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

Binomial Theorem Class 11 Maths MCQs Questions with Answers

Check the below NCERT MCQ Questions for Class 11 Maths Chapter 8 Binomial Theorem with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have Provided Binomial Theorem Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.

Class 11 Maths Chapter 8 Quiz

Class 11 Maths Chapter 8 MCQ Online Test

You can refer to NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

Binomial Theorem Class 11 Maths MCQ online test

Q1.	The number (101)¹⁰⁰ – 1 is divisible by
	A.100
	B.1000
	C.10000
	D.All the above

Ans: All the above
Given, (101)¹⁰⁰ – 1 = (1 + 100)¹⁰⁰ – 1
= [¹⁰⁰C₀ + ¹⁰⁰C₁ × 100 + ¹⁰⁰C₂ × (100)² + ……….+ ¹⁰⁰C₁₀₀ × (100)¹⁰⁰] – 1
= 1 + [¹⁰⁰C₁ × 100 + ¹⁰⁰C₂ × (100)² + ……….+ ¹⁰⁰C₁₀₀ × (100)¹⁰⁰] – 1
= ¹⁰⁰C₁ × 100 + ¹⁰⁰C₂ × (100)² + ……….+ ¹⁰⁰C₁₀₀ × (100)¹⁰⁰
= 100 × 100 + ¹⁰⁰C₂ × (100)² + ……….+ ¹⁰⁰C₁₀₀ × (100)¹⁰⁰
= (100)² + ¹⁰⁰C₂ × (100)² + ……….+ ¹⁰⁰C₁₀₀ × (100)¹⁰⁰
= (100)² [1 + ¹⁰⁰C₂ + ……….+ ¹⁰⁰C₁₀₀ × (100)⁹⁸]
Which is divisible by 100, 1000 and 10000

Q2.	The value of -1° is
	A.1
	B.-1
	C.0
	D.None of these

Ans: -1
First we find 10
So, 10 = 1
Now, -10 = -1

Q3.	If the fourth term in the expansion (ax + 1/x)ⁿ is 5/2, then the value of x is
	A.4
	B.6
	C.8
	D.5

Ans: 6
Given, T₄ = 5/2
⇒ T₃₊₁ = 5/2
⇒ ⁿC₃ × (ax)^n-3 × (1/x)³ = 5/2
⇒ ⁿC₃ × a^n-3 × x^n-3 × (1/x)² = 5/2
Clearly, RHS is independent of x,
So, n – 6 = 0
⇒ n = 6

Q4.	The number 111111 ………….. 1 (91 times) is
	A.not an odd number
	B.none of these
	C.not a prime
	D.an even number

Ans: not a prime
111111 ………….. 1 (91 times) = 91 × 1 = 91, which is divisible by 7 and 13.
So, it is not a prime number.

Q5.	In the expansion of (a + b)ⁿ, if n is even then the middle term is
	A.(n/2 + 1)^th term
	B.(n/2)^th term
	C.n^th term
	D.(n/2 – 1)^th term

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

Q6.	The number of terms in the expansion (2x + 3y – 4z)ⁿ is
	A.n + 1
	B.n + 3
	C.{(n + 1) × (n + 2)}/2
	D.None of these

Ans: {(n + 1) × (n + 2)}/2
Total number of terms in (2x + 3y – 4z)ⁿ is
= ^n+3-1C_3-1
= ⁿ⁺²C₂
= {(n + 1) × (n + 2)}/2

Q7.	If A and B are the coefficient of xⁿ in the expansion (1 + x)²ⁿ and (1 + x)^2n-1 respectively, then A/B equals
	A.1
	B.2
	C.1/2
	D.1/n

Ans: 2
A/B = ²ⁿC_n/ ^2n-1C_n
= {(2n)!/(n! × n!)}/{(2n – 1)!/(n! × (n – 1!))}
= {2n(2n – 1)!/(n(n – 1)! × n!)}/{(2n – 1)!/(n! × (n – 1!))}
= 2
So, A/B = 2

Q8.	The coefficient of y in the expansion of (y² + c/y)⁵ is
	A.29c
	B.10c
	C.10c³
	D.20c²

Ans: 10c³
We have,
T_r+1 = ⁵C_r ×(y²)^5-r × (c/y)^r
⇒ T_r+1 = ⁵C_r × y^10-3r × c^r
For finding the coefficient of y,
⇒ 10 – 3r = 1
⇒ 33r = 9
⇒ r = 3
So, the coefficient of y = ⁵C₃ × c³
= 10c³

Q9.	The coefficient of x^-4 in (3/2 – 3/x²)¹⁰ is
	A.405/226
	B.504/289
	C.450/263
	D.None of these

Ans: None of these
Let x^-4 occurs in (r + 1)th term.
Now, T_r+1 = ¹⁰C_r × (3/2)^10-r ×(-3/x²)^r
⇒ T_r+1 = ¹⁰C_r × (3/2)^10-r ×(-3)^r × (x)^-2r
Now, we have to find the coefficient of x^-4
So, -2r = -4
⇒ r = 2
Now, the coefficient of x^-4 = ¹⁰C₂ × (3/2)^10-2 × (-3)²
= ¹⁰C₂ × (3/2)⁸ × (-3)²
= 45 × (3/2)⁸ × 9
= (3¹² × 5)/2⁸

