Check the below NCERT MCQ Questions for Class 11 Maths Chapter 8 Binomial Theorem with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have provided Binomial Theorem Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.

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## Binomial Theorem Class 11 MCQs Questions with Answers

MCQ On Binomial Theorem Class 11 Question 1.
The coefficient of y in the expansion of (y² + c/y)5 is
(a) 10c
(b) 10c²
(c) 10c³
(d) None of these

Hint:
Given, binomial expression is (y² + c/y)5
Now, Tr+1 = 5Cr × (y²)5-r × (c/y)r
= 5Cr × y10-3r × Cr
Now, 10 – 3r = 1
⇒ 3r = 9
⇒ r = 3
So, the coefficient of y = 5C3 × c³ = 10c³

Binomial Theorem MCQ Question 2:
(1.1)10000 is _____ 1000
(a) greater than
(b) less than
(c) equal to
(d) None of these

Answer: (a) greater than
Hint:
Given, (1.1)10000 = (1 + 0.1)10000
10000C0 + 10000C1 × (0.1) + 10000C2 ×(0.1)² + other +ve terms
= 1 + 10000×(0.1) + other +ve terms
= 1 + 1000 + other +ve terms
> 1000
So, (1.1)10000 is greater than 1000

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MCQ On Binomial Theorem Question 3.
The fourth term in the expansion (x – 2y)12 is
(a) -1670 x9 × y³
(b) -7160 x9 × y³
(c) -1760 x9 × y³
(d) -1607 x9 × y³

Answer: (c) -1760 x9 × y³
Hint:
4th term in (x – 2y)12 = T4
= T3+1
= 12C3 (x)12-3 ×(-2y)³
= 12C3 x9 ×(-8y³)
= {(12×11×10)/(3×2×1)} × x9 ×(-8y³)
= -(2×11×10×8) × x9 × y³
= -1760 x9 × y³

MCQs On Binomial Theorem Question 4.
If n is a positive integer, then (√3+1)2n+1 + (√3−1)2n+1 is
(a) an even positive integer
(b) a rational number
(c) an odd positive integer
(d) an irrational number

Answer: (d) an irrational number
Hint:
Since n is a positive integer, assume n = 1
(√3+1)³ + (√3−1)³
= {3√3 + 1 + 3√3(√3 + 1)} + {3√3 – 1 – 3√3(√3 – 1)}
= 3√3 + 1 + 9 + 3√3 + 3√3 – 1 – 9 + 3√3
= 12√3, which is an irrational number.

Binomial Theorem Class 11 MCQ Question 5.
If the third term in the binomial expansion of (1 + x)m is (-1/8)x² then the rational value of m is
(a) 2
(b) 1/2
(c) 3
(d) 4

Hint:
(1 + x)m = 1 + mx + {m(m – 1)/2}x² + ……..
Now, {m(m – 1)/2}x² = (-1/8)x²
⇒ m(m – 1)/2 = -1/8
⇒ 4m² – 4m = -1
⇒ 4m² – 4m + 1 = 0
⇒ (2m – 1)² = 0
⇒ 2m – 1 = 0
⇒ m = 1/2

Binomial Theorem MCQ Pdf Question 6.
The greatest coefficient in the expansion of (1 + x)10 is
(a) 10!/(5!)
(b) 10!/(5!)²
(c) 10!/(5! × 4!)²
(d) 10!/(5! × 4!)

Hint:
The coefficient of xr in the expansion of (1 + x)10 is 10Cr and 10Cr is maximum for
r = 10/ = 5
Hence, the greatest coefficient = 10C5
= 10!/(5!)²

Binomial Theorem MCQs With Answers Pdf Question 7.
The coefficient of xn in the expansion of (1 – 2x + 3x² – 4x³ + ……..)-n is
(a) (2n)!/n!
(b) (2n)!/(n!)²
(c) (2n)!/{2×(n!)²}
(d) None of these

Hint:
We have,
(1 – 2x + 3x² – 4x³ + ……..)-n = {(1 + x)-2}-n
= (1 + x)2n
So, the coefficient of xnC3 = 2nCn = (2n)!/(n!)²

Binomial Theorem MCQ With Solution Question 8.
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

Hint:
Given that the first three terms of the expansion are 729, 7290 and 30375 respectively.
Now T1 = nC0 × an-0 × b0 = 729
⇒ an = 729 ……………. 1
T2 = nC1 × an-1 × b1 = 7290
⇒ n
an-1 × b = 7290 ……. 2
T3 = nC2 × an-2 × b² = 30375
⇒ {n(n-1)/2}
an-2 × b² = 30375 ……. 3
Now equation 2/equation 1
n
an-1 × b/an = 7290/729
⇒ n×b/n = 10 ……. 4
Now equation 3/equation 2
{n(n-1)/2}
an-2 × b² /n
an-1 × b = 30375/7290
⇒ b(n-1)/2a = 30375/7290
⇒ b(n-1)/a = (30375×2)/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 × 5/3 = 10
⇒ 5n = 30
⇒ n = 30/5
⇒ n = 6
So, the value of n is 6

Maths MCQs For Class 11 With Answers Pdf Question 9.
If α and β are the roots of the equation x² – x + 1 = 0 then the value of α2009 + β2009 is
(a) 0
(b) 1
(c) -1
(d) 10

