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

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.

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

Answer

General term in (x + 3)⁸ = ⁸C_r x^{8 – r}. 3^r
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 a⁵b ⁷ in (a – 2b)¹².

Answer

(a – 2b) ¹² = [a + (– 2b)]¹²
General term T_{r + 1} = C(12, r) a ^{12 – r} (–2b) ^r .
Putting 12 – r = 5 or 12 – 5 = r => r = 7
T_{7 +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 = T_{r + 1} = ⁶C_r (x ²)^{6 – r} (– y) ^r
= (–1)^r 6!/(r!(6 – r)!) .x^12-2r .y^r 

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

Answer

Binomial expansion is (x ² – yx) ¹²
General term T_{r +1} = ¹²C_r (x ²) ^{12 – r}. (–yx) ^r
= 12!/(r!(12 – r)!) . x^{24 – 2r} .(-1)^r . y^r x^r
= ((-1)^r 12!)/(r!(12 – r)!) . x^24-r . y^r

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

Answer

4^th term = T₃₊₁ in the expansion of (x+(-2y))¹²
= ¹²C₃.x^{12 – 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

13^th term, T₁₃ = T_{12 +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., 4^th & 5^th
Hence, we have to find T₄ and T₅ in the given expansion
(3 – (x³/6))⁷ = [3 + (-(x³/6)]⁷
T_r+1 = C(7, r)3^7-r (-(x³/6)^r) ………(i)
Now T_{r + 1} = T₄ or r + 1 = 4 ∴ r = 3
Putting r = 3, we have
T₃₊₁ = C(7, 3)3^{7 – 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 T_r+1 = T₅ or r + 1= 5 or r = 4
Putting r = 4 in (i), we have
T_{4+ 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 = 6^th term
T₆ = T_{5 + 1} = C(10, 5)(x/3)^{10 – 5} (9y)⁵
= C(10, 5) x⁵/3⁵ . 9⁵y⁵ = 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
T_{r + 1} = ^{m + n}C_r a ^r
Putting r = m
T_m+1 = ^m+nC_ma^m ……(i)
∴ Coefficient of a^m = ^{m + n}C_m
Again putting r = n
T_n+1 = ^m+nC_naⁿ
Coefficient of aⁿ = ^m+nC_n = ^m+nC_m  ..… (ii)
[∵ ⁿC_r = ⁿC_{n – r}]
From (i) and (ii) coefficient of a^m 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
T_{r – 1} = T_{(r – 2)} = ⁿC_{r – 2}. x^r-2
T_r = T_{(r – 1)} = ⁿC_r-1 . x ^{r -1}
T_r+1 = ⁿC_r.x^r
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
T_{r + 1} = C (2n, r) x ^r
Putting r = n, we have
T_{n + 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 T_{r + 1} = C (2n –1, r) x ^r
Putting r = n, we have
T_{n + 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 N^r and D^r 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 = T_r+1 = ^mC_rx^r
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