Multiple Choice Questions on Binomial Theorem Class 11 OP Malhotra Exe-13D ISC Maths Solutions Ch-13. In this article you would learn to solve hard mcq questions easily on Binomial Theorem. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11.

Binomial Theorem Class 11 OP Malhotra Multiple Choice Questions ISC Maths Solutions Ch-13

Board	ICSE
Publications	S Chand
Subject	Maths
Class	11th
Chapter-13	Binomial Theorem
Writer	OP Malhotra
Exe-13(D)	Multiple Choice Questions.

Multiple Choice Questions on Binomial Theorem

OP Malhotra ISC Class 11 Maths Solutions

Que-1: The number of terms in the expansion of (a² − 2ab + b²)¹⁰ is

(a) 20      (b) 15      (c) 21      (d) 11

Sol: (c) 21
(a² − 2ab + b²)¹⁰ = (a − b)²⁰
Number of terms = 20 + 1 = 21

Que-2: The 6th term in the expansion of (2x² − ¹/_3x²)¹⁰ is

(a) −3196/27      (b) −896/27      (c) 4580/17      (d) 5580/17 

Sol: (b) −896/27
General term: T_r+1 = ¹⁰C_r(2x²)^10−r(−1/3x²)^r
For 6th term r = 5
T₆ = −896/27

Que-3: (i) If the coefficients of (r+1)th and (r+3)th terms are equal in the expansion of (1 + x)²⁰, then the value of r will be

(a) 7      (b) 8      (c) 9      (d) 10

Sol: (c) 9
If coefficients equal: ²⁰C_r = ²⁰C_r+2
⇒ r = 9

(ii) If the integers r > 1, n > 2 and the coefficients of (3r)th and (r+2)th terms in the binomial expansion of (1 + x)²ⁿ are equal, then

(a) n = 2r   
(b) n = 3r   
(c) n = 2r + 1   
(d) None of these

Sol: (a) n = 2r
T_k+1 = ²ⁿC_k x^k
Equal coefficients ⇒ ²ⁿC_3r-1 = ²ⁿC_r+1
Using property ⁿC_r = ⁿC_n-r ⇒ n = 2r

Que-4: The coefficient of the middle term in the expansion of (x + 2y)⁶ is

(a) 4( ⁶C₄ )      (b) 8( ⁶C₃ )      (c) 4( ⁶C₅ )       (d) 8( ⁶C₄ )

Sol: (b) 8( ⁶C₃ )
Middle term = T_(6/2)+1 = T₄
Coefficient = ⁶C₃(2y)³
= 8( ⁶C₃ )

Que-5: Sum of the coefficients in the binomial expansion of (¹/_x + 2x)⁶ is equal to

(a) 1024       (b) 729        (c) 243       (d) 512

Sol: (b) 729
Put x = 1
(1 + 2)⁶ = 3⁶ = 729

Que-6: If (1 + ax)ⁿ = 1 + 6x + 27/2 x² + …… + aⁿxⁿ, then the values of a and n are respectively

(a) 2, 3      (b) 3, 2       (c) 3/2, 4       (d) 3/4, 8

Sol: (c) 3/2, 4
Coefficient of x = na = 6
Coefficient of x² = n(n−1)a²/2 = 27/2
Solving ⇒ a = 3/2 and n = 4

Que-7: The 7th term from the end in the expansion of (x + ¹/_x)¹⁵ is

(a) ¹⁵C₇ · 1/x   
(b) ¹⁵C₆ · x³   
(c) ¹⁵C₆ · 1/x³   
(d) None of these

Sol: (c) ¹⁵C₆ · 1/x³ .
General term: T_r+1 = ¹⁵C_rx^15−2r
7th from end ⇒ r = 6
Term = ¹⁵C₆ / x³

Que-8: In the binomial expansion of (a − b)ⁿ, n ≥ 9, the sum of 5th and 6th terms is zero, then a/b equals

(a) (n − 5)/6   
(b) (n − 4)/5   
(c) 5/(n − 4)   
(d) 6/(n − 5)

Sol: (b) (n − 4)/5
General term: T_r+1 = ⁿC_r a^n−r(−b)^r
5^th term → T₅ = ⁿC₄aⁿ⁻⁴(−b)⁴
6^th term → T₆ = ⁿC₅aⁿ⁻⁵(−b)⁵
Given: T₅ + T₆ = 0
ⁿC₄aⁿ⁻⁴b⁴ − ⁿC₅aⁿ⁻⁵b⁵ = 0
Divide by aⁿ⁻⁵b^4
nC₄a = ⁿC₅b
a/b = ⁿC₅ / ⁿC₄
= (n−4)/5

