NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Exercise 8.2

Chapter 8 Binomial Theorem Exercise 8.2 Class 11 Maths NCERT Solutions is provided here that will essential for development of problem solving skills. NCERT Solutions for Class 11 Maths will help every students in completing their home assignments on time and easily. These NCERT Solutions are updated according to the latest pattern released by CBSE which will save your precious time.

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Exercise 8.2


1. Find the coefficient of x 5 in (x + 3)8. 

Answer

General term in (x + 3)8 = 8Cr x8 – r. 3r
We have to find the coefficient of x 5
8 – r = 5, r = 8 – 5 = 3
∴ Coefficient of x 5 (putting r = 3)
8C3. 33 = (8.7.6)/( 1.2.3) . 27 = 56 . 27 = 1512.

2. Find the coefficient of a57 in (a – 2b)12.

Answer

(a – 2b) 12 = [a + (– 2b)]12
General term Tr + 1 = C(12, r) a 12 – r (–2b) r .
Putting 12 – r = 5 or 12 – 5 = r => r = 7
T7 +1 = C (12, 7) a 12 – 7 (–2b) 7
= C(12, 7) a 5 (–2b) 7= C(12, 7) (–2)7 a 5b 7
Hence required coefficient is C(12, 7) (–2)7
= 12!/ (7! 5!) . 27
= (-12 × 11 × 10 × 9 × 8 × 7!)/(7! × 5 × 4 × 3 × 2 × 1) × 27
= 8 × -11 × 9 × 27
= -99 × 8 × 128 = -101376.

3. Write the general term in the expansion of (x2 – y)6

Answer

General term = Tr + 1 6Cr (x 2)6 – r (– y) r
= (–1)r 6!/(r!(6 – r)!) .x12-2r .y

4. Write the general term in the expansion of (x2 – yx)12, x ≠ 0.

Answer

Binomial expansion is (x 2 – yx) 12
General term Tr +1 12Cr (x 212 – r. (–yx) r
= 12!/(r!(12 – r)!) . x24 – 2r .(-1)r . yr xr
= ((-1)r 12!)/(r!(12 – r)!) . x24-r . yr

5. Find the 4th term in the expansion of (x – 2y)12

Answer

4th term = T3+1 in the expansion of (x+(-2y))12
12C3.x12 – 3 [-2y]3
= (12 . 11. 10)/(1. 2. 3) x9 (-1)3.23. y3
= -220 × 8x9.y3 = -1760 x9 y3

6. Find the 13th term in the expansion of
(9x – 1/3√x)18 , x ≠ 0

Answer

13th term, T13 = T12 +1
18C12 (9x)18 – 12 (-(1/3√x))12
18C696x6 (-1)12 . 1/312 × 1/x6
= 18564 × (32)6. 1/312 × x6/x6
= 18564 × 312/312 = 18564

7. Find the middle term in the expansion of (3 = x3/6)7.

Answer

Number of terms in the expansion is 7 + 1 = 8
There are two middle terms which are
(8/2)th & ((7 + 3)/2)th i.e., 4th & 5th
Hence, we have to find T4 and T5 in the given expansion
(3 – (x3/6))7 = [3 + (-(x3/6)]7
Tr+1 = C(7, r)37-r (-(x3/6)r) ………(i)
Now Tr + 1 = T4 or r + 1 = 4 ∴ r = 3
Putting r = 3, we have
T3+1 = C(7, 3)37 – 3(-(x3/6)3)
= C(7, 3)34 (-1)3 x9/63 = -7!/(3!4!) . 3/23 . x9
= (- 7 ×6 ×5 ×4!)/(3 × 2 × 1 × 4!) . 3/23 . x9 = -105/8 x9
Again Tr+1 = T5 or r + 1= 5 or r = 4
Putting r = 4 in (i), we have
T4+ 1 = T5 = C(7, 4) 3 7 – 4 (-1)4 x12/64
= 7/4!3! . 33x12/3424 = (7× 6 × 5 × 4!)/(3 × 2 × 1 × 4!) × x12/3 × 24
35/48 .x12