Hint:
Given, x² – x + 1 = 0
Now, by Shridharacharya formula, we get
x = {1 ± √(1 – 4×1×1) }/2
⇒ x = {1 ± √(1 – 4) }/2
⇒ x = {1 ± √(-3)}/2
⇒ x = {1 ± √(3 × -1)}/2
⇒ x = {1 ± √3 × √-1}/2
⇒ x = {1 ± i√3}/2 {since i = √-1}
⇒ x = {1 + i√3}/2, {–1 – i√3}/2
⇒ x = -{-1 – i√3}/2, -{-1 + i√3}/2
⇒ x = w, w² {since w = {-1 + i√3}/2 and w² = {-1 – i√3}/2 }
Hence, α = -w, β = w²
Again we know that w³ = 1 and 1 + w + w² = 0
Now, α2009 + β2009 = α2007 × α² + β2007 × β²
= (-w)2007 × (-w)² + (-w²)2007 × (-w²)² {since 2007 is multiple of 3}
= -(w)2007 × (w)² – (w²)2007 × (w4)
= -1 × w² – 1 × w³ × w
= -1 × w² – 1 × 1 × w
= -w² – w
= 1 {since 1 + w + w² = 0}
So, α2009 + β2009 = 1

MCQ Questions For Class 11 Maths With Answers Pdf Question 10.
The general term of the expansion (a + b)n is
(a) Tr+1 = nCr × ar × br
(b) Tr+1 = nCr × ar × bn-r
(c) Tr+1 = nCr × an-r × bn-r
(d) Tr+1 = nCr × an-r × br

Answer: (d) Tr+1 = nCr × an-r × br
Hint:
The general term of the expansion (a + b)n is
Tr+1 = nCr × an-r × br

MCQ Questions For Class 11 Maths With Answers Question 11.
The coefficient of xn in the expansion (1 + x + x² + …..)-n is
(a) 1
(b) (-1)n
(c) n
(d) n+1

Hint:
We know that
(1 + x + x² + …..)-n = (1 – x)-n
Now, the coefficient of x = (-1)n × nCn
= (-1)n

Question 12.
If n is a positive integer, then (√5+1)2n + 1 − (√5−1)2n + 1 is
(a) an odd positive integer
(b) not an integer
(c) none of these
(d) an even positive integer

Answer: (b) not an integer
Hint:
Since n is a positive integer, assume n = 1
(√5+1)² + 1 − (√5−1)² + 1
= (5 + 2√5 + 1) + 1 – (5 – 2√5 + 1) + 1 {since (x+y)² = x² + 2xy + y²}
= 4√5 + 2, which is not an integer

Question 13.
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) nth term
(d) (n/2 – 1)th term

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 14.
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

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 15.
if n is a positive ineger then 23nn – 7n – 1 is divisible by
(a) 7
(b) 9
(c) 49
(d) 81

Hint:
Given, 23n – 7n – 1 = 23×n – 7n – 1
= 8n – 7n – 1
= (1 + 7)n – 7n – 1
= {nC0 + nC1 7 + nC2 7² + …….. + nCn 7n} – 7n – 1
= {1 + 7n + nC2 7² + …….. + nCn 7n} – 7n – 1
= nC2 7² + …….. + nCn 7n
= 49(nC2 + …….. + nCn 7n-2)
which is divisible by 49
So, 23n – 7n – 1 is divisible by 49

Question 16.
In the binomial expansion of (71/2 + 51/3)37, the number of integers are
(a) 2
(b) 4
(c) 6
(d) 8

Hint:
Given, (71/2 + 51/3)37
Now, general term of this binomial Tr+1 = 37Cr × (71/2)37-r × (51/3)r
⇒ Tr+1 = 37Cr × 7(37-r)/2 × (5)r/3
This General term will be an integer if 37Cr is an integer, 7(37-r)/2 is an integer and (5)r/3 is an integer.
Now, 37Cr will always be a positive integer.
Since 37Cr denotes the number of ways of selecting r things out of 37 things, it can not be a fraction.
So, 37Cr is an integer.
Again, 7(37-r)/2Cr 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 (71/2 + 51/3)37

Question 17.
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

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+3-1C3-1
= 99C2
= (99 × 98)/2
= 4851

Question 18.
The greatest coefficient in the expansion of (1 + x)10 is
(a) 10!/(5!)
(b) 10!/(5!)²
(c) 10!/(5! × 4!)²
(d) 10!/(5! × 4!)

Hint:
The coefficient of xr in the expansion of (1 + x)10 is 10Cr and 10Cr is maximum for
r = 10/2 = 5
Hence, the greatest coefficient = 10C5
= 10!/(5!)²

Question 19.
If the third term in the binomial expansion of (1 + x)m is (-1/8)x² then the rational value of m is
(a) 2
(b) 1/2
(c) 3
(d) 4

Hint:
(1 + x)m = 1 + mx + {m(m – 1)/2}x² + ……..
Now, {m(m – 1)/2}x² = (-1/8)x²
⇒ m(m – 1)/2 = -1/8
⇒ 4m² – 4m = -1
⇒ 4m² – 4m + 1 = 0
⇒ (2m – 1)² = 0
⇒ 2m – 1 = 0
⇒ m = 1/2

Question 20.
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