## MP Board Class 11th Maths Important Questions Chapter 9 Sequences and Series Important Questions

### Sequences and Series Important Questions

Sequences and Series Objective Type Questions

(A) Choose the correct answer of the following:

Question 1.
The sum of the cube of first n positive integer is :
(a) $$\frac {n(n + 1)}{2}$$
(b) $$\frac {n(n + 1)(2n + 1)}{6}$$
(c) $$\frac {n(n + 1)(n + 2)}{6}$$
(d) {$$\frac {n(n + 1)}{2}$$}2
(d) {$$\frac {n(n + 1)}{2}$$}2

Make use of this free Radius of Convergence Calculator to get the radius of convergence of a power series. Online calculator tool gives output in no time.

Question 2.
The sum of n term of arithmetic progression is 2n + 3n2, its second term will be :
(a) 10
(b) 12
(c) 16
(d) 11
(c) 16

Question 3.
If arithmetic mean of a and b is $$\frac { { a }^{ n }+{ n }^{ n } }{ { a }^{ n – 1 }+{ b }^{ n – 1 } }$$, then the value of n will be :
(a) 1
(b) 0
(c) – 1
(d) $$\frac {1}{2}$$
(a) 1

Question 4.
Which term of the series 8, 4,0, ………….. is – 24 :
(a) 7th
(b) 28th
(c) 8th
(d) 9th
(d) 9th

Question 5.
If the first term of arithmetic progression be a and last term is l, then the sum of n terms will be :
(a) $$\frac {n}{2}$$[2a – (n – 1)d]
(b) $$\frac {n}{2}$$[2a + (n – 1)d]
(c) $$\frac {n}{2}$$(a + l)
(d ) $$\frac {n}{2}$$(a – l).
(c) $$\frac {n}{2}$$(a + l)

Question 6.
The next term of the sequence 2$$\sqrt { 2 }$$, $$\sqrt { 2 }$$, 0, …………… is :
(a) – $$\sqrt { 3 }$$
(b) $$\frac { 1 }{ \sqrt { 2 } }$$
(c) – $$\sqrt { 2 }$$
(d) $$\sqrt { 2 }$$
(c) – $$\sqrt { 2 }$$

Question 7.
If 2x, x + 8, 3x + 1 are in A.P, then value of x will be :
(a) 3
(b) 7
(c) 5
(d) 2
(c) 5

Question 8.
The 15th term from the end of the A.P. 2, 6,10, …………., 86 is :
(a) 30
(b) 32
(c) 46
(d) 48
(a) 30

Question 9.
If first 15th term of an A.P. is a, second term is b and nth term is 2 a, then sum of the n terms is :
(a) $$\frac {ab}{2(b – a)}$$
(b) $$\frac {2ab}{3(b – a)}$$
(c) $$\frac {3ab}{2(b – a)}$$
(d) $$\frac {3ab}{(b – a)}$$
(c) $$\frac {3ab}{2(b – a)}$$

Question 10.
In an A.P. Sn = 3n2 + 5n and Tm = 164, then m equals to :
(a) 26
(b) 27
(c) 28
(d) None of these.
(b) 27

Question 11.
A.M. of two number is 10 and GM. is 8, the numbers are
(a) a = 4, b = 16
(b) a = 2, b = 8
(c) a = 4, b = 9
(d) a = 2, b = 18.
(a) a = 4, b = 16

Question 12.
The A.M. of two numbers is A and G M is G, then relation between them is :
(a) A < G (b) A = G (c) A > G
(d) None of these.
(c) A > G

Question 13.
21/4.41/8.81/16 ……………. ∞ =
(a) 1
(b) 2
(c) 3/2
(d) 4
(b) 2

Question 14.
If y = x – x2 + x3 – x4 + ………. ∞, then value of x be (- 1 < x < 1) :
(a) y + $$\frac {1}{y}$$
(b) $$\frac {y}{1 + y}$$
(c) y – $$\frac {1}{y}$$
(d) $$\frac {y}{1 – y}$$
(d) $$\frac {y}{1 – y}$$

Question 15.
If the second, third and sixth terms of an A.P. are in GP, then the common ratio of the GP. is :
(a) 2
(b) 5
(c) 4
(d) 3
(d) 3

Question 16.
Sum up to infinity of 1 + $$\frac {4}{5}$$ + $$\frac { 7 }{ { 5 }^{ 2 } }$$ + $$\frac { 10 }{ { 5 }^{ 2 } }$$ + …………. :
(a) $$\frac {35}{16}$$
(b) $$\frac {37}{16}$$
(c) $$\frac {39}{16}$$
(d) 3
(a) $$\frac {35}{16}$$

(B) Match the following :

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

(C) Fill in the blanks :

1. – 7$$\sqrt { 3 }$$ will be term of the series 5$$\sqrt { 3 }$$, 3$$\sqrt { 3 }$$, $$\sqrt { 3 }$$, …………….
2. 116 is sum of the term of the series 25,22, 19 …………….
3. If sum of n terms of series is n2 + 4n, then 15th term of the series is …………….
4. 0 will be the term of the series 27, 24,21, 18 …………….
5. 729 will be the term of the series – $$\frac {1}{27}$$, $$\frac {1}{9}$$, – $$\frac {1}{3}$$ …………….
6. The first term of GP. is a, common ratio r < 1 and its last term l, then its sum will be …………….
7. The ratio of the sum of first three term to the sum of first six term is 125 : 152, then common ratio will be …………….
8. First term 16 and fifth term $$\frac {1}{16}$$, then its 4th term will be …………….
9. Sum of n terms of the series x + 2x2 + 4x3 + 8x4 + ……………. will be …………….
10. If a = 2, d = 2 and n = 50, then the last term of the series will be …………….

1. 7
2. 12
3. 33
4. 10
5. 10
6. $$\frac {a – rl}{1 – r}$$
7. $$\frac {3}{5}$$
8. $$\frac {1}{4}$$
9. $$\frac { 1-({ 2x }^{ n })x }{ 1 – 2x }$$
10. 100

(D) Write true / false :

1. 1, 3, 5, 8, ………….. are inA.P.
2. nth term of a G.P. is a + (n – 1 )d.
3. If 2x, x + 5 and x + 11 are in A.P., then value of x will be – 1.
4. Arithmetic mean of a and b is $$\sqrt { ab }$$
5. Sum of 9 terms of a sequence 24 + 20 + 16 + will be.
6. Four consicutive term of G.P. are $$\frac { a }{ { r }^{ 3 } }$$ $$\frac {a}{r}$$, ar, ar3.

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

(E) Write answer in one word / sentence :

1. If a, b, care inA.P., then find the value of ab + ac?
2. Arithmetic mean of (a + b)2and (a – b)2 will be?
3. Sum of n arithmetic mean between x and 3x will be.
4. Find the sum of infinite terms of series : 91/3.91/3.91/27 ………….. up to ∞.
5. If a, b, c are in GP., then find the value $$\frac {1}{b}$$ + $$\frac {1}{a – b}$$ –$$\frac {1}{b – c}$$

1. 2b2
2. a2 + b2
3. 3nx
4. 3
5. 0

Sequences and Series Short Answer Type Questions

Question 1.
Which term of the sequence 27,24,21,18, …………. is zero? (NCERT)
Solution:
Here a = 27 and d= T2 – T1 = 24 – 27 = – 3.
Let the nth term of series be 0.
∴ Tn = a + (n – 1)d
⇒ 0 = 27 + (n – 1)(- 3)
⇒ 3n – 3 = 27
⇒ 3n = 30
n = 10 or 10th term.

Question 2.
The last term of the series 8,4,0, ……………. is – 24. Find the total number of terms. (NCERT)
Solution:
Given series 8,4, 0, … (1)
First term = a = 8 Common difference d = 4 – 8 = -4
d = 0 – 4 = – 4
∵ The common difference is same the series is in A.P.
last term l = – 24,
l = a + (n – 1)d,
= – 24 = 8 + (n – 1)(-4)
⇒ – 24 – 8 = (n – 1)(-4)
⇒ – 32 = (n – 1)(- 4)
⇒ (n – 1) = $$\frac {32}{4}$$
⇒ n – 1 = 8
⇒ n = 8 + 1
⇒ n = 9
∴ Number of terms n = 9

Question 3.
Seven times the 7th term of a series is equal to eleven times of its 11th term. Find the 18th term of the series. (NCERT)
Solution:
Let first term = a and common difference = d of A.P.
∴ 7th term = a + 6d and 11th term = a + 10 d.
∴ According to question,
7(a + 6d) = 11 + (a + 10d)
⇒ 7a + 42d = 11a + 110d
⇒ 7a – 11a = 110d – 42d
⇒ – 4a = 68 d
⇒ a = – 17d
Hence 18th term = a + 17d
= – 17d + 17d [∵ a = – 17d]
= 0.

Question 4.
Prove that the sum of (m + n)th term and (m – n)th terms of an A.P. is twice of its mth term.
Solution:
Let the first term = a and common difference = d.
Tn = a+(n – 1)d
Tm + n= a + (m + n – 1)d … (1)
Tm – n = a + (m – n – 1)d … (2)
Tm = a + (m – 1)d … (3)
Tm+n + Tm-n = a + (m + n – 1)d + a(m – n – 1)d
= 2a + (m + n – 1 + m – n – 1)d
= 2a + (2m – 2)d
= 2a + 2(m – 1)d
= 2[a+(m – 1)d]
Tm+n + Tm-n = 2Tm.

Question 5.
Insert three Arithmetic means between 3 and 19. (NCERT)
Solution:
Let three Arithmetic mean be A1, A2, A3,
then 3, A1, A2, A3, 19 are in A.P.
∴ 3 = a, 19 = T5, let common difference = d
T5 = a + 4d
T5 = a + 4d
⇒ 19 = 3 + 4d
⇒ 16 = 4d
⇒ d = 4
Hence A1 = 3 + 4 = 7, A2 = 7 + 4 = 11, A3 = 11 + 4 = 15.

Question 6.
If – 8, A1, A2 are in Arithmetic progression (A.P.), then find the value of A1, A2. (NCERT)
Solution:
– 8, A1, A2, 9 are in A.P.
∴ a = – 8, T4 = 9, let common difference = d
T4 = a + 3d
⇒ 9 = – 8 + 3 d
⇒ 3d = 11
⇒ d = $$\frac {17}{3}$$

Instruction :
Write the first five terms of each of the sequence and obtain the corresponding series.

Question 7.
(a) a1 = 3, an = 3an – 1 + 2, where n > 1. (NCERT)
Solution:
Given : a1 = 3,
a2 = 3an – 1 + 2, where n > 1
a2 = 3a2 – 1 + 2
⇒ a2 = 3a1 + 2
⇒ a2 = 3 x 3 + 2 = 9 + 2 = 11
a3 = 3a1 + 2
= 3a2 + 2
⇒ a3 = 3(11)+ 2 = 33+ 2 = 35
a4 = 3a4 – 1 + 2
= 3a3 + 2
⇒ a4 = 3(35) + 2 = 105 + 2 = 107
a5 = 3a5 – 1 + 2
= 3a4 + 2
= 3(107) + 2 = 321 + 2 = 323
Hence series is 3, 11, 35, 107, 323.

Question 7.
(b) a1 = -1, an = $$\frac { { a }_{ n-1 } }{ n }$$ where n ≥ 2.
Solution:

Question 8.
A person pays first instalment of Rs. 100 towards his loan. If he increases his instalment every month by Rs. 5, then what will be his 30th instalment.
Solution:
Given : a = Rs. 100, d= Rs. 5, n = 30,
Tn = a + (n – 1)d
Amount of 30th instalment
J30 = 100 + (30 – 1) x 5
= 100 + 29 x 5 = 100 + 145
= Rs. 245.
Hence the 30th instalment = Rs. 245.

Question 9.
Which term of the sequence $$\sqrt { 3 }$$, 3, 3$$\sqrt { 3 }$$ ……. is 729?
Solution:

Question 10.
How many terms are required in GP. 3,32,33 …………, so that their sum would be 120? (NCERT)
Solution:
Given : a = 3, r = $$\frac {9}{3}$$ = 3, Sn = 120,
Sn = $$\frac { a({ r }^{ n }-1) }{ r-1 }$$
⇒ $$\frac { 3({ 3 }^{ n }-1) }{ 3-1 }$$ = 120
⇒ 3n – 1 = $$\frac {120 × 2}{3}$$
⇒ 3n – 1 = 80
⇒ 3n = 81
⇒ 3n = 34
⇒ n = 4

Question 11.
Show that ratio between sum of n terms and sum of (n +1)th term to (2n)th terms in a G.P. is $$\frac { 1 }{ { r }^{ n } }$$
Solution:
Let 1st term of G.P. = a and common ratio = r
∴ n terms of GP. is a, ar, ar2, ………….. , arn – 1
Let the sum to n terms be S1
S1 = $$\frac { a({ r }^{ n }-1) }{ r-1 }$$
GP. from (n + 1)th term to (2n)th term
arn, arn + 1, …………. , ar2n – 1
Let the sum of this G.P. up to n terms be S2

Question 12.
If A.M. and GM. of roots of a quadratic equation are 8 and 5 respectively, then find the quadratic equation. (NCERT)
Solution:
Let the roots of equation be α and β
A.M. of roots = $$\frac {α + β}{2}$$ = 8
⇒ α +β = 16
G.M. of roots = $$\sqrt {αβ }$$ = 5
⇒ αβ = 25
If α, β are the roots of equation
then, x2 – (α + β)x + αβ = 0
⇒ x2 -16x + 25 = 0.

Question 13.
If the first and nth term of GP. are a and b respectively and if p is the product of n terms, then prove that
p2 = (ab)n. (NCERT)
Solution:
Let the common ratio of G.P. = r
Given: arn – 1 = b, …(1)
then a.ar.ar2……. arn – 1 = p
⇒ anr1+2+3+……+(n – 1) = p
⇒ anr$$\frac {1}{2}$$(n – 1)n = p
⇒ a2n.r(n – 1)n = p2
⇒ p2 = (a2rn – 1n
⇒ p2 = (a.arn – 1)n
⇒ p2 = (ab)n, [fromeqn. (1)]

Question 14.
If the 4th term of a GP. is square of its second term and the 1st term is -3, then find the 7th term. (NCERT)
Solution:
Let a and r be the 1st term and common ratio of GP.
∴ Tn = arn-1
T4 = ar4-1 = ar3 … (1)
T2 = ar2-1 = ar … (2)
Given : T4 = (T2)2
⇒ ar3 = (ar)2
⇒ ar3 = a2r2
⇒ a = r = – 3 (given, a = -3)
T7 = ar7-1 = ar6
= (-3)(-3)6 = (-3)7
∴ T7 = – 2187
Hence 7th term of G.P. = – 2187.

Question 15.
Find the value of $$\sum _{ k=1 }^{ 11 }{ (2+{ 3 }^{ k }) }$$.
Solution:

Sequences and Series Long Answer Type Questions

Question 1.
If m times of the mth term of an A.P. is equal to n times the nth term, then prove that (m + n)th term of the series is zero.
Solution:
Let the 1st term = a and common difference = d.
then tm = a + (m – 1)d
tn = a + (n – 1)d
Given: m[a + (n – 1)d] = n[a + (n – 1)d]
⇒ ma + m(m – 1 )d = na + n(n – 1 )d
⇒ (m2 – m)d + (m – n)a = (n2 – n)d
⇒ (m2 – m – n2 + n) d + (m – n)a = 0
⇒ [m2 – n2 – (m – n)] d + (m – n)a = 0
⇒ [(m – n)(m + n) – (m – n)] d+(m – n)a = 0
⇒ (m – n)[(m + n) – 1] d + (m – n)a = 0
⇒ a + (m + n – 1 )d = 0
⇒ tm+n=0.

Question 2.
If the 6th term of A.P. is 12 and 9th term is 27, then find its rth term.
Solution:
Let the 1st term = a and common difference = d.
Given : t6 = 12
⇒ a + 5d = 12 … (1)
t9 = 27
⇒ a + 8d = 27 … (2)

Question 3.
If pth term of an A.P. is $$\frac {1}{q}$$ and qth term is $$\frac {1}{p}$$, then prove that (pq)th
Solution:

Question 4.
If pth term of an A.P. is $$\frac {1}{q}$$ and qth term is $$\frac {1}{p}$$, then prove that (pq)th terms is $$\frac {1}{2}$$ (pq + 1), where p ≠ q.
Solution:

Question 5.
The sum of the series 25, 22, 19, ………….. of A.P. is 116, then find its last term. (NCERT)
Solution:
Given : a = 25, d = 22 – 25 = – 3, Sn = 116
Sn = $$\frac {n}{2}$$ [2a + (n – 1)d]
⇒ 116 = $$\frac {n}{2}$$[2 x 25 + (n -1) x (-3)]
⇒ 232 = n[50 – 3n + 3]
⇒ 232 = n[53 – 3n]
⇒ 232 = 53n – 3n2
⇒ 3n2 – 53n +232 = 0
⇒ 3n2 – 24n – 29n + 232 = 0
⇒ 3n(n – 8) – 29(n – 8) = 0
⇒ (n – 8)(3n – 29) = 0
n = 8, or n = $$\frac {29}{3}$$, (which is impossible)
∴ Last term l = a + (n – 1 )d
= 25 + (8 – 1) x (- 3)
⇒ l = 25 – 21 = 4.

Question 6.
If the A.M. of a and b is $$\frac { { a }^{ n }+{ b }^{ n } }{ { a }^{ n – 1 }+{ b }^{ n – 1 } }$$, then find the value of it. (NCERT)
Solution:
A.M. of a and b = $$\frac {a+b}{2}$$
According to question,
$$\frac { { a }^{ n }+{ b }^{ n } }{ { a }^{ n – 1 }+{ b }^{ n – 1 } }$$ = $$\frac {a+b}{2}$$
⇒ 2an + 2bn = (a + b) (an – 1 + bn – 1)
⇒ 2an +2bn = an + bn + abn – 1 + ban – 1
⇒ an + bn = abn – 1 + ban – 1
⇒ an – ban – 1 = abn – 1 – bn
⇒ an – 1(a – b) = bn – 1(a – b)
⇒ an – 1 = bn – 1
⇒ $$\frac {a}{b}$$n – 1 = $$\frac {a}{b}$$0
⇒ n – 1 = 0
⇒ n = 1.

Question 7.
Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7th and (m – 1)th mean is 5 : 9 then, find the value of m.
Solution:
Let A1, A2, A3, ……… ,Am are m A.M. respectively inserted between 1 and 31.
Then 1, A1, A2, A3, ……… Am 31 are in A.P. Here, 1st term = 1 and (m + 2)th term = 31 and let the common difference of sequence be d.

Question 8.
Show that the sum of (m + n)th and (m – n)th term of an A.P. is equal to twice the mth term. (NCERT)
Solution:
Let the first term = a and common difference = d.
Tn = a + (n – 1)d
Tm + n =a +(m + n – 1)d … (1)
Tm – n = a + (m – n – 1)d … (2)
Tm = a + (m – 1)d … (3)
Tm + n + Tm – n = a + (m + n – 1)d + a + (m – n – 1)d
= 2a + (m + n – 1 + m – n – 1)d
= 2a + (2m – 2 )d
= 2a + 2 (m – 1)d
= 2[a + (m – 1)d]
∴ Tm + n + Tm – n = 2Tm [From equation (3)]

Question 9.
If the sum of three numbers in A.P. is 24 and their product is 440, then find the numbers? (NCERT)
Solution:
Let the three numbers of A.P. are a – d, a, a + d.
Given : a – d + a + a + d = 24
3a = 24
⇒ a = 8
and (a – d) × a × (a + d) = 440
a(a2 – d2) = 440
⇒ 8(64 – d2) = 440
⇒ 64 – d2 = $$\frac {440}{8}$$
⇒ 64 – d2 = 55
⇒ d2 = 64 – 55
⇒ d2 = 9
⇒ d = ± 3
When a = 8 and d = 3
Then, a – d = 8 – 3 = 5, a = 8, a + d= 8 + 3 = 11
When a = 8 and d = – 3
Then, a – d = 8 + 3 = 11, a = 8, a + d = 8 – 3 = 5
Hence required numbers are 5, 8, 11 or 11, 8, 5.

Question 10.
If the sum of n, 2n and 3n terms in A.P. are S1 S2 and S3, then show that S3 = 3(S2 – S1).
Solution:

Question 11.
Find the sum of all natural numbers lying between 200 and 400, which are divisible by 7. (NCERT)
Solution:
Numbers divisible by 7 are 203, 210, 217, …………. , 399
Here a = 203, d = 210 – 203 = 7, l = 399
l = a + (n – 1)d
399 = 203 + (n – 1) x 7
⇒ 399 – 203 = (n – 1) x 7
⇒ (n – 1)7 = 196
⇒ (n – 1) = $$\frac {196}{7}$$
⇒ (n – 1) = 28
∴ n = 29.
Hence required sum, S29 = $$\frac {n}{2}$$[a+l]
⇒ S29 = $$\frac {29}{2}$$[203 + 399]
⇒ S29 = $$\frac {29}{2}$$[602] = $$\frac {17458}{2}$$ = 8729.

Question 12.
If the 5th, 8th and 11th terms of a GP. are p, q and s respectively, then show
that q2 = ps. (NCERT)
Solution:
Let the 1st term of GP. = a and common ratio = r.
Then, Tn = arn – 1
T5 = ar5 – 1 =p
⇒ ar4 = P … (1)
T8 = ar8-1 = q
⇒ ar7 = q … (2)
T11 = ar11 – 1 = s
⇒ ar10 = s
ps = ar4.ar10, [from equation (1) And (3)]
⇒ ps = a2r14
⇒ ps = (ar7)2
⇒ ps = q2, [from equation (2)]
∴ q2 = ps.

Question 13.
If the first term of G.P. a = 729 and 7th term is 64, then find S7. (NCERT)
Solution:

Question 14.
If the 4th terms of a GP is square of its 2nd term and first term is – 3, then find its 7th (NCERT)
Solution:
Let the 1st term of GP. = a and common ratio = r.
Then, Tn = arn – 1
T4 = ar4 – 1 = ar3 … (1)
T2 = ar2-1 = ar … (2)
Given : T4 = T22
ar3 = (ar)2
ar3 = a2r2
⇒ ar3 = a2r2
⇒ a = r = – 3 (given a = – 3)
T7 = ar7 – 1 = ar6
= (-3)(-3)6 = (-3)7
∴ T7 = – 2187
Hence, 7thterm of G. P. = – 2187.