Que-9: The constant term in the expansion of (x² + 1/x²)¹⁶ is

(a) ¹⁶C₈      (b) ¹⁶C₇       (c) ¹⁶C₉       (d) ¹⁶C₁₀

Sol: (a) ¹⁶C₈
Power of x = 0 ⇒ r = 8

Que-10: The coefficient of x in the expansion of (1 − 3x + 7x²)(1 − x)¹⁶ is

(a) 17       (b) 19        (c) −17        (d) −19

Sol: (d) -19
Coefficient of x = (1)(−16) + (−3)(1)
= −16 −3 = −19

Que-11: In the expansion of (2 − 3x³)²⁰, if the ratio of 10th term to 11th term is 45/22, then x =

(a) 2/3       (b) 3/2        (c) −2/3         (d) −3/2

Sol: (c) -2/3
General term: T_r+1 = ²⁰C_r (2)^20−r(−3x³)^r
10^th term → r = 9
11^th term → r = 10
T₁₀ / T₁₁ = 45/22
After simplification → x = −2/3

Que-12: The two successive terms in the expansion of (1 + x)²⁴ whose coefficients are in the ratio 1 : 4 are

(a) 3rd and 4th   
(b) 4th and 5th   
(c) 5th and 6th   
(d) 6th and 7th

Sol: (c) 5th and 6th
General term of (1+x)²⁴ is
T_r+1 = ²⁴C_rx^r
Ratio of successive coefficients:
²⁴C_r : ²⁴C_r+1 = 1 : 4
Using property of combinations and solving, we get r = 4.
Therefore the terms are 5th and 6th.

Que-13: The coefficient of xⁿ in the expansion (1 + x)²ⁿ and (1 + x)^{2n − 1} are in the ratio

(a) 1 : 2       (b) 1 : 3        (c) 3 : 1        (d) 2 : 1

Sol: (d) 2 : 1
Coefficient of xⁿ in (1+x)²ⁿ = ²ⁿC_n
Coefficient of xⁿ in (1+x)²ⁿ⁻¹ = ²ⁿ⁻¹C_n
Ratio =
²ⁿC_n : ²ⁿ⁻¹C_n
= 2 : 1

Que-14: The coefficient of x⁻¹⁰ in the expansion of (x² + 1/x³)¹⁰ is

(a) −252       (b) 210       (c) −(5!)        (d) 120

Sol:  (b) 210
General term: T_r+1 = ¹⁰C_r(x²)^10−r(x⁻³)^r
= ¹⁰C_r x^{20 − 5r}
For x⁻¹⁰ :
20 − 5r = −10
r = 6
Coefficient = ¹⁰C₆ = 210

Que-15: The total number of terms in the expansion of (x + y)¹⁰⁰ + (x − y)¹⁰⁰ after simplification is

(a) 100        (b) 50        (c) 51        (d) 502

Sol: (c) 51
In (x+y)¹⁰⁰ + (x−y)¹⁰⁰, odd power terms cancel.
Only even powers remain → terms = 100/2 + 1 = 51.

Que-16: The number of terms in the expansion of (3y + x)¹⁰ − (3y − x)¹⁰ is

(a) 5        (b) 4          (c) 9        (d) 11

Sol: (a) 5
Even powers cancel in subtraction.
Only odd powers remain → 5 terms.

Que-17: The value of (²¹C₁ − ¹⁰C₁) + (²¹C₂ − ¹⁰C₂) + (²¹C₃ − ¹⁰C₃) + … + (²¹C₆ − ¹⁰C₆) is

(a) 2²¹ − 2¹⁰   
(b) 2²⁰ − 2⁹   
(c) 2²⁰ − 2¹⁰   
(d) 2²¹ − 2¹¹

Sol: (c) 2²⁰ − 2¹⁰
Using the identity:
C(n,1) + C(n,2) + … + C(n,n) = 2ⁿ − 1
So,
Σ (²¹C_r − ¹⁰C_r)
= Σ ²¹C_r − Σ ¹⁰C_r
= (2²¹ − 1) − (2¹⁰ − 1)
= 2²¹ − 2¹⁰

Que-18: The value of (⁷C₀ − ⁷C₁) + (⁷C₁ − ⁷C₂) + … + (⁷C₆ − ⁷C₇) is

(a) 2⁸ − 2   
(b) 2⁸   
(c) 2⁸ − 1   
(d) 2⁸ − 2

Sol: (a) 2⁸ − 2
(⁷C₀ − ⁷C₁) + (⁷C₁ − ⁷C₂) + … + (⁷C₆ − ⁷C₇)
= ⁷C₀ − ⁷C₇
Using identity: Σ ⁿC_r = 2ⁿ
= 2⁸ − 2

–: End Binomial Theorem Class 11 OP Malhotra Exe-13D ISC Maths Ch-13 Solutions :–

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