8. Find the middle term in the expansion of (x/3 + 9y)10

Answer

Number of terms in the expansion is
10 + 1 = 11 (odd)
Middle term of the expansions is (n/2 + 1)th term
= (5 + 1)th term = 6th term
T6 = T5 + 1 = C(10, 5)(x/3)10 – 5 (9y)5
= C(10, 5) x5/35 . 95y= C(10, 5) 35x5y5
= 10!/(5! (10 – 5)!) 35x5y5 = 10!/5!5! . 35x5y5
= (10 × 9 ×8 ×7 × 6 × 5!)/(5 × 4 ×3 ×2 ×1× 5!) 35x5y5
= 61236x5y5

9. In the expansion of (1 + a) m + n, prove that coefficients of a m and a n are equal.

Answer

General term in the expansion of (1 + a) m + n is
Tr + 1 m + nCr a r
Putting r = m
Tm+1 m+nCmam ……(i)
∴ Coefficient of am = m + nCm
Again putting r = n
Tn+1 = m+nCnan
Coefficient of an = m+nCn = m+nC ..… (ii)
[∵ nCr = nCn – r]
From (i) and (ii) coefficient of am is equal to coefficient of an .

10. The coefficients of the (r – 1)th, r th and (r + 1)th terms in the expansion of (x + 1)n are in the ratio of 1: 3: 5. Find n and r.

Answer

General term in the expansion of (x + 1)n is
Tr – 1 = T(r – 2) = nCr – 2. xr-2
Tr = T(r – 1) = nCr-1 . x r -1
Tr+1 = nCr.xr
C(n, r -2) : C(n, r -1): C(n, r) = 1 : 3 : 5
Or (C(n, r – 2))/1 = (C(n, r – 1))/3 = (C(n,r))/5
If (C(n, r -2))/1 = (C(n, r – 1))/3
Or 3C(n, r- 2) = C(n, r – 1)
Or 3. n!/((r – 2) ! (n + 2 – r) !) = n!/((r – 1) ! (n + 1 – r) !)
Or 3/((r – 2)!(n + 2 – r)(n + 1 – r) !)
= 1/((r – 1)(r – 2)!(n + 1 – r)!)
Or 3/(n + 2 – r) = 1/(r – 1)
Or 3r – 3 = n + 2 – r
Or 4r = n + 5 ….(i)
Again if (C(n, r – 1))/3 = (C(n,r))/5
Or 5C(n, r – 1) = 3C(n, r)
Or 5. n!/((r – 1) !(n + 1 – r) !) = 3. n!/(r!(n – r) !)
Or 5/((r – 1) !(n + 1 – r)(n – r) ! = 3/(r(r – 1)!(n – r)!)
Or 5r = 3(n + 1 – r) or 8r = 3n + 3 ….(ii)
From (i) & (ii) 2n + 10 = 3n + 3
Or 3n – 2n = 10 -3 => n = 7
From (ii) 8r = 21 + 3 = 24
∴ r = 3
∴ n = 7, r = 3

11. Prove that the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n –1 .

Answer

General term in the expansion of (1 + x) 2n is
Tr + 1 = C (2n, r) x r
Putting r = n, we have
Tn + 1 = C(2n, n) x n
Coefficient of x n = C(2n, n)
Again general term in the expansion of
(1 + x)2n –1 is Tr + 1 = C (2n –1, r) x r
Putting r = n, we have
Tn + 1 = C(2n – 1, n) x n
Coefficient of x n in the expansion of (1+ x)2n – 1 is C(2n – 1, n)
According to the problem, we have to prove that
C(2n, n) = 2 × C(2n – 1, n)
Or 2n!/(n!(2n – n)!) = 2. ((2n – 1)!)/(n!(2n – 1 – n)!)
Or 2n!/n!n! = 2. ((2n – 1)!)/(n!(n – 1)!)
Multiplying Nr and Dr by n on RHS, we have
2n!/n!n! = (2n(2n -1)!)/(n!n(n – 1)!)
i.e 2n!/n!n! = 2n!/n!n!, Which is true.
Hence proved.

12. Find a positive value of m for which the coefficient of x 2 in the expansion (1 + x) m is 6. 

Answer

Given expansion is (1 + x)m. Now,
General term = Tr+1 mCrxr
Put r = 2, we have
T3 = mC2.x2
According to the question C(m, 2) = 6
or (m(m – 1))/2! = 6
=> m2 – m = 12
Or m2 – m – 12 = 0
=> m2 – 4m + 3m – 12 = 0
Or (m – 4)(m + 3) = 0
∴ m = 4, since m ≠ –3
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