probability = \(\frac{13}{49}\)
Therefore, by compound probability principle, required probability
= \(\frac{1}{4}\) × \(\frac{13}{51}\) × \(\frac{13}{50}\) × \(\frac{13}{49}\)

Question 7.
A problem of Maths is given to three students whose chances of solving it are \(\frac{1}{2}\), \(\frac{1}{3}\) and \(\frac{1}{4}\). What is the probability that the problem is solved by all?
Solution:
Let the probabilities of solving the question by the three students be P₁, P₂, P₃ respectively.
Then given that P₁ = \(\frac{1}{2}\), P₂ = \(\frac{1}{3}\), P₃ = \(\frac{1}{4}\)
∴ The probabilities of not solving the problem by them are
q₁ = 1 – p₁ = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
q₂ = 1 – p₂ = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
and q₃ = 1 – p₃ = 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\)
∴ Probability that all of the three students do not solve the problem
= q₁q₂q₃
= 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\).

Question 8.
A bag contains 6 red, 4 white and 5 blue balls. If balls are drawn one by one from bag and they are not replaced, then what is the probability of 1st red, 2nd white and 3rd blue?
Solution:
A bag contains 6 red (R), 4 white (W), 5 blue (B) balls.
Total No. of balls = 6 + 4 + 5 = 15
∴ Probability of 1st ball is red,
P(R) = \(\frac{6}{15}\)
Now total balls in the bag 15 – 1 = 14
∴ Probability of 2nd ball is white,
P(W) = \(\frac{4}{14}\)
Now the total No. of balls in bag 14 – 1 = 13
∴ Probability of 3rd ball is blue,
P(B) = \(\frac{5}{13}\)
Hence, required probability
= \(\frac{6}{15}\) × \(\frac{4}{14}\) × \(\frac{5}{13}\)
= \(\frac{2}{5}\) × \(\frac{2}{7}\) × \(\frac{5}{13}\) = \(\frac{4×1}{91}\) = \(\frac{4}{91}\)

Question 9.
A bag contains 6 red, 4 white and 6 blue balls. If balls are drawn one by one from bag without replacement, then what is the probability of drawing 1st ball red, 2nd ball white and 3rd ball blue?
Solution:
Solve as Q. No. 8.

Question 10.
Two cubical dice are thrown simultaneously. Find the probability that the first dice shows an even number or both the dice show the sum 9?
Solution:
Here (S) = 6 × 6 = 36
The events of getting an even number on the first dice is
E₁ = {(2, 1), (2,-2), (2, 3), (2,4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4,4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6,4), (6, 5), (6, 6)}
∴ n(E₁) = 18
The event of getting a total of 9 on the two dice is
E₂ = {(3,6), (4, 5), (5,4), (6, 3)}
∴ n(E₂) = 4 .
Also, E₁∩E₂
⇒ n(E₁∩E₂) = 2
∴ P(E₁) = \(\frac{18}{36}\), P(E₂) = \(\frac{4}{36}\) and P(E₁∩E₂) = \(\frac{2}{36}\)
∴ The required probability = P(E₁∪E₂)
= P(E₁) + P(E₂) – P(E₁∩E₂)
= \(\frac{18}{36}\) + \(\frac{4}{36}\) – \(\frac{2}{36}\) = \(\frac{20}{36}\) = \(\frac{5}{9}\)

Question 11.
Out of two bags one contains 3 black and 4 red balls and the second bag contains 8 black and 10 red balls. If one bag is chosen and a ball is drawn from it, then find the probability that is a red ball?
Solution:
I bag
3B + 4R = 7 balls total

II bag
8B + 10R = 18 balls total
(i) Selecting the I bag:
Probability that one bag is chosen (out of two bags) = \(\frac{1}{2}\)
1 red ball is drawn from I bag probability = \(\frac{4}{7}\)
Hence, the probability that the I bag is chosen as well as 1 red ball is drawn from it
(Compound event) P₁ = \(\frac{1}{2}\) × \(\frac{4}{7}\) = \(\frac{2}{7}\) ………………… (1)

