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.

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## Binomial Theorem Class 11 MCQs Questions with Answers

**MCQ On Binomial Theorem Class 11 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

**Binomial Theorem MCQ 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

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**MCQ On Binomial Theorem 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

**MCQs On Binomial Theorem 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.

**Binomial Theorem Class 11 MCQ 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

**Binomial Theorem MCQ Pdf 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

**Binomial Theorem MCQs With Answers Pdf 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

**Binomial Theorem MCQ With Solution 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

**Maths MCQs For Class 11 With Answers Pdf 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

**MCQ Questions For Class 11 Maths With Answers Pdf 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}

**MCQ Questions For Class 11 Maths With Answers 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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