Binomial Theorem Class 11 MCQs Questions with Answers
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
Answer
Answer: (c) 10c³
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
Question 2:
(1.1)10000 is _____ 1000
(a) greater than
(b) less than
(c) equal to
(d) None of these
Answer
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
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
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
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
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.
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
Answer
Answer: (b) 1/2
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 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!)
Answer
Answer: (b) 10!/(5!)²
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
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
Answer
Answer: (b) (2n)!/(n!)²
Hint:
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.
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
Answer
Answer: (c) 6
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
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
Answer
Answer: (b) 1
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 × Î±² + β2007 × Î²²
= (-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 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
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
Question 11.
The coefficient of xn in the expansion (1 + x + x² + …..)-n is
(a) 1
(b) (-1)n
(c) n
(d) n+1
Answer
Answer: (b) (-1)n
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
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
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
Answer
Answer: (c) 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
Answer
Answer: (c) 49
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
Answer
Answer: (c) 6
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
Answer
Answer: (b) 4851
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!)
Answer
Answer: (b) 10!/(5!)²
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
Answer
Answer: (b) 1/2
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
Answer
Answer: (b) 15
Hint:
Given, in the binomial expansion of (a + b)n, the coefficient of fourth and thirteenth terms are equal to each other
⇒ nC3 = nC12
This is possible when n = 15
Because 15C13 = 15C12
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