(ii) If selecting II bag:
Probability that Il bag is chosen = \(\frac{1}{2}\)
I red ball is drawn from II bag probability = \(\frac{10}{18}\) = \(\frac{5}{9}\)
Hence, the probability that the II bag is chosen as well as 1 red ball is drawn from it
(Compound event) P₂ = \(\frac{1}{2}\) × \(\frac{5}{9}\) = \(\frac{5}{18}\) ……………………… (2)
The above two event are mutually exclusive.
Hence, only one event out of these two will happen.
Hence, the required probability P = P₁ + P₂
= \(\frac{2}{7}\) + \(\frac{5}{18}\)
= \(\frac{36+35}{126}\) = \(\frac{71}{126}\).

Question 12.
There are two bags, one contains 5 red and 7 white balls and second bag contains 3 red and 12 white balls. One ball is drawn from any of these bags at random? Find the probability that the ball is red?
Solution:
Solve same as Q. No. 11.

Question 13.
A bag contains 50 bolts and 150 nuts. Half of the bolts and nuts are rusted. If one item is taken out at random find out the probability that ¡ts rusted o is a bolt?
Solution:

Question 14.
A can hit a target 4 times in 5 shots; B, 3 times in 4 shots and C, 2 times 3 shoots at one time. Find the probability of at least two shots hit?
Solution:
Probability of A hitting the target = \(\frac{4}{5}\)
Probability of B hitting the target = \(\frac{3}{4}\)
Probability of C hitting the target = \(\frac{2}{3}\)
At least 2 can hit the target in the following ways:
(i) Probability that A, B,C all hit the target = \(\frac{4}{5}\) × \(\frac{3}{4}\) × \(\frac{2}{3}\) = \(\frac{2}{5}\)

(ii) Probability that B and C hit the target and not A
= ( 1 – \(\frac{4}{5}\) ) × \(\frac{3}{4}\) × \(\frac{2}{3}\) = \(\frac{1}{5}\) × \(\frac{3}{4}\) × \(\frac{2}{3}\) = \(\frac{1}{10}\)

(iii) Probability that C and A hit the target and not B
= \(\frac{4}{5}\) × (1 – \(\frac{3}{4}\) × \(\frac{2}{3}\) = \(\frac{4}{5}\) × \(\frac{1}{4}\) × \(\frac{2}{3}\)
= \(\frac{2}{15}\)

(iv) Probability that A and B hit the target but not C
= \(\frac{4}{5}\) × \(\frac{3}{4}\) × (1 – \(\frac{2}{3}\)) = \(\frac{4}{5}\) × \(\frac{3}{4}\) × \(\frac{1}{3}\) = \(\frac{1}{5}\)
Since, all these events are mutually exclusive,
∴ Required probability = \(\frac{2}{5}\) + \(\frac{1}{10}\) + \(\frac{2}{15}\) + \(\frac{1}{5}\) = \(\frac{5}{6}\)

Question 15.
In a class 30% students fail in physics, 25% fails in maths and 10% fails in both. If one student selects at random, then And the probability of:

(i) Fail in maths when fail in physics also,
(ii) Fail in physics when fail in maths also,
(iii) Fail in maths or physics.

Solution:
Probability of fail in physics is
P(A) = \(\frac{30}{100}\)
Probability of fail in maths is
P(B) = \(\frac{25}{100}\)
Probability of fail in physics and maths both is
P(A∩B) = \(\frac{10}{100}\)

Question 16.
Two bags A and B contain 8 green and 9 white balls and 5 green and 4 white balls respectively. One ball is drawn at random from one of the bags and it is found to be green. Find the probability that it is drawn from bag B? (NCERT)
Solution:
Let E₁ : Event of choosing bag A
∴ P(E₁) = \(\frac{1}{2}\) ………………………. (1)
E₂ = \(\frac{1}{2}\) ………………………… (2)
Again, let C: Event of drawing a green ball
∴ By defnition of conditional probability

The required probability (i.e; the probability of obtaining a ball from bag B when it is green.)

