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

Do you need some help in preparing for your upcoming Class 11 Maths exams? We’ve compiled a list of MCQ on Binomial Theorem Class 11 MCQs Questions with Answers to get you started with the subject. You can download NCERT MCQ Questions for Class 11 Maths Chapter 8 Binomial Theorem with Answers Pdf free download, and learn how smart students prepare well ahead with MCQ Questions for Class 11 Maths with Answers.

## Binomial Theorem Class 11 MCQs Questions with Answers

Solving the MCQ Questions of Binomial Theorem Class 11 with answers can help you understand the concepts better.

Question 1.
The number (101)100 – 1 is divisible by
(a) 100
(b) 1000
(c) 10000
(d) 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

Question 2.
The value of -1° is
(a) 1
(b) -1
(c) 0
(d) None of these

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

Question 3.
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

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

Question 4.
The number 111111 ………….. 1 (91 times) is
(a) not an odd number
(b) none of these
(c) not a prime
(d) an even number

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

Question 5.
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) nth term
(d) (n/2 – 1)th term

Answer: (a) (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

Question 6.
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

Answer: (c) {(n + 1) × (n + 2)}/2
Total number of terms in (2x + 3y – 4z)ⁿ is
= n+3-1C3-1
= n+2C2
= {(n + 1) × (n + 2)}/2

Question 7.
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

A/B = ²ⁿCn/ 2n-1Cn
= {(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

Question 8.
The coefficient of y in the expansion of (y² + c/y)5 is
(a) 29c
(b) 10c
(c) 10c³
(d) 20c²

We have,
Tr+1 = 5Cr ×(y²)5-r × (c/y)r
⇒ Tr+1 = 5Cr × y10-3r × cr
For finding the coefficient of y,
⇒ 10 – 3r = 1
⇒ 33r = 9
⇒ r = 3
So, the coefficient of y = 5C3 × c³
= 10c³

Question 9.
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

Let x-4 occurs in (r + 1)th term.
Now, Tr+1 = 10Cr × (3/2)10-r ×(-3/x²)r
⇒ Tr+1 = 10Cr × (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

Question 10.
If n is a positive integer, then 9n+1 – 8n – 9 is divisible by
(a) 8
(b) 16
(c) 32
(d) 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

Question 11.
The general term of the expansion (a + b)ⁿ is
(a) Tr+1 = ⁿCr × ar × br
(b) Tr+1 = ⁿCr × ar × bn-r
(c) Tr+1 = ⁿCr × an-r× bn-r
(d) Tr+1 = ⁿCr × an-r × br

Answer: (d) Tr+1 = ⁿCr × an-r × br
The general term of the expansion (a + b)ⁿ is
Tr+1 = ⁿCr × an-r × br

Question 12.
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) nth term
(d) (n/2 – 1)th term

Answer: (a) (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

Question 13.
The smallest positive integer for which the statement 3n+1 < 4ⁿ is true for all
(a) 4
(b) 3
(c) 1
(d) 2

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

Question 14.
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

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+3-1C3-1
= 99C2
= (99 × 98)/2
= 4851

Question 15.
if n is a positive ineger then 2³ⁿ – 7n – 1 is divisible by
(a) 7
(b) 9
(c) 49
(d) 81

Given, 2³ⁿ – 7n – 1 = 23×n – 7n – 1
= 8ⁿ – 7n – 1
= (1 + 7)ⁿ – 7n – 1
= {ⁿC0 + ⁿC1 7 + ⁿC2 7² + …….. + ⁿCn 7ⁿ} – 7n – 1
= {1 + 7n + ⁿC2 7² + …….. + ⁿCn 7ⁿ} – 7n – 1
= ⁿC2 7² + …….. + ⁿCn 7ⁿ
= 49(ⁿC2 + …….. + ⁿCn 7n-2)
which is divisible by 49
So, 2³ⁿ – 7n – 1 is divisible by 49

Question 16.
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!)

The coefficient of xr in the expansion of (1 + x)10 is 10Cr and 10Cr is maximum for r = 10/ = 5
Hence, the greatest coefficient = 10C5
= 10!/(5!)²

Question 17.
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

A/B = ²ⁿCn/2n-1Cn
= {(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

Question 18.
(1.1)10000 is _____ 1000
(a) greater than
(b) less than
(c) equal to
(d) None of these

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

Question 19.
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

Answer: (c) 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.

Question 20.
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