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

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.
Answer:
(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)^{–\(\frac { 1 }{ 3 }\)}
Answer:
(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
Answer:
(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
Answer:
(a) – ¹⁰C₅

An online remainder theorem calculator allows you to determine the remainder of given polynomial expressions by remainder theorem.

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.
Answer:
(c) Even positive integer

Question 6.
Expansion of (a + x)ⁿ is :
(a) aⁿ + ⁿC₁a^n-1x + ⁿC₂a^n-2x² + ……….. + ⁿC_ra^n-rx^r + ………… + aⁿ
(b) xⁿ + ⁿC₁x^n-1a +ⁿC₂x^n-2a² + ……….. + ⁿC_rx^n-ra^r + ………… + aⁿ
(c) aⁿ – ⁿC₁a^n-1x + ⁿC₂a^n-2x² + ……….. + (-1)^rnC_ra^n-rx^r + ………… + (-1)ⁿaⁿ
(d) xⁿ – ⁿC₁a^n-1a + ⁿC₂a^n-2a² + ……….. + (-1)^rnC_ra^n-rx^r + ………… + (-1)ⁿxⁿ
Answer:
(a) aⁿ + ⁿC₁a^n-1x +ⁿC₂a^n-2x² + ……….. + ⁿC_ra^n-rx^r + ………… + 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) \(\frac { ^{ 10 }{ C }_{ 6 } }{ { x }^{ 10 } }\)
Answer:
(a) \(\frac { ^{ 10 }{ C }_{ 6 } }{ x }\)
[Hint: The total number of terms be 11 in the expansion of it.
∵ The 5^th term from end = (11 – 5)^th = 7^th 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
Answer:
(b) 2

Question 9.
The value of ⁿC₀ + ⁿC₁ + ⁿC₂ + …………….. + ⁿC_n :
(a) 2ⁿ + 1
(b) 2^{n – 1}
(c) 2ⁿ – 1
(d) 2ⁿ
Answer:
(d) 2ⁿ

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

The Chebyshev’s Theorem calculator, above, will allow you to enter any value of k greater than 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) \(\frac { (n+1)(n+2) }{ 6 }\)
Answer:
(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.
Answer:
(a) | x | < 1

Question 13.
The general term in the expansion of (x + a)ⁿ will be :
(a) r^th
(b) (r+1)^thterm
(c) (r-1)^th
(d) (r+2)^thterm
Answer:
(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) \(\frac { 256 }{ 243 }\)
Answer:
(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
Answer:
(b) 21

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

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

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

(B) Match the following :

Answer:

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 5^th 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 + ………….. + nC_n will be ……………..
6. The value of ⁿC₀ – ⁿC₁ + ⁿC₂ – ⁿC₃ + ………….. 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 (3^rd) 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 ……………..

Answer:

1. 08243216
2. \(\frac { { x }^{ 3 } }{ 16 }\)
3. \(\frac { 105 }{ 32{ x }^{ 2 } }\)
4. 924
5. n.2^n-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 + ………….. + \(\frac { (- 1)(r + 1)(r +2) }{ 2! }\)x^r + ……………..
2. The expansion of (1 – x)^-3 is 1 + 3x + 6x² + 10x³ + ………….. + \(\frac { (- 1)^r(r + 1)(r +2) }{ 2! }\)x^r + ……………..
3. The expansion of (1 – x)^-2 is 1 + 2x + 3x² + (r + 1) x^r+ ………….. +
4. The (r + 1 )^th term in the expansion of (1 – x)^-2 is (- 1)^r(r + 1) x^r+ ………….. +
5. The (r + 1 )^th term in the expansion of (1 – x)ⁿ will be x^r
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.

Answer:

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 x^r and x^{r – 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 r^th term, then the value of r will be.
7. The 5^th term in the expansion of (1 – 2x)^{– 1} will be.

Answer:

1. 997002999
2. – 20 x⁵
3. ⁿC_r x^{n – r} a^r
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 13^th 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 T_r+1 = nC₁x^n-ra^r
Here n = 8, x = x, a = 3
T_r+1 = ⁸C_r(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 17^th and 18^th 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
T_r+1 = ⁿC_r x^{n – r} a^r
Here n = 50, x = 2, a = a
T₁₇ = T_{16 + 1} = ⁵⁰C₁₆ (2)^{50 – 16} (a)¹⁶
⇒ T₁₇ = ⁵⁰C₁₆ (2)³⁴ (a)¹⁶
and T₁₈ = T_{17 + 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, r^th 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,
T_{r + 1} = ⁿC_rx^{n – r}(1)^r
T_{r – 1} = T_{r – 2 + 1} = ⁿC_{r – 2}(x)^{n – (r – 2)}(1)^{r – 2}
Coefficient of T_{r – 1}^th term = ⁿC_{r – 2}
T_r = T_{r – 1 + 1} = ⁿC_{r – 1}(x)^{n – (r – 1)}(1)^{r – 1}
Coefficient of T_r^thterm = ⁿC_{r – 1}
T_{r + 1} = ⁿC_r x^{n – r} (1)^r
Coefficient of T_r+1^th term = ⁿC_r

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
T_r+1 = ⁿC_r x^{n – r} a^r
Here x = 1, a = x, n = 2n
T_r+1 = ²ⁿC_r(1)^{2n – r}(x)^r
For the coefficient of xⁿ, put r = n,
T_n+1 = ²ⁿC_n(a)^{2n – n}(x)ⁿ
and T₁₈ = T_{17 + 1} = ⁵⁰C₁₇ (2)^{50 – 17} (a)¹⁷
= (²ⁿC_n) xⁿ
∴ In the expansion of (1 + x)²ⁿ, the coefficient of xⁿ = ²ⁿC_n …. (1)
and in the expansion of (1 + x)^{2n – 1}, x = 1, a = x, n = 2n – 1
∴ T_r+1 = ²ⁿC_r (1)^{2n – 1  -r} (x)ⁿ
For the coefficient of xⁿ, put r = n, ‘
We get T_n+1 = ^{2n – 1}C_n xⁿ
The coefficient of xⁿ in the expansion of (1 + x)^{2n – 1} = ^{n – 1}C_n

∴ 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