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## Binomial Theorem Class 11 MCQs Questions with Answers

Question 1.

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

(a) 10c

(b) 10c²

(c) 10c³

(d) None of these

## Answer

Answer: (c) 10c³

Hint:

Given, binomial expression is (y² + c/y)^{5}

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

= ^{5}C_{r} × y^{10-3r} × C^{r}

Now, 10 – 3r = 1

⇒ 3r = 9

⇒ r = 3

So, the coefficient of y = ^{5}C_{3} × c³ = 10c³

Question 2:

(1.1)^{10000} is _____ 1000

(a) greater than

(b) less than

(c) equal to

(d) None of these

## Answer

Answer: (a) greater than

Hint:

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 3.

The fourth term in the expansion (x – 2y)^{12} is

(a) -1670 x^{9} × y³

(b) -7160 x^{9} × y³

(c) -1760 x^{9} × y³

(d) -1607 x^{9} × y³

## Answer

Answer: (c) -1760 x^{9} × y³

Hint:

4th term in (x – 2y)^{12} = T_{4}

= T_{3+1}

= ^{12}C_{3} (x)^{12-3} ×(-2y)³

= ^{12}C_{3} x^{9} ×(-8y³)

= {(12×11×10)/(3×2×1)} × x^{9} ×(-8y³)

= -(2×11×10×8) × x^{9} × y³

= -1760 x^{9} × y³

Question 4.

If n is a positive integer, then (√3+1)^{2n+1} + (√3−1)^{2n+1} is

(a) an even positive integer

(b) a rational number

(c) an odd positive integer

(d) an irrational number

## Answer

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.

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Question 5.

If the third term in the binomial expansion of (1 + x)^{m} is (-1/8)x² then the rational value of m is

(a) 2

(b) 1/2

(c) 3

(d) 4

## Answer

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

Question 6.

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

Hint:

The coefficient of x^{r} 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 7.

The coefficient of x^{n} in the expansion of (1 – 2x + 3x² – 4x³ + ……..)^{-n} is

(a) (2n)!/n!

(b) (2n)!/(n!)²

(c) (2n)!/{2×(n!)²}

(d) None of these

## Answer

Answer: (b) (2n)!/(n!)²

Hint:

We have,

(1 – 2x + 3x² – 4x³ + ……..)^{-n} = {(1 + x)^{-2}}^{-n}

= (1 + x)^{2n}

So, the coefficient of x^{n}C_{3} = ^{2n}C_{n} = (2n)!/(n!)²

Question 8.

The value of n in the expansion of (a + b)^{n} if the first three terms of the expansion are 729, 7290 and 30375, respectively is

(a) 2

(b) 4

(c) 6

(d) 8

## Answer

Answer: (c) 6

Hint:

Given that the first three terms of the expansion are 729, 7290 and 30375 respectively.

Now T_{1} = ^{n}C_{0} × a^{n-0} × b^{0} = 729

⇒ a^{n} = 729 ……………. 1

T_{2} = ^{n}C_{1} × a^{n-1} × b^{1} = 7290

⇒ n

a^{n-1} × b = 7290 ……. 2

T_{3} = ^{n}C_{2} × 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/a^{n} = 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

Question 9.

If α and β are the roots of the equation x² – x + 1 = 0 then the value of α^{2009} + β^{2009} is

(a) 0

(b) 1

(c) -1

(d) 10

## Answer

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, α^{2009} + β^{2009} = α^{2007} × α² + β^{2007} × β²

= (-w)^{2007} × (-w)² + (-w²)^{2007} × (-w²)² {since 2007 is multiple of 3}

= -(w)^{2007} × (w)² – (w²)^{2007} × (w^{4})

= -1 × w² – 1 × w³ × w

= -1 × w² – 1 × 1 × w

= -w² – w

= 1 {since 1 + w + w² = 0}

So, α^{2009} + β^{2009} = 1

Question 10.