Q10.	If n is a positive integer, then 9ⁿ⁺¹ – 8n – 9 is divisible by
	A.8
	B.16
	C.32
	D.64

Ans: 64
Let n = 1, then
9ⁿ⁺¹ – 8n – 9 = 9¹⁺¹ – 8 × 1 – 9 = 9² – 8 – 9 = 81 – 17 = 64
which is divisible by 64
Let n = 2, then
9ⁿ⁺¹ – 8n – 9 = 9²⁺¹ – 8 × 2 – 9 = 9³ – 16 – 9 = 729 – 25 = 704 = 11 × 64
which is divisible by 64
So, for any value of n, 9ⁿ⁺¹ – 8n – 9 is divisible by 64

Q11.	The general term of the expansion (a + b)ⁿ is
	A.T_r+1 = ⁿC_r × a^r × b^r
	B.T_r+1 = ⁿC_r × a^r × b^n-r
	C.T_r+1 = ⁿC_r × a^n-r× b^n-r
	D.T_r+1 = ⁿC_r × a^n-r × b^r

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

Q12.	In the expansion of (a + b)ⁿ, if n is even then the middle term is
	A.(n/2 + 1)^th term
	B.(n/2)^th term
	C.n^th term
	D.(n/2 – 1)^th term

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

Q13.	The smallest positive integer for which the statement 3ⁿ⁺¹ < 4ⁿ is true for all
	A.4
	B.3
	C.1
	D.2

Ans: 4
Given statement is: 3ⁿ⁺¹ < 4ⁿ is
Let n = 1, then
3¹⁺¹ < 4¹ = 3² < 4 = 9 < 4 is false
Let n = 2, then
3²⁺¹ < 4² = 3³ < 4² = 27 < 16 is false
Let n = 3, then
3³⁺¹ < 4³ = 3⁴ < 4³ = 81 < 64 is false
Let n = 4, then
3⁴⁺¹ < 4⁴ = 3⁵ < 4⁴ = 243 < 256 is true.
So, the smallest positive number is 4

Q14.	The number of ordered triplets of positive integers which are solution of the equation x + y + z = 100 is
	A.4815
	B.4851
	C.8451
	D.8415

Ans: 4851
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

Q15.	if n is a positive ineger then 2³ⁿ – 7n – 1 is divisible by
	A.7
	B.9
	C.49
	D.81

Ans: 49
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

Q16.	The greatest coefficient in the expansion of (1 + x)¹⁰ is
	A.10!/(5!)
	B.10!/(5!)²
	C.10!/(5! × 4!)²
	D.10!/(5! × 4!)

Ans: 10!/(5!)²
The coefficient of xr 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

Q17.	If A and B are the coefficient of xn in the expansion (1 + x)²ⁿ and (1 + x)^2n-1 respectively, then A/B equals
	A.1
	B.2
	C.1/2
	D.1/n

Ans: 2
A/B = ²ⁿC_n/^2n-1C_n
= {(2n)!/(n! × n!)}/{(2n – 1)!/(n! × (n – 1!))}
= {2n(2n – 1)!/(n(n – 1)! × n!)}/{(2n – 1)!/(n! × (n – 1!))}
= 2
So, A/B = 2

Q18.	(1.1)¹⁰⁰⁰⁰ is _____ 1000
	A.greater than
	B.less than
	C.equal to
	D.None of these

Ans: greater than
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

Q19.	If n is a positive integer, then (√3+1)²ⁿ + (√3−1)²ⁿ is
	A.an odd positive integer
	B.none of these
	C.an even positive integer
	D.not an integer

Ans: an even positive integer
Since n is a positive integer, assume n = 1
(√3 + 1)² + (√3 – 1)²
= (3 + 2√3 + 1) + (3 – 2√3 + 1) {since (x + y)² = x² + 2xy + y²}
= 8, which is an even positive number.

Q20.	if y = 3x + 6x² + 10x³ + ………. then x =
	A.4/3 – {(1 × 4)/(3² × 2)}y² + {(1 × 4 × 7)/(3² ×3)}y³ – ………..
	B.-4/3 + {(1 × 4)/(3² × 2)}y² – {(1 × 4 × 7)/(3² ×3)}y³ + ………..
	C.4/3 + {(1 × 4)/(3² × 2)}y² + {(1 × 4 × 7)/(3² ×3)}y³ + ………..
	D.None of these

Ans: None of these
Given, y = 3x + 6x² + 10x³ + ……….
⇒ 1 + y = 1 + 3x + 6x² + 10x³ + ……….
⇒ 1 + y = (1 – x)^-3
⇒ 1 – x = (1 + y)^-1/3
⇒ x = 1 – (1 + y)^-1/3
⇒ x = (1/3)y – {(1 × 4)/(3² × 2)}y² + {(1 × 4 × 7)/(3² × 3!)}y³ – ………..