Question 15.
Find the sum of the sequence 8,88,888,8888 ……….. up to n terms. (NCERT)
Solution:
Let the sum of the n terms be S

Question 16.
Find the sunt of the numbers 7,77,777, 7777 up to n terms. (NCERT)
Solution:
Let the sum of n terms is S.

Question 17.
If the A.M. and GM. between two positive numbers a and b are 10 and 8 respectively, then find the numbers. (NCERT)
Solution:
A.M. A = $$\frac {a+b}{2}$$
a + b = 20 … (1)
G.M. G = $$\sqrt {ab}$$
ab = 64 … (2)
(a – b)2 = (a + b)2 – 4ab
= (20)2 – 4 × 64, [From equation (2)]
= 400 – 256
(a – b)2 = 144 = (12)2 … (3)

Put a = 16 in equation (1), we get,
16 + b = 20
∴ b = 4
Numbers are 16 and 4.
When a – b = – 12, then
a + b = 20
a – b = – 12
⇒ a = 4
Put a = 4 in equation (1), we get,
4 + 6 = 20
∴ b = 16
Numbers are 4 and 16.
Hence numbers a and b are 4, 16 or 16,4.

Question 18.
The sum of two numbers is 6 times their geometric mean, show that the . numbers are in the ratio (3 + 2$$\sqrt {2}$$) ; (3 – 2$$\sqrt {2}$$). (NCERT)
Solution:
Let the numbers are a and 6.
Given: a + b = 6 $$\sqrt {ab}$$
$$\frac { a+b }{ 2\sqrt { ab } }$$ = $$\frac {3}{1}$$
⇒ a + b = 3K …. (1)
and 2 $$\sqrt {ab}$$ = K ⇒ 4ab = K2
(a – b)2 = (a + b)2 – 4ab
= (3K)2 – (K)2
= 9K2 – K2 = 8K2
⇒ a – b = 2$$\sqrt {2}$$K …. (2)

Sequences and Series Very Long Answer Type Questions

Question 1.
If the ratio of the sum of n terms of two A.P. is 5n + 4 : 9n + 6, then find the ratio of their 18th term.
Solution:
Let the two A.P. are :
a, a + d, a + 2d, ………….
and A, A + D, A +2D …………..

Question 2.
The ratio of the sum of m and n terms of an A.P. is m2 : n2. Show that the ratio of and u* term is (2m – 1) : (2n – 1). (NCERT)
Solution:
Let the A.P. are a, a + d, a + 2d, ……………
∴ mth term of A.P. Tm = a + (m – 1)d
nth term of A.P. Tn= a + (n – 1)d

Question 3.
If the sum of first three terms of a G.P. is $$\frac {39}{10}$$ and their product is 1, then find the common ratio and the terms. (NCERT)
Solution:
Let the three terms of G.P. are $$\frac {a}{r}$$, a, ar.
Given: $$\frac {a}{r}$$ × a × ar = 1
⇒ a3 = 1 ⇒ a = 1
and $$\frac {a}{r}$$ + a + ar = $$\frac {39}{10}$$
⇒ a($$\frac {a}{r}$$ + r + 1) = $$\frac {39}{10}$$
⇒ 1 x $$\frac { 1+{ r }^{ 2 }+r }{ r }$$ = $$\frac {39}{10}$$
⇒ 10r2 + 10r + 10 = 39r
⇒ 10r2 – 29r + 10 = 0
⇒ 10r2 – 25r – 4r + 10 = 0
⇒ 5r(2r – 5) – 2(2r – 5) = 0
⇒ (5r – 2)(2r – 5) = 0
∴ r = $$\frac {2}{5}$$ and r = $$\frac {5}{2}$$
When a = 1 and r = $$\frac {2}{5}$$

Question 4.
Find four numbers forming a G.P. in which the third term is greater than the lint term by 9 and the second term is greater than 4th term by 18. (NCERT)
Solution:
Let the four terms of G.P. be a, ar, ar2,ar3.
Given: T3 = T1 + 9
⇒ T3 = T1 + 9
⇒ T3 = a + 9
⇒ ar2 = a + 9 …. (1)
According to question,
T2 = T4 + 18
⇒ ar = a3 + 18
⇒ ar – ar3 = 18 …. (2)

Put r = – 2 in equation (1), we get
a(- 2)2 – a = 9
⇒ 4a – a = 9
⇒ 3a = 9
⇒ a = 3
∴ Numbers are 3, 3(- 2), 3(- 2)2, 3(-2)3, ……………..
3, – 6, 12, – 24, ………………

Question 5.
If S be the sum of it terms of a GP., P be the product and R be the sum of reciprocal of n terms, then prove that
P2Rn = Sn. (NCERT)
Solution:
Let n terms of GP. be a, ar, ar2, …………… arn – 1.
According to the question,

Question 6.
If x = 1 + a + a2 + ………….∞ ($$\left| a \right|$$<1)
y = 1 + b + b2 + …………….∞ ($$\left| b \right|$$<1)
then prove that
1 + ab + a2b2 + …………….∞ = $$\frac {xy}{x + y – 1}$$
Solution:

Question 7.
The sum of infinite terms of a Geometric Progression is 15 and sum of the square of its terms is 45. Find the Geometric Progression. (NCERT)
Solution:
Let a be the first term and r be the common ratio.
∵ $$\left| r \right|$$<1,
Then, $$\frac {a}{1 – r}$$ = 1.5 …. (1)
Squaring the terms of GP. the new G.P. is
a2, a2 r2, a2 r4, a2 r6, ……….
Sum of infinity of G.P. = $$\frac { { a }^{ 2 } }{ { 1 – r }^{ 2 } }$$.
$$\frac { { a }^{ 2 } }{ { 1 – r }^{ 2 } }$$ = 45, (given) … (2)
Squaring both sides of equation (1) and the result is divided by equation (2),
$$\frac { { a }^{ 2 } }{ { 1 – r }^{ 2 } }$$2 × $$\frac { 1-{ r }^{ 2 } }{ { a }^{ 2 } }$$ = $$\frac {15×15}{45}$$
⇒ $$\frac {1 + r}{1 – r}$$ = 5
⇒ 1 + r = 5 – 5r
⇒ 6r = 4
⇒ r = $$\frac {2}{3}$$
Put r =$$\frac {2}{3}$$ in equation (1), we get

Hence required progression is 5, 5 × $$\frac {2}{3}$$, 5 × ($$\frac {2}{3}$$)2, …………….
Hence 5, $$\frac {10}{3}$$, $$\frac {20}{9}$$, ………………

Question 8.
A farmer buys a used tractor of Rs. 12,000. He pays Rs. 6,000 cash and agrees to pay the balance in annual instalment of Rs. 500 plus 12% interest on theunpaid amount How much the tractor cost him?
Solution:
Cost of tractor = Rs. 12, 000, down payment = Rs. 6,000
Balance amount = 12,000 – 6,000 = Rs. 6,000

Actual cost of tractor = 12,000 + 4,680 = Rs. 16,680. Ans.

Question 9.
Shamshad Ali buys a scooter for Rs. 22,000. He pays Rs. 4,000 cash and agrees to pay the balance in annual instalment of Rs. 1,000 plus 10% interest on the unpaid amount How much will scooter cost him? (NCERT)
Solution:
Cost of scooter = Rs. 22,000, Cash down payment = Rs. 4,000.
Remaining amount = 22,000 – 4,000 = Rs. 18,000

Actual cost = 22,000 + 17,100 = Rs. 39,100.
Total amount paid for scooter = Rs. 39,100.

Question 10.
A person writes a letter to four of his friends. He asks each one of them, to copy the letter and mail to four different persons with instructions that they move the chain similarly. Assuming that the chain is not broken and that is costs 50 paisa to mail one letter. Find the amount on postage when 8th set of letter is mailed. (NCERT)
Solution:
First person sends 4 letters.
2nd step, he sends 4 x 4= 16 letters
3rd step, he sends 4 x 4 x 4 = 64 letters
Hence 4, 16, 64, 256, …………… is a Geometric series.
Here a = 4, r = $$\frac {16}{4}$$ = 4
Sn = $$\frac { { a(r }^{ n } – 1) }{ r – 1 }$$ [∵r>1]
∴ Total number of letters till 8th set = S8 = $$\frac { { 4(r }^{ 4 } – 1) }{ 4 – 1 }$$
= $$\frac {4}{3}$$ (65536 – 1) = $$\frac {4}{3}$$ x 65535
Cost for one letter = Rs. 0.50
∴ Hence total cost = $$\frac {4}{3}$$ x 65535 x 0.50 = Rs. 43690.

Question 11.
150 workers were engaged to finish a job in a certain number of days, 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed. (NCERT)
Solution:
150, 146, 142, 138, …………, it is a Geometric series.
Let the number of days required complete the work be n.

If the workers not dropped then the work would have complited in (n – 8) days with 150 workers working on each day.
Hence the total workers worked for n days = 150 (n – 8)
= 150n – 1200 ….. (2)
From equation (1) and (2),
150n – 1200 = 152n – 2n2
2n2 + 150n – 152n – 1200 = 0
2n2 – 2n – 1200 = 0
n2 – n – 600 =0
n2 – 25n + 24n – 600 = 0
n(n – 25) + 24(n – 25) = 0
(n – 25)(n + 24) = 0
n =25 and n = – 24 (not possible)
∴ Work is completed in 25 days.

## MP Board Class 11th Maths Important Questions Chapter 16 Probability

### Probability Important Questions

Probability Objective Type Questions

(A) Choose the correct option :

Question 1.
A leap year is selected at random. Then the probability that this year will contain 53 Sunday is :
(a) $$\frac { 1 }{ 7 }$$
(b) $$\frac { 2 }{ 7 }$$
(c) $$\frac { 3 }{ 7 }$$
(d) $$\frac { 4 }{ 7 }$$
(b) $$\frac { 2 }{ 7 }$$

Question 2.
A and B are two events such that P(A) = 0.4, P(A + B) = 0.7, P(AB) = 0.2, then
P(B) =
(a) 0.1
(b) 0.3
(c) 0.5
(d) 0.2
(c) 0.5

Question 3.
Three letters are to be sent to different persons and address on the three envelope are also written. Without looking at the address, the probability that the letters go into right envelope is equal to :
(a) $$\frac { 1 }{ 27 }$$
(b) $$\frac { 1 }{ 9 }$$
(c) $$\frac { 4 }{ 27 }$$
(d) $$\frac { 1 }{ 6 }$$
(d) $$\frac { 1 }{ 6 }$$

Question 4.
If A and B are two events such that P(A ∪B) = $$\frac { 3 }{ 4 }$$, P(A ∩ B) = $$\frac { 1 }{ 4 }$$, P($$\bar { A }$$) = $$\frac { 2 }{ 3 }$$, then P(A ∩ B) is:
(a) $$\frac { 5 }{ 12 }$$
(b) $$\frac { 3 }{ 8 }$$
(c) $$\frac { 5 }{ 8 }$$
(d) $$\frac { 1 }{ 4 }$$
(a) $$\frac { 5 }{ 12 }$$

Question 5.
The probability that A speaks truth is $$\frac { 4 }{ 5 }$$ while that this probability for B is $$\frac { 3 }{ 4 }$$. The probability that they contradict each other when asked to speak on a fact:
(a) $$\frac { 5 }{ 12 }$$
(b) $$\frac { 3 }{ 8 }$$
(c) $$\frac { 5 }{ 8 }$$
(d) $$\frac { 1 }{ 4 }$$
(c) $$\frac { 5 }{ 8 }$$

Question 6.
A player drawn one card from a pack of 52 cards, then the probability that it is not a diamond card :
(a) $$\frac { 1 }{ 4 }$$
(b) $$\frac { 2 }{ 4 }$$
(c) $$\frac { 3 }{ 4 }$$
(d) $$\frac { 1 }{ 13 }$$
(c) $$\frac { 3 }{ 4 }$$

Question 7.
Probability of getting number greater than 3 when a dice is thrown :
(a) 1
(b) $$\frac { 1 }{ 3 }$$
(c) $$\frac { 1 }{ 2 }$$
(d) $$\frac { 1 }{ 6 }$$
(c) $$\frac { 1 }{ 2 }$$

Question 8.
A bag contain 6 white, 5 black and 4 yellow balls. The probability of drawing a white or a black ball is :
(a) $$\frac { 6 }{ 11 }$$
(b) $$\frac { 6 }{ 15 }$$
(c) $$\frac { 5 }{ 15 }$$
(d) $$\frac { 11 }{ 15 }$$
(d) $$\frac { 11 }{ 15 }$$

Question 9.
Probability of getting an even number on throwing a dice is :
(a) 1
(b) $$\frac { 1 }{ 3 }$$
(c) $$\frac { 1 }{ 2 }$$
(d) $$\frac { 1 }{ 6 }$$
(c) $$\frac { 1 }{ 2 }$$

Question 10.
If P(A) = $$\frac { 1 }{ 4 }$$, P(B) = $$\frac { 1 }{ 2 }$$ and P(A ∩ B) = $$\frac { 1 }{ 8 }$$ then P(A’ ∩ B’)=
(a) $$\frac { 1 }{ 8 }$$
(b) $$\frac { 3 }{ 8 }$$
(c) $$\frac { 5 }{ 8 }$$
(d) $$\frac { 7 }{ 8 }$$
(b) $$\frac { 3 }{ 8 }$$

(B) Fill in the blanks :

1. a : b is odds in favour of an event A, then probability that it happen ………………..
2. The probability of getting odd number when a cubical dice is thrown ………………..
3. Total number of sample points when three coins are thrown are ………………..
4. It is given that the event A and B 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) = ………………..
5. If P(A) = 0.42, P(B) = 0.48, P(A ∩ B) = 0-16, then P(A’ ∩ B’) = ………………..
6. If two events A and B related to a random experiment are not mutually exclusive, then P(A ∪ B) + P(A ∩ B) = ………………..
7. Probability of happening of an event and not happening are P (A) and P(A) respectively P(A) + P($$\bar { A }$$) = ………………..
8. The number of sample space when two cubical dice are thrown is ………………..
9. If A and B are two mutually exclusive events and P(B) = 2 P(A) then the value of P(A) will be ………………..
10. If two dice are thrown simultaneously, then the probability of getting sum is 7 will be ………………..

1. $$\frac { a }{ a + b }$$
2. $$\frac { 1 }{ 2 }$$
3. 8
4. $$\frac { 1 }{ 3 }$$
5. 0.26
6. P(A) + P(B)
7. 1
8. 36
9. $$\frac { 1 }{ 3 }$$
10. $$\frac { 1 }{ 6 }$$

(C) Write true / false :

1. If two coins are tossed together, then the probability of getting atleast one head is $$\frac { 3 }{ 4 }$$.
2. From a pack of cards one card is drawn at random then the chance of getting knave $$\frac { 1 }{ 52 }$$
3. If tossing a coin once, then the probability of getting tail on the upper face is $$\frac { 1 }{ 2 }$$
4. If two dice are thrown simultaneously, then the probability of getting product 6 is $$\frac { 1 }{ 9 }$$
5. The odds favour in event is $$\frac { 3 }{ 4 }$$, then the probability of happening is $$\frac { 4 }{ 7 }$$

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

(D) Write answer in one word / sentence :

1. If two coins are tossed together, then the sample space will be?
2. From a well shuffled pack of 52 cards one card is drawn at random; find the probability it is a king.
3. Two dice are thrown simultaneously, find the probability getting sum is multiple of 3.
4. The odds in favour of an event is 3 : 5, then find the probability of not happening.
5. If A, B and C are three mutually exclusive even, then P(A ∪ B ∪ C) =

1. {HT, HH, TH, TT}
2. $$\frac { 1 }{ 13 }$$
3. $$\frac { 1 }{ 3 }$$
4. $$\frac { 5 }{ 8 }$$
5. P(A) + P(B) + P(C )

Question 1.
A coin is tossed twice, what is the probability that at least one tail occurs? (NCERT)
Solution:
Let S is the sample space, then
S = {(H, H), (H, T), (T, H), (T, T)}
n(S) = 4
Let E be the event of getting at least one tail, then
E = {(H, T), (T, H), (T, T)}
∴ n(E) = 3
Hence, required probability P(E) = $$\frac { n(E) }{ n(S) }$$ = $$\frac { 3 }{ 4 }$$

Question 2.
There are four men and six women on the city council. If one council member is selected for a committee at random, then how likly is it that it is a woman?
Solution:
Let S is the sample space, then
n (S) = 10
Let E be the event that woman is selected, then
n(E) = 6
Hence, required probability P(E) = $$\frac { n(E) }{ n(S) }$$
P(E) = $$\frac { 6 }{ 10 }$$ = $$\frac { 3 }{ 5 }$$

Question 3.
If $$\frac { 2 }{ 11 }$$ is the probability of an event A, then find what is the probability of the event is ‘not A’.
Solution:
Given that :
P(A) = $$\frac { 2 }{ 11 }$$
We know that,
P(A) + P($$\bar { A }$$) = 1
$$\frac { 2 }{ 11 }$$ + P($$\bar { A }$$) = 1
P($$\bar { A }$$) = 1 – $$\frac { 2 }{ 11 }$$ = $$\frac { 11 – 2 }{ 11 }$$
P($$\bar { A }$$) = $$\frac { 9 }{ 11 }$$.

Question 4.
A box contain 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn from the box. What is the probability that: (NCERT)

1. All marbles will be blue.
2. At least one marble will be green.

Solution:
In the box, there are 10 red, 20 blue and 30 green marbles.
The No. of marbles = 10 + 20 + 30 = 60
1. Total ways of choosing 5 marbles out of 60 marbles,
n(S) = 60C5
Let E is the event of choosing blue marbles, then
n(E) = 20C5
Probability = $$\frac { n(E) }{ n(S) }$$ = $$\frac{^{20} C_{5}}{^{60} C_{5}}$$

2. P(At least one green marble)
= 1 – P(No green)
= 1 – $$\frac{^{30} C_{5}}{^{60} C_{5}}$$

Question 5.
4 cards are drawn from a well shuffled deck of 52 cards. What is the probability of obtaining in card of 3 diamond and one spade?
Solution:
Let S is the sample space.
Total number of selecting 4 cards out of 52 cards, n(S) = 52C4
If E is the event obtaining card of 3 diamond and 2 spade, then
n(E) = 13C3 x 13C1
Probability P(E) = $$\frac { n(E) }{ n(S) }$$ = $$\frac{^{13} C_{3} \times^{13} C_{1}}{^{52} C_{4}}$$

Question 6.
Given : P(A) = $$\frac { 3 }{ 5 }$$ and P(B) = $$\frac { 1 }{ 5 }$$. If A and B are mutually exclusive events, then find P(A or B). (NCERT)
Solution:
A and B are mutually exclusive events.
∴ P (A or B) = P(A ∪ B) = P(A) + P(B)
P(A ∪ B) = $$\frac { 3 }{ 5 }$$ + $$\frac { 1 }{ 5 }$$ = $$\frac { 3 + 1 }{ 5 }$$ = $$\frac { 4 }{ 5 }$$

Question 7.
In a lottery, a person choose six different natural numbers at random from 1 to 20 and if there six numbers match with the six numbers already fixed by lottery committee, he wins the prize, what is the probability of winning the prize in the game? (Hint: Order of the numbers is not important) (NCERT)
Solution:
Let S is the sample space, then
n(S) = 20C6
= $$\frac { 20 × 19 × 18 × 17 × 16 × 15 }{ 6 × 5 × 4 × 3 × 2 × 1 }$$ = 38760
Only one prize can be win.
∴ n(E) = 1
Hence, required probability P(E) = $$\frac { n(E) }{ n(S) }$$ = $$\frac { 1 }{ 38760 }$$

Question 8.
Two dice are thrown simultaneously. Find the probability of getting a sum 9 in a single throw.
Solution:
Total number of ways in which two dice may be thrown
= 6 x 6 = 36
∴ n(S) = 36
Event of getting sum 9 is A = {(3, 6), (4, 5), (5, 4), (6, 3)}
∴ n(A) = 4
∴ Required probability P(A) = $$\frac { n(A) }{ n(S) }$$ = $$\frac { 1 }{ 9 }$$

Question 9.
A bag contains 8 black, 6 white and 5 red balls. Find the probability of drawing a black or a white ball from it
Solution:
Total number of balls = 8 + 6 + 5 = 19
∴ n(S) = 19
Event A of drawing 1 black or 1 white ball
n(A) = 8 + 6 = 14
∴ n(A) = 14
∴ Required probability P(A) = $$\frac { n(A) }{ n(S) }$$ = $$\frac { 14 }{ 19 }$$

Question 10.
From a pack of well shuffled cards two cards drawn simultaneously. Find the probability that both the cards are ace.
Solution:
Total number of ways of drawing two cards out of 52
= 52C2
Number of ways drawing two ace out of 4 ace
= 4C2
∴ Required Probability

Question 11.
A pair of dice are thrown. Find the probability that the sum is 9 or 11.
Solution:
Let the sample space be S, then
∴ n(S) = 36
Let E be the event that sum is 9 or 11, then
E = {(5, 4), (4, 5), (6, 3), (3, 6), (6, 5), (5, 6)}
∴ n (E) = 6
The probability of getting sum 9 or 11 is
P(E) = $$\frac { n(E) }{ n(S) }$$ = $$\frac { 6 }{ 36 }$$ = $$\frac { 1 }{ 6 }$$
Hence, probability that the sum is not 9 or 11 is
P($$\bar { E }$$) = 1 – P(E)
= 1 – $$\frac { 1 }{ 6 }$$ = $$\frac { 5 }{ 6 }$$

Question 12.
In a city 20% persons read English newspaper, 40% persons read Hindi newspaper and 5% persons read both, then find the probability that the newspapers are not read.
Solution:
P(E) = The probability that persons read English newspapers = $$\frac { 20 }{ 100 }$$
p(H) = The probability that persons read Hindi newspapers = $$\frac { 40 }{ 100 }$$
P(E ∩ H) = The probability that persons read both the papers = $$\frac { 5 }{ 100 }$$
The probability that the newspapers are not read

Question 13.
In class XI of a school, 40% of the students study Mathematics and 30% study Biology. 10% of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology. (NCERT)
Solution:
Let M and B denote the students of Mathematics and Biology respectively. Then, as given:
P(M) = 40% = $$\frac { 40 }{ 100 }$$
= P(B) = 30% = $$\frac { 30 }{ 100 }$$
P(M ∩ 5) = 10% = $$\frac { 10 }{ 100 }$$
∴ P(M ∪ B) = P(M) + P(B) – P(M ∩ B)
$$\frac { 40 }{ 100 }$$ + $$\frac { 30 }{ 100 }$$ – $$\frac { 10 }{ 100 }$$
= $$\frac { 60 }{ 100 }$$ = 60% = 0.6.