The general term of the expansion (a + b)^{n} is

(a) T_{r+1} = ^{n}C_{r} × a^{r} × b^{r}

(b) T_{r+1} = ^{n}C_{r} × a^{r} × b^{n-r}

(c) T_{r+1} = ^{n}C_{r} × a^{n-r} × b^{n-r}

(d) T_{r+1} = ^{n}C_{r} × a^{n-r} × b^{r}

## Answer

Answer: (d) T_{r+1} = ^{n}C_{r} × a^{n-r} × b^{r}

Hint:

The general term of the expansion (a + b)^{n} is

T_{r+1} = ^{n}C_{r} × a^{n-r} × b^{r}

Question 11.

The coefficient of x^{n} in the expansion (1 + x + x² + …..)^{-n} is

(a) 1

(b) (-1)^{n}

(c) n

(d) n+1

## Answer

Answer: (b) (-1)^{n}

Hint:

We know that

(1 + x + x² + …..)^{-n} = (1 – x)^{-n}

Now, the coefficient of x = (-1)^{n} × ^{n}C_{n}

= (-1)^{n}

Question 12.

If n is a positive integer, then (√5+1)^{2n} + 1 − (√5−1)^{2n} + 1 is

(a) an odd positive integer

(b) not an integer

(c) none of these

(d) an even positive integer

## Answer

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

Question 13.

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

Hint:

In the expansion of (a + b)^{n},

if n is even then the middle term is (n/2 + 1)^{th} term

Question 14.

In the expansion of (a + b)^{n}, if n is odd then the number of middle term is/are

(a) 0

(b) 1

(c) 2

(d) More than 2

## Answer

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

Question 15.

if n is a positive ineger then 2^{3n}n – 7n – 1 is divisible by

(a) 7

(b) 9

(c) 49

(d) 81

## Answer

Answer: (c) 49

Hint:

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

= 8^{n} – 7n – 1

= (1 + 7)^{n} – 7n – 1

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

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

= ^{n}C_{2} 7² + …….. + ^{n}C_{n} 7^{n}

= 49(^{n}C_{2} + …….. + ^{n}C_{n} 7^{n-2})

which is divisible by 49

So, 2^{3n} – 7n – 1 is divisible by 49

Question 16.

In the binomial expansion of (7^{1/2} + 5^{1/3})^{37}, the number of integers are

(a) 2

(b) 4

(c) 6

(d) 8

## Answer

Answer: (c) 6

Hint:

Given, (7^{1/2} + 5^{1/3})^{37}

Now, general term of this binomial T_{r+1} = ^{37}C_{r} × (7^{1/2})^{37-r} × (5^{1/3})^{r}

⇒ T_{r+1} = ^{37}C_{r} × 7(^{37-r})^{/2} × (5)^{r/3}

This General term will be an integer if ^{37}C_{r} is an integer, 7(^{37-r})^{/2} is an integer and (5)^{r/3} is an integer.

Now, ^{37}C_{r} will always be a positive integer.

Since ^{37}C_{r} denotes the number of ways of selecting r things out of 37 things, it can not be a fraction.

So, ^{37}C_{r} is an integer.

Again, 7(^{37-r})^{/2}C_{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}

Question 17.

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

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+3-1}C_{3-1}

= ^{99}C_{2}

= (99 × 98)/2

= 4851

Question 18.

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

Hint:

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

r = 10/2 = 5

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

= 10!/(5!)²

Question 19.

If the third term in the binomial expansion of (1 + x)^{m} is (-1/8)x² then the rational value of m is

(a) 2

(b) 1/2

(c) 3

(d) 4

## Answer

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

Question 20.

In the binomial expansion of (a + b)^{n}, the coefficient of fourth and thirteenth terms are equal to each other, then the value of n is

(a) 10

(b) 15

(c) 20

(d) 25

## Answer

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

⇒ ^{n}C_{3} = ^{n}C_{12}

This is possible when n = 15

Because ^{15}C_{13} = ^{15}C_{12}

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