MCQ Questions for Class 11 Maths Chapter 8 Binomial Theorem with Answers

Answer: (c) 10c³
Hint:
Given, binomial expression is (y² + c/y)⁵
Now, T_r+1 = ⁵C_r × (y²)^5-r × (c/y)^r
= ⁵C_r × y^10-3r × C^r
Now, 10 – 3r = 1
⇒ 3r = 9
⇒ r = 3
So, the coefficient of y = ⁵C₃ × c³ = 10c³

Answer: (a) greater than
Hint:
Given, (1.1)¹⁰⁰⁰⁰ = (1 + 0.1)^10000
10000C₀ + ¹⁰⁰⁰⁰C₁ × (0.1) + ¹⁰⁰⁰⁰C₂ ×(0.1)² + other +ve terms
= 1 + 10000×(0.1) + other +ve terms
= 1 + 1000 + other +ve terms
> 1000
So, (1.1)¹⁰⁰⁰⁰ is greater than 1000

Answer: (c) -1760 x⁹ × y³
Hint:
4th term in (x – 2y)¹² = T₄
= T₃₊₁
= ¹²C₃ (x)^12-3 ×(-2y)³
= ¹²C₃ x⁹ ×(-8y³)
= {(12×11×10)/(3×2×1)} × x⁹ ×(-8y³)
= -(2×11×10×8) × x⁹ × y³
= -1760 x⁹ × y³

Answer: (d) an irrational number
Hint:
Since n is a positive integer, assume n = 1
(√3+1)³ + (√3−1)³
= {3√3 + 1 + 3√3(√3 + 1)} + {3√3 – 1 – 3√3(√3 – 1)}
= 3√3 + 1 + 9 + 3√3 + 3√3 – 1 – 9 + 3√3
= 12√3, which is an irrational number.

Answer: (b) 1/2
Hint:
(1 + x)^m = 1 + mx + {m(m – 1)/2}x² + ……..
Now, {m(m – 1)/2}x² = (-1/8)x²
⇒ m(m – 1)/2 = -1/8
⇒ 4m² – 4m = -1
⇒ 4m² – 4m + 1 = 0
⇒ (2m – 1)² = 0
⇒ 2m – 1 = 0
⇒ m = 1/2

Answer: (b) 10!/(5!)²
Hint:
The coefficient of x^r in the expansion of (1 + x)¹⁰ is ¹⁰C_r and ¹⁰C_r is maximum for
r = 10/ = 5
Hence, the greatest coefficient = ¹⁰C₅
= 10!/(5!)²

Answer: (b) (2n)!/(n!)²
Hint:
We have,
(1 – 2x + 3x² – 4x³ + ……..)^-n = {(1 + x)^-2}^-n
= (1 + x)²ⁿ
So, the coefficient of xⁿC₃ = ²ⁿC_n = (2n)!/(n!)²

Answer: (c) 6
Hint:
Given that the first three terms of the expansion are 729, 7290 and 30375 respectively.
Now T₁ = ⁿC₀ × a^n-0 × b⁰ = 729
⇒ aⁿ = 729 ……………. 1
T₂ = ⁿC₁ × a^n-1 × b¹ = 7290
⇒ n
a^n-1 × b = 7290 ……. 2
T₃ = ⁿC₂ × a^n-2 × b² = 30375
⇒ {n(n-1)/2}
a^n-2 × b² = 30375 ……. 3
Now equation 2/equation 1
n
a^n-1 × b/aⁿ = 7290/729
⇒ n×b/n = 10 ……. 4
Now equation 3/equation 2
{n(n-1)/2}
a^n-2 × b² /n
a^n-1 × b = 30375/7290
⇒ b(n-1)/2a = 30375/7290
⇒ b(n-1)/a = (30375×2)/7290
⇒ bn/a – b/a = 60750/7290
⇒ 10 – b/a = 6075/729 (60750 and 7290 is divided by 10)
⇒ 10 – b/a = 25/3 (6075 and 729 is divided by 243)
⇒ 10 – 25/3 = b/a
⇒ (30-25)/3 = b/a
⇒ 5/3 = b/a
⇒ b/a = 5/3 …………….. 5
Put this value in equation 4, we get
n × 5/3 = 10
⇒ 5n = 30
⇒ n = 30/5
⇒ n = 6
So, the value of n is 6

Answer: (b) 1
Hint:
Given, x² – x + 1 = 0
Now, by Shridharacharya formula, we get
x = {1 ± √(1 – 4×1×1) }/2
⇒ x = {1 ± √(1 – 4) }/2
⇒ x = {1 ± √(-3)}/2
⇒ x = {1 ± √(3 × -1)}/2
⇒ x = {1 ± √3 × √-1}/2
⇒ x = {1 ± i√3}/2 {since i = √-1}
⇒ x = {1 + i√3}/2, {–1 – i√3}/2
⇒ x = -{-1 – i√3}/2, -{-1 + i√3}/2
⇒ x = w, w² {since w = {-1 + i√3}/2 and w² = {-1 – i√3}/2 }
Hence, α = -w, β = w²
Again we know that w³ = 1 and 1 + w + w² = 0
Now, α²⁰⁰⁹ + β²⁰⁰⁹ = α²⁰⁰⁷ × α² + β²⁰⁰⁷ × β²
= (-w)²⁰⁰⁷ × (-w)² + (-w²)²⁰⁰⁷ × (-w²)² {since 2007 is multiple of 3}
= -(w)²⁰⁰⁷ × (w)² – (w²)²⁰⁰⁷ × (w⁴)
= -1 × w² – 1 × w³ × w
= -1 × w² – 1 × 1 × w
= -w² – w
= 1 {since 1 + w + w² = 0}
So, α²⁰⁰⁹ + β²⁰⁰⁹ = 1

