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MCQ Class 11 Mathematics Binomial Theorem with Answers - Set - 1
Question 1:
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
Correct Answer – (D)
The general term of the expansion (a + b)n is
Tr+1 = nCr × an-r × br
Question 2 :
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
Correct Answer – (C)
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
Question 3 :
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!)
Correct Answer – (B)
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
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
Correct Answer – (D)
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.
Question 5 :
(1.1)10000 is _____ 1000
(a) greater than
(b) less than
(c) equal to
(d) None of these
Correct Answer – (A)
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
MCQ Class 11 Mathematics Binomial Theorem with Answers
Question 6 :
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
Correct Answer – (B)
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
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
Correct Answer – (B)
We have,
(1 – 2x + 3x² – 4x³ + ……..)-n = {(1 + x)-2}-n
= (1 + x)2n
So, the coefficient of xnC3 = 2nCn = (2n)!/(n!)²
Question 8 :
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
Correct Answer – (B)
(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 9 :
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
Correct Answer – (C)
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
Question 10 :
The coefficient of y in the expansion of (y² + c/y)5 is
(a) 10c
(b) 10c²
(c) 10c³
(d) None of these
Correct Answer – (C)
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
- NCERT Solutions Class 11 Mathematics Binomial Theorem with Answers : Exercise 8.1
- NCERT Solutions Class 11 Mathematics Binomial Theorem with Answers : Exercise 8.2
- NCERT Solutions Class 11 Mathematics Binomial Theorem with Answers : Exercise 8 Misc
- NCERT Solutions Class 11 Mathematics Textbook download
