NCERT Solutions Class 11th Maths Chapter – 8 Binomial Theorem Miscellaneous Exercise
| Textbook | NCERT |
| class | Class – 11th |
| Subject | Mathematics |
| Chapter | Chapter – 8 |
| Chapter Name | Binomial Theorem |
| grade | Class 11th Maths solution |
| Medium | English |
| Source | last doubt |
NCERT Solutions Class 11th Maths Chapter – 8 Binomial Theorem Miscellaneous Exercise
?Chapter – 8?
✍Binomial Theorem✍
?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
2. Find a if the coefficients of x2 and x3 in the expansion of (3 + a x)9 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 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
?♂️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 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 of
?♂️solution –
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)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 in the expansion of is √6: 1
?♂️solution –
9. Expand using Binomial Theorem
?♂️solution – Using binomial theorem the given expression can be expanded as
Again by using binomial theorem to expand the above terms we get
From equation 1, 2 and 3 we get
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