Question 17.
A company has two plants to manufacture bicycles. The first plant manufacture 60% of the bicycles and second plant 40%. Also 80% of the bicycles are rated standard quality at the first plant and 90% of standard quality at the second plant. A bicycle is picked up at random and found to be standard quality? Find the probability that it comes from the second plant? (CBSE 2003)
Solution:
Let E₁: The event of choosing a bicycle from first plant.
∴ P(E₁) = \(\frac{60}{100}\) ………………….. (1)
Let E₂: The event of choosing a bicycle from first plant.
P(E₂) = \(\frac{40}{100}\) ……………………….. (2)
Let E be event of choosing a bicycycle of standard quality, then
P ( \(\frac { E }{ E_{ 1 } } \) ) = Probability of choosing a bicycle of standard quality, given that it is produced by the first plant.
⇒ P ( \(\frac { E }{ E_{ 1 } } \) ) = \(\frac{80}{100}\), (given 80%)
P( \(\frac { E }{ E_{ 2 } } \) ) = Probability of choosing a bicycle of standard quality, given that it is produced by the second plant.
⇒ P( \(\frac { E }{ E_{ 2 } } \) ) = \(\frac{90}{100}\) (given 90%)
∴ The required probability (i.e., probability of choosing a bicycle from the second plant, given that it is of standard quality,

Question 18.
A company manufactures T.V., machine A, B and C manufatures 30%, 20% and 50% T.V. of the total production of their outputs 7%, 5% and 2% are defec¬tive T.V. respectively. A T.V. is selected at random from the production and is found to be defective?
Find the probability that defective T.V. which is manufactured by machine A? (CBSE 2015)
Solution:
Let the events T₁, T₂ and T₃ are the following:
Event T₁ : T.V. manufactured by machined A.
Event T₂ : T.V. manufactured by mechine B.
Event T₃: T.V. manufactured by machine C.
It is noted that these events are mutually exclusive.
Again, let E : The event of defective T.V.
Given, P(T₁) = 30% = \(\frac{30}{100}\) ………………….. (1)
P(T₂) = 20% = \(\frac{20}{100}\) ………………….. (2)
P(T₃) = 50% = \(\frac{50}{100}\) ………………….. (3)
Again P( \(\frac { E }{ T_{ 1 } } \) ) = The probability that T.V. is defective when it is manufactured by machine A.
⇒ P(\(\frac { E }{ T_{ 1 } } \) ) = 7% = \(\frac{7}{100}\)
P( \(\frac { E }{ T_{ 2 } } \) ) = The probability that T.V. is defective when it is manufactured by machine B.
⇒ P( \(\frac { E }{ T_{ 3 } } \) ) = 5% = \(\frac{5}{100}\)
Similarly,
P( \(\frac { E }{ T_{ 3 } } \) ) = The probability that T.V. is defective when it is manufactured by machine C.
⇒ P( \(\frac { E }{ T_{ 3 } } \) ) = 2% = \(\frac{2}{100}\)
Now, from Bayes’s theorem
P( \(\frac { T_{ 1 } }{ E } \) ) = The probability that T.V. is defective when it is manufactured by machine A.

Question 19.
The probability that Mohan does not speak the truth is \(\frac{1}{5}\) Mohan speaks the head when a coin is thrown, find the probability that really gettting head by the throw of a coin? (NCERT)
Solution:
Let A: The event of getting head.
P(A) = \(\frac{1}{2}\)
B: The event of not getting head.
∴ P(B) = 1 – P(B) = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
Again, Let E: The event that really getting head by the throw of a coin and Mohan speaks.
P( \(\frac{E}{B}\) ) = \(\frac{1}{5}\)
and P( \(\frac{E}{A}\) ) = 1 – \(\frac{1}{5}\) = \(\frac{4}{5}\)
Now, from Bayes’s thoerem
P( \(\frac{A}{E}\) ) = Probability that the report of the Mohan that head has occured is actually head.

