**CBSE Class 11 Maths – MCQ and Online Tests – Unit 8 – Binomial Theorem**

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**CBSE Class 11 Maths – MCQ and Online Tests – Unit 8 – Binomial Theorem**

Question 1.

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 2.

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 3.

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 з 98)/2

= 4851

Question 4.

The greatest coefficient in the expansion of (1 + x)^{10} is

(a) 10!/(5!)

(b) 10!/(5!)В

(c) 10!/(5! з 4!)В

(d) 10!/(5! з 4!)

## Answer

Answer: (b) 10!/(5!)В

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!)В

Question 5.

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

(a) -1670 x^{9} з yГ

(b) -7160 x^{9} з yГ

(c) -1760 x^{9} з yГ

(d) -1607 x^{9} з yГ

## Answer

Answer: (c) -1760 x^{9} з yГ

Hint:

4th term in (x – 2y)^{12} = T_{4}

= T_{3+1}

= ^{12}C_{3} (x)^{12-3} з(-2y)Г

= ^{12}C_{3} x^{9} з(-8yГ)

= {(12з11з10)/(3з2з1)} з x^{9} з(-8yГ)

= -(2з11з10з8) з x^{9} з yГ

= -1760 x^{9} з yГ

Question 6.

If n is a positive integer, then (v3+1)^{2n+1} + (v3-1)^{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

(v3+1)Г + (v3-1)Г

= {3v3 + 1 + 3v3(v3 + 1)} + {3v3 – 1 – 3v3(v3 – 1)}

= 3v3 + 1 + 9 + 3v3 + 3v3 – 1 – 9 + 3v3

= 12v3, which is an irrational number.

Question 7.

If the third term in the binomial expansion of (1 + x)^{m} is (-1/8)xВ 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}xВ + ……..

Now, {m(m – 1)/2}xВ = (-1/8)xВ

? m(m – 1)/2 = -1/8

? 4mВ – 4m = -1

? 4mВ – 4m + 1 = 0

? (2m – 1)В = 0

? 2m – 1 = 0

? m = 1/2

Question 8.

The greatest coefficient in the expansion of (1 + x)^{10} is

(a) 10!/(5!)

(b) 10!/(5!)В

(c) 10!/(5! з 4!)В

(d) 10!/(5! з 4!)

## Answer

Answer: (b) 10!/(5!)В

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!)В

Question 9.

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

(a) (2n)!/n!

(b) (2n)!/(n!)В

(c) (2n)!/{2з(n!)В}

(d) None of these

## Answer

Answer: (b) (2n)!/(n!)В

Hint:

We have,

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

= (1 + x)^{2n}

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

Question 10.

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} з a^{n-0} з b^{0} = 729

? a^{n} = 729 ……………. 1

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

? n

a^{n-1} з b = 7290 ……. 2

T_{3} = ^{n}C_{2} з a^{n-2} з bВ = 30375

? {n(n-1)/2}

a^{n-2} з bВ = 30375 ……. 3

Now equation 2/equation 1

n

a^{n-1} з b/a^{n} = 7290/729

? nзb/n = 10 ……. 4

Now equation 3/equation 2

{n(n-1)/2}

a^{n-2} з bВ /n

a^{n-1} з b = 30375/7290

? b(n-1)/2a = 30375/7290

? b(n-1)/a = (30375з2)/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 з 5/3 = 10

? 5n = 30

? n = 30/5

? n = 6

So, the value of n is 6

Question 11.

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

(a) 10c

(b) 10cВ

(c) 10cГ

(d) None of these

## Answer

Answer: (c) 10cГ

Hint:

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

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

= ^{5}C_{r} з y^{10-3r} з C^{r}

Now, 10 – 3r = 1

? 3r = 9

? r = 3

So, the coefficient of y = ^{5}C_{3} з cГ = 10cГ

Question 12:

(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} з (0.1) + ^{10000}C_{2} з(0.1)В + other +ve terms

= 1 + 10000з(0.1) + other +ve terms

= 1 + 1000 + other +ve terms

> 1000

So, (1.1)^{10000} is greater than 1000

Question 13.

If n is a positive integer, then (v5+1)^{2n} + 1 – (v5-1)^{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

(v5+1)В + 1 – (v5-1)В + 1

= (5 + 2v5 + 1) + 1 – (5 – 2v5 + 1) + 1 {since (x+y)В = xВ + 2xy + yВ}

= 4v5 + 2, which is not an integer

Question 14.

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^{3зn} – 7n – 1

= 8^{n} – 7n – 1

= (1 + 7)^{n} – 7n – 1

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

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

= ^{n}C_{2} 7В + …….. + ^{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 15.

If the third term in the binomial expansion of (1 + x)^{m} is (-1/8)xВ 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}xВ + ……..

Now, {m(m – 1)/2}xВ = (-1/8)xВ

? m(m – 1)/2 = -1/8

? 4mВ – 4m = -1

? 4mВ – 4m + 1 = 0

? (2m – 1)В = 0

? 2m – 1 = 0

? m = 1/2

Question 16.

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}

Question 17.

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} з (7^{1/2})^{37-r} з (5^{1/3})^{r}

? T_{r+1} = ^{37}C_{r} з 7(^{37-r})^{/2} з (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 18.

If a and п are the roots of the equation xВ – x + 1 = 0 then the value of a^{2009} + п^{2009} is

(a) 0

(b) 1

(c) -1

(d) 10

## Answer

Answer: (b) 1

Hint:

Given, xВ – x + 1 = 0

Now, by Shridharacharya formula, we get

x = {1 Б v(1 – 4з1з1) }/2

? x = {1 Б v(1 – 4) }/2

? x = {1 Б v(-3)}/2

? x = {1 Б v(3 з -1)}/2

? x = {1 Б v3 з v-1}/2

? x = {1 Б iv3}/2 {since i = v-1}

? x = {1 + iv3}/2, {–1 – iv3}/2

? x = -{-1 – iv3}/2, -{-1 + iv3}/2

? x = w, wВ {since w = {-1 + iv3}/2 and wВ = {-1 – iv3}/2 }

Hence, a = -w, п = wВ

Again we know that wГ = 1 and 1 + w + wВ = 0

Now, a^{2009} + п^{2009} = a^{2007} з aВ + п^{2007} з пВ

= (-w)^{2007} з (-w)В + (-wВ)^{2007} з (-wВ)В {since 2007 is multiple of 3}

= -(w)^{2007} з (w)В – (wВ)^{2007} з (w^{4})

= -1 з wВ – 1 з wГ з w

= -1 з wВ – 1 з 1 з w

= -wВ – w

= 1 {since 1 + w + wВ = 0}

So, a^{2009} + п^{2009} = 1

Question 19.

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

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

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

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

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

## Answer

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

Hint:

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

T_{r+1} = ^{n}C_{r} з a^{n-r} з b^{r}

Question 20.

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

(a) 1

(b) (-1)^{n}

(c) n

(d) n+1

## Answer

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

Hint:

We know that

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

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

= (-1)^{n}