NCERT Solutions for Class 11 Maths Chapter 8- Binomial Theorem Miscellaneous Exercise

NCERT Solutions are provided to help the students in understanding the steps to solve mathematical problems that are provided in the textbook. The Miscellaneous Exercise of NCERT Solutions for Class 11 Maths Chapter 8- Binomial Theorem is based on the following topics:

  1. Binomial Theorem for Positive Integral Indices
  2. General and Middle Terms

The NCERT Solutions for Class 11 Maths enhance topics with frequent, focused, engaging math challenges and activities that strengthen math concepts. Each question of the exercises have been carefully solved for the students to understand, keeping the examination point of view in mind.

Download PDF of NCERT Solutions for Class 11 Maths Chapter 8- Binomial Theorem Miscellaneous Exercise

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Solutions for Class 11 Maths Chapter 8 – Miscellaneous Exercise

1. Find a, b and n in the expansion of (a + b)n if the first three terms of the expansion are 729, 7290 and 30375, respectively.

Solution:

We know that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by

Tr+1 = nCr an-t br

The first three terms of the expansion are given as 729, 7290 and 30375 respectively. Then we have,

T1 = nC0 an-0 b0 = an = 729….. 1

T2 = nC1 an-1 b1 = nan-1 b = 7290…. 2

T3 = nC2 an-2 b2 = n (n -1)/2 an-2 b2 = 30375……3

Dividing 2 by 1 we get

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 31

From 4 and 5 we have

n. 5/3 = 10

n = 6

Substituting n = 6 in 1 we get

a6 = 729

a = 3

From 5 we have, b/3 = 5/3

b = 5

Thus a = 3, b = 5 and n = 76

2. Find a if the coefficients of x2 and x3 in the expansion of (3 + a x)9 are equal.

Solution:

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 32

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 33

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 34

3. Find the coefficient of x5 in the product (1 + 2x)6 (1 – x)7 using binomial theorem.

Solution:

(1 + 2x)6 = 6C0 + 6C1 (2x) + 6C2 (2x)2 + 6C3 (2x)3 + 6C4 (2x)4 + 6C5 (2x)5 + 6C6 (2x)6

= 1 + 6 (2x) + 15 (2x)2 + 20 (2x)3 + 15 (2x)4 + 6 (2x)5 + (2x)6

= 1 + 12 x + 60x2 + 160 x3 + 240 x4 + 192 x5 + 64x6

(1 – x)7 = 7C07C1 (x) + 7C2 (x)27C3 (x)3 + 7C4 (x)47C5 (x)5 + 7C6 (x)6 7C7 (x)7

= 1 – 7x + 21x2 – 35x3 + 35x4 – 21x5 + 7x6 – x7

(1 + 2x)6 (1 – x)7 = (1 + 12 x + 60x2 + 160 x3 + 240 x4 + 192 x5 + 64x6) (1 – 7x + 21x2 – 35x3 + 35x4 – 21x5 + 7x6 – x7)

192 – 21 = 171

Thus, the coefficient of x5 in the expression (1+2x)6(1-x)7 is 171.

4. If a and b are distinct integers, prove that a – b is a factor of an – bn, whenever n is a positive integer. [Hint write an = (a – b + b)n and expand]

Solution:

In order to prove that (a – b) is a factor of (an – bn), it has to be proved that

an – bn = k (a – b) where k is some natural number.

a can be written as a = a – b + b

an = (a – b + b)n = [(a – b) + b]n

= nC0 (a – b)n + nC1 (a – b)n-1 b + …… + n C n bn

an – bn = (a – b) [(a –b)n-1 + nC1 (a – b)n-1 b + …… + n C n bn]

an – bn = (a – b) k

Where k = [(a –b)n-1 + nC1 (a – b)n-1 b + …… + n C n bn] is a natural number

This shows that (a – b) is a factor of (an – bn), where n is positive integer.

5. Evaluate 

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 35

Solution:

Using binomial theorem the expression (a + b)6 and (a – b)6, can be expanded

(a + b)6 = 6C0 a6 + 6C1 a5 b + 6C2 a4 b2 + 6C3 a3 b3 + 6C4 a2 b4 + 6C5 a b5 + 6C6 b6

(a – b)6 = 6C0 a66C1 a5 b + 6C2 a4 b26C3 a3 b3 + 6C4 a2 b46C5 a b5 + 6C6 b6

Now (a + b)6 – (a – b)6 =6C0 a6 + 6C1 a5 b + 6C2 a4 b2 + 6C3 a3 b3 + 6C4 a2 b4 + 6C5 a b5 + 6C6 b6 – [6C0 a66C1 a5 b + 6C2 a4 b26C3 a3 b3 + 6C4 a2 b46C5 a b5 + 6C6 b6]

Now by substituting a = √3 and b = √2 we get

(√3 + √2)6 – (√3 – √2)6 = 2 [6 (√3)5 (√2) + 20 (√3)3 (√2)3 + 6 (√3) (√2)5]

= 2 [54(√6) + 120 (√6) + 24 √6]

= 2 (√6) (198)

= 396 √6

6. Find the value of 

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 36

Solution:

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 37

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 38

7. Find an approximation of (0.99)5 using the first three terms of its expansion.

Solution:

0.99 can be written as

0.99 = 1 – 0.01

Now by applying binomial theorem we get

(o. 99)5 = (1 – 0.01)5

= 5C0 (1)55C1 (1)4 (0.01) + 5C2 (1)3 (0.01)2

= 1 – 5 (0.01) + 10 (0.01)2

= 1 – 0.05 + 0.001

= 0.951

8. Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 39 is √6: 1

Solution:

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 40

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 41

9. Expand using Binomial Theorem 

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 42

Solution:

Using binomial theorem the given expression can be expanded as

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 43

Again by using binomial theorem to expand the above terms we get

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 44

From equation 1, 2 and 3 we get

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 45

10. Find the expansion of (3x2 – 2ax + 3a2)3 using binomial theorem.

Solution:

We know that (a + b)3 = a3 + 3a2b + 3ab2 + b3

Putting a = 3x2 & b = -a (2x-3a), we get

[3x2 + (-a (2x-3a))]3

= (3x2)3+3(3x2)2(-a (2x-3a)) + 3(3x2) (-a (2x-3a))2 + (-a (2x-3a))3

= 27x6 – 27ax4 (2x-3a) + 9a2x2 (2x-3a)2 – a3(2x-3a)3

= 27x6 – 54ax5 + 81a2x4 + 9a2x2 (4x2-12ax+9a2) – a3 [(2x)3 – (3a)3 – 3(2x)2(3a) + 3(2x)(3a)2]

= 27x6 – 54ax5 + 81a2x4 + 36a2x4 – 108a3x3 + 81a4x2 – 8a3x3 + 27a6 + 36a4x2 – 54a5x

= 27x6 – 54ax5+ 117a2x4 – 116a3x3 + 117a4x2 – 54a5x + 27a6

Thus, (3x2 – 2ax + 3a2)3

= 27x6 – 54ax5+ 117a2x4 – 116a3x3 + 117a4x2 – 54a5x + 27a6


Access other exercise solutions of Class 11 Maths Chapter 8- Binomial Theorem

Exercise 8.1 Solutions 14 Questions

Exercise 8.2 Solutions 12 Questions

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