Question 20.
Bag A contains 3 white and 4 red balls and bag B contains 5 white and 6 red balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it is drawn from bag B?
Solution:
Let E₁ = Event of choosing bag A.
E₂ = Event of choosing bag B.
P(E₁) = \(\frac{1}{2}\); P(E₂) = \(\frac{1}{2}\) ………………….. (1)
⇒ P(E₁) = P(E₂) = \(\frac{1}{2}\)
Again, let R = Event of drawing a red ball.
According to the question.
P( \(\frac { R }{ E_{ 1 } } \) ) = The probability that a red ball is drawn when it is selected from bag A.
⇒ P( \(\frac { R }{ E_{ 1 } } \) ) = \(\frac{4}{3+4}\) = \(\frac{4}{7}\) …………………. (2) [since bag A contains 3 white and 4 red balls]
P( \(\frac { R }{ E_{ 2 } } \) ) = \(\frac{6}{5+6}\) = \(\frac{6}{11}\) ……………….. (3) [since bag B contains 5 white and 6 red balls]
Now, from Bayes’s theorem,
P( \(\frac { E_{ 2 } }{ R } \) ) = The probability that a ball is drawn from bag B when it is red.

Question 21.
Three different bags A, B and C are given, each bag contains 2-2 books. Bag A contains both 2 mathematics books, bag B contains 2 chemistry books and bag C contains, one mathematics and one chemistry book. The student select one bag ran¬domly and from that bag he randomly take out one book. The book he has taken out is that of mathematics. Find the probability that the second book he draws out from that bag is also of mathematics or that he is selected bag A given? (NCERT)
Solution:
Let events E₁, E₂ and E₃ represents the selection of bag A, B and C respctively.
P(E₁) = P(E₂) = P(E₃) = \(\frac{1}{3}\) …………………….. (1)
D: The event of mathematics book happen
P( \(\frac { D }{ E_{ 1 } } \) ) = P(A mathematics book chosen from bag A)
= \(\frac{2}{2}\) = 1 ……………….. (2)
P( \(\frac { D }{ E_{ 2 } } \) ) = P(A mathematics book chosen from bag B)
= \(\frac{0}{2}\) = 0 ………………… (3)
[∵according to the questions, both chemistry books are in bag B]
P( \(\frac { D }{ E_{ 3 } } \) ) = P(A mathematics book chosen from bag C)
= \(\frac{1}{2}\)
Now, the required probability
P( \(\frac { E_{ 1 } }{ D } \) )
Now, by Bayes’ thoerem,
P( \(\frac { E_{ 1 } }{ D } \) )

Question 22.
By examining about a student that he is known to speak the truth 2 out of 5 times. He throws a die and reports that it is a six. Find the probability that it is actually a six?
Solution:
Let E: Event that the student reports that it is a six.
A : Event of getting a six on the upper face of die.
B : Event of not getting a six on the upper face of die.
∴ P(A) = \(\frac{1}{6}\) ………………….. (1)
P(B) = P(Not A) = P( \(\bar { A } \) )
⇒ P(B) = 1 – P(A) = 1 – \(\frac{1}{6}\)
⇒ P(B) = \(\frac{5}{6}\) ………………… (2)
P( \(\frac{E}{A}\) ) = Probability that the student reports that six occurs, when six has actually occured.
P ( \(\frac{E}{A}\) ) = Probability that the student speaks the truth
= \(\frac{2}{5}\) …………………… (3)
Now, P ( \(\frac{E}{B}\) ) = Probability that the student reports that six occurs, when six that has not actually occured.
P ( \(\frac{E}{B}\) ) = Probability that the student does not speak the truth.
P( \(\frac{E}{B}\) ) = 1 – \(\frac{2}{5}\) = \(\frac{3}{5}\)
Now, by Bayes’s thoerem
P( \(\frac{A}{E}\) ) = Probability of getting six, given that the student reports it to be six.

MP Board Class 12 Maths Important Questions