Question 14.
In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and the probability of passing the second examination is 0.7. The probability of passing at least one of them is 0.95. What is the probability of passing both? (NCERT)
Solution:
Let probability of passing in first examination is A and passing in the second examination is B.
P(A) = 0.8, P(B) = 0.7, P(A ∪ B) = 0.95, P(A ∩ B) = ?
We know that,
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
0.95 = 0.8 + 0.7 – P(A ∩ B)
⇒ P(A ∩ B) = 1.5 – 0.95
∴ P(A ∩ B) = 0.55.

Question 1.
A letter is selected at random from the word ‘ASSASSINATION’. Find the probability that letter is

1. a vowel
2. a consonant

Solution:
Number of letters is 13 in which there are 6 vowels and 7 consonants.
1. Let sample space is S, then
∴ n(S) = 13
E1 is the event of choosing a vowel, then
n(E1) = 6
Hence, required probability P(E1) = $$\frac{n\left(E_{1}\right)}{n(S)}$$
P(E1) = $$\frac { 6 }{ 13 }$$

2. E2 is the event of choosing a consonant, then
n(E1) = 7
Hence, required probability P(E2) = $$\frac{n\left(E_{2}\right)}{n(S)}$$

Question 2.
Check whether the following probabilities P{A) and P(B) are consistently defined: (NCERT)

1. P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6
2. P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8.

Solution:
1. Given : P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6
If P(A ∩ B) ≤ P(A) and P(A ∩ B) ≤ P(B)
Then, P(A) and P(B) are consistent.
Here

P(A) and P(B) are not consistent.

2. Given : P(A) = 0.5, P(B) = 0 4, P(A ∪ B) = 0.8

∴ P(A) and P(B) are consistent.

Question 3.
In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS. If one of these students is selected at random, find the probability that:

1. The student opted for NCC or NSS.
2. The student has opted for neither NCC nor NSS.
3. The student has opted NSS but not NCC.

Solution:
Let A and B denote the students in NCC and NSS respectively.
Given : n(A) = 30, n(B) = 32, n(S) = 60, n(A ∩ S) = 24

1. P (Student opted for NCC or NSS)

2. P (Student has opted neither NCC nor NSS)

3. P (The student has opted NSS but not NCC)

Question 4.
From the employees of a company, 5 persons are selected to represent them in managing committee of the company. Particulars of five persons are as follows :

A person is selected at random from this group to act as a spokesperson. What is the probability that the spokesperson will be either male or over 35 years? (NCERT)
Solution:
Let A is the event that selected person is male and B be the event that selected person is over 35 years.

Hence, required probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= $$\frac { 3 }{ 5 }$$ + $$\frac { 2 }{ 5 }$$ – $$\frac { 1 }{ 5 }$$ = $$\frac { 3 + 2 – 1 }{ 5 }$$ = $$\frac { 4 }{ 5 }$$

Question 5.
A fair coin with 1 marked on one face and 6 on the other and a fair dice are both tossed. Find the probability that the sum of the numbers that turn up is: (NCERT)

1. 3
2. 12.

Solution:
The number of possible cases in coin = 2 Number of possible cases in dice = 6 Let S is the sample space, then
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(S) = 12
1. Let E is the event of coming sum of the numbers as 3, then
E = { 1, 2 }
∴ n(E) = 1
Required probability P(E) = $$\frac { n(E) }{ n(S) }$$ = $$\frac { 1 }{ 12 }$$

2. Let E’ is the event of coming sum of the numbers as 12, then E’ = (6, 6)
∴ n(E’) = 1
Required probability P(E’) = $$\frac { n(E’) }{ n(S) }$$ = $$\frac { 1 }{ 12 }$$

Question 6.
In a certain lottery 10,000 tickets are sold and ten equal prizes are awarded. What is the probability of not getting a prize if you buy?

1. One ticket
2. Two tickets
3. 10 tickets.

Solution:
Total tickets = 10,000
The tickets getting not prize = 10,000 – 10 = 9,990
Let S is the sample space, then
n(S) = 10,000C1 = 10,000
Let E1 is the event of not getting prize on buying one ticket, then
n(E1) = 9,990C1 = 9,990
∴ Required probability P(E1) = $$\frac{n\left(E_{1}\right)}{n(S)}$$ = $$\frac { 9, 990 }{ 1, 000 }$$ = $$\frac { 999 }{ 1, 000 }$$

2. On buying two tickets, then
n(S) = 10,000C2
Let E2 is the event of not getting prize on buying two tickets, then
n(E2) = 9,990C2
∴ Required probability P(E2) = $$\frac{n\left(E_{2}\right)}{n(S)}$$ = $$\frac { 9, 990 }{ 1, 000 }$$

3. On buying 10 tickets, then
n(S) = 10,000C10
Let E3 is the event of not getting prize on buying ten tickets, then
n(E3) = 9,990C10
∴ Required probability P(E3) = $$\frac{n\left(E_{3}\right)}{n(S)}$$

Question 7.
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 it is rusted or is a bolt.
Solution:
Let S be the sample space, then
∴ n (S) = 200, [∵ 50 bolt + 150 nuts = 200]
According to the question half of the bolts and nuts are rusted.
∴ Total rusted article = 25 + 75 = 100
Let E1 is the event of drawing a rusted article.
Then, P(E1) = $$\frac{n\left(E_{1}\right)}{n(S)}$$ = $$\frac { 100 }{ 200 }$$
Let E2 is the event of drawing nuts.
Then, P(E2) = $$\frac{n\left(E_{2}\right)}{n(S)}$$ = $$\frac { 50 }{ 200 }$$
In both the events E1, and E2, 25 rusted bolts are common.

Hence, required probability

## MP Board Class 11th Maths Important Questions Chapter 15 Statistics

### Statistics Important Questions

Question 1.
Find the mean deviation about the mean for the data in following table: (NCERT)

Solution:

Mean $$\bar { x }$$ = $$\frac{\sum f_{i} x_{i}}{\sum f_{i}}$$ = $$\frac { 4000 }{ 80 }$$ = 50
Mean deviation about mean = $$\frac{\sum f_{i}\left|x_{i}-\bar{x}\right|}{\sum f_{i}}$$
= $$\frac { 1280 }{ 80 }$$ = 16.

Question 2.
Find the mean deviation about the median for the data in following table: (NCERT)

Solution:

$$\frac { N }{ 2 }$$ = $$\frac { 26 }{ 2 }$$ = 13
Which lies on cumulative frequency (C.F.) 14.
∴ Median = Md = 7.
Mean deviation about mean = $$\frac{\sum f_{i}\left|x_{i}-M_{d}\right|}{\sum f_{i}}$$
= $$\frac { 1128.8 }{ 100 }$$

Question 3.
Find the mean deviation about the mean for the data in following table: (NCERT)

Solution:
Let A = 130

Mean $$\bar { x }$$ = A + $$\frac { Σfd }{ Σf }$$ × i = 130 + $$\frac { (- 47) }{ 10 }$$ = 10
$$\bar { x }$$ = 130 – 4.7 = 125.3
Mean deviation about mean = $$\frac{\sum f_{i}\left|x_{i}-\bar{x}\right|}{\sum f_{i}}$$
= $$\frac { 1128.8 }{ 100 }$$ = 11.228.

Question 4.
Find the mean deviation about the median for the data in following table: (NCERT)

Solution:

$$\frac { N }{ 2 }$$ = $$\frac { 50 }{ 2 }$$ = 25
20 – 30 is median class.
Here F = 14, f = 14, l = 20, i = 10

Question 5.
Calculate the mean deviation about median for the age distribution of 100 persons given below :

Solution:
First make the class interval uniform:

$$\frac { N }{ 2 }$$ = $$\frac { 100 }{ 2 }$$ = 50
Median class = 35.5 – 40.5
where l = 35.5, F = 37, f = 26, i = 5

Question 6.
Find the mean and variance for the following frequency distribution in following table:
(A)

Solution:

Let assumed mean A = 105

(B)

Solution:

Let assumed mean A = 25

Question 7.
Find the mean and variance and standard deviation using short cut method:

Solution:

Let assumed mean A = 92.5

Standard deviation σ = $$\sqrt {variance}$$
= $$\sqrt {105.58}$$ = 10.27

Question 8.
The diameter of circle (in mm) drawn in design are given below :

Calculate the standard deviation and mean diameter of the circles. (NCERT)
Solution:
First make the class interval uniform :

Let assume mean A = 42.5

Question 9.
From the prices of shares X and Y below, find out which is more stable in value:

Solution:
For first share X : Let assumed mean A = 58, n = 10

Foe second share Y : Let assumed mean A = 107, n = 10

Coefficient of variation for X :
Mean M = A + $$\frac { Σd }{ n }$$
= 58 + $$\frac { ( – 70) }{ 10 }$$ = 58 – 7
= 51
∴ Coefficient of variance = $$\frac { σ }{ M }$$ x 100
= $$\frac { 6 }{ 51 }$$ x 100 = 11.8
Coefficient of variance of Y :
Mean M = A + $$\frac { Σd }{ n }$$
= 107 + $$\frac { – 20 }{ 10 }$$ = 107 – 2
= 105
∴ Coefficient of variation = $$\frac { σ }{ M }$$ x 100
= $$\frac { 2 }{ 105 }$$ x 100 = 1.9
Coefficient of variance of Y is less than the coefficient of variance of X.
Share Y is more stable than share X.

Question 10.
An analysis of monthly wages paid to the workers in two firms A and B belonging to the same industry. Give the following results: (NCERT)

1. Which firm A or B pays out larger amount as monthly wages.
2. Which firm A or B shows greater variability in individual wages.

Solution:
For firm A :
Number of wages earners = 586
Mean of monthly wages x = Rs. 5,253
Amount paid by firm A = 586 x 5,253 = Rs. 30, 78, 258
Variance of distribution of wages = 100
Standard deviation σ = $$\sqrt {100}$$ = 10
Coefficient of variation = $$\frac{\sigma}{\bar{x}}$$ x 100 = $$\frac { 10 }{ 5253 }$$ = 0.19

For firm B :
Number of wages earners = 648
Mean $$\bar { x }$$ = Rs. 5253
Amount paid by firm B = 648 x 5253 = Rs. 34, 03, 944
Coefficient of variation = 121
Standard deviation σ = $$\sqrt {121}$$ = 11
Coefficient of variation = $$\frac{\sigma}{\bar{x}}$$ x 100 = $$\frac { 11 }{ 5253 }$$ x 100 = 0.21
Thus, firm B pays out larger amount as monthly wages.
∵ C.V. of firm B > C.V. of firm A.
∴ Firm B shows greater variability in individual wages.

Question 11.
The following is the record of goals scored by team A in a football session:

For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent? (NCERT)
Solution:

For team B :

C.V of team A < C.V of team B. Hence team A is more consistent.

Question 12.
The mean and standard deviation of marks obtained by 50 students of three subjects Mathematics, Physics and Chemistry are given below :

Which of these three subjects shows the highest variability in marks and which shows lowest? (NCERT)
Solution:
C.V. of Mathematics (C.V.) = $$\frac{\sigma}{\bar{x}}$$ x 100
Where σ = 12, $$\bar { x }$$ = 42
∴ C.V. of Mathematics = $$\frac { 12 }{ 42 }$$ x 100 = 28.57
C.V. of Physics = $$\frac{\sigma}{\bar{x}}$$ x 100
Where σ = 15, $$\bar { x }$$ = 32
∴ C.V. of Physics = $$\frac { 15 }{ 32 }$$ x 100 = 46.87
C.V. of Chemistry = $$\frac{\sigma}{\bar{x}}$$ x 100
Where σ = 20, $$\bar { x }$$ = 40.9
∴ C.V. of Chemistry = $$\frac { 20 }{ 40.9 }$$ x 100
= 48.89
∵ C.V. of Chemistry > C.V. of Physics > C.V. of Mathematics.
∴ Chemistry shows the highest variability and Mathematics shows the least variability.

Question 13.
The mean and standard deviation of a group of 100 observations were found to be 20 and 3 respectively. Later on it was found that three observations were incorrect Which are recorded by 21,21, and 18. Find the mean and standard deviation, if the incorrect observations are omitted. (NCERT)
Solution:
Given: n = 100, $$\bar { x }$$ = 20, σ = 3
$$\bar { x }$$ = $$\frac{\sum x}{n}$$ ⇒ 20 = $$\frac{\sum x}{n}$$
Σx = 20 x 100 = 2000
If incorrect observations 21,21 and 18 are omitted, then correct sum
Σx = 2000 – 21 – 21 – 18 = 2000 – 60 = 1940
Now correct mean of remaining 97 observations are
$$\bar { x }$$ = $$\frac{\sum x}{n}$$ = $$\frac { 1940 }{ 97 }$$ = 20
Given : σ = 3

correct Σx2 = 40900 – (21)2 – (21)2 – (18)2
= 40900 – 441 – 441 – 324 = 39694
Now correct S.D. of remaining 97 observations are

Question 14.
The mean and variance of eight observations are 9 and 9.25 respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, then And the remaining two observations. (NCERT)
Solution:
Let the two remaining observations are x1 and x2
Mean M = $$\frac{6 + 7 + 10 + 12 + 12 + 13 + x_{1} + x_{2}}{8}$$
9 = $$\frac{60 + x_{1} + x_{2}}{8}$$
⇒ x1 + x2 + 60 = 72
⇒ x1 + x2 = 12 …. (1)

⇒ x21 + x22 = 722 – 642
⇒ x21 + x22 = 80 …. (2)
From eqn. (1),
x2 = 12 – x1
Put the value of x2 in equation (2),
x21 + (12 – x1)2 = 80
⇒ x2 + 144 + x21 – 24x1 = 80
⇒ 2x21 – 24x1 + 64 = 0
⇒ x21 – 12x1 + 32 = 0
⇒ x21 – 4x1 – 8x1 + 32 = 0
⇒ x1(x1 – 4) – 8(x1 – 4) = 0
⇒ (x1 – 8)(x1 – 4) = 0
⇒ x1 = 4, 8
When x1 = 4, then x2 = 12 – 4 = 8
When x1 = 8, then x2 = 12 – 8 = 4
Hence remaining observations are 4 and 8.

## MP Board Class 11th Maths Important Questions Chapter 14 Mathematical Reasoning

### Mathematical Reasoning Important Questions

Mathematical Reasoning Objective Type Questions

(A) Choose the correct option :

Question 1.
If p and H are statements then determine of equivalance statement is :
(a) p ⇔ q
(b) p ∨ q
(c) p ∧ q
(d) None of these.
(a) p ⇔ q

Question 2.
Examine of the following sentence is not statement:
(a) The sum of interior angles of a quadrilateral is 360°
(b) There are only three sides in a triangle
(c) You have long live
(d) The sum of three and three is 6.
(c) You have long live

Question 3.
Examine that which of the following is correct part of a negative statement:
(a) 3 + 5 > 9
(b) Each square is a rectangle
(c) Equilateral triangle is an isosceles triangle
(d) None of these.
(a) 3 + 5 > 9

Question 4.
Which of the following is a statement:
(a) n is a real numbers
(b) Let us go
(c) Switch of the fan
(d) 3 is a natural number.
(d) 3 is a natural number.

Question 5.
The connective in the statement “3 + 5 > 9 or 3 + 5 < 9” is :
(a) >
(b) <
(c) or
(d) and.
(c) or

Question 6.
Examine which of the following sentence statement:
(a) Each square is a rectangle
(b) Door closed
(c) God bless you
(d) Oh ! I passed.
(a) Each square is a rectangle

Question 7.
The opposite statement of the statement p => q is :
(a) ~ q ⇔ p
(b) q ⇒ P
(c) ~ p ⇒ q
(d) None of these.
(b) q ⇒ P

Question 8.
The negative of the statement “51 is not a multiple of 3” is :
(a) 51 is an odd number
(b) 51 is not an odd number
(c) 51 is a multiple of 2
(d) 51 is a multiple of 3.
(d) 51 is a multiple of 3.

Question 9.
The contrapositive of the statement “ If p, then q ” is :
(a) If q, then ~ P
(b) If ~ p, then ~ q
(c) If p, then ~ q
(d) If ~ q, then ~ p
(d) If ~ q, then ~ p

(B) Write answer in one word / sentence :

1. Write the component statements of the compound statement:
“13 is an odd number and a prime number.”
p : Number 13 is prime; q: Number 13 is odd.

2. Write the negation of the statement:
“Everyone who lives in India is an Indian.”
Every one who live in India is an Indian

3. Identify the connective in the following compound statement:
“It is raining or the Sun is shining.”
or.

4. From the biconditional statement p ⇔ q, where :
p ≡ A triangle is an equilateral.
q ≡ All three sides of a triangle are equal.
A triangle is an equilateral triangle if and only if all three sides of the triangle are equal.

5. If p ≡ Mathematics is hard, q ≡ 4 is even number, then write the formula p ∨ q in logical sentences.
Mathematics is hard or 4 is even number.

6. If p ≡ the question paper is hard, q ≡ I will fail in the examination, then write the statement in symbolically, “The question paper is hard if and only if I will fail in the examination.”
p ⇔ q.

7. State whether the statement is exclusive or inclusive :
“All integers are positive or negative.”
Exclusive.

Mathematical Reasoning Very Short Answer Type Questions

Question 1.
Which of the following sentences are statements? Give reasons for your answer: (NCERT)

1. There are 35 days in a month.
2. Mathematics is difficult.
3. The sum of 5 and 7 is greater than 10.
4. The square of a number is an even number.
5. The sides of a quadrilateral have equal lengths.
7. The product of (- 1) and 8 is 8.
8. The sum of all interior angles of a triangle is 180°.
9. Today is a windy day.
10. All real numbers are complex numbers.

1. A month has 30 or 31 days. It is false to say that a month has 35 days, hence it is a statement.
2. Mathematics may be difficult for one but may be easy for the others. Hence it is not a statement.
3. It is true that sum of 5 and 7 is greater than 10. Hence it is a statement.
4. The square of a number may be even or it may be odd. Squaring of an odd number is always odd and square of even number is always even. Hence it is not a statement.
5. A quadrilateral may have equal lengths as it may be a rhombus or a square or the quadrilateral may have unequal sides
6. like parallelogram, hence it is not a statement.
7. It is an order, hence it is not a statement.
8. It is false because the product of (- 1) and 8 is – 8, hence it is a statement. It is true because sum of three angles of a triangles is 180°. Hence it is a statement.
9. It is a windy day. It is not clear that about which day it is said. Thus it can’t be concluded whether it is true or false. Hence it is not a statement.
10. It is true that all real numbers are complex numbers. All real numbers can be expressed as a + ib, hence it is a statement.

Question 2.
State whether the ‘or’ used in the following statements is exclusive or inclusive. Give reasons for your answer: (NCERT)

1. Sun rises or Moon sets.
2. To apply for driving licence, you should have a ration card or a passport.
3. All integers are positive or negative.

1. When Sun rises, the Moon sets. One of happening will take place, hence ‘or’ is exclusive.
2. To apply for a driving licence either a ration card or a passport or both can be used, hence ‘or’ is inclusive.
3. All integers are positive or negative. An integers cannot be both positive or negative, hence ‘or’ is exclusive.

Question 3.
Rewrite the following statement ‘if then’ in five different ways conveying the same meaning:
If a natural number is odd than its square is odd. (NCERT)

1. A natural number is odd implies that its square is odd.
2. A natural number is odd only if its square is odd.
3. If the square of a natural number is not odd, then the natural number is also not odd.
4. For a natural number to be odd, its necessary that its square is odd.
5. For a square of a natural number to be odd, if it is sufficient that the number is odd.

Question 4.
Give statement in (a) and (b) identify the statements given below as contra – positive or converse of each other: (NCERT)
(a) If you live in Delhi, then you have winter clothes.
(i) If you do not have winter clothes, then you do not live in Delhi.
(ii) If you have winter clothes, then you live in Delhi.

(b) If a quadrilateral is parallelogram, then its diagonal bisect each other.
(i) If the diagonals of a quadrilateral do not bisect each other, then the quadrilateral is not a parallelogram.
(ii) If the diagonals of a quadrilateral bisect each other, then it is a parallelogram
(a) (i) Contrapositive statement, (ii) Converse statement.
(b) (i) Contrapositive statement, (ii) Converse statement.

Question 5.
Which of the following sentence is a statement:

1. New Delhi is a capital of India.
2. How are you?

1. Its a statement.
2. Its not a statement.

Question 6.
Find the component statement of the following statements :
1. The sky is blue and grass is green.
2. All the rational numbers are real number and all the real numbers are complex number.
Solution:
1. The components are:
p : The sky is blue.
q : The grass is green.

2. The component statement are as follows:
p : All the rational numbers are real number.
q : All the real numbers are complex number.

Question 7.
If p = It is 7 o’clock
q = The train is late,
then write the following symbols in the form of statement:

1. q ∨ ~ p
2. p ∧ q
3. ~ (p ∧ q)
4. p ∧ ~ q

1. Either the train is late or it is not 7 o’clock.
2. It is 7 o’clock, but train is late.
3. It is not true that it is 7 o’clock and the train is late.
4. It is 7 o’clock, but train is not late.

Question 8.
If p ≡ He is intelligent and q ≡ He is strong, then write the following statement symbolically with the help of logical connectives:

1. He is intelligent and strong.
2. He is intelligent but not strong.
3. He is neither intelligent nor strong.

1. p ∧ q
2. p ∧ ~ q
3. ~ q ∧ ~ q

Mathematical Reasoning Short Answer Type Questions

Question 1.
Show that the statement p : If x is a real number such that x3 + 4x = 0, then x is ‘0’ is true by :

1. Direct method
3. Method of contrapositive

Solution:
1. Direct method :
Given :
x3 + 4x = 0 ⇒ x(x2 + 4) = 0
x = 0 or x3 + 4 = 0
x is real number.
∴ x2 + 4 ≠ 0
∴ x = 0.

Let x ≠ 0
Let x = p, (where p is a real number)
p is one the root of equation x3 + 4x = 0.
∴ p3 + 4p = 0
⇒ P(P2 + 4) = 0
⇒ P = o
and p2 + 4 ≠ 0, (It is a real number)
∴ p = 0.

