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## Binomial Theorem Class 11 MCQs Questions with Answers

Students are advised to solve the Binomial Theorem Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Binomial Theorem Class 11 with answers will boost your confidence thereby helping you score well in the exam.

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Question 1.

The number (101)^{100} – 1 is divisible by

(a) 100

(b) 1000

(c) 10000

(d) All the above

## Answer

Answer: (d) All the above

Given, (101)^{100} – 1 = (1 + 100)^{100} – 1

= [^{100}C_{0} + ^{100}C_{1} × 100 + ^{100}C_{2} × (100)² + ……….+ ^{100}C_{100} × (100)^{100}] – 1

= 1 + [^{100}C_{1} × 100 + ^{100}C_{2} × (100)² + ……….+ ^{100}C_{100} × (100)^{100}] – 1

= ^{100}C_{1} × 100 + ^{100}C_{2} × (100)² + ……….+ ^{100}C_{100} × (100)^{100}

= 100 × 100 + ^{100}C_{2} × (100)² + ……….+ ^{100}C_{100} × (100)^{100}

= (100)² + ^{100}C_{2} × (100)² + ……….+ ^{100}C_{100} × (100)^{100}

= (100)² [1 + ^{100}C_{2} + ……….+ ^{100}C_{100} × (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

## Answer

Answer: (b) -1

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

## Answer

Answer: (b) 6

Given, T_{4} = 5/2

⇒ T_{3+1} = 5/2

⇒ ⁿC_{3} × (ax)^{n-3} × (1/x)³ = 5/2

⇒ ⁿC_{3} × a^{n-3} × x^{n-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

## Answer

Answer: (c) not a prime

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) n^{th} term

(d) (n/2 – 1)^{th} term

## Answer

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

Answer: (c) {(n + 1) × (n + 2)}/2

Total number of terms in (2x + 3y – 4z)ⁿ is

= ^{n+3-1}C_{3-1}

= ^{n+2}C_{2}

= {(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

## Answer

Answer: (b) 2

A/B = ²ⁿC_{n}/ ^{2n-1}C_{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

Question 8.

The coefficient of y in the expansion of (y² + c/y)^{5} is

(a) 29c

(b) 10c

(c) 10c³

(d) 20c²

## Answer

Answer: (c) 10c³

We have,

T_{r+1} = ^{5}C_{r} ×(y²)^{5-r} × (c/y)^{r}

⇒ T_{r+1} = ^{5}C_{r} × y^{10-3r} × c^{r}

For finding the coefficient of y,

⇒ 10 – 3r = 1

⇒ 33r = 9

⇒ r = 3

So, the coefficient of y = ^{5}C_{3} × 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

## Answer

Answer: (d) None of these

Let x^{-4} occurs in (r + 1)th term.

Now, T_{r+1} = ^{10}C_{r} × (3/2)^{10-r} ×(-3/x²)^{r}

⇒ T_{r+1} = ^{10}C_{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} = ^{10}C_{2} × (3/2)^{10-2} × (-3)^{2}

= ^{10}C_{2} × (3/2)^{8} × (-3)^{2}

= 45 × (3/2)^{8} × 9

= (3^{12} × 5)/2^{8}

Question 10.

If n is a positive integer, then 9^{n+1} – 8n – 9 is divisible by

(a) 8

(b) 16

(c) 32

(d) 64

## Answer

Answer: (d) 64

Let n = 1, then

9^{n+1} – 8n – 9 = 9^{1+1} – 8 × 1 – 9 = 9² – 8 – 9 = 81 – 17 = 64

which is divisible by 64

Let n = 2, then

9^{n+1} – 8n – 9 = 9^{2+1} – 8 × 2 – 9 = 9³ – 16 – 9 = 729 – 25 = 704 = 11 × 64

which is divisible by 64

So, for any value of n, 9^{n+1} – 8n – 9 is divisible by 64

Question 11.

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}

## Answer

Answer: (d) 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}

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) n^{th} term

(d) (n/2 – 1)^{th} term

## Answer

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 3^{n+1} < 4ⁿ is true for all

(a) 4

(b) 3

(c) 1

(d) 2

## Answer

Answer: (a) 4

Given statement is: 3^{n+1} < 4ⁿ is

Let n = 1, then

3^{1+1} < 4^{1} = 3² < 4 = 9 < 4 is false

Let n = 2, then

3^{2+1} < 4² = 3³ < 4² = 27 < 16 is false

Let n = 3, then

3^{3+1} < 4³ = 3^{4} < 4³ = 81 < 64 is false

Let n = 4, then

3^{4+1} < 4^{4} = 3^{5} < 4^{4} = 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

## Answer

Answer: (b) 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+3-1}C_{3-1}

= ^{99}C_{2}

= (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

## Answer

Answer: (c) 49

Given, 2³ⁿ – 7n – 1 = 2^{3×n} – 7n – 1

= 8ⁿ – 7n – 1

= (1 + 7)ⁿ – 7n – 1

= {ⁿC_{0} + ⁿC_{1} 7 + ⁿC_{2} 7² + …….. + ⁿC_{n} 7ⁿ} – 7n – 1

= {1 + 7n + ⁿC_{2} 7² + …….. + ⁿC_{n} 7ⁿ} – 7n – 1

= ⁿC_{2} 7² + …….. + ⁿC_{n} 7ⁿ

= 49(ⁿC_{2} + …….. + ⁿC_{n} 7^{n-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!)

## Answer

Answer: (b) 10!/(5!)²

The coefficient of xr in the expansion of (1 + x)^{10} is ^{10}C_{r} and ^{10}C_{r} is maximum for r = 10/ = 5

Hence, the greatest coefficient = ^{10}C_{5}

= 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

## Answer

Answer: (b) 2

A/B = ²ⁿC_{n}/^{2n-1}C_{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

Question 18.

(1.1)^{10000} is _____ 1000

(a) greater than

(b) less than

(c) equal to

(d) None of these

## Answer

Answer: (a) greater than

Given, (1.1)^{10000} = (1 + 0.1)^{10000}

= ^{10000}C_{0} + ^{10000}C_{1} ×(0.1) + ^{10000}C_{2} × (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

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

## Answer

Answer: (d) 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 – ………..

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