This set of Class 11 Maths Chapter 8 Multiple Choice Questions & Answers (MCQs) focuses on “Binomial Theorem for Positive Integral Index”.

1. What is the coefficient of x^{2}y^{2} in (x + 1)^{2} . (x + 1)^{3}?

a) 1

b) 5

c) 2

d) 10

View Answer

Explanation: We know that (a + b)

^{2}= a

^{2}+ 2ab + b

^{2}

(a + b)

^{3}= a

^{3}+ 3ab

^{2}+ 3a

^{2}b + b

^{2}

Using these formulae, we get

P(x) = (x

^{2}+ 2xy + y

^{2})(x

^{3}+ 3xy

^{2}+ 3x

^{2}y + y

^{2})

P(x) = 3xy

^{4}+ 9x

^{2}y

^{3}+ 10x

^{3}y

^{2}+ 5x

^{4}y + x

^{5}+ y

^{4}+ 2xy

^{3}+ x

^{2}y

^{2}

The coefficient of x

^{2}y

^{2}in (x + 1)

^{2}. (x + 1)

^{3}is 1.

2. What is the remainder when 8^{48} is divided by 63?

a) 4

b) 2

c) 1

d) 7

View Answer

Explanation: 8

^{58}can be written as (8

^{2})

^{24}.

8

^{48}= (64)

^{24}

8

^{48}= (63 + 1)

^{24}

We know that (60 + 1)

^{24}= \(\Sigma_{r = 0}^{r = 24}\)(24Cr 63

^{24 – r}1

^{r})

= 24C

_{0}63

^{24}4

^{0}+ 24C

_{1}63

^{23}4

^{1}+….+24C

_{23}63

^{1}4

^{23}+ 24C

_{24}63

^{0}1

^{24}

= 63 x k + 1

Therefore, the remainder will be 1.

3. What is the remainder when 4^{103} is divided by 17?

a) 10

b) 14

c) 13

d) 16

View Answer

Explanation: 4

^{103}= 4 x 4

^{102}

4

^{103}= 4 x (4

^{2})

^{51}

4

^{103}= 4 x (16)

^{51}

4

^{103}= 4 x (17 – 1)

^{51}

4

^{103}= 4 x \(\Sigma_{r = 0}^{r = 51}\)(51Cr 17

^{24 – r}(-1)

^{r}

4

^{103}= 4 x [51C

_{0}17

^{51}(-1)

^{0}+ 51C

_{1}17

^{51}(-1)

^{1}+….+ 51C

_{50}17

^{1}(-1)

^{50}+ 51C

_{51}17

^{0}(-1)

^{51}]

4

^{103}= 4 x 17 x k – 1

The remainder = 17 – 1

Remainder = 16.

4. What is the integral part of (√3 + 1)^{8}?

a) 1558

b) 1551

c) 1552

d) 1556

View Answer

Explanation: By binomial expansion,

(√3 + 1)

^{7}= \(\Sigma_{r = 0}^{r = 7}\)(7Cr √3

^{7 – r}(1)

^{r})

Whenever, r is an even number, 8 – r will also be even. Then √3 will also have an even power and thereby be integral.

Integral parts = 8C

_{0}(√3)

^{0}+ 8C

_{2}(√3)

^{2}+ 8C

_{4}(√3)

^{4}+ 8C

_{6}(√3)

^{6}+ 8C

_{8}(√3)

^{8}

Integral parts = 1 + 28 x 3 + 70 x 9 + 28 x 27 + 1 x 81

Integral part = 1552.

5. What is the expansion of (x + y)^{1000}?

a) \(\Sigma_{r = 0}^{r = 1000}\)(1000Cr x^{r – 1000} y^{r})

b) \(\Sigma_{r = 0}^{r = 1000}\)(100Cr x^{1000 – r} y^{r})

c) \(\Sigma_{r = 0}^{r = 999}\)(1000Cr x^{r – 1000} y^{r})

d) \(\Sigma_{r = 0}^{r = 999}\)(1000Cr x^{1000 – r} y^{r})

View Answer

Explanation: The expansion can be done using binomial theorem.

(x + y)

^{1000}= 1000C

_{0}x

^{1000}y

^{0}+ 1000C

_{1}x

^{999}y

^{1}+….+ 1000C

_{999}x

^{1}y

^{999}+ 1000C

_{1000}x

^{0}y

^{1000}

This can also be written as,

(x + y)

^{1000}= \(\Sigma_{r = 0}^{r = 1000}\)(100Cr x

^{1000 – r}y

^{r}).

6. What is the real part of (11 + i)^{3}?

a) 1331

b) 1332

c) 1328

d) 1329

View Answer

Explanation: (11 + i)

^{3}= 11

^{3}+ 3.11

^{2}.i +3.i

^{2}.11 +i

^{3}

= 1331 + 363i – 3 – i

= 1328 + 365i.

7. What are the coefficients of the first and the last term of (a + b)^{n}?

a) 2

b) 1

c) Coefficients depend on n

d) 3

View Answer

Explanation: The coefficient of the first term and last term is same. The first term is nC

_{1}a

^{n}and the last term is nC

_{0}b

^{n}unless, a and b are numbers that change the value of the coefficient.