3. Method of contrapositive :
Let x = 0 is not true and x = p ≠ 0
∴p being the root of x3 + 4x = 0
∴ p3 + 4 p = 0
⇒ P(p2 + 4) = 0
p = 0 and p2 + 4 = 0
⇒ P(p2 + 4) ≠ 0, if p is not true
∴ x = 0 is one root of equation x3 + 4x = 0.

Question 2.
Show that the statement ‘for any real number a and b, a2 = b2 implies that a = b’ is not true by giving a counter example.
Solution:
Let a = 3 and b = -3 then a, b are real numbers.
Here a2 = b2 but a b
Hence a, b ∈ R and a2 = b2
⇒ a = b, statement is not true.

Question 3.
Find the components of the following compound statement and check whether they are true or false.
(i) The number 3 is prime or odd.
(ii) All the integers are positive or negative.
(iii) Number 100 is divisible by the numbers 3, 11 and 5.
(i) p : Number 3 is prime.
q : Number 3 is odd.
Here p and q both are true.

(ii) p : All the integers are positive. q : All the integers are negative.
Clearly p and q both are false.

(iii) p : The number 100 is divisible by 3.
q : The number 100 is divisible by 11.
r : The number 100 is divisible by 5.
p is false, q is false and r is true. p, q and r are false statement.

Question 4.
Show that the following statement is true by the method of contrapositive :
p : If x is an integer and x2 is even, then x is also even.
Solution:
P – If x is an integer and x2 is even, then x is also even.
Let q : x is an integer and x2 is even.
r : x is even.
To prove that P is true by contrapositive method, we assume that r is false and prove that q is also false.
Let x is not even.
To prove that q is false, it has to be proved that
x is not an integer or x2 is not even.
x is not even implies that x2 is also not even.
Therefore, statement q is false.
Thus, the given statement p is true.

Question 5.
For each of the following compound statements first identify the connecting words and then break it into compound statements: (NCERT)
(i) All rational numbers are real and all real numbers are not complex.
(ii) x = 2 and x = 3 are the roots of the equation 3x2 – x – 10 = 0.
Solution:
(i) The connecting word in the compound statement is ‘and’.
p : All the rational numbers are real number.
q: AH the real numbers are not complex number.

(ii) The connecting word in the compound statement is ‘and’.
p : x = 2 is the root of equation 3x2 – x – 10 = 0.
q : x = 3 is the root of equation 3x2 – x – 10 = 0.

## MP Board Class 11th Maths Important Questions Chapter 13 Limits and Derivatives

### Limits and Derivatives Important Questions

Limits and Derivatives Short Answer Type Questions

Evaluate the following limits :

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.
If = 405, then find the value of n.
Solution:

Question 10.
Find the value of :

Solution:

Question 11.

Solution:

Question 12.
If y = ex cos x, then find the value of $$\frac { dy }{ dx }$$
Solution:
y = ex.cos x
I II
$$\frac { dy }{ dx }$$ = $$\frac { d }{ dx }$$(ex.cos x)
= ex.$$\frac { d }{ dx }$$cos x + cos x$$\frac { d }{ dx }$$ex
= ex( – sin x) + cos x$$\frac { d }{ dx }$$ex
= ex[cos x – sin x].

Question 13.
If y = ex cos x, then find the value of $$\frac { dy }{ dx }$$
Solution:
y = ex.sin x
I II
$$\frac { dy }{ dx }$$ = $$\frac { d }{ dx }$$(ex.sin x)
= ex.$$\frac { d }{ dx }$$sin x + sin x$$\frac { d }{ dx }$$ex
= excos x + x.ex
= ex(cos x + sin x).

Question 14.
Differentiate sin(x + a) w.r.t. x.
Solution:
Let y = sin(x + a)
y = sin x cos a + cos x sin a
∴$$\frac { dy }{ dx }$$ = $$\frac { d }{ dx }$$(sin x cos a + cos x sin a)
= cos a$$\frac { d }{ dx }$$ sin x + sin$$\frac { d }{ dx }$$cos x
= cos a cos x – sin a sin x
⇒ $$\frac { dy }{ dx }$$ = cos(x + a).

Question 15.
Differentiate cosecx.cotx w.r.t. x.
Solution:
Let y = cosec x. cot x
⇒ $$\frac { dy }{ dx }$$ = $$\frac { d }{ dx }$$[cosec x. cot x]
⇒ $$\frac { dy }{ dx }$$ = cot x$$\frac { d }{ dx }$$ cosec x + cosec x$$\frac { d }{ dx }$$ cot x
= – cot xcosec x cot x – cosec x.cosec2 x
$$\frac { dy }{ dx }$$ = – cosec x[cot2 x + cosec2 x].

Question 16.
Differentiate $$\frac { cos x }{ 1 + sin x }$$ w.r.t. x.
Solution:
Let y = $$\frac { cos x }{ 1 + sin x }$$

Question 17.
Differentiate $$\frac { sec c – 1 }{ sec x + 1 }$$ w.r.t. x.
Solution:
Let y = $$\frac { sec c – 1 }{ sec x + 1 }$$

Question 18.
Differentiate sinn x w.r.t. x.
Solution:
Let y = sinn x
$$\frac { dy }{ dx }$$ = $$\frac { d }{ dx }$$(sinn x )
Put sin x = t,
$$\frac { dy }{ dx }$$ = $$\frac { d }{ dx }$$tn = $$\frac { d }{ dt }$$tn. $$\frac { dt }{ dx }$$ = n sinn – 1$$\frac { d }{ dx }$$(sin x)
⇒ $$\frac { dy }{ dx }$$ = n sinn – 1x cos x.

Limits and Derivatives Long Answer Type Questions

Question 1.
Let f(x) =
, then find the value of a and b.
Solution:

Question 2.
Find the value of , where

Solution:

Question 3.
f(x) is defined such that
and is exist x = 2, then find the value of k.
Solution:

Question 4.
If the function f(x) satisfies= π, then find the value of
Solution:

Question 5.
Find the differential coefficient of the following functions by using first principle method :
(i) sin(x + 1), (ii) cos(x – $$\frac { π }{ 8 }$$,
Solution:
(i) Let f(x) = sin(x + 1)
∴ f(x + h) = sin[x + h + 1]
By definition of first principle,

By definition of first principle,

Find the differential coefficient of the following functions

Question 6.
(ax2 + sin x)(p + q cos x). (NCERT)
Solution:
Let y = (ax2 + sin x)(p + q cos x).
∴ $$\frac { dy }{ dx }$$ = $$\frac { d }{ dx }$$[(ax2 + sin x)(p + q cos x)]
⇒ $$\frac { dy }{ dx }$$ = (p + q cos x)$$\frac { d }{ dx }$$(ax2 + sin x) + (ax2 + sin x)$$\frac { d }{ dx }$$ (p + q cos x)
⇒ $$\frac { dy }{ dx }$$ = (p + q cos x)(2a + cos x) + (ax2 + sin x)(0 – q sin x)
⇒ $$\frac { dy }{ dx }$$ = – q sin x(ax2 + sin x) + (p + q cos x)(2a + cos x)

Question 7.
(x + cos x)(x – tan x) (NCERT)
Solution:
Let y = (x + cos x)(x – tan x)
∴ $$\frac { dy }{ dx }$$ = $$\frac { d }{ dx }$$[(x + cos x)(x – tan x)]
⇒ $$\frac { dy }{ dx }$$ = (x – tan x)$$\frac { d }{ dx }$$(x + cos x) + (x + cos x)$$\frac { d }{ dx }$$(x – tan x)
⇒ $$\frac { dy }{ dx }$$ = (x – tan x)(1 – sinx) + (x + cos x)(1 – sec2 x)
= (x – tan x)(1 – sinx) + (x + cos x)(sec2 x – 1)
⇒ $$\frac { dy }{ dx }$$ = (x – tan x)(1 – sinx) – tan2 x(x + cos x)

Question 8.
$$\frac { 4x + 5sin x }{ 3x + 7 cos x}$$
Solution:
Let y = $$\frac { 4x + 5sin x }{ 3x + 7 cos x}$$

Question 9.
$$\frac{x^{2} \cos \frac{\pi}{4}}{\sin x}$$
Solution:
Let y = $$\frac{x^{2} \cos \frac{\pi}{4}}{\sin x}$$

Question 10.
$$\frac { x }{ 1 + tan x}$$
Solution:
Let y = $$\frac { x }{ 1 + tan x}$$

Question 11.
Find the differential coefficient of $$\frac { sec x – 1 }{ sec x + 1}$$
Solution:
Let y = $$\frac { sec x – 1 }{ sec x + 1}$$

Question 12.
If y = $$\frac { x }{ x + 4 }$$, then prove that:
x$$\frac { dy }{ dx}$$ = y(1 – y)
Solution:
Given : y = $$\frac { x }{ x + 4 }$$

Question 13.
If y = $$\sqrt {x}$$ + $$\frac{1}{\sqrt{x}}$$, then prove that : 2x$$\frac { dy }{ dx}$$ + y – 2 $$\sqrt {x}$$ = 0
Solution:
Given : y = $$\sqrt {x}$$ + $$\frac{1}{\sqrt{x}}$$

Question 14.
Find the differential coefficient of $$\frac { sin(x + a) }{ cos x }$$
Solution:
Let y = $$\frac { sin(x + a) }{ cos x }$$
⇒ y = $$\frac { sin x + cos a + cos x sin a }{ cos x }$$
⇒ y = $$\frac { sin x + cos a }{ cos x }$$ + $$\frac { cos x + sin a }{ cos x }$$
⇒ y = cos a tan x + sin a
∴ $$\frac { dy }{ dx}$$ = $$\frac { d }{ dx}$$(cos a tan x + sin a)
= cos a $$\frac { d }{ dx}$$ tan x + $$\frac { d }{ dx}$$ sin a
= cos a x sec2 x + 0
= sec2x. cos a.

Question 15.
If f(x) = $$\frac { { x }^{ 100 } }{ 100 } +\frac { { x }^{ 99 } }{ 100 }$$+……..$$\frac { { x }^{ 2 } }{ 2 }$$ + x +1, then prove that:
f'(1) = 100 f'(0). (NCERT)
Solution:
Put x = 1, we get
f'(1) = 1 + 1 + ………. 1 + 1 (100 times)
f’(1) = 100 …. (1)
Put x = 0, we get
f'(0) = 0 + 0 + ……… 0 + 1
f’(0) = 1 …. (2)
From equation (1) and (2),
f'(1) = 100 f'(0).

Question 16.
Find the differential coefficient of cos x by first principle method. (NCERT)
Solution:
Let f(x) = cosx
∴ f(x + h) = cos(x + h)
By definition of first principle

Question 17.
Find the differential coefficient of f(x) = $$\frac { x + 1 }{ x – 1 }$$ by first principle method.
Solution:
Given : f(x) = $$\frac { x + 1 }{ x – 1 }$$
f(x + h) = $$\frac { x + h + 1}{ x + h – 1 }$$
By definition of first principle,

## MP Board Class 11th Maths Important Questions Chapter 12 Introduction to Three Dimensional Geometry

### Introduction to Three Dimensional Geometry Important Questions

Introduction to Three Dimensional Geometry Objective Type Questions

(A) Choose the correct option :

Question 1.
Distance of a point (3,4, 5) from origin:
(a) 3
(b) 4
(c) 5
(d) 5$$\sqrt {2}$$
(d) 5$$\sqrt {2}$$

Question 2.
The perpendicular distance of the point P (x, y, z) from X – axis is:
(a) $$\sqrt { { x }^{ 2 } + { z }^{ 2 } }$$
(b) $$\sqrt { { y }^{ 2 } + { z }^{ 2 } }$$
(c) $$\sqrt { { x }^{ 2 } + { y }^{ 2 } }$$
(d) $$\sqrt { { x }^{ 2 } + { y }^{ 2 } + { z }^{ 2 } }$$
(b) $$\sqrt { { y }^{ 2 } + { z }^{ 2 } }$$

Question 3.
Distance of a point (3, 2, 5) from X – axis is:
(a) $$\sqrt {28}$$
(b)$$\sqrt {30}$$
(c) $$\sqrt {29}$$
(d) 3
(c) $$\sqrt {29}$$

Question 4.
The YZ – plane divides the line segment joining the points (2, 3, 4) and (3, 5, – 4) in the ratio:
(a) 2 : 3 internally
(b) 1 : 2
(c) 2 : 3 externally
(d) 1 : 3
(c) 2 : 3 externally

Question 5.
Distance between the points (1, 2,3) and (1,3, -2) is :
(a) $$\sqrt {- 26}$$
(b) $$\sqrt {26}$$
(c) 26
(d) ± $$\sqrt {24}$$
(b) $$\sqrt {26}$$

Question 6.
Coordinate of a point on Z – axis, equidistant from the points (1, 5, 7) and (5, 1, – 4) is :
(a) (0, 0, $$\frac { 3 }{ 2 }$$
(b) (o, $$\frac { 3 }{ 2 }$$, 0)
(c) ($$\frac { 3 }{ 2 }$$, 0, 0)
(d) (4, – 4, – 11)
(a) (0, 0, $$\frac { 3 }{ 2 }$$

Question 7.
Distance of a point (2, 1, 4) from Y – axis is:
(a) $$\sqrt {20}$$
(b) 1
(c) $$\sqrt {12}$$
(d) $$\sqrt {10}$$
(a) $$\sqrt {20}$$

(B) Match the following :

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

(C) Fill in the blanks :

1. If the points (- 1, 3, 2), (- 4, 2, – 2) and (5, 5, λ ) are collinear, then value of λ is …………….
2. The perpendicular distance of point P(x, y, z) from YZ – plane is …………….
3. XY – plane divides the line segment joining the points (- 3, 4, 8) and (5, – 6, 4) in the ratio …………….
4. Distance between the points (1, – 3, 4) and (3, 11, – 6) is …………….
5. Perpendicular distance of a point P (3,4, 5) from YZ – plane is …………….

1. 10
2. x
3. 2 : 1
4. 10$$\sqrt {3}$$
5. 3

(D) Write true / false :

1. Distance of a point (4, 3, 5) from Y – axis is $$\sqrt {40}$$.
2. Distance of a point (5, 12, 13) from YZ – axis is $$\sqrt {313}$$.
3. The coordinate of a point where YZ – plane divides the join of the points (3, 5, – 7) and (- 2, 1, 8) are (0, $$\frac { 13 }{ 5 }$$, 2).
4. Coordinate of mid point of the line segment joining the points (- 3, 4, – 8) and (5, – 6, 4) are (1, – 1, 2).
5. Points A(1, 2, 3), B(4, 0, 4) and C (- 2, 4, 2) are collinear.

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

(E) Write answer in one word / sentence :

1. A point R lies on the line segment joining the points P (2, – 3, 4) and Q(8, 0, 10) whose x coordinate is 4, then the coordinate of R will be.
2. In what ratio does the plane 2x + y – z = 3 divide the line segment joining the points A (2, 1, 3) and 5(4, – 2, 5)?
3. If the origin is the centroid of the triangle ABC with vertices A (a, 1, 3), B (- 2, b, – 5), C (4, 7, c) then write the value of a, b and c.
4. Find the locus of the point, which is equidistant to the points (3, 4, – 5) and (- 2, 1, 4).
5. In which ratio the XY – plane divides the line joining the points (2, 4, 2) and (2, 5, – 4)?

1. (4, – 2, 6)
2. 1 : 2 externally
3. – 2, – 8, – 2
4. 10x + 6y – 18z – 29 = 0
5. 1 : 2 internally.

Introduction to Three Dimensional Geometry Very Short Answer Type Questions

Question 1.
A point is on X – axis. What are its y – co – ordinate and z – co – ordinate?
y – co – ordinate and z – co – ordinate of point on X – axis is 0, 0 is (x, 0, 0).

Question 2.
A point is in the xz – plane. What can you say about is y – co – ordinate?
On xz – plane the y – co – ordinate will be zero.

Question 3.
The X – axis and Y – axis taken together to form a plane, name of the plane is.
The plane is known as xy – plane.

Question 4.
Co – ordinate of any point on xy – plane is.
The co – ordinate of any point on xy – plane is (x, y, 0).

Question 5.
Length of perpendicular of point P(x, y, z) from X – axis is.
Length of perpendicular = $$\sqrt { { y }^{ 2 } + { z }^{ 2 } }$$

Question 6.
Length of perpendicular of point P(x, y, z) from yz – plane is.
Length of perpendicular of point P(x, y, z) from yz – plane is x.

Question 7.
Find the distance of point (3, 4, 5) from xz – plane.
Distance of point (3, 4, 5) from xz – plane M(3, 0, 5) = $$\sqrt {0 + 16 + 0}$$ = 4.

Question 8.
What is the centroid of ∆ABC whose vertices A(x1, y1, z1), B(x2, y2, Z2) and C(x3, y3, z3)?
Centroid of ∆ABC is

Question 9.
Find distance between points (2,3, 5) and (4,3,1).
Solution:
Distance between two points A(x1, y1, z1) and B(x2, y2, Z2):

Question 10.
Find the distance between points (- 3, 7, 2) and (2, 4, – 1).
Solution:

Question 11.
Find the distance of the point (3, 2, 5) from X – axis.
Solution:
Point on X – axis (3, 0, 0)

Question 12.
Find the distance of the point (2, 1, 4) from Y – axis.
Solution:
Point on Y – axis (0, 1, 0)

Question 13.
The three dimensional planes divides the space in how many octants?
Three dimensional plane divides the space in eight octants.

Introduction to Three Dimensional Geometry Short Answer Type Questions

Question 1.
Prove that the points (- 2, 3, 5) (1, 2, 3) and (7, 0, – 1) are collinear. (NCERT)
Solution:
Given point are A (- 2, 3, 5), B(1, 2, 3) and C (7, 0, – 1).

Hence points A, B, Care collinear.

Question 2.
Find the co – ordinates of the point which divides the line segment joining the points (- 2, 3, 5) and (1, – 4, 6) in the ratio 2 : 3 internally. (NCERT)
Solution:
Required co – ordinates are :

Question 3.
Given that P (3, 2, – 4), Q (5, 4, – 6) and R (9, 8, – 10) are collinear. Find the ratio in which Q divides PR. (NCERT)
Solution:

⇒ 5k + 5 = 9k + 3
⇒ 4k = 2
⇒ k = $$\frac { 1 }{ 2 }$$
Hence, point Q divides PR internally in ratio 1 : 2.

Question 4.
Find the ratio in which the YZ – plane divides the line segment formed by joining the points (- 2, 4, 7) and (3, – 5, 8).
Solution:
Let the point R (0, y, z) on YZ – plane divides the points P (- 2, 4, 7) and Q (3, – 5, 8) in the ratio m : n.
x = $$\frac{m x_{2}+n x_{1}}{m+n}$$
⇒ 0 = $$\frac { m × 3 + n( – 2) }{ m + n }$$
⇒ 3m – 2n = 0
⇒ 3m = 2n
⇒ $$\frac { m }{ n }$$ = $$\frac { 2 }{ 3 }$$
⇒ m : n = 2 : 3
YZ – plane divides the line PQ internally in ratio 2:3.

Question 5.
If the origin is the centroid of the ∆PQR with vertices P(2a, 2, 6), Q(- 4, 3b, – 10) and R (8, 14, 2c) then find the value of a, b and c. (NCERT)
Solution:
Centroid of APQR is :
x = $$\frac{x_{1} + x_{2} + x_{3}}{3}$$, y = $$\frac{y_{1} + y_{2} + y_{3}}{3}$$, z = $$\frac{z_{1} + z_{2} + z_{3}}{3}$$
Co – ordinate of centroid are (0, 0, 0),

Question 6.
Find the co – ordinates of a point on Y – axis which are at a distance of 5$$\sqrt {2}$$ from the point P(3, – 2, 5).
Solution:
Let A (0, y, 0) be any point on Y – axis.
Given : PA = 5$$\sqrt {2}$$
PA2 = 50
⇒ (3 – 0)2 + (- 2 – y)2 + (5 – 0)2 = 50
⇒ 9 + 4 + y2 + 4y + 25 = 50
⇒ y2 + 4y + 38 – 50 = 0
⇒ y2 + 4y – 12 = 0
⇒ y2 + 6y – 2y – 12 = 0
⇒ y(y + 6) – 2(y + 6) = 0
⇒ (y – 2)(y + 6) = 0
⇒ y = 2, – 6
The co – ordinate on Y – axis is (0, 2, 0) or (0, – 6, 0).

Question 7.
A point R with x co – ordinate 4 lies on the line segment joining the points P (2, – 3, 4) and Q (8, 0, 10). Find the co – ordinate of the point R.
Solution:
Let the point R (x, y, z) divides PQ in ratio m : n.

Co – ordinate of R (4, – 2, 6).

Question 8.
Find the equation of set of points which are equidistant from the points (1, 2, 3) and (3, 2, – 1). (NCERT)
Solution:
Let point P (x, y, z) be equidistant from points A (1, 2, 3) and B (3, 2, – 1).
∴ PA = PB

⇒ PA2 = PB2
⇒ (x – 1)2 + (y – 2)2 + (z – 3)2 = (x – 3)2 + (y – 2)2 + (z +1)2
⇒ x2 – 2x + 1 + z2 – 6z + 9
– 2x – 6z + 10 = – 6x + 2z + 10
⇒ 4x – 8z = 0
⇒ x – 2z = 0.
This is the required equation.

Question 9.
In which ratio the point (1, 1, 1) divides the line joining the points (3, – 2, 4) and (- 1, 4, 2)?
Solution:
Let point R (1,1,1), divides the line joining the points P (3, – 2, 4) and Q(- 1, 4, 2) in ratio λ : 1.
By formula:
x = $$\frac{m x_{2}+n x_{1}}{m+n}$$
Here x1 = 3, x2 = – 1, x = 1, m = λ and n = 1.
∴ 1 = $$\frac { λ(- 1) + 1 × 3 }{ λ + 1}$$
⇒ λ + 1 = – λ + 3
⇒ 2λ = 2
∴ λ = 1
Hence, required ratio is 1 : 1.

