This set of Class 11 Maths Chapter 8 Multiple Choice Questions & Answers (MCQs) focuses on “Binomial Theorem”. These MCQs are created based on the latest CBSE syllabus and the NCERT curriculum, offering valuable assistance for exam preparation.

1. What is the general term of (x – y)^{xy}?

a) ^{x – y}C_{r} (x^{xy – r} . y^{r})

b) ^{xy}C_{r} (x^{x – y – r} . -y^{r})

c) ^{xy}C_{r} (x^{xy – r} . -y^{r})

d) ^{x – y}C_{r} (x^{x – y – r} . y^{r})

View Answer

Explanation: The general term of a binomial series is given by

^{n}C

_{r}a

^{n – r}b

^{r}.

Here a = x, b = -y and n = xy

Therefore the general term is given by

^{xy}C

_{r}(x

^{xy – r}. -y

^{r}).

2. What is the value of n, if the coefficients of the second term of (x – y)^{3} is equal to the third term of the expansion (x + y)^{n}?

a) –2

b) 3

c) 4

d) 5

View Answer

Explanation: Coefficient of the second term of (x – y)

^{3}is

^{3}C

_{1}and the coefficient of the third term of the expansion (x + y)

^{n}is

^{n}C

_{2}.

^{3}C

_{1}=

^{n}C

_{2}

3 = \(\frac{n!}{2!(n – 2)!}\)

6 = \(\frac{n(n-1)(n-2)!}{(n – 2)!}\)

6 = n

^{2}– n

n

^{2}– n – 6 = 0

n

^{2}– 3n + 2n – 6 = 0

(n – 3) (n + 2) = 0

n = 3, – 2

Since n cannot be negative, n = 3.

3. Which term will be the middle term of (xyz – x)^{2n}?

a) (n + 1)^{th} term

b) (n + 2)^{th} term

c) n^{th} term

d) (n – 1)^{th} term

View Answer

Explanation: Clearly 2n is an even number and the binomial has 2n + 1 terms.

The middle term for a binomial with even power, is the term equal to (n/2 + 1) where n is number of terms.

In this case, (2n/2 + 1) = n + 1.

4. What is the middle term of (4 + 2x)^{6}?

a) 11240 x^{2}

b) 10240 x^{3}

c) 12240 x^{4}

d) 10340 x^{4}

View Answer

Explanation: The middle term will be the 4

^{th}term

4

^{th}term =

^{6}C

_{3}(4)

^{6 – 3}(2x)

^{3}

= 20 (64) (8x

^{3})

= 10240 x

^{3}

5. What is the middle term of (x^{2} + x)^{3}?

a) 3x^{4}

b) 6x^{4}

c) 4x^{4}

d) 3x^{6}

View Answer

Explanation:

Since the power is odd, there will be even number of terms and two middle terms.

r = \(\frac{n + 1}{2}\) and r = \(\frac{n – 1}{2}\) Here n = 3

Therefore, r = 2 and r = 1.

When r = 2,

^{3}C

_{2}(x

^{2})

^{3 – 2}(x)

^{2}= 3x

^{4}

When r = 1,

^{3}C

_{1}(x

^{2})

^{3 – 1}(x)

^{1}= 3x

^{5}

6. Which of the following values of n are possible, if the middle term of (x + 3y)^{n} is the fifth term.

a) 6, 7 or 8

b) 7, 8 or 10

c) 7, 8 or 9

d) 8, 9 or 10

View Answer

Explanation: If n is the number of terms and is even, then the middle term is the \((\frac{n}{2} + 1)^{th}\) term.

Else if n is the number of terms and is odd, there are two middle terms which are the \((\frac{n + 1}{2})^{th}\) term and the \((\frac{n + 3}{2})^{th}\) term.

Case 1: \(\frac{n}{2} + 1\) = 5 ; n = 8

Case 2: \(\frac{n + 1}{2}\) = 5 ; n = 9

Case 3: \(\frac{n + 3}{2}\) = 5 ; n = 7

7. What is the even value of n, if the middle term of (a + b)^{2n – 3} is 11?

a) 12

b) 10

c) 20

d) 22

View Answer

Explanation: If 2n – 3 is even, then the middle term is the \((\frac{2n-3}{2}+1)^{th}\) term.

Else if 2n – 3 is odd, there are two middle terms which are the \((\frac{2n-3+1}{2})^{th}\) term and the \((\frac{2n-3+3}{2})^{th}\) term.

Case 1: \(\frac{2n-3}{2}\) + 1= 11 ; n = 11.5

Case 2: \(\frac{2n-2}{2}\) = 11 ; n = 12

Case 3: \(\frac{2n}{2}\) = 11 ; n = 11

8. What is the value of n if the middle term (x + 2y)^{2n + 1} is the 19^{th} term?

a) 33

b) 34

c) 35

d) 38

View Answer

Explanation: Clearly (2n + 1) is an odd number. Therefore this is a case of binomial with an odd power.

For a binomial expansion with odd power, there are two middle terms.

Case 1: \(\frac{n+1}{2}\) = 19

Therefore n = 37

Case 2: \(\frac{n+3}{2}\) = 19

Therefore n = 35

9. If the general term is ^{91}C_{2} x^{89}, what is the expansion?

a) (x)^{91}

b) (x – 2)^{90}

c) (x – 1)^{91}

d) (x + 1)^{90}

View Answer

Explanation: The general term of an expansion is

^{n}C

_{r}x

^{n – r}y

^{r}.

Clearly here n is 91 and the first term is x raised to the power 89.

The second term is raised to power 2.

y

^{2}= 1

y = +1 or -1

Therefore the expansion can either be (x + 1)

^{91}or (x – 1)

^{91}.

10. What is the middle term of (xyz + 3)^{80}?

a) ^{80}C_{41} (xyz)^{41} (3)^{39}

b) ^{80}C_{40} (xyz)^{40} (3)^{40}

c) ^{80}C_{39} (xyz)^{39} (3)^{40}

d) ^{80}C_{41} (xyz)^{41} (3)^{40}

View Answer

Explanation: Since the power is even, there are odd number of terms.

The middle term is the \((\frac{n}{2} + 1)^{th}\) term.

= \((\frac{80}{2} + 1)^{th}\) term

= 41

^{st}term

The 41

^{st}term =

^{80}C

_{40}(xyz)

^{40}(3)

^{40}

11. What is the coefficient of the middle term of (z + y)^{3x}, if 3x is considered to be even and the middle term is the 4^{th} term?

a) ^{7}C_{3}

b) ^{6}C_{2}

c) ^{6}C_{3}

d) ^{7}C_{2}

View Answer

Explanation: Since the middle term is the fourth term

Considering 3x to be even, \((\frac{3x + 1}{2})^{th}\) term = 4

x = 7/3

Therefore, the fourth term coefficients are

^{7}C

_{3}

12. What is the fourth term of (x – 5y)^{96}?

a) 125 ^{96}C_{3} x^{93} y^{3}

b) 625 ^{96}C_{3} x^{93} y^{4}

c) 625 ^{96}C_{4} x^{92} y^{4}

d) 125 ^{96}C_{4} x^{92} y^{4}

View Answer

Explanation: T

_{r + 1}=

^{n}C

_{r}x

^{n – r}y

^{r}

Here first term is 4 and second term is 5y.

n = 96

r = 3

Therefore, T

_{r + 1}=

^{96}C

_{3}x

^{96 – 3}(5y)

^{3}

= 125

^{96}C

_{3}x

^{93}y

^{3}

**More MCQs on Class 11 Maths Chapter 8:**

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