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:
- 5!
- 6
- 216
- 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 Tr+1 =nCr xn-r ar
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
- 5.011
- 5.012
- 5.013
- 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:
- nth term
- (n/2)th term
- [(n/2)-1]th term
- [(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:
- 10! / (5!)2
- 10! / 5!
- 10! / (5!×4!)2
- 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 Tr+1 =nCr xn-r ar
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
- 360
- 720
- 240
- 1080
Answer: (b) 720
Explanation:
Given: (2x+3y2)5
Therefore, the general form for the expression (2x+3y2)5 is Tr+1 = 5Cr. (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
- 6th term
- 7th term
- 8th term
- 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:
- 10c
- 29c
- 10c3
- 20c3
Answer: (c) 10c3
Explanation: Given: (y2+(c/y))5
We know that general term of (x+a)n is Tr+1 =nCr xn-r ar
Here, n=5, r=?
(y2+(c/y))5 = 5Cr.(y2)r.(c/y)5-r
(y2+(c/y))5 = 5Cr. y2r. (c5-r/y5-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:
- -1760 x9 × y3
- -1670 x9 × y3
- -7160 x9 × y3
- -1607 x9 × y3
Answer: (a) -1760 x9 × y3
Explanation: We know that the general term of an expansion (a+b)n is Tr+1 = nCr an-r br.
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 x12-3 (-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
- n=6, p=6
- n=8, p=6
- n=8, p= ½
- 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
- 7
- 10
- 49
- 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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