## MP Board Class 11th Maths Important Questions Chapter 11 Conic Sections

### Conic Sections Important Questions

Conic Sections Objective Type Questions

(A) Choose the correct option :

Question 1.
Coordinates of the focus of the parabola y = 2x2 + x are:
(a) (0, 0)
(b) ($$\frac { 1 }{ 2 }$$, $$\frac { 1 }{ 4 }$$)
(c) (- $$\frac { 1 }{ 4 }$$, 0)
(d) ( – $$\frac { 1 }{ 4 }$$, $$\frac { 1 }{ 8 }$$)
(c) (- $$\frac { 1 }{ 4 }$$, 0)

Question 2.
In a ellipse $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}$$ = 1 a > b, the relation between a, b an eccentricity e is:
(a) b2 = a2(1 – e2)
(b) b2 = a2(e2 – 1)
(c) a2 = b2(1 – e2)
(d) a2 = b2(e2 – 1)
(a) b2 = a2(1 – e2)

Question 3.
The length of latus rectum of ellipse $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}$$ = 1, represent a circle then its eccentricity will be:
(a) $$\frac { { 2a }^{ 2 } }{ b }$$
(b) $$\frac { { 2b }^{ 2 } }{ a }$$
(c) $$\frac { { a }^{ 2 } }{ b }$$
(d) $$\frac { { b }^{ 2 } }{ a }$$
(b) $$\frac { { 2b }^{ 2 } }{ a }$$

Question 4.
The eccentricity of the parabola is:
(a) Less than 1
(b) Greater than 1
(c) 0
(d) 1
(d) 1

Question 5.
The eccentricity of the ellipse is:
(a) Less than 1
(b) Greater than 1
(c) 0
(d) 1
(a) Less than 1

Question 6.
The eccentricity of the hyperbola is:
(a) Less than 1
(b) Greater than 1
(c) 0
(d) 1
(b) Greater than 1

Question 7.
In a ellipse $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}$$ = 1, represent a circle then its eccentricity will be:
(a) Less than 1
(b) Greater than 1
(c) 0
(d) 1
(c) 0

Question 8.
The sum of focal distances from any point on the ellipse is:
(a) Equal to major axis
(b) Equal to minor axis
(c) The distance between two foci
(d) Equal to latus rectum.
(a) Equal to major axis

Question 9.
The differecne of the focal distances from any point on the hyperbola is:
(a) Equal to its conjugate axis
(b) Equal to its transverse axis
(c) The distance between two foci
(d) Equal to its latus rectum.
(b) Equal to its transverse axis

Question 10.
The value of the eccentricity of ellipse 25x2 + 16y2 = 400 is:
(a) $$\frac { 3 }{ 5 }$$
(b) $$\frac { 1 }{ 3 }$$
(c) $$\frac { 2 }{ 5 }$$
(d) $$\frac { 1 }{ 5}$$
(a) $$\frac { 3 }{ 5 }$$

Question 11.
Equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represent a circle if:
(a) a = b, c = 0
(b) f = g, h = 0
(c) a = b, h = 0
(d) f = g, c = 0
(a) a = b, c = 0

Question 12.
Area of triangle whose centre (1,2) and which is passes through the point (4,6) will be:
(a) 5π
(b) 10π
(c) 25π
(d) 25π2
(c) 25π

Question 13.
The circle passing through (1, – 2) and touching the X – axis at (3,0), also passes through the point:
(a) (2, – 5)
(b) (5, – 2)
(c) (- 2, 5)
(d) (- 5, 2)
(a) (2, – 5)

Question 14.
The length of the diameter of the circle which touches the X – axis at the point (1,0) and passes through the point (2,3) is:
(a) $$\frac { 10 }{ 3 }$$
(b) $$\frac { 3 }{ 5 }$$
(c) $$\frac { 6 }{ 5 }$$
(d) $$\frac { 5 }{ 3 }$$
(a) $$\frac { 10 }{ 3 }$$

Question 15.
Eccentricity of the hyperbola 3x2 – y2 = 4 :
(a) 1
(b) 2
(c) – 2
(d) $$\sqrt {2}$$
(b) 2

(B) Match the following :

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

(C) Fill in the blanks :

1. The length of the latus rectum of the parabola y2 = 4ax is ……………
2. The centre of the ellipse $$\frac { { (x-1) }^{ 2 } }{ 9 } +\frac { { (y-2) }^{ 2 } }{ 4 }$$ = 1 will be ……………
3. The vertex of the parabola (y – 2)2 = 4a(x -1) is ……………
4. The lines $$\frac { x }{ a }$$ – $$\frac { y }{ b }$$ = m and $$\frac { x }{ a }$$ + $$\frac { y }{ b }$$ = $$\frac { 1 }{ m }$$ meets always at ……………
5. If an ellipse $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}$$ = 1, a > b and its eccentricity is e, then the foci will be
6. Standard form of equation of parabola is ……………
7. Parametric equation of a circle x2 + y2 = 4 is ……………
8. A line y = x + a$$\sqrt {2}$$ touches the circle x2 + y2 = a2 point ……………
9. A line y = mx + c touches the circle x2 + y2 = a2 if c = ……………
10. Vertex of the parabola 3y2 + 6y – 4x + 11 = 0 is ……………
11. Equation 2x2 + 2y2 – 12x – 16y + 4 = 0 represent a point circle if k = ……………
12. Radius of circle 3x2 + 3y2 – 5x – 6y + 4 = 0 is ……………

1. 4a
2. (1, 2)
3. (1, 2)
4. Hyperbola
5. (± ae, 0)
6. y2 = 4ax
7. x = 2cosθ
8. (- $$\frac { a }{ \sqrt { 2 } }$$, $$\frac { a }{ \sqrt { 2 } }$$ )
9. ±a$$\sqrt { 1+{ m }^{ 2 } }$$
10. (5, 1)
11. 50, 12
12. $$\sqrt {61}$$

(D) Write true / false :

1. Conic section is a locus of the point whose the ratio between the distance from the fixed point and distance from the fixed line, this ratio is called eccentricity of the conic section.
2. The ellipse $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}$$ = 1 has two directrics, the equation of directrics are x = ± $$\frac { a }{ e }$$; Where a > b and y = ± $$\frac { b }{ e }$$ ; where b > a.
3. The foci of the ellipse $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}$$ = 1 are (0, ± be) where a < b.
4. A circle drawn by taking major axis of the ellipse as diameter is called auxiliary circle of the ellipse.
5. The locus of intersection point of the lines bx + ay = abt and bx – ay = $$\frac { ab}{ t }$$ will be a ellipse.
6. The focus of a parabola x2 = – 16y will be (0, – 4).
7. Equation x2 + y2 – 6x + 8y + 50 = 0 represent a circle.
8. Circle x2 + y2 = 9 and x2 + y2 + 8y + c = 0 touches externally if c = 15 .
9. Eccentricity of hyperbola is 1.
10. Minimum distance between line y – x = 1 and curve x = y2 is $$\frac { 3\sqrt { 2 } }{ 8 }$$

1. True
2. True
3. True
4. True
5. False
6. True
7. False
8. True
9. False
10. True

(E) Write answer in one word / sentence :

1. If the circle x2 + y2 + 2ax + 8y +16 = 0, touches X – axis, then the value of α will be.
2. Coordinate of focus of parabola x2 = – 10y will be.
3. Write the equation of a circle whose centre is (2,2) and passes through the point (4, 5).
4. The centre of a circle is (5, 7) and touches Y – axis, then its radius will be.
5. If the radius of a circle x2 + y2 – 6x + ky – 25 = 0 is $$\sqrt {38}$$ the value of k will be.
6. Vertex of the parabola y = x2 – 2x + 3 will be.
7. Equation of a parabola whose vertex (0, 0) and focus (0, 3) will be.
8. Length of major axis of ellipse 9x2 + 16y2 = 144 will be.
9. Eccentricity of an ellipse whose latus rectum in half of its minor axis will be.
10. Equation of hyperbola whose one focus in (4, 0 ) and corresponding equation of directrix x = 1 will be.

1. ± 4
2. (0, $$\frac { – 5 }{ 2 }$$)
3. x2 + y2 – 4x – 4y – 5 = 0
4. 7, 5
5. ± 4
6. (1, 2)
7. x2 = 12y
8. 6
9. $$\frac { \sqrt { 3 } }{ 2 }$$
10. $$\frac { x^{ 2 } }{ 4 } -\frac { y^{ 2 } }{ 12 }$$ = 1

Conic Sections Long Answer Type Questions

Question 1.
Find the equation of circle which touches the X – axis at a distance of 4 units in the negative direction and makes intercept of 6 units on positive direction of Y – axis.
Solution:
Here OA = CM = 4, BD = 6.
Length of perpendicular drawn from centre C on BD.
Then, BM = MD = 3
In right angled ∆ CMB,
CB2 = CM2 + BM2
= 42 + 32
= 16 + 9 =25
⇒ CB = 5

∴ CA = Radius of circle = CB = 5
∴ Centre of circle (- 4, 5) and radius = 5
Hence, equation of circle :
(x + 4)2 + (y – 5)2 = 52
⇒ x2 + 8x + 16 + y2 – 10y + 25 = 25
⇒ x2 + y2 + 8x – 10y + 16 = 0

Question 2.
Find the equation of circle which touches Y – axis at a distance of 4 units and makes intercept of 6 units on Y – axis?
Solution:
Given : OP = 4, AB = 6, PC = AC = radius.
CM ⊥ AB ∴ AM = BM = $$\frac { 6 }{ 2 }$$ = 3
OP = CM = 4
In right angled ∆ AMC,

AC2 = AM2 + CM2
= (3)2 + (4)2 = 9 + 16 = 25
∴ AC = 5
From figure PC – OM= 5 = radius
Centre of circle is (5, 4) and radius = 5.
Hence, required equation of circle :
(x – 5)2 + (y – 4)2 = (5)2
x2 – 10x + 25 + y2 – 8y + 16 = 25
x2 + y2 – 10x – 8y + 16 = 0.

Question 3.
ABCD is a square. Supposing AB and AD as the coordinate axes. Find the equation of the circle circumscribing the square if each side of square is of length l.
Solution:
Taking AB and AD as X – axis and Y – axis respectively
Given : AB = BC = CD = DA = 1
M is mid point of AB.
N is mid point of AD.
AM = $$\frac { l }{ 2 }$$, AN = $$\frac { l }{ 2 }$$ = OM
In ∆OAM,
OA2 = AM2 + OM2
= $$\frac { l }{ 2 }$$2 + $$\frac { l }{ 2 }$$2
= $$\frac{l^{2}}{4}+\frac{l^{2}}{4} = \frac{l^{2}}{2}$$
∴ Radius = OA = $$\frac{l}{\sqrt{2}}$$
Centre of circle (AM, OM) = ( $$\frac { l }{ 2 }$$, $$\frac { l }{ 2 }$$ )

Required equation of circle is :
(x – $$\frac { l }{ 2 }$$ )2 + (y – $$\frac { l }{ 2 }$$ )2 = $$\frac{l^{2}}{2}$$
Centre of circle (AM, OM) = ($$\frac { l }{ 2 }$$, $$\frac { l }{ 2 }$$)
Required equation of circle is :
(x – $$\frac { l }{ 2 }$$)2 + (y – $$\frac { l }{ 2 }$$)2 = $$\frac{l^{2}}{2}$$
⇒ x2 – lx + $$\frac{l^{2}}{4}$$ + y2 – ly + $$\frac{l^{2}}{4}$$ = $$\frac{l^{2}}{2}$$
⇒ x2 + y2 – l(x + y) = 0
⇒ x2 + y2 = l(x + y)

Question 4.
Find the equation of the circle passing through the points (4, 1) and (6, 5). Whose centre lies on line 4x + y = 16. (NCERT)
Solution:
Let the equation of circle be
x2 + y2 + 2gx + 2 fy + c = 0 …. (1)
It passes through points (4, 1) and (6, 5).
∴ 8g + 2f + c + 17 = 0 …. (2)
and 12g + 10f + c + 61 = 0 …. (3)
Centre of circle (1) is (- g, – f) which lies on line 4x + y = 16.
∴ – 4g – f – 16 = 0
⇒ 4g + f + 16 – 0 …. (4)
Subtracting equation (2) from equation (3), we get
4g + 8f + 44 = 0
⇒ g + 2f + 11 = 0 …. (5)
On solving equation (4) and (5), g = – 3, f = – 4
Put g = – 3 and f = – 4 in equation (2),
– 24 – 8 + C + 17 = 0
⇒ c = 15
Put values of g, f and c in equation (1), then required equation of circle is :
x2 + y2 – 6x – 8y + 15 = 0.

Question 5.
Find the equation of the circle which passes through the points (2, 3) and (- 1, 1) whose centre lies on line x – 3y – 11 = 0. (NCERT)
Solution:
Let the equation of circle is :
x2 + y2 + 2gx + 2fy + c = 0 …. (1)
∵Points (2, 3) and (- 1, 1) lies on equation (1),
∴ (2)2 + (3)2 + 2g(2) + 2f(3) + c = 0
⇒ 4 + 9 + 4g + 6f + c + 13 = 0
4g + 6f + c + 13 = 0 …. (2)
and (-1)2 + (l)2 – 2g + 2f + c = 0
⇒ 1 + 1 – 2g + 2f + c = 0
⇒ – 2g + 2f + c + 2 = 0 …. (3)

Putting the value of g, f and c in equation (1), then required equation of circle will be :
x2 + y2 + 2($$\frac { -7 }{ 2 }$$)x + 2($$\frac { 5 }{ 2 }$$)y + c = 0 …. (1)
x2 + y2 – 7x + 5y -14 = 0.

Question 6.
Find the equation of circle whose radius is 5, centre is on Y – axis and which passes through point (2, 3).
Solution:
Centre of circle is on X – axis, so k = 0.
Let the equation of circle be :
(x – h)2 + (y – k)2 = a2
Here a = 5
(x – h)2 + (y – 0)2 = (5)2
(x – h)2 + y2 = 25
Circle (1) passes through point (2, 3),
∴ (2 – h)2 + (3)2 = 25
⇒ (2 – h)2 = 25 – 9 = 16 = (4)2
⇒ 2 – h = ± 4

Question 7.
y = mx is a chord of the circle whose radius is ‘a’ and its diameter is X – axis. Origin is one of the limiting points of the chord. Show that the equation to a circle whose diameter is the given chord is given by the equation (1 + m2) (x2 + y2 ) – 2a (x + my) = 0. Solution:
Equation of circle whose radius is a and centre is (a, 0) will be
(x – a)2 + y2 = a2
⇒ x2 – 2ax + y2 + a2 = a2
⇒ x2 – 2ax + y2 = 0 …. (1)
Equation of given line is :
y = mx …. (2)
Now, equation of circle passing through the intersection of eqns. (1) and (2) will be :
x2 + y2 – 2ax + λ(y – mx) = 0 … (3)
Centre of co – ordinate of circle (3) are ($$\frac { λm + 2a }{ 2 }$$, $$\frac { λ }{ 2 }$$)
∵ Centre lies on line y = mx.
∴ – $$\frac { λ }{ 2 }$$ = m$$\frac { λm + 2a }{ 2 }$$
⇒ λ = $$\frac{-2 a m}{1+m^{2}}$$

Put the value of λ in Equation (3),
x2 + y2 – 2ax + $$\frac{-2 a m}{1+m^{2}}$$ (y – mx) = 0
⇒ (l + m2)(x2 + y2) = 2ax + 2am2x + 2amy – 2am2x
⇒ (l + m2)(x2 + y2) = 2a(x + my)
⇒ (l + m2)(x2 + y2) – 2a(x + my) = 0
Which is required equation of circle.

Question 8.
If the straight line x cos α + y sin α = p cuts a circle x2 + y2 = a2 in two points M and N, then show that the equation of the circle whose diameter is MN will be x2 + y2 – a2 = 2p(x cos α + y sin α – p).
Solution:
Given : Equation of line is :
x cos α + y sin α = p …. (1)
and Equation of circle is
x2 + y2 = a2 …. (2)
Now, equation of circle passing through the intersection of line (1) and circle (2) at points M and N is :
x2 + y2 – a2 + λ(x cos α + y sin α – p) = 0 …. (3)
If MN is diameter of above circle then centre is :
(- $$\frac { λ }{ 2 }$$cos α, – $$\frac { λ }{ 2 }$$sin α)
Which is lies on line x cos α + y sin α = p.
– ( $$\frac { λ }{ 2 }$$cos α )cos α + (- $$\frac { λ }{ 2 }$$sin α)sin α = p
⇒ – $$\frac { λ }{ 2 }$$[cos2 α + sin2 α] = p
⇒ λ = – 2p
Put the value of λ in equation (3), then required equation of circle is
x2 + y2 – a2 – 2p(x cos α + y sin α – p) = 0
⇒ x2 + y2 – a2 = 2p(x cos α + y sin α – p)

Question 9.
Find the following equation of parabola : (i) co – ordinates of focus, (ii) axis, (iii) equation of directrix, (iv) length of Iatus rectum. (NCERT)
(A) y2 = 12x
Solution:
Equation of parabola : y2 = 12x
Comparing with y2 = 4 ax,
4a = 12 ⇒ a = 3
∴Co – ordinates of focus (a, 0) = (3, 0).
Axis of parabola = X – axis.
Equation of directrix is x = – a ⇒ x = – 3.
Length of latus rectum = 4a = 4 x 3 = 12.

(B) x2 = 6y.
Solution:
Equation of parabola: x2 = 6y
Comparing with x2 = 4ay
4a = 6 a ⇒ 3/2
∴ Co – ordinate of focus (0, a) = (0, 3/2).
Axis of parabola = Y – axis.
Equation of directrix is y = – a ⇒ y = – 3/2.

(C) y2 = – 8x
Solution:
Equation of parabola : y2 = – 8x
Comparing with y2 = – 4ax
– 4a – = – 8 ⇒ a = 2
∴ Co – ordinates of focus (- a, 0) = (- 2, 0).
Axis of parabola = X – axis.
Equation of directrix is x = a ⇒ x = 2.
Length of latus rectum 4a = 4 x 2 = 8.

(D) x2 = – 16y
Solution:
Equation of parabola : x2 = – 16y
Comparing with x2 = – 4ay
– 4a = – 16 ⇒ a = 4
∴ Co – ordinate of focus (0, – a) = (0, – 4)
Axis of parabola = Y – axis
Equation of directrix is y = a ⇒ y = 4
Length of latus rectum = 4a = 16.

Question 10.
An equilateral triangle inscribed in the parabola y2 = 4ax, where one vertex is at the vertex of parabola. Find the length of the side of triangle. (NCERT)
Solution:
Let the equation of parabola is y2 = 4ax.
Let APQ be the equilateral triangle whose vertex A(0, 0), P(h, k) and Q(h, – k).
AP2 = (h – 0)2 + (k – 0)2
= h2 + k2
⇒ AP = $$\sqrt{h^{2}+k^{2}}$$
Similarly, AQ = $$\sqrt{h^{2}+k^{2}}$$

Again, PQ = $$\sqrt{(h-k)^{2}+(k+h)^{2}}$$
= $$\sqrt{(2k)^{2}}$$ = 2k
∴ AP = PQ
⇒ $$\sqrt{h^{2}+k^{2}}$$ = 2k
⇒ h2 + k2 = 4k2
⇒ h2 = 3k2
⇒ h = $$\sqrt {3}$$.k
∵ Point P(h, k) lies on parabola y2 = 4ax.
k2 = 4ah = 4a.$$\sqrt {3}$$k
⇒ k = 4a$$\sqrt {3}$$, [∵ k ≠ 0]
Hence, length of side PQ = 2k = 2.(4a$$\sqrt {3}$$) = 8a$$\sqrt {3}$$.

Question 11.
If a parabola reflector is 20 cm in diameter and 5 cm deep. Find the focus. (NCERT)
Solution:
Taking vertex of parabola reflector at origin and X – axis along the axis of parabola.

Equation of parabola y2 = 4ax …. (1)
Given : OS = 5 cm, AB = 20 cm, AS = 10 cm
∴ Co – ordinate of A will be (5, 10).
∴ (10)2 = 4a x 5
⇒ 100 = 20a
⇒ a = 5
∴ OS = 5 cm
Co – ordinates of focus S will be (5, 0).

Question 12.
An arch is in the form of a parabola with its vertical axis. The arch is 10 m high and 5 m wide at the base. How wide it is 2 m vertex of the parabola. (NCERT)
Solution:
Let the equation of parabola is :
x2 = 4 ay …. (1)
Given : AB = 5 metre
AF = BF = $$\frac { 5 }{ 2 }$$ metre
OE = 2 metre
OF = 10 metre
Co – ordinate of A will be ($$\frac { 5 }{ 2 }$$, 10)
This point lies on parabola, hence it will be satisfy equation (1),
∴ $$\frac { 5 }{ 2 }$$2 = 4a x 10
⇒ $$\frac { 25 }{ 4 }$$ = 4 x a x 10
⇒ a = $$\frac { 25 }{ 4 × 4 × 10 }$$
⇒ a = $$\frac { 5 }{ 32 }$$

Put the value of a in Equation (1),
∴ x2 = 4 x $$\frac { 5 }{ 32 }$$y
⇒ x2 = $$\frac { 5 }{ 8 }$$y
Let EC = k
OE = 2
Co – ordinate of C will be (k, 2) and it will satisfy equation of parabola.
We get k2 =$$\frac { 5 }{ 8 }$$ x 2
⇒ k2 = $$\frac { 5 }{ 4 }$$
⇒ k = $$\frac{\sqrt{5}}{2}$$
DE = 2EC
2 x $$\frac{\sqrt{5}}{2}$$ = $$\sqrt {5}$$
= 2.23 metre (approx.)

Question 13.
In each of the following ellipse. Find the co – ordinates of the foci and vertices, the length of major axis and minor axis, the eccentricity and the length of latus rectum of the ellipse. (NCERT)
(A) = $$\frac{x^{2}}{36}+\frac{y^{2}}{16}$$ = 1.
Solution:
Comparing with standard form of ellipse, $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}$$ = 1
We get, a2 = 36 ⇒ a = 6, b2 = 16 ⇒ b = 4
Here a > b.
∴ b2 = a2(1 – e2)
⇒ 16 = 36(1 – e2)
⇒ 1 – e2 = $$\frac { 16 }{ 36 }$$ = $$\frac { 4 }{ 9 }$$
⇒ e2 = 1 – $$\frac { 4}{9 }$$ = $$\frac { 5 }{ 9 }$$
∴ Eccentricity e = $$\frac{\sqrt{5}}{3}$$
Foci (± ae, o) = (± 6 × $$\frac{\sqrt{5}}{3}$$, o)
= (± 2$$\sqrt {5}$$, 0)
Vertices (± a, 0) = (± 6, 0)
Length of major axis = 2a = 2 x 6 = 12.
Length of minor axis = 2b = 2 x 4 = 8.
Length of latus rectum = $$\frac{2 b^{2}}{a}$$ = $$\frac { 2 × 16 }{ 6 }$$ = $$\frac { 16 }{ 3 }$$.