8. What is the remainder when (4)^{2n + 1} is divided by 5?

a) 4

b) 1

c) 2

d) 3

View Answer

Explanation: The powers of four follow the given order:

4

^{1}= 4

4

^{2}= 16

4

^{3}= 64

4

^{4}= 256

4

^{5}= 1024 and so on.

Odd powers of 4, have the number 4 in the units place. When 5 divides the nearest ten, 4 will be obtained as the remainder each time.

9. What is the expansion of the series (xy + 2)^{2}?

a) x^{2} + y^{2} + 4

b) xy^{2} + 4 +2xy

c) x^{2}y^{2} + 2xy + 4

d) x^{2}y^{2} + 4xy + 4

View Answer

Explanation: (a + b)

^{2}can be expanded using binomial theorem to get:

(a + b)

^{2}= a

^{2}+ 2ab + b

^{2}

Here, a = xy and b = 2

Therefore, (xy + 2)

^{2}= (xy)

^{2}+ 2(xy)(2) + (2)

^{2}

(xy + 2)

^{2}= x

^{2}y

^{2}+ 4xy + 4.

10. What is the answer of \(\frac{x^2+y^2+2xy}{x^2-y^2}\)?

a) (x – y) (x + y)^{-2}

b) (x + y) (x – y)^{-2}

c) (x + y) (x – y)^{-1}

d) (x – y) (x + y)^{-1}

View Answer

Explanation: x

^{2}+ y

^{2}+ 2xy is the expansion of (x + y)

^{2}

x

^{2}– y

^{2}can be written as (x – y)(x + y)

Substituting in the fraction we get, \(\frac{(x + y)^2}{(x – y)(x + y)}\).

After cancelling the terms we get, \(\frac{x^2+y^2+2xy}{x^2-y^2}\) = (x + y) (x – y)

^{-1}.

11. What is the value of \(\frac{7^3+2^3+84}{7^2-2^2}\) ?

a) 9 \(\frac{1}{2}\)

b) 9 \(\frac{2}{3}\)

c) 9 \(\frac{1}{3}\)

d) 9 \(\frac{1}{4}\)

View Answer

Explanation: Using binomial theorem we know that (a + b)

^{3}= a

^{3}+ 3ab

^{2}+ 3a

^{2}b + b

^{3}

Therefore, (7 + 2)

^{3}= 7

^{3}+ 2

^{3}+ (3 x 7 x 2

^{2}) + (3 x 2 x 7

^{2})

(9)

^{3}= 7

^{3}+ 2

^{3}+ 84 + (3 x 2 x 7

^{2})

729 = 7

^{3}+ 2

^{3}+ 84 + 294

7

^{3}+ 2

^{3}+ 84 = 435

Also 7

^{2}– 2

^{2}= (7 – 2)(7 + 2)

7

^{2}– 2

^{2}= (5)(9)

7

^{2}– 2

^{2}= 45

So \(\frac{7^3 + 2^3 + 84}{7^2-2^2}\) = 435 / 45

= 9 \(\frac{30}{45}\)

= 9 \(\frac{2}{3}\).

12. What is the value of \(\frac{101^3-99^3+2969703–3029697}{ 101^2 – 99^2}\)?

a) 1

b) 1/200

c) 1/100

d) 1/50

View Answer

Explanation: The numerator when simplified is of the form (101 – 99)

^{3}

The denominator can be simplified as (101 – 99)(101 + 99)

When we substitute in the numerator and denominator we get (2 x 2 x 2) / (2 x 200)

= 1/50.

13. What is the quotient when x^{4} + 4x^{3}y + 6x^{2}y + 4xy^{3} + y^{4} is divided by (x + y)?

a) (x + y)^{3}

b) x^{2} + y^{2}

c) (x + y)^{2}

d) (x + y)

View Answer

Explanation: Using binomial expansions properties, x

^{4}+ 4x

^{3}y + 6x

^{2}y + 4xy

^{3}+ y

^{4}can be written as

= 4C

_{0}x

^{4}y

^{0}+ 4C

_{1}x

^{3}y

^{1}+ 4C

_{2}x

^{2}y

^{2}+ 4C

_{3}x

^{1}y

^{2}+ 4C

_{4}x

^{0}y

^{4}

= (x + y)

^{4}

When divided by (x + y), we get (x + y)

^{3}.

14. What is the real part of (9 + 3i)^{2}?

a) 81

b) 90

c) 54

d) 72

View Answer

Explanation: Using binomial theorem (9 + 3i)

^{2}= 81 + 54i + 9i

^{2}

We know that i

^{2}= –1

Therefore, (9 + 3i)

^{2}= 81 + 54i – 9

(9 + 3i)

^{2}= 72 + 54i

Real part = 72.

**Sanfoundry Global Education & Learning Series – Mathematics – Class 11**.

To practice all chapters and topics of class 11 Mathematics, __ here is complete set of 1000+ Multiple Choice Questions and Answers__.

**Related Posts:**

- Practice Class 11 - Chemistry MCQs
- Practice Class 11 - Biology MCQs
- Practice Class 11 - Physics MCQs
- Practice Class 12 - Mathematics MCQs
- Check Class 11 - Mathematics Books