*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 7.
NCERT Solutions are provided to help the students understand the steps to solve mathematical problems presented in the textbook. The Miscellaneous Exercise of NCERT Solutions for Class 11 Maths Chapter 8 – Binomial Theorem is based on the following topics:
- Binomial Theorem for Positive Integral Indices
- General and Middle Terms
The solutions enhance topics with frequent, focused, engaging challenges and activities that strengthen Maths concepts. Each question of the exercises has been carefully solved in NCERT Class 11 Maths Solutions for the students to understand and learn for the examination.
NCERT Solutions for Class 11 Maths Chapter 8 – Binomial Theorem Miscellaneous Exercise
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
Dividing 3 by 2, we get
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 x2Â and x3Â in the expansion of (3 + a x)9Â are equal.
Solution:
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 = 7C0 – 7C1 (x) + 7C2 (x)2 – 7C3 (x)3 + 7C4 (x)4 – 7C5 (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 x5Â 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 a positive integer.
5. EvaluateÂ
Solution:
Using the 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 a6 – 6C1 a5 b + 6C2 a4 b2 – 6C3 a3 b3 + 6C4 a2 b4 – 6C5 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 a6 – 6C1 a5 b + 6C2 a4 b2 – 6C3 a3 b3 + 6C4 a2 b4 – 6C5 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 ofÂ
Solution:
7. Find an approximation of (0.99)5Â using the first three terms of its expansion.
Solution:
0.99 can be written as
0.99 = 1 – 0.01
Now, by applying the binomial theorem, we get
(o. 99)5 = (1 – 0.01)5
= 5C0 (1)5 – 5C1 (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 of the expansion of 
 is √6: 1
Solution:
9. Expand using the Binomial Theorem.Â
Solution:
Using the binomial theorem, the given expression can be expanded as
Again, by using the binomial theorem to expand the above terms, we get
From equations 1, 2 and 3, we get
10. Find the expansion of (3x2 – 2ax + 3a2)3 using binomial theorem.
Solution:
We know that (a + b)3Â = a3Â + 3a2b + 3ab2Â + b3
Putting a = 3x2Â & b = -a (2x-3a), we get
[3x2Â + (-a (2x-3a))]3= (3x2)3+3(3x2)2(-a (2x-3a)) + 3(3x2) (-a (2x-3a))2Â + (-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
Also explore – NCERT Class 11 Solutions
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