(B) $$\frac{x^{2}}{4}+\frac{y^{2}}{25}$$ = 1.
Solution:
Comparing with standard form of ellipse, $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}$$ = 1
We get, a2 = 36 ⇒ a = 6, b2 = 16 ⇒ b = 4
Here a < b.
∴ a2 = b2(1 – e2)
⇒ 4 = 25(1 – e2)
⇒ 1 – e2 = $$\frac { 4 }{ 25 }$$
⇒ e2 = 1 – $$\frac { 4}{25 }$$ = $$\frac { 21 }{ 25 }$$
∴ Eccentricity e = $$\frac{\sqrt{21}}{5}$$
Foci (0, ± b) = (0, ± 5 × $$\frac{\sqrt{21}}{5}$$)
= (0, ± $$\sqrt {21}$$)
Vertices (0, ± b) = (0, ± 5)
Length of major axis = 2b = 2 x 5 = 12.
Length of minor axis = 2a = 2 x 2 = 8.
Length of latus rectum = $$\frac{2 a^{2}}{b}$$ = $$\frac{2 \times 2^{2}}{5}$$ = $$\frac { 2 × 4 }{ 5 }$$ = $$\frac { 8 }{ 5 }$$.

Question 14.
Find the equation of hyperbola whose foci is (± 4, 0) and length of latus rectum is 12. (NCERT)
Solution:
Foci of hyperbola (± 4, 0).
Hence equation of hyperbola will be :
$$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}$$ = 1
Foci (±ae, 0) = (± 4, 0)
∴ ae = 4
Length of latus rectum = $$\frac{2 b^{2}}{a}$$ = 12
⇒ b2 = 6a
We know, b2 = a2(e2 – 1)
⇒ 6a = a2e2 – a2
⇒ 6a = 42 – a2
⇒ 6a = 16 – a2
⇒ 6a = 16 – a2
⇒ a2 + 6a – 16 = 0
⇒ a2 + 8a – 2a – 16 = 0
⇒ a(a – 2)(a + 8) = 0
⇒ a = 2, a = – 8, (∵ a cannot be negative)
∴ a = 2
b2 = 6a = 6 x 2 = 12
⇒ b = $$\sqrt {12}$$
Putting values of a and b in equation (1), then required equation of hyperbola will be :
$$\frac{x^{2}}{2^{2}}-\frac{y^{2}}{(\sqrt{12})^{2}}$$ = 1
⇒ $$\frac{x^{2}}{4}+\frac{y^{2}}{12}$$ = 1
⇒ 3x2 – y2 = 12.

Question 15.
Find the axis, foci, directrix, eccentricity and the latus rectum of the ellipse 9x2 + 4y2 = 36.
Solution:
Given equation of ellipse
9x2 + 4y2 = 36
⇒ $$\frac{x^{2}}{4}+\frac{y^{2}}{9}$$ = 1
Here, b2 > a2 or b > a
∴ Major axis = 2.3 = 6
Minor axis = 2.2 = 4
Now,
a2 = b2(1 – e2)
(2)2 = (3)2(1 – e2)
⇒ 4 = 9(1 – e2)
⇒ $$\frac { 4 }{ 9 }$$ = 1 – e2
⇒ e2 = 1 – $$\frac { 4 }{ 9 }$$ = $$\frac { 9 – 4 }{ 9 }$$ = $$\frac { 5 }{ 9 }$$
∴ e = $$\frac{\sqrt{5}}{3}$$
Co – ordinate of foci = (0, ± be )
= (0, ± 3. $$\frac{\sqrt{5}}{3}$$)
= (0, ± $$\sqrt {5}$$).
Co – ordinate of vertex = (0, ± b )
= (0 ± 3 ).
Length of latus rectum = $$\frac{2 a^{2}}{b}$$
= $$\frac { 2.4 }{ 3 }$$ = $$\frac { 8 }{ 3 }$$.

Question 16.
(A) Find the equation of ellipse whose vertices are (± 5, 0) and foci (± 4, 0).
Solution:
Given : Vertices are (± 5, 0) and foci are (± 4, 0)
∴ a = 5
and ae = 4
⇒ 5e = 4
⇒ e = $$\frac { 4 }{ 5 }$$
Let the equation of ellipse is :
$$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}$$ = 1, where a > b …. (1)
∴ From b2 = a2(1 – e2),
b2 = 52(1 – ($$\frac { 4 }{ 5 }$$)2)
b2 = 25(1 – $$\frac { 16 }{ 25 }$$)
b2 = 25 x $$\frac { 9 }{ 25 }$$ = 9
Putting values of a and b in equation (1), the required equation of ellipse will be :
$$\frac{x^{2}}{25}+\frac{y^{2}}{9}$$ = 1
⇒ 9x2 + 25y2 = 225

(B) Find the equation of ellipse whose vertices are (0, ± 13) and foci is (0, ± 5).
Solution:
Let the equation of ellipse is :
$$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}$$ = 1, where a < b …. (1)
Vertices of ellipse = (0, ± b) = (0, ± be)
∴ b = 13
Foci = (0, ± 5) = (0, ± be)
∴ be = 15
⇒ 13 x 3 = 5
⇒ e = $$\frac { 5 }{ 13 }$$
Now, a2 = b2(1 – e2)
⇒ a2 = 132[1 – ( $$\frac { 5 }{ 13 }$$)2 ]
⇒ a2 = 169[1 – $$\frac { 25 }{ 169 }$$ ]
⇒ a2 = 169$$\frac { 169 – 25 }{ 169 }$$
⇒ a2 = 144
⇒ a = 12.
Putting values of a and b in equation (1), then required equation of ellipse will be :
$$\frac{x^{2}}{(12)^{2}}-\frac{y^{2}}{(13)^{2}}$$ = 1
⇒ $$\frac{x^{2}}{144}+\frac{y^{2}}{169}$$ = 1

Question 17.
Find the equation of ellipse whose centre is at (0, 0), major axis on the Y – axis passing through the points (3, 2) and (1, 6).
Solution:
Let the equation of ellipse is :
$$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}$$ = 1, where a < b …. (1)
∵ Equation (1) passes through points(3, 2) and (1, 6)
∴ $$\frac{9}{a^{2}}+\frac{4}{b^{2}}$$ = 1 …. (2)
and $$\frac{1}{a^{2}}+\frac{36}{b^{2}}$$ = 1 …. (3)
Multiply equation (2) by 9, we get

Putting value of a2 in equation (2), we get
$$\frac { 9 }{ 10 }$$ + $$\frac{4}{b^{2}}$$ = 1
⇒ $$\frac{4}{b^{2}}$$ 1 – $$\frac { 9 }{ 10 }$$
⇒ $$\frac{4}{b^{2}}$$ = $$\frac { 1}{ 10 }$$
⇒ b2 = 40
Putting values of a2 and b2 in equation (1), then required equation of ellipse will be :
$$\frac{x^{2}}{10}+\frac{y^{2}}{40}$$ = 1

Question 18.
Find the equation of ellipse whose major axis on the X – axis which passes through the points (4, 3) and (6, 2).
Solution:
Let the equation of ellipse is :
$$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}$$ = 1 …. (1)
∵ It passes through points(4, 3) and (6, 2)
∴$$\frac{36}{a^{2}}+\frac{4}{b^{2}}$$ = 1 …. (2)
and $$\frac{16}{a^{2}}+\frac{9}{b^{2}}$$ = 1 …. (3)
Subtracting equation (3) from equation (2), we get
$$\frac{20}{a^{2}}-\frac{5}{b^{2}}$$ = 1
⇒ $$\frac{4}{a^{2}}-\frac{1}{b^{2}}$$ = 1
⇒ $$\frac{4}{a^{2}} = \frac{1}{b^{2}}$$
⇒ a2 = 4b2
Putting value of a2 in equation (2), we get
$$\frac{36}{4b^{2}}+\frac{4}{b^{2}}$$ = 1
⇒ $$\frac{9}{b^{2}}+\frac{4}{b^{2}}$$ = 1
⇒ 9 + 4 = b2
⇒ b2 = 13
Putting values of a2 and b2 in equation (1), hence
Required equation of ellipse $$\frac{x^{2}}{52}+\frac{y^{2}}{13}$$ = 1

Question 19.
An arch is the form of a semi ellipse. It is 8 m wide and 2 m high of the centre. Find the height of the arch at a point 1-5 m from one end. (NCERT)
Solution:
Let the equation of ellipse is :
$$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}$$ = 1 …. (1)
Here 2a = 8 ⇒ a = 4, b = 2
Putting the values of a and b, we get

$$\frac{x^{2}}{4^{2}}-\frac{y^{2}}{2^{2}}$$ = 1 …. (1)
$$\frac{x^{2}}{16}+\frac{y^{2}}{4}$$ = 1
Given :
AP = 1.5m, OA = $$\frac { 8 }{ 2 }$$ = 4m,
OP = OA – AP = 4 – 1.5 = 2.5m.
Let PQ = k
∴ Co – ordinate of Q will be (2.5, k), which will satisfy ellipse’s equation.
Hence,
$$\frac{(2.5)^{2}}{16}+\frac{k^{2}}{4}$$ = 1
⇒ $$\frac { 6.25}{ 16 }$$ + $$\frac{k^{2}}{4}$$ = 1
⇒ $$\frac{k^{2}}{4}$$ = 1 – $$\frac { 6.25 }{ 16 }$$
⇒ $$\frac{k^{2}}{4}$$ = $$\frac { 16 – 6.25 }{ 16 }$$
⇒ k2 = $$\frac { 9.75 }{ 4 }$$
⇒ k2 = 2.437
⇒ k = 1.56metre (approx).

Question 20.
A rod of length 12 cm moves with its ends always touching the co – ordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the X – axis. (NCERT)
Solution:
Let AB be the rod of length 12 cm which make an angle θ with X – axis.
∴ ∠BAO = θ
AB = 12 cm
AP = 3 cm, then PB = 9 cm

In ∆PNA,
sin θ = $$\frac {PN }{ PA }$$ = $$\frac { y }{ 3 }$$
In ∆PMB,
cos θ = $$\frac { PM }{ PB }$$ = $$\frac { x }{ 9 }$$
sin2θ + cos2θ = $$\frac { y }{ 3 }$$2 + $$\frac { x }{ 9 }$$2
⇒ 1 = $$\frac{y^{2}}{9}+\frac{x^{2}}{81}$$
Hence required equation is :
$$\frac{x^{2}}{81}+\frac{y^{2}}{9}$$ = 1

Question 21.
Find the eccentricity, co – ordinate of foci, equation of directrix and length of Iatus rectum of ellipse 4x2 + y2 – 8x + 2y + 1 = 0.
Solution:
4x2 + y2 – 8x + 2y + 1 = 0
⇒ 4x2 – 8x + y2 + 2y +1 = 0
⇒ 4x2 – 8x + (y + 1)2 = 0
⇒ 4(x2 – 2x) + (y + 1)2 = 0
⇒ 4(x2 – 2x + 1) + (y + 1)2 = 4
⇒ 4(x2 – 2x + 1) + (y + 1)2 = 4
⇒ $$\frac{(x-1)^{2}}{1}+\frac{(y+1)^{2}}{4}$$ or $$\frac{X^{2}}{1}+\frac{Y^{2}}{4}$$ = 1
Here, b > a
a2 = b2(1 – e2)
⇒ 1 = (1 – e2)
⇒ $$\frac { 1 }{ 4}$$ = 1 – e2
⇒ e2 = 1 – $$\frac { 1 }{ 4}$$ = $$\frac { 3 }{ 4}$$
⇒ Eccentricity e = $$\frac{\sqrt{3}}{2}$$
Co – ordinate of foci (0, ± be)
= (0, ±2. $$\frac{\sqrt{3}}{2}$$
= (0, ± $$\sqrt {3}$$ )
Here X = 0, Y = ± $$\sqrt {3}$$
∴ x – 1 = 0, y + 1 = ± $$\sqrt {3}$$
⇒ x = 1, y = – 1 ± $$\sqrt {3}$$
foci = (1 ± $$\sqrt {3}$$ – 1)
Equation of directrix Y = ± $$\frac { b }{ e}$$
⇒ Y = ± $$\frac{2}{\sqrt{3}}$$.2
⇒ Y = ± $$\frac{4}{\sqrt{3}}$$
Y + 1 = $$\frac{4}{\sqrt{3}}$$, (∵ Y = y+1)
⇒ y = ± $$\frac{4}{\sqrt{3}}$$ – 1
Length of latus rectum = $$\frac{2 a^{2}}{b}$$
= 2. $$\frac { 1 }{ 2 }$$ = 1.

Question 22.
Find the vertices, co-ordinate of foci, eccentricity and length of latus rectum of hyperbola :
(A) 9y2 – 4x2 = 36.
Solution:
Given : 9y2 – 4x2 = 36
⇒ $$\frac{9 y^{2}}{36}-\frac{4 x^{2}}{36}$$ = 1
⇒ $$\frac{y^{2}}{4}-\frac{x^{2}}{9}$$ = 1
Comparing the above equation with the standard form of hyperbola
$$\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}$$ = 1 …. (1)
b2 = 4 ⇒ b = 2, a2 = 9 ⇒ a = 3
Let e is the eccentricity of hyperbola.
Then, a2 = b2(e2 – 1)
⇒ 9 = (e2 – 1)
⇒ $$\frac { 9 }{ 4 }$$ = e2 – 1
⇒ e2 = $$\frac { 9 }{ 4 }$$ + 1 = $$\frac { 13 }{ 4 }$$
∴ Eccentricity e = $$\frac{\sqrt{3}}{2}$$
Vertices = (0, ± b) = (0, ± 2)
Foci = (0, ± be) = (0, ± 2 x $$\frac{\sqrt{3}}{2}$$) = (0, ±$$\sqrt {3}$$)
Length of latus rectum = $$\frac{2 a^{2}}{b}$$ = $$\frac { 2 x 9 }{ 2 }$$ = 9

(B) 16x2 – 9y2 = 576.
Solution:
Given : 16x2 – 9y2 = 576
⇒ $$\frac{16 x^{2}}{576}-\frac{9 y^{2}}{576}$$ = 1
⇒ $$\frac{x^{2}}{36}-\frac{y^{2}}{64}$$ = 1
Comparing the above equation with the standard form of hyperbola
$$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}$$ = 1 …. (1)
Here, a2 = 36 ⇒ a = 6, b2 = 64 ⇒ b = 3
We Know that, b2 = a2(e2 – 1)
64 = 36(e2 – 1)
⇒ $$\frac { 64 }{ 36 }$$ = e2 – 1
⇒ $$\frac { 16 }{ 9 }$$ = e2 – 1
⇒ e2 = $$\frac { 16 }{ 9 }$$ + 1 = $$\frac { 25 }{ 9 }$$
∴ Eccentricity e = $$\frac{5}{3}$$
Vertices = (± a, 0) = (± 6, 0)
Foci = (± ae, 0) = (± 6 x $$\frac { 5 }{ 3 }$$) = (± 10, 0)
Length of latus rectum = $$\frac{2 b^{2}}{a}$$ = $$\frac { 2 x 64 }{ 6 }$$ = $$\frac { 64 }{ 3 }$$.

(c) 5y2 – 9x2 = 36.
Solution:
Given : 5y2 – 9x2 = 36
⇒ $$\frac{5 y^{2}}{36}-\frac{9 x^{2}}{36}$$ = 1
$$\frac{y^{2}}{\frac{36}{5}}-\frac{x^{2}}{4}$$ = 1
Comparing the above equation with the standard form of hyperbola
$$\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}$$ = 1 …. (1)
Here b2 = $$\frac { 36 }{ 5 }$$ ⇒ b = $$\frac{\sqrt{6}}{5}$$, a2 = 4 ⇒ a = 2
We know that, a2 = b2(e2 – 1)
⇒ 4 = $$\frac{\sqrt{36}}{5}$$(e2 – 1)
⇒ e2 – 1 = $$\frac { 20 }{ 36 }$$ = $$\frac { 5 }{ 9 }$$
⇒ e2 = 1 + $$\frac { 5 }{ 9 }$$ = $$\frac { 14 }{ 9 }$$

Question 23.
Find the equation of hyperbola whose foci are (0, ± $$\sqrt {10}$$) and which passes through point (2,3).
Solution:
Foci of hyperbola are (0, ±$$\sqrt {10}$$).
∴Form of hyperbola is :
$$\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}$$ = 1 …. (1)
Foci (0, ± be) = (0, ± $$\sqrt {10}$$)
be = $$\sqrt {10}$$
Equation (1) passes through point (2, 3).
∴ $$\frac{9}{b^{2}}-\frac{4}{a^{2}}$$
⇒ 9a2 – 4b2 = a2b2
We know that, a2 = b2 (e2 – 1)
⇒ a2 = b2e2 – b2
a2 = ( $$\sqrt {10}$$)2 – b2
⇒ a2 = 10 – b2
⇒ b2 = 10 – a2
Putting value of b2 in equation (2),
9a2 – 4(10 – a2) = a2 (10 – a2)
⇒ 9a2 – 40 + 4a2 = 10a2 – a4
⇒ 13a2 – 40 = 10a2 – a4
⇒ a4 + 13a2 – 10a2 – 40 = 0
⇒ a4 + 3a2 – 40 = 0
⇒ a4 + 8a2 – 5a2 – 40 = 0
⇒ a2(a2 + 8) – 5(a2 + 8) = 0
⇒ (a2 – 5)(a2 + 8) = 0
a2 = 5, a2 = – 8
∵ The value of a cannot be negative.
∴ a2 = 5
b2 = 10 – a2
⇒ b2  = 10 – 5
⇒  b2 = 5
Putting values of a2 and b2 in equation (1), then required equation of hyperbola will be :
$$\frac{y^{2}}{5}-\frac{x^{2}}{5}$$ = 1
⇒ y2 – x2 = 5.

Question 24.
Find the equation of hyperbola in which the distance between foci is 8 and distance between directrix is 6.
Solution:
Let the equation of hyperbola is :
$$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}$$ … (1)
and Eccentricity of hyperbola is e and foci are (ae, 0) and (- ae, 0) and latus rectum
are x = $$\frac { a }{ e }$$ and x = – $$\frac { a }{ e }$$
Distance between foci = 2ae
Distance between latus rectum = $$\frac { 2a }{ e }$$
According to question, 2ae = 8
and $$\frac { 2a }{ e }$$ = 6
Multiplying equation (2) and (3),
4a2 = 48 ⇒ a2 = 12
⇒ a = 2$$\sqrt {3}$$
Putting value of a in equation (2),
2.2$$\sqrt {3}$$e = 8 ⇒ e = $$\frac{2}{\sqrt{3}}$$
b2 = a2(e2 – 1)
= 12( $$\frac { 4 }{ 3 }$$ – 1) = 4
Putting values of a2 and b2 in equation (1), the required equation of hyperbola is :
$$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}$$
⇒ x2 – 3y2 = 12.

Question 25.
Find the centre, eccentricity, foci and length of latus rectum of hyperbola 9x2 – 16y2 + 18x + 32y – 151 = 0.
Solution:
Equation of hyperbola is :
9x2 – 16y2 + 18x + 32y – 151 = 0
⇒ 9x2 + 18x – 16y2 + 32y = 151
⇒ 9(x2 + 2x) – 16(y2 – 2y) = 151
⇒ 9(x + 1)2 – 16(y – 1)2 = 151 – 16 + 9
⇒ 9(x + 1)2 – 16(y – 1)2 = 144
⇒ $$\frac{9(x+1)^{2}}{144}-\frac{16(y-1)^{2}}{144}$$
⇒ $$\frac{(x+1)^{2}}{16}-\frac{(y-1)^{2}}{9}$$ = 1
Let x + 1 = X and y – 1 = Y, then equation of hyperbola is
$$\frac{X^{2}}{16} – \frac{Y^{2}}{9}$$ = 1
∴ a2 = 16 ⇒ a = 4 and b2 = 9 ⇒ b = 3.
∴ Centre is (- 1, 1).
For eccentricity = e
b2 = a2(e2 – 1)
⇒ 9 = 16(e2 – 1)
⇒ $$\frac { 9 }{ 16 }$$ = e2 – 1
⇒ e2 = 1 + $$\frac { 9 }{ 16 }$$ = $$\frac { 25 }{ 16 }$$
⇒ e = $$\frac { 5 }{ 4 }$$
For foci, X = ± ae, Y = 0
⇒ x +1 = ± 4 x $$\frac { 5 }{ 4 }$$, y – 1 = 0
⇒ x + 1 = ± 5, y = 1
⇒ x = 4, – 6, y = 1
∴ Foci are (4, 1) and (6, 1).
Equation of directrix is X = ± $$\frac { a }{ e }$$
⇒ x +1 = ± $$\frac { 4 }{ 5/4 }$$
⇒ x = ± $$\frac { 16 }{ 5 }$$ – 1
⇒ x = ± $$\frac { 11 }{ 5 }$$ and x = – $$\frac { 21 }{ 5 }$$
⇒ 5x = 11 and 5x + 21 = 0.

Question 26.
If e and ex are the eccentricity of hyperbola $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}$$ and $$\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}$$ = 1, then prove that: $$\frac{1}{e^{2}}+\frac{1}{e_{1}^{2}}$$ = 1.
Solution:
Equation of hyperbola is
$$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}$$ = 1
and
$$\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}$$ = 1
For eccentricity e of equation (1),
b2 = a2(e2 – 1)
⇒ $$\frac{b^{2}}{a^{2}}$$ = (e2 – 1)
⇒ e 2 = 1 + $$\frac{b^{2}}{a^{2}}$$ = $$\frac{a^{2}+b^{2}}{b^{2}}$$
⇒ $$\frac{1}{e_{1}^{2}}$$ = $$\frac{a^{2}}{a^{2}+b^{2}}$$
Again for eccentricity e of equation (2),

Question 27.
On a level plain the crack of the rifle and the thud of the ball striking the target are heard at the same instant, prove that the locus of the hearer is a hyperbola.
Solution:
Let P be the situation of hearer and T be the situation of the rifle and S is target. Let the velocity of the ball be v1 and the velocity of sound be v2.