Answer: (d) T_r+1 = ⁿC_r × a^n-r × b^r
Hint:
The general term of the expansion (a + b)ⁿ is
T_r+1 = ⁿC_r × a^n-r × b^r

Answer: (b) (-1)ⁿ
Hint:
We know that
(1 + x + x² + …..)^-n = (1 – x)^-n
Now, the coefficient of x = (-1)ⁿ × ⁿC_n
= (-1)ⁿ

Answer: (b) not an integer
Hint:
Since n is a positive integer, assume n = 1
(√5+1)² + 1 − (√5−1)² + 1
= (5 + 2√5 + 1) + 1 – (5 – 2√5 + 1) + 1 {since (x+y)² = x² + 2xy + y²}
= 4√5 + 2, which is not an integer

Answer: (a) (n/2 + 1)^th term
Hint:
In the expansion of (a + b)ⁿ,
if n is even then the middle term is (n/2 + 1)^th term

Answer: (c) 2
Hint:
In the expansion of (a + b)ⁿ,
if n is odd then there are two middle terms which are
{(n + 1)/2}^th term and {(n+1)/2 + 1}^th term

Answer: (c) 49
Hint:
Given, 2³ⁿ – 7n – 1 = 2^3×n – 7n – 1
= 8ⁿ – 7n – 1
= (1 + 7)ⁿ – 7n – 1
= {ⁿC₀ + ⁿC₁ 7 + ⁿC₂ 7² + …….. + ⁿC_n 7ⁿ} – 7n – 1
= {1 + 7n + ⁿC₂ 7² + …….. + ⁿC_n 7ⁿ} – 7n – 1
= ⁿC₂ 7² + …….. + ⁿC_n 7ⁿ
= 49(ⁿC₂ + …….. + ⁿC_n 7^n-2)
which is divisible by 49
So, 2³ⁿ – 7n – 1 is divisible by 49

Answer: (c) 6
Hint:
Given, (7^1/2 + 5^1/3)³⁷
Now, general term of this binomial T_r+1 = ³⁷C_r × (7^1/2)^37-r × (5^1/3)^r
⇒ T_r+1 = ³⁷C_r × 7(^37-r)^/2 × (5)^r/3
This General term will be an integer if ³⁷C_r is an integer, 7(^37-r)^/2 is an integer and (5)^r/3 is an integer.
Now, ³⁷C_r will always be a positive integer.
Since ³⁷C_r denotes the number of ways of selecting r things out of 37 things, it can not be a fraction.
So, ³⁷C_r is an integer.
Again, 7(^37-r)^/2C_r will be an integer if (37 – r)/2 is an integer.
So, r = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37 …………. 1
And if (5)^r/3 is an integer, then r/3 should be an integer.
So, r = 0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36 ………….2
Now, take intersection of 1 and 2, we get
r = 3, 9, 15, 21, 27, 33
So, total possible value of r is 6
Hence, there are 6 integers are in the binomial expansion of (7^1/2 + 5^1/3)³⁷

Answer: (b) 4851
Hint:
Given, x + y + z = 100;
where x ≥ 1, y ≥ 1, z ≥ 1
Let u = x – 1, v = y – 1, w = z – 1
where u ≥ 0, v ≥ 0, w ≥ 0
Now, equation becomes
u + v + w = 97
So, the total number of solution = ^97+3-1C_3-1
= ⁹⁹C₂
= (99 × 98)/2
= 4851

Answer: (b) 10!/(5!)²
Hint:
The coefficient of x^r in the expansion of (1 + x)¹⁰ is ¹⁰C_r and ¹⁰C_r is maximum for
r = 10/2 = 5
Hence, the greatest coefficient = ¹⁰C₅
= 10!/(5!)²

Answer: (b) 1/2
Hint:
(1 + x)^m = 1 + mx + {m(m – 1)/2}x² + ……..
Now, {m(m – 1)/2}x² = (-1/8)x²
⇒ m(m – 1)/2 = -1/8
⇒ 4m² – 4m = -1
⇒ 4m² – 4m + 1 = 0
⇒ (2m – 1)² = 0
⇒ 2m – 1 = 0
⇒ m = 1/2

Answer: (b) 15
Hint:
Given, in the binomial expansion of (a + b)ⁿ, the coefficient of fourth and thirteenth terms are equal to each other
⇒ ⁿC₃ = ⁿC₁₂
This is possible when n = 15
Because ¹⁵C₁₃ = ¹⁵C₁₂