Class 11 Maths Chapter 8 Binomial Theorem MCQs

Class 11 Maths Chapter 8 Binomial Theorem MCQs are provided here to assist students in achieving the highest possible score in the board exam in 2022-23. All of these multiple-choice questions for the Class 11 Maths exam are based on the NCERT curriculum and the most recent CBSE standards. As binomial theorem is one of the most significant topics in mathematics, understanding them will help you not only with board examinations but also with other exams. As a result, MCQs on binomial theorem is a great way to refresh your memory and practise at any time. To get all chapter-by-chapter MCQs for Class 11 Maths, click here.

MCQs on Class 11 Maths Chapter 8 Binomial Theorem

Check out the multiple-choice questions for Class 11 Maths Chapter 8 binomial theorem. Each MCQ contains four possible answers, but only one is correct.

Download PDF – Chapter 8 Binomial Theorem MCQs

Students must choose the best choice and compare their results to the ones provided on this page. This form of preparation can help you gain confidence to attempt any type of question in the exam. Also, make sure to look at the important questions for class 11 Maths.

1) The coefficient of the middle term in the expansion of (2+3x)⁴ is:

1. 5!
2. 6
3. 216
4. 8!

Answer: (c) 216

Explanation:

If the exponent of the expression is n, then the total number of terms is n+1.

Hence, the total number of terms is 4+1 = 5.

Hence, the middle term is the 3rd term.

We know that general term of (x+a)ⁿ is T_r+1 =ⁿC_r x^n-r a^r

Here, n=4, r=2

Therefore, T₃ = ⁴C₂.(2)².(3x)²

T₃ = (6).(4).(9x²)

T₃ = 216x².

Therefore, the coefficient of the middle term is 216.

2) The value of (126)^1/3 up to three decimal places is

1. 5.011
2. 5.012
3. 5.013
4. 5.014

Answer: (c) 5.013

Explanation:

(126)^⅓ can also be written as the cube root of 126.

Hence, (126)^⅓ is approximately equal to 5.013.

Hence, option (c) 5.013 is the correct answer.

3) If n is even in the expansion of (a+b)ⁿ, the middle term is:

1. n^th term
2. (n/2)^th term
3. [(n/2)-1]^th term
4. [(n/2)+1]^th term

Answer: (d) [(n/2)+1]^th term

Explanation:

In general, if “n” is the even in the expansion of (a+b)ⁿ, then the number of terms will be odd. (i.e) n+1.

Hence, the middle term of the expansion (a+b)ⁿ is [(n/2)+1]^th term.

4) The largest coefficient in the expansion of (1+x)¹⁰ is:

1. 10! / (5!)²
2. 10! / 5!
3. 10! / (5!×4!)²
4. 10! / (5!×4!)

Answer: (a) 10! / (5!)²

Explanation:

Given: (1+x)¹⁰

The greatest coefficient will always occur in the middle term.

Hence, the total number of terms in an expansion is 11. (i.e. 10+1 = 11)

Therefore, middle term = [(10/2) + 1] = 5+1 = 6th term.

We know that general term of (x+a)ⁿ is T_r+1 =ⁿC_r x^n-r a^r

Here, n=10, r=5

So, T6 = ¹⁰C₅.x⁵

Therefore, the coefficient of the greatest term = ¹⁰C₅ = 10!/(5!)².

So, option (a) 10!/(5!)² is the correct answer.

5) The coefficient of x³y⁴ in (2x+3y²)⁵ is

1. 360
2. 720
3. 240
4. 1080

Answer: (b) 720

Explanation:

Given: (2x+3y²)⁵

Therefore, the general form for the expression (2x+3y²)⁵ is T_r+1 = ⁵C_r. (2x)^r.(3y²)^5-r

Hence, T₃₊₁ = ⁵C₃ (2x)³.(3y²)^5-3

T₄ = ⁵C₃ (2x)³.(3y²)²

T₄ = ⁵C₃.8x³.9y⁴

On simplification, we get

T₄ = 720x³y⁴

Therefore, the coefficient of x³y⁴ in (2x+3y²)⁵ is 720.

6) The largest term in the expansion of (3+2x)⁵⁰, when x = ⅕ is

1. 6th term
2. 7th term
3. 8th term
4. None of the above

Answer: (a) 6th term

Explanation:

The greatest term in the expansion of (x+y)ⁿ is the kth term. Where k= [(n+1)y]/[x+y]..(1)

On comparing the given expression with the general form, x = 3, y=2x, n=50

Now, substitute the values in the given expression, we get

Hence, kth term = [(50+1)(2x)]/[3+2x]

When x = ⅕,

Kth term = [(51)(2(⅕))]/[3+2(⅕)] = 6

Hence, the 6th term is the largest term in the expansion of (3+2x)⁵⁰, when x = ⅕.

7) The coefficient of y in the expansion of (y²+(c/y))⁵ is:

1. 10c
2. 29c
3. 10c³
4. 20c³

Answer: (c) 10c³

Explanation: Given: (y²+(c/y))⁵

We know that general term of (x+a)ⁿ is T_r+1 =ⁿC_r x^n-r a^r

Here, n=5, r=?

(y²+(c/y))⁵ = ⁵C_r.(y²)^r.(c/y)^5-r

(y²+(c/y))⁵ = 5Cr. y^2r. (c^5-r/y^5-r)

On solving this, we get r = 2.

Hence, the coefficient of y = ⁵C₃.c³ = 10c³.

Therefore, option (c) 10c³ is the correct answer.

8) The fourth term in the expansion of (x-2y)¹² is:

1. -1760 x⁹ × y³
2. -1670 x⁹ × y³
3. -7160 x⁹ × y³
4. -1607 x⁹ × y³

Answer: (a) -1760 x⁹ × y³

Explanation: We know that the general term of an expansion (a+b)ⁿ is T_r+1 = ⁿC_r a^n-r b^r.

Now, we have to find the fourth term in the expansion (x-2y)¹²

Hence, r = 3, a = x, b = -2y, n= 12.

Now, substitute the values in the formula, we get

T₃₊₁ = ¹²C₃ x^12-3 (-2y)³.

On solving this, we get

T₄ = -1760x⁹y³.

9) If the fourth term of the binomial expansion of (px+(1/x))ⁿ is 5/2, then

1. n=6, p=6
2. n=8, p=6
3. n=8, p= ½
4. n=6, p=½

Answer: (d) n=6, p=½

Explanation:

Given: (px+(1/x))ⁿ

Hence, the fourth term, T₃₊₁ = ⁿC₃(px)^n-3(1/x)³

Given that the fourth term of the binomial expansion of (px+(1/x))ⁿ is 5/2, which is independent of x.

Hence, (5/2)= ⁿC₃(px)^n-3(1/x)³ …(1)

On solving this, we get n=6.

Now, substitute n=6 in (5/2)= ⁿC₃(p)³

20p³ = 5/2

p³=⅛

p=½.

Therefore, n=6 and p=½.

10) If n is the positive integer, then 2³ⁿ – 7n -1 is divisible by

1. 7
2. 10
3. 49
4. 81

Answer: (c) 49

Explanation:

Given: 2³ⁿ – 7n -1. It can also be written as 8ⁿ – 7n – 1

Let 8ⁿ – 7n – 1 =0

So, 8ⁿ = 7n+1

8ⁿ = (1+7)ⁿ

By applying binomial theorem, we get

8n – 1 – 7n = 49 (or) 2³ⁿ – 7n -1 = 49

Hence, 2³ⁿ – 7n -1 is divisible by 49.

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