Then,
Time to reach the ball from T to S = $$\frac{T S}{v_{1}}$$
Time to reach the sound from S to P = $$\frac{S P}{v_{2}}$$
and Time to reach the sound from T to P = $$\frac{T P}{v_{2}}$$
∴ The crack of the rifle and the thud of the ball are heard at the same instant.
∴ $$\frac{T S}{v_{1}}$$ + $$\frac{S P}{v_{2}}$$ = $$\frac{T P}{v_{2}}$$
⇒ $$\frac{T P}{v_{2}}$$ – $$\frac{S P}{v_{2}}$$ = $$\frac{T S}{v_{1}}$$
⇒ TP – SP = $$\frac{v_{2}}{v_{1}}$$
⇒ PT – PS = A constant (∵ v2, v2, TS are constant)
Hence locus of point P is hyperbola whose foci is T and S.

## MP Board Class 11th Maths Important Questions Chapter 8 Binomial Theorem

### Binomial Theorem Important Questions

Binomial Theorem Objective Type Questions

(A) Choose the correct option :

Question 1.
The total number of terms in the expansion of
(a) 7
(b) 12
(c) 13
(d) 6.
(c) 13

Question 2.
If y = 3x + 6x2 + 10x3 + …………… ∞, then the correct relation will be :
(a) x = 1 – (1 + y)–$$\frac { 1 }{ 3 }$$
(b) x = (1 + y)–$$\frac { 1 }{ 3 }$$
(c) y = 1 – (1 – x)-3
(d) x = 1 + (1 + y)–$$\frac { 1 }{ 3 }$$
(a) x = 1 – (1 + y)–$$\frac { 1 }{ 3 }$$

Question 3.
The total number of terms in the expansion of (1+ x)-1 will be :
(a) 0
(b) ∞
(c) 2
(d) It can not be expand
(b) ∞

Question 4.
The mid – term in the expansion of (x – $$\frac { 1 }{ x }$$)10 will be :
(a) – 10C5
(b) 10C5
(c) 251
(d) 252
(a) – 10C5

Question 5.
For all positive integer of n, n(n – 1) is :
(a) Integer
(b) Natural number
(c) Even positive integer
(d) Odd positive integer.
(c) Even positive integer

Question 6.
Expansion of (a + x)n is :
(a) an + nC1an-1x + nC2an-2x2 + ……….. + nCran-rxr + ………… + an
(b) xn + nC1xn-1a +nC2xn-2a2 + ……….. + nCrxn-rar + ………… + an
(c) annC1an-1x + nC2an-2x2 + ……….. + (-1)rnCran-rxr + ………… + (-1)nan
(d) xnnC1an-1a + nC2an-2a2 + ……….. + (-1)rnCran-rxr + ………… + (-1)nxn
(a) an + nC1an-1x +nC2an-2x2 + ……….. + nCran-rxr + ………… + an

Question 7.
The fifth term in the expansion of ( x – $$\frac { 1 }{ x }$$ )10 from the end, will be :
(a) $$\frac { ^{ 10 }{ C }_{ 6 } }{ x }$$
(b) $$\frac { 105 }{ 32{ x }^{ 2 } }$$
(c) $$\frac { ^{ 10 }{ C }_{ 6 } }{ { x }^{ 2 } }$$
(d) $$\frac { ^{ 10 }{ C }_{ 6 } }{ { x }^{ 10 } }$$
(a) $$\frac { ^{ 10 }{ C }_{ 6 } }{ x }$$
[Hint: The total number of terms be 11 in the expansion of it.
∵ The 5th term from end = (11 – 5)th = 7th term from the beginning.]

Question 8.
The number of mid – terms in the expansion of ( x – $$\frac { 1 }{ x }$$ )10
(a) 1
(b) 2
(c) – $$\frac { ^{ 13 }{ C }_{ 7 } }{ x }$$
(d) 1716x
(b) 2

Question 9.
The value of nC0 + nC1 + nC2 + …………….. + nCn :
(a) 2n + 1
(b) 2n – 1
(c) 2n – 1
(d) 2n
(d) 2n

Question 10.
The value of nC0 + nC2 + nC4 + …………….. = nC1 + nC3 + ……………. will be :
(a) 2n + 1
(b) 2n – 1
(c) 2n – 1
(d) 2n
(b) 2n – 1

Question 11.
The total number of terms in the expansion of (a + b + c + d)n will be :
(a) $$\frac { (n+1)(n+2) }{ 2 }$$
(b) $$\frac { n(n+1) }{ 2 }$$
(c) $$\frac { (n+1)(n+2)(n+3) }{ 6 }$$
(d) $$\frac { (n+1)(n+2) }{ 6 }$$
(c) $$\frac { (n+1)(n+2)(n+3) }{ 6 }$$

Question 12.
The necessary condition for expansion of (1 + x)-1 is :
(a) | x | < 1
(b) | x | > 1
(c) | x | = 1
(d) | x | = – 1.
(a) | x | < 1

Question 13.
The general term in the expansion of (x + a)n will be :
(a) rth
(b) (r+1)thterm
(c) (r-1)th
(d) (r+2)thterm
(b) (r+1)thterm

Question 14.
In the expansion of ( 2x + $$\frac { 1 }{ { 3x }^{ 2 } }$$ )9, then term independent of x will be :
(a) $$\frac { 8 }{ 127 }$$
(b) $$\frac { 124 }{ 81 }$$
(c) $$\frac { 1792 }{ 9 }$$
(d) $$\frac { 256 }{ 243 }$$
(c) $$\frac { 1792 }{ 9 }$$

Question 15.
The coefficient of x3 in the expansion of ( x – $$\frac { 1 }{ x }$$ )15 is :
(a) 14
(b) 21
(c) 28
(d) 35
(b) 21

Question 16.
The independent term in the expansion of (x2 – $$\frac { 2 }{ { x }^{ 3 } }$$)15 is :
(a) 5th
(b) 6th
(c) 7th
(d) 8th
(c) 7th

Question 17.
The value of nC0 + nC1 + nC2 + …………….. = nCn the expansion of (l + x)n is :
(a) 2n – 1
(b) 2n – 2
(c) 2n
(d) 2n-1
(c) 2n

Question 18.
The value of 15C0 + 15C2 + 15C4 + 15C6 + …………….. = 15C14 is :
(a) 214
(b) 215
(c) 215 – 1
(d) None of these
(a) 214

(B) Match the following :

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

(C) Fill in the blanks :

1. By binomial theorem the value of (102)4 is ……………..
2. The value of second term in the expansion of (1 – x)-3/2 is ……………..
3. The 5th term from the end in the expansion of (x – $$\frac { 1 }{ 2x }$$ )10 is ……………..
4. The constant term will be …………….. in the expansion of ( x2 – 2 + $$\frac { 1 }{ { x }^{ 2 } }$$ )6
5. The value of C1 + 2C2 + 3C3 + ………….. + nCn will be ……………..
6. The value of nC0nC1 + nC2nC3 + ………….. will be ……………..
7. The value of is ……………..
8. The value of is ……………..
9. The value of is ……………..
10. (2x + 3y)5 = …………….. up to three terms.
11. The coefficient of x7 in the expansion of (x2 + $$\frac { 1 }{ x }$$ )11 will be …………….
12. In the expansion of (1 – x)10 the value of middle term is ……………..
13. Third (3rd) term in the expansion of e-3x will be ……………..
14. If n is odd, in the expansion of (x + a)n, then number of middle terms are ……………..
15. The middle term in the expansion of ($$\frac { x }{ a }$$ + $$\frac { a }{ x }$$ )10
16. The coefficient of xn in the expansion of (1 + x) (1 – x)n will be ……………..

1. 08243216
2. $$\frac { { x }^{ 3 } }{ 16 }$$
3. $$\frac { 105 }{ 32{ x }^{ 2 } }$$
4. 924
5. n.2n-1
6. 0
7. $$\frac { n(n+1) }{ 2 }$$
8. $$\frac { n(n+1)(2n+1) }{ 6 }$$
9. [ $$\frac { n(n+1) }{ 2 }$$ ]2
10. 32x5 + 240x4y + 720x3y2
11. 462
12. – 252 x-5
13. $$\frac { 1 }{ 2 }$$
14. Two
15. 252
16. (- 1)n(1 – n)

(D) Write true / false :

1. The expansion of (1 + x)-3 is 1 – 3x + 6x2 – 10x3 + ………….. + $$\frac { (- 1)(r + 1)(r +2) }{ 2! }$$xr + ……………..
2. The expansion of (1 – x)-3 is 1 + 3x + 6x2 + 10x3 + ………….. + $$\frac { (- 1)r(r + 1)(r +2) }{ 2! }$$xr + ……………..
3. The expansion of (1 – x)-2 is 1 + 2x + 3x2 + (r + 1) xr+ ………….. +
4. The (r + 1 )th term in the expansion of (1 – x)-2 is (- 1)r(r + 1) xr+ ………….. +
5. The (r + 1 )th term in the expansion of (1 – x)n will be xr
6. The total number of terms in the expansion of (a + b + c)n is $$\frac { (n + 1)(n + 1) }{ 2 }$$
7. In the expansion of ( 3x – $$\frac { { x }^{ 3 } }{ 9 }$$ )9, No . of terms is 9.
8. The number of term in the expansion of ( 3x – $$\frac { { x }^{ 3 } }{ 9 }$$ )9 is 8.
9. In the expansion of (x + a)n then sum of powers of x and α in any term is n.
10. The coefficient of x in the expansion of (1 – 2x)-3 is 6.
11. The second term in the expansion of (2x + 3y)5 is 240x4y.
12. The value of second term in the expansion of (1 – x)-3/2 is $$\frac { 3 }{ 2 }$$x.

1. True
2. True
3. True
4. True
5. False
6. True
7. False
8. False
9. True
10. False
11. True
12. True.

(E) Write answer in one word / sentence :

1. Find the value of 9993 by the binomial theorem
2. Find the middle term in expansion of (x2 – $$\frac { 1 }{ x }$$)6
3. General term in the expansion of (x + a)n will be.
4. If in the expansion of (1+x)51 the coefficient of xr and xr – 5 are equal, then the value of r will be.
5. Find the coefficient of xn in the expansion of (1 + x + x2 + …………… ∞)2, if | x | < 1.
6. In the expansion of ($$\frac { x }{ 3 }$$ – $$\frac { 2 }{ { x }^{ 2 } }$$)10, x4 comes in rth term, then the value of r will be.
7. The 5th term in the expansion of (1 – 2x)– 1 will be.

1. 997002999
2. – 20 x5
3. nCr xn – r ar
4. 28
5. (n + 1)
6. 3
7. 16x2

Binomial Theorem Long Answer Type Questions

Question 1.
Expand : ($$\frac { 2 }{ x }$$ – $$\frac { x }{ 2 }$$)5 (NCERT)
Solution:

Question 2.
Expand : (2x – 3)6 (NCERT)
Solution:

Question 3.
Expand : ($$\frac { x }{ 3 }$$ + $$\frac { 1 }{ x }$$)5 (NCERT)
Solution:

Question 4.
Expand : (x + $$\frac { 1 }{ x }$$)6. (NCERT)
Solution:

Question 5.
find 13th term in the expansion of ( 9x – $$\frac { 1 }{ 3\sqrt { x } }$$ )18. (NCERT)
Solution:

Question 6.
Find the middle term of (3 – $$\frac { { x }^{ 3 } }{ 6 }$$)7
Solution:

Question 7.
Find the middle term in the expansion of ($$\frac { x }{ 3 }$$ + 9y)10
Solution:
Here n =10
Total number of terms = n + 1 = 10 + 1 = 11 (odd)
Here, the term will be middle term.

Question 8.
If coefficient of x2 and x3 in the expansion of (3 + ax)9 are equal, the value of a.
Solution:

Question 9.
Find the coefficient of x5 in the expansion of (x + 3)8
Solution:
Suppose x5 appears in (r + 1)th term Tr+1 = nC1xn-rar
Here n = 8, x = x, a = 3
Tr+1 = 8Cr(x)8 – r(3)r
For the coefficient of x5,
8 – r = 5
=> r = 3
T3+1 = 8C3(3)3
= $$\frac { 8 × 7 × 6 }{ 3 × 2 × 1}$$ × 3 × 3 × 3 × x5
= 1512 × x5
Hence coefficient of x5 is 1512.

Question 10.
Find the coefficient of a5b7 in the expansion of (a – 2b)12.
Solution:

Question 11.
If the 17th and 18th terms in the expansion of (2 + a)50 are equal, then find the value of a. (NCERT)
Solution:
In the expansion of (x + a)n
Tr+1 = nCr xn – r ar
Here n = 50, x = 2, a = a
T17 = T16 + 1 = 50C16 (2)50 – 16 (a)16
⇒ T17 = 50C16 (2)34 (a)16
and T18 = T17 + 1 = 50C17 (2)50 – 17 (a)17
= 50C17 (2)33 (a)17

Question 12.
Prove that the value of the middle term in the expansion of (1+x)2n is $$\frac { { 1.3.5 …….. (2n – 1)} }{ n! }$$.2n xn
Solution:

Question 13.
In the expansion of (x + 1)n, the coefficient of the (r – 1)th, rth and (r + 1)th terms are in the ratio 1 : 3 : 5, then find the value of n and r.
Solution:
In the expansion of (x + 1)n,
Tr + 1 = nCrxn – r(1)r
Tr – 1 = Tr – 2 + 1 = nCr – 2(x)n – (r – 2)(1)r – 2
Coefficient of Tr – 1th term = nCr – 2
Tr = Tr – 1 + 1 = nCr – 1(x)n – (r – 1)(1)r – 1
Coefficient of Trthterm = nCr – 1
Tr + 1 = nCr xn – r (1)r
Coefficient of Tr+1th term = nCr

Put n = 4r – 5 from equation (1) in equation (2),
3(4r – 5) – 8r = – 3
⇒ 12r – 15 – 8r = – 3
⇒ 4r = 12
∴ r = 3
Put r = 3 in equation (2),
n – 4 x 3 = – 5
⇒ n = 12 – 5
⇒ n = l
n = 7, r = 3

Question 14.
Prove that the coefficient of xn in the expansion of (1 + x)2n of in the expansion of (1 + x)2n – 1.
Solution:
In the expansion of (x + a)n
Tr+1 = nCr xn – r ar
Here x = 1, a = x, n = 2n
Tr+1 = 2nCr(1)2n – r(x)r
For the coefficient of xn, put r = n,
Tn+1 = 2nCn(a)2n – n(x)n
and T18 = T17 + 1 = 50C17 (2)50 – 17 (a)17
= (2nCn) xn
∴ In the expansion of (1 + x)2n, the coefficient of xn = 2nCn …. (1)
and in the expansion of (1 + x)2n – 1, x = 1, a = x, n = 2n – 1
∴ Tr+1 = 2nCr (1)2n – 1  -r (x)n
For the coefficient of xn, put r = n, ‘
We get Tn+1 = 2n – 1Cn xn
The coefficient of xn in the expansion of (1 + x)2n – 1 = n – 1Cn

∴ The coefficient of xn in the expansion of (1 + x)2n
= 2 x The coefficient of xn in the expansion of (1 + x)2n, [from equation (1) and (2)]

Question 15.
Find the constant term in the expansion of ($$\frac { 3 }{ 2 }$$x2 – $$\frac { 1 }{ 3x }$$)6
Solution:

## MP Board Class 11th Maths Important Questions Chapter 7 Permutations and Combinations

### Permutations and Combinations Important Questions

Permutations and Combinations Objective Type Questions

(A) Choose the correct option :

Question 1.
The correct relation between nPr and nCr is :
(a) nCr = $$\frac{[r}{^{n} P_{r}}$$
(b) nCr.nPr= 1
(c) nCr = $$^{n} P_{r} \times\lfloor r$$
(d) $$^{n} C_{r} \times\left\lfloor r=^{n} P_{r}\right.$$
(d) $$^{n} C_{r} \times\left\lfloor r=^{n} P_{r}\right.$$

Question 2.
If nC2:nC4 = 2 : 1, then n = …………..
(a) 3
(b) 4
(c) 5
(d) 6
(c) 5

Question 3.
How many words can be made by the letter of the word ‘COMMITEE’:

Question 4.
If nP4 = 20 x nP3, then n =
(a) 20
(b) 21
(c) 18
(d) 23
(d) 23

Question 5.
The number of different words that can be formed from the letter of the word ‘TRIANGLE’:
(a) 7200
(b) 36000
(c) 144000
(d) 1240.
(b) 36000

Question 6.
In how many ways three person can be seated in 4 chairs :
(a) 42
(b) 20
(c) 24
(d) 12.
(c) 24

Question 7.
If 15Cr = 15Cr – 3 , then the value of rC6 will be :
(a) 84
(b) 83
(c) 82
(d) 80.
(a) 84

Question 8.
5C0 + 5C1 + 5C2 + 5C3 + 5C4 + 5C5 = ……………..
(a) 30
(b) 32
(c) 23
(d) 42
(b) 32

Question 9.
The value of : $$\frac{1}{\lfloor- 4}$$ will be :
(a) 0
(b) ∞
(c) 4
(d) 1.
(a) 0

Question 10.
If 10Pr = 720, then the value of r will be :
(a) 4
(b) 5
(c) 6
(d) 3
(d) 3

(B) Match the following :

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

(C) Fill in the blanks :

1. The number of words can be made by the letters of the word ‘INDIA’ ……………
2. In order to invite 6 friends, the number of ways to send 3 servant is ……………
3. If 20Cr = 20Cy; x ≠ y, then the value of x + y will be ……………
4. The numbers of ways greater than 4000 can be made by the digits 2, 4, 5, 7 is ……………
5. If 12Pr = 1320, then r will be ……………
6. The number of ways can 8 friends be seated in a circular table is ……………
7. If nC8 = nC12, then n will be ……………

1. 60
2. 46
3. 20
4. 48
5. 3
6. 5040
7. 20

(D) Write true / false :

1. If nCr + nCr + 1 = n + 1Cx, then x will be 6.
2. 5 letters be posted in 4 letter boxes in 54 ways.
3. Fruit can be chosen from 5 guavas, 3 apples and 4 mangoes in 60 ways.
4. The number of ways can 8 friends stands in a row is 8.
5. The value of nCr ÷ n – 1Cris $$\frac { n }{ r – 1 }$$

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

(E) Write answer in one word / sentence :

1. How many four digit numbers can be formed with the digits 2, 3, 4, 5 in these number divisible by 2?
2. How many ways can a garland be prepared with 6 flowers?
3. If 2.nC5 = 9.n – 2C5, then the value of n will be
4. Five points out of 15 points are collinear. If these points be joined, then how many triangles can be formed?
5. The value of will be?

1. 12
2. 60
3. 10
4. 30
5. 56C4

Permutations and Combinations Short Answer Type Questions

Question 1.
If $$\frac{1}{\lfloor6}$$ + $$\frac{1}{\lfloor7}$$ = $$\frac{x}{\lfloor8}$$, then find the value of x.
Solution:

Question 2.
Find the value of

(i) n = 6, r = 2 (ii) n = 9, r = 5
Solution:

Question 3.
If n-1P3 : nP4 = 1 : 9, then find the value of n.
Solution:

Question 4.
IF nP5 : 42.nP3, then find the value of n.
Solution:
IF nP5 = 42.nP3, then find the value of n.
Solution:
Given:  nP5 = 42.nP3
⇒ n(n – 1)(n – 2)(n – 3)(n – 4) = 42n(n – 1)(n – 2)
⇒ (n – 3)(n – 4) = 42
⇒ n2 – 7n – 30 = 0
⇒ n2 – 10n + 3n – 30 = 00
⇒ (n – 10) (n + 3) = 0
n = 10, n ≠ – 3.

Question 5.
IF $$\frac{^{n} P_{4}}{^{n-1} P_{4}}$$ = $$\frac { 5 }{ 3 }$$, then find the value of n.
Solution:
Given : $$\frac{^{n} P_{4}}{^{n-1} P_{4}}$$ = $$\frac { 5 }{ 3 }$$
⇒ 3.nP3 = 5.n-1P4
⇒ 3n(n – 1)(n – 2)(n – 3) = 5 (n – 1)(n – 2)(n – 3)(n – 4)
⇒ 3n = 5 (n – 4)
⇒ n = 10.

Question 6.
If nC9 = nC8, then find the value of nC17.
Solution:

Question 7.
If nPn-2 = 60, then find the value of n.
Solution:

Question 8.
If 12Pr = 1320, then find the value of r.
Solution:

Question 9.
If nC14 = nC16, then find the value of 32Cn.
Solution:
Given : nC14 = nC16
nC14 = nC16-1
⇒ 14 = n – 16
⇒ n = 30
32Cn = 32C30 = 32C2
= $$\frac { 32.31 }{ 2 }$$ = 496.

Question 10.
If 20Cx = 20Cy, then find the value of x + y.
Solution:
Given :
20Cx = 20Cy
20Cx = 20C20-y
⇒ x = 20 – y
⇒ x + y = 20

Question 11.
In how many ways can 8 friends sit around a round table?
Solution:
Total number of ways of sitting 8 friends around table = $$\lfloor 8 – 1$$ = $$\lfloor 7$$
= 7 x 6 x 5 x 4 x 3 x 2 x 1
= 5040.

Question 12.
In how many ways can a garland be prepared with 6 flowers?
Solution:
Ways of preparing the garlands with 6 flowers = $$\frac { 1 }{ 2 }$$ $$\lfloor 6 – 1$$
= $$\frac { 1 }{ 2 }$$ $$\lfloor 5$$
= $$\frac { 1 }{ 2 }$$.5.4.3.2.1 = 60.

Question 13.
How many numbers between 100 and 1000 can be formed with the digits 2, 3, 0, 4, 8, 9 if repetition of digits is not allowed in the same number?
Solution:
The numbers between 100 and 1000 are of three digits.
∴Number of three digits made from 6 digits will be 6P3.
In this number which begins with 0 will be 5P2.
∴Required number = 20Cy5P2 = 120 – 20 = 100.

Question 14.
How many words, with or without meaning can be formed using all the letters of the word EQUATION? (NCERT)
Solution:

Question 15.
How many chords can be drawn through 21 points on a circle? (NCERT)
Solution:
By joining two points we get one chord.
∴ No. of chords = 21C2
= $$\frac { 21 x 20 }{ 2 x 1 }$$
= 210.

Question 16.
In how many ways can a team of 3 boys and 3 girls can be selected from 5 boys and 4 girls? (NCERT)
Solution:
Number of ways of selecting 3 boys from 5 boys and 3 girls from 4 girls will be
= 5C3 × 4C3
= $$\frac { 5 × 4 × 3 }{ 3 × 2 × 1}$$ × $$\frac { 4 × 3 × 2 }{ 3 × 2 × 1 }$$
= 10 x 4 = 40.

Permutations and Combinations Long Answers Type Questions

Question 1.
Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour. (NCERT)
Solution:
Since, 3 red balls out of 6 red balls can be selected
= 6C3 = $$\frac { 6 × 5 × 4 }{ 3 × 2 × 1}$$ = 20
3 white balls out of 5 white balls = 5C3 = $$\frac { 5 × 4 × 3 }{ 3 × 2 × 1}$$ = 10
and 3 blue balls out of 5 blue balls = 5C3 = $$\frac { 5 × 4 × 3 }{ 3 × 2 × 1}$$ = 10
Total number of ways of selecting 9 balls = 20 x 10 x 10

Question 2.
Determine the number of 5 cards combinations out of a deck of 52 cards, if there is exactly one ace in each combination (NCERT)
Solution:
Number of ways of selecting one ace from 4 ace = 4C1
Number of selecting 4 cards from 48 cards = 48C4
= $$\frac { 48 × 47 × 46 × 45 }{ 4 × 3 × 2 × 1}$$
Total number of ways = 4C1 x 48C4
= 4 × = $$\frac { 48 × 47 × 46 × 45 }{ 4 × 3 × 2 × 1}$$
= 7, 78, 320.

Question 3.
In how many ways can one select a cricket team of eleven players from 17 players. In which only 5 players can bowl, if each cricket team of 11 must include exactly 4 bowlers? (NCERT)
Solution:
Out of 17 players, to select 11 players in which 5 bowlers and 12 batsman.
Out of 5 bowlers to select 4 bowlers = 5C4 = 5
Out of 12 batsman to select 7 batsman = 12C7
= $$\frac { 12 × 11 × 10 × 9 × 8 × 7 × 6 }{ 7 × 6 × 5 × 4 × 3 × 2 × 1 }$$
= 11 × 9 × 8 = 792
∴ Number of ways selecting 11 players = 5 x 792 = 3960.

Question 4.
A bag contains 5 black and 6 red balls, determine the number of ways in which 2 black and 3 red balls can be selected. (NCERT)
Solution:
There are 5 black and 6 red balls.
Out of 5 black balls, 2 black balls can be =5C2 = $$\frac { 5 × 4 }{ 2 × 1 }$$ = 10
Out of 6 red balls, 3 red balls can be = 6C3 = $$\frac { 6 × 5 × 4 }{ 3 × 2 × 1 }$$ = 20
∴ Total No. of ways = 10 x 20 = 200.

Question 5.
In how many ways can a student choose a program of 5 courses, if 9 courses are available and 2 specific courses are compulsory for each students? (NCERT)
Solution:
Number of available courses = 9.
Number of courses to choose = 5.
Compulsory courses = 2.
∴ Required number of ways = 9-2C5-2 = 7C3
= $$\frac { 7 × 6 × 5 }{ 3 × 2 × 1 }$$ = 35.

Question 6.
How many words, with or without meaning each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER? (NCERT)
Solution:
There are 3 vowels and 5 consonants in DAUGHTER.
2 vowels out of 3 vowels = 3C2 = $$\frac { 3! }{ 1!2! }$$ = $$\frac { 3 × 2 }{ 2 × 1 }$$
3 consonants out of 5 consonats = 5C3 = $$\frac { 5! }{ 2!3! }$$ = $$\frac { 5 × 4 × 3 }{ 3 × 2 × 1 }$$ = 10
Arrangement of 2 vowels and 3 consonants = 5!
= 5 × 4 × 3 × 2 × 1 = 120

Question 7.
How many words, with or without menning can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?
Solution:
Given word is EQUATION.
Vowels – E, U, A, I, O
Consonants – T, Q, N.
Arrangement of 5 vowels = 5! = 5 x 4 x 3 x 2 x 1 = 120
Arrangement of 3 consonants = 3! = 3 x 2 x 1 = 6
Arrangements of vowels and consonants = 2!
∴ Total No. of ways = 120 x 6 x 2! = 720 x 2 = 1440.

Question 8.
A committee of 7 members has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of: (NCERT)

1. Exactly 3 girls
2. At least 3 girls
3. At most 3 girls.

Solution:
Total members in committee = 7
1. Selection of 3 girls and 4 boys in committee
= 4C3 x 9C4
= $$\frac { 4 × 3 × 2 × 1 }{ 3 × 2 × 1 }$$ x $$\frac { 9 × 8 × 7 × 6 }{ 4 × 3 × 2 × 1 }$$
= 9 x 8 x 7 = 504.

2. Selection of minimum 3 girls in committee
= 4C3 x 9C4 + 4C4 x 9C3
= 504 + 1 x $$\frac { 9 × 8 × 7 }{ 3 × 2 × 1 }$$ = 504 + 84 = 588.

3. Selection of maximum 3 girls in a committee

Question 9.
How many different words can be formed by using the letters of the word ALLAHABAD, out of which in how many words vowels will come in even places? (NCERT)
Solution:
Total number of letters in ALLAHABAD = 9.
In which A = 4 times, L = 2 times.

= 9.8.7.3.5 = 7560.
Number of even places are = 4 and odd places = 5. Hence we can put A only in 4 places and rest 5 letters L, L, H, B, D in 5 odd places inways.

Question 10.
In how many ways can a cricket team of eleven can be formed from 14 players? Which win always have :

1. Captain is always taken.
2. When two fast bowlers are always taken?

Solution:
1. When captain is always taken (11 – 1) from (14 – 1) players.
= 13C10 = 13C3
= $$\frac { 13.12.11 }{ 3.2.1 }$$ = 286.

2. When two fast bowlers are always taken = 12C9 = 12C3
= $$\frac { 12.11.10 }{ 3.2.1}$$

Question 11.
In how many ways committee of 5 members can be formed out of 6 men and 4 women ? If the committee has : (i) Only one woman, (ii) At least one woman.
Solution:
(i) Only one woman out of 4 women and 4 men out of 6 men.
No. of ways = 6C4 x 4C1
= 6C2 x 4
= $$\frac { 6 × 5 × 4 }{ 2.1 }$$ = 60

(ii) Atleast one woman :

Hence, total No. of ways = 60 + 120 + 60 + 6 = 246.

Question 12.
If 2nC3 : nC2 = 12 : 1,then find the value of it. (NCERT)
Solution:

Question 13.
If 2nC3 : nC3 = 11 : 1, then find the value of it. (NCERT)
Solution:

Question 14.
Prove that:

Solution:

Question 15.
Prove that:
2nPn = 2n{1.3.5 …………. (2n – 1)}.
Solution:

## MP Board Class 11th Maths Important Questions Chapter 6 Linear Inequalities

### Linear Inequalities Important Questions

Linear Inequalities Objective Type Questions

(A) Choose the correct option :

Question 1.
If x is a real number, then the solution of inequality 3x + 1 < 5x + 7 is :
(a) (- ∞, 3)
(b) (- 3,∞)
(c) (3, ∞)
(d) None of these.
(b) (- 3,∞)

Question 2.
If x is a real number, then the solution of inequality $$\frac { 1 }{ 2 }$$ (3x – 1) ≥ $$\frac { 1 }{ 3 }$$(4x + 3) – 1 is :
(a) (- ∞, 3)
(b) (3, ∞)
(c) (3, ∞)
(d) None of these.
(c) (3, ∞)

Question 3.
The graph of x ≤ 2 and y ≥ 2 is situated in the :
(d) None of these.

Question 4.
In linear programming problems, the objective function will be :
(a) One of the constraints
(b) A linear function which gives optimum solution
(c) Relation between varivable
(d) None of these.
(b) A linear function which gives optimum solution

Question 5.
In linear constraints the maximum value of objective function will be :
(a) At the centre of feasible region
(b) At (0, 0)
(c) At one of the vertices of the feasible region
(d) At the vertex which is situated at maximum distance from (0, 0)
(c) At one of the vertices of the feasible region

Question 6.
Which word is not used in linear programming problems :
(a) Slack variable
(b) Objective function
(c) Concave region
(d) Feasible solution.
(a) Slack variable

Question 7.
The minimum value of p = 6x + 16y when constraints are x ≤ 40 and y ≥ 20 and x, y ≥ 0 is :
(a) 240
(b) 320
(c) 0
(d) None of these.
(b) 320

Question 8.
At which point the value of 3x + 2y is maximum under the constraints x + y ≤ 2, x ≥ 0 y ≥ 0 :
(a) (0, 0)
(b) (1.5, 1.5)
(c) (2, 0)
(d) (0, 2).
(c) (2, 0)

Question 9.
Under constraints x ≥ 0, y ≥ 0, x + y ≤ 4, maximum value of P = 3x + y is :
(a) 8
(b) 12
(c) 6
(d) 10.
(b) 12

Question 10.
Under constraints x – 2y ≥ 6, x + 2y ≥ 0, x ≤ 6 , maximum value of P = x + 3y is :
(a) 16
(b) 17
(c) 18
(d) 19.
(c) 18

Question 11.
Feasible solution of linear programming is in :

Question 12.
If x is a real number and |x| < 4, then :
(a) x ≥ 4
(b) – 4 < x < 4
(c) x ≤ – 4
(d) – 4 ≤ x ≤ 4
(b) – 4 < x < 4

Question 13.
Solution of inequality 3x – 2 < 0 will be :
(a) [3, ∞]
(b) [- ∞, $$\frac { 2 }{ 3 }$$]
(c) [3, 2]
(d) [2, 3]
(b) [- ∞, $$\frac { 2 }{ 3 }$$]

Question 14.
The set of solutions of inequality 4x – 12 ≥ 0 is :
(a) (4, 2)
(b) [4, 12]
(c) [3, ∞]
(d) [3, ∞]
(c) [3, ∞]

Question 15.
The solution of inequality |4x – 3| < 27 is :
(a) (- 6, $$\frac { 15 }{ 2 }$$)
(b) [ – 6, $$\frac { 15 }{ 2 }$$ ]
(c) [ – 6, $$\frac { 15 }{ 2 }$$ )
(d) ( – 6, $$\frac { 15 }{ 2 }$$ ]
(a) (- 6, $$\frac { 15 }{ 2 }$$)

Question 16.
Solution of inequality x > 2 is :

Question 17.
Sawant has a space to store at most 13 boxes of apples and oranges. If he bought x boxes of oranges and y boxes of apples, then correct inequality will be :
(a) x + y < 13
(b) x + y > 13
(c) x + y < 13
(d) x + y = 13.
(c) x + y < 13

Question 18.
In which point the maximum of objective function P = 10x + 6y in the following points:
(a) (0, 0)
(b) (8, 4)
(c) (10, 0)
(d) (0, 0).
(b) (8, 4)

Question 19.
The graph of inequality x +y > 2 is :

Question 20.
The solution of inequality 3x – 7 ≥ x + 1 is :
(a) [4, ∞]
(b) (4, ∞]
(c) (4, ∞)
(d) [4, ∞).
(d) [4, ∞).

Question 21.
The solution of inequality 3x – 6 > 0; 2x – 6 > 0 is :
(a) [3, ∞]
(b) (3, ∞)
(c) [2, ∞)
(d) (2, ∞).
(b) (3, ∞)

Question 22.
The solution of inequality 2x + 6 = 0; 4x – 8 < 0 is :
(a) [- 3, ∞)
(b) (3, 2)
(c) [- 3, 2)
(d)[- 3, 2].
(c) [- 3, 2)

Question 23.
The set of solution of inequality x ≥ 0 is :
(a) All points of X – axis and first, fourth quadrants
(b) All points of first and fourth quadrants
(c) All points of first and second quadrants
(d) All points of Y – axis and first, fourth quadrants.
(d) All points of Y – axis and first, fourth quadrants.

Question 24.
Variables of the objective function of linear Programming problem are :
(a) Negative
(b) Zero or Negative
(c) Zero
(d) Zero or Positive.
(d) Zero or Positive.

(B) Match the following :

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

(C) Fill in the blanks :

1. The maximum or minimum value of objective function is called ……………….
2. The graph of x ≥ 0 is situated in the ………………. quadrant.
3. The graph of y ≤ 0 is situated in the ………………. quadrant.
4. Solution of inequality 3x + 1 < 5x + 7 is (where x is an integer) ……………….
5. Solution of inequality 3x – 4 < 5 is (where x is an integer) ……………….
6. Solution of inequality 4x + 3 > – 13 is (where x is a real number) ……………….
7. Solution of inequality 7x – 2 < 5x + 4 is (where x is a real number) ……………….
8. Solution of inequality 20x < 90 is (where x is a natural number) ……………….
9. Solution of inequality 3x – 2 < x + 4 is ……………….
10. The inequality of one variable is ……………….
11. The set of solutions of y > 0 are in the ………………. and ………………. quadrant.
12. x ≥ 2 and y ≥ 2 will be situated in the graph ……………….
13. The domain of inequality 3x – 15 ≤ 0 is ……………….
14. A function whose maximum and minimum value is to be found subject to the given constrains is known as ……………….

1. Optimum value
2. First and fourth
3. Third and fourth
4. (- 3, ∞)
5. {… – 3, – 2, -1, 0, 1, 2}
6. (- 4, ∞)
7. (- ∞, 3)
8. {1, 2, 3, 4}
9. (- ∞, 3)
10. ax + b ≥ c or ax + b ≤ c; where a ≠ 0
11. First and second
13. (- ∞, $$\frac { 15 }{ 3 }$$)
14. Objective function

(D) Write true / false :

1. If the variable x is such that its value lies between two fixed point a and b, then {x : a < x < b} is called a closed interval.
2. The function whose maximum or minimum value is to be found is called objective function.
3. The set of values of the variable satisfying all constraint is called feasible solution of the problem.
4. The process of doing certain specified steps in a given order is called programming.
5. The set {x : a < x < b} which consists of both a and b is called open interval.
6. The solution of inequality 6x – 30 ≥ 0 will be x ≤ 5
7. The solution of inequality – 2x + 7 < – 13 will be (10, ∞).
8. The linear inequality of one variable is ax + b = 0; a ≠ 0, ∀a, b ∈
9. R. The line inequality of two variables is ax + by + c = 0; a ≠ 0, b ≠ 0, ∀a, b, c ∈ R.
10. The graph of inequality x > 0 is :
11. The graph of inequality y ≤ 0 is :
12. The solution of inequality 5x – 30 ≤ 0 is :

1. False
2. True
3. True
4. True
5. False
6. False
7. True
8. True
9. True
10. True
11. True
12. False.

(E) Write answer in one word / sentence :

1. In which quadrant lies the solution of x ≥ 0 and y ≥ 0?
2. In which quadrant lies the solution of x ≤ 2 and y ≥ 2?
3. Write the solution of 3(2 – x) ≥ 2(1 – x) .
4. Write the solution of $$\frac { x }{ 3 }$$ > $$\frac { x }{ 2 }$$ + 1.
5. Write the solution of $$\frac { x – 4 }{ x + 2 }$$ ≤ 2.

3. (- ∞, 4]
4. (- ∞, – 6)
5. (- ∞, – 8]∪(- 2, ∞)

Linear Inequalities Long Answer Type Questions

Question 1.
Draw the graph of inequality 2x + y ≥ 6 in two – dimensional plane. (NCERT)
Solution:
Given inequality is 2 x + y ≥ 6
Consider this as an equation 2x + y = 6.
For this equation following values of x and y are given in the table :

Plot the graph using the above table. Take a point (0, 0) and put in given inequality,
2x + y ≥ 6
⇒ 2 x 0 + 0 ≥ 6
⇒ 0 ≥ 6 Which is false.
Hence, shaded portion will be opposite of origin.

∴ The shaded region represents the inequality 2x + y ≥ 6.

Question 2.
Draw the graph of inequality 3x + 4y ≤ 12 in two – dimensional plane. (NCERT)
Solution:
Given inequality : 3x + 4y = 12
Consider this as an equation 3x + 4y = 12.
For this equation the following values of x and y are given in the table :

Plot the above points in a graph paper and join them.
Put (0, 0) in given inequality,
⇒ 3x + 4y ≤ 12
⇒ 3 x 0 + 4 x 0 ≤ 12
⇒ 0 ≤ 12
Which is true.
Hence, shaded portion will be towards origin as shown in figure.

∴ The shaded region represents the inequality 3x + 4y ≤ 12

Question 3.
Draw the graph of inequality y + 8 ≥ 2x in two – dimensional plane. (NCERT)
Solution:
Given inequality : y+8 > 2x Consider this as an equation 2x – y = 8
For this equation following values of x and y are given in the table :

Plot the above points in graph paper and join them.
Put (0, 0) in the given inequality,
y + 8 ≥ 2x
⇒ 0 + 8 ≥ 2 x 0
⇒ 8 ≥ 0 Which is true.
Hence, the shaded portion will be towards origin.

∴ Shaded portion represents the inequality y + 8 ≥ 2x.

Draw the graph of following inequality in two – dimensional plane :

Question 4.
2x – 3y > 6.
Solution:
Given inequality is :
2x – 3y > 6
Consider this as an equation 2x – 3y = 6
Following table is prepared for different values of x and y :

Plot the above points on xy – plane and join them.
Put x = 0, y = 0 in given inequality,
2x – 3y > 6
⇒ 2 x 0 – 3 x 0 > 6
⇒ 0 > 6 Which is false.
Hence, the shaded portion will be opposite of origin.

∴The shaded region represents the inequality 2x – 3y > 6

Question 5.
– 3x + 2y ≥ – 6.
Solution:
Given inequality is : – 3x + 2y ≥ – 6.
Consider this as an equation – 3x + 2y = – 6.
Following table is prepared for different values of x and y :

Plotting the above points on xy – plane and join them. Put x = 0, y = 0 in given inequality,
– 3x + 2y ≥ – 6
⇒ – 3(0) + 2(0) ≥ – 6
⇒ 0 ≥ – 6 Which is true.
∴ Shaded portion will be towards origin.

Hence, the shaded portion represents the inequality – 3x + 2y ≥ – 6.

Question 6.
3y – 5x < 30.
Solution:
Given inequality: y – 5x < 30.
Consider this as an equation 3y – 5x = 30.
Following table is prepared for different values of x and y :

Ploting the above points on xy – plane and join them.
Put x = 0, y = 0 in given inequality,
3y – 5x < 30
⇒ 3(0) – 5(0) < 30
⇒ 0 < 30
Which is true.
Hence, the shaded region will be towards origin.

∴ Shaded region represents the inequality 3y – 5x < 30

Question 7.
2x + y ≥ 6, 3x + 4y ≤ 12. (NCERT)
Solution:
Given inequalities are :
2x + y ≥ 6
and 3x + 4y ≤ 12
For the equation 2x+y = 6.
Following table is prepared for different values of x and y :

Plotting the above points on xy – plane to get straight line.
Put x = 0, y = 0 in the inequality,
2x + y ≥ 6
⇒ 2 x 0 + 0 ≥ 6
⇒ 0 ≥ 6
Which is false.
∴ The shaded region will be opposite of origin.
For the equation 3x + 4y = 12.
Following table is prepared for different values of x and y :

Plotting the above points on xy – plane to get straight line.
Put x = 0, y = 0 in given inequality,
3x + 4y ≤ 12
⇒ 3 x (0) + 4(0) ≤ 12
⇒ 0 ≤ 12
Which is true.
Hence, the shaded portion will be towards origin.

∴ The common shaded region is the required solution of the given inequalities.

Question 8.
Solve the following inequalities graphically in two – dimensional plane : x + y ≥ 4, 2x – y >0. (NCERT)
Solution:
Given inequalities are :
x + y ≥ 4
and 2x – y > 0
For the equation x + y = 4, following table is prepared for different values of x and y:

Plotting the above points on xy – plane to get straight line.
Put x = 0, y = 0 in the inequality,
x + y ≥ 4
⇒ 0 + 0 ≥ 4
⇒ 0 ≥ 4
Which is false.
Hence, the shaded region will be opposite of origin.
Preparing table for different value of x and y of the equation 2x – y = 0 :

Plotting the above points in xy – plane to get straight line.
Put x = 1 and y = 0 in given inequality,
2x – y > 0
⇒ 2(1) – 0 > 0
⇒ 2 > 0
Which is true.
Hence, the shaded region will be towards of point (1,0).

The common shaded region is the required solution of the given inequalities.

Question 9.
Solve the following inequalities graphically in two – dimensional plane :
5x + 4y ≤ 20, x ≥ 1, y ≥ 2 (NCERT)
Solution:
The given inequalities are :
5x + 4y ≤ 20
x ≥ 1
y ≥ 2
Prepared table for different values of x and y of the equation 5x +4y = 20 :

Plotting the points of the above table on xy – plane to get straight line.
Then, put x = 0, y = 0 in the inequality,
5x + 4y ≤ 20
⇒ 0 + 0 ≤ 20
⇒ 0 < 20
Which is true.
∴Shaded region will be towards origin.
For equation x + O.y = 1 to get different values of x and y are in the table :

Plotting the points from table on xy plane to get straight line.
Putting x = 0, y = 0 in the inequality,
x ≥ 1
⇒ 0 ≥ 1
Which is false.
Hence, the shaded region will be opposite side of origin.
For equation 0.x + y = 2, the different values of x and y are given in the table :

Plotting the points on xy plane to get straight line.
Then, put x = 0, y = 0 (0,0) in the inequality,
y ≥ 2
⇒ 0 ≥ 2
Which is false.
Hence, the shaded region will be opposite side of origin.

∴ The common shaded region is the required solution of the given inequalities.

Question 10.
Solve the following inequalities graphically :
3x + 4y ≤ 60, x + 3y ≤ 30, x ≥ 0, y ≥ 0. (NCERT)
Solution:
The given inequalities are :
3x + 4y ≤ 60
x + 3y ≤ 30
x ≥ 0
y ≥ 0
Prepared table for values of x and y of equation 3x + 4y = 60 :

Plotting the above points on xy – plane to get straight line.
Then, put x = 0, y = 0 in the inequality,
3x + 4y ≤ 60
⇒ 0 + 0 ≤ 60
⇒ 0 ≤ 60
Which is true.
Hence, the shaded portion will be towards origin.
Prepared table for values of x and y of equation x + 3y = 30:

Plotting the above points on xy – plane to get straight line.
Then, put x = 0, y = 0 in the inequality,
x + 3y ≤ 30
⇒ 0 + 0 ≤ 30
⇒ 0 ≤ 30
Which is true.
Hence, the shaded region will be towards origin.
The graph of x = 0 is y – axis and graph of y = 0 is x – axis.

The common shaded region is the required solution set of the given inequalities.