Revision and Self Assessment on Mathematical Induction Class 11 OP Malhotra Exe-8C ISC Maths Solutions Ch-8. In this article you would learn to solve all mcq questions on Mathematical Induction. 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

Mathematical Induction Class 11 OP Malhotra Revision and Self Assessment ISC Maths Solutions Ch-8

Board	ICSE
Publications	S Chand
Subject	Maths
Class	11th
Chapter-8	Mathematical Induction
Writer	OP Malhotra
Exe-8(C)	Revision and Self Assessment.

Revision and Self Assessment on Mathematical Induction

OP Malhotra ISC Class 11 Maths Solutions

Que-1: If P(n): 2²ⁿ − 1 is divisible by k for all n ∈ N is true, then the value of k is

(a) 6        (b) 3          (c) 7         (d) 2

Sol: (b) 3
If P(n): 2²ⁿ − 1 is divisible by k for all n ∈ N, then find k.
2²ⁿ − 1 = (4ⁿ − 1)
Since 4 ≡ 1 (mod 3), therefore
4ⁿ − 1 ≡ 0 (mod 3)
Hence divisible by 3 for all n.

Que-2: If x³ − 1 is divisible by x − k, then the least value of k is

(a) 1          (b) 2         (c) 3        (d) 4

Sol: (a) 1
If x³ − 1 is divisible by x − k, find the least value of k.
By Remainder Theorem,
Put x = k:
k³ − 1 = 0
⇒ k³ = 1
⇒ k = 1

Que-3: If 10ⁿ + 3·4ⁿ⁺² + k is divisible by 9, for all n ∈ N, then the positive integral value of k is

(a) 5         (b) 3         (c) 7        (d) 1

Sol: (a) 5
10 ≡ 1 (mod 9) ⇒ 10ⁿ ≡ 1
4 ≡ 4 (mod 9) and 4² = 16 ≡ 7 (mod 9)
3·4ⁿ⁺² ≡ 3×7 ≡ 21 ≡ 3 (mod 9)
Expression ≡ 1 + 3 + k ≡ 4 + k (mod 9)
For divisibility by 9 ⇒ 4 + k ≡ 0 (mod 9)
k ≡ 5 (mod 9)

Que-4: For all n ∈ N, 3·5²ⁿ⁺¹ + 2³ⁿ⁺¹ is divisible by

(a) 19          (b) 17          (c) 23          (d) 25

Sol: (b) 17
5² = 25 ≡ 8 (mod 17)
Using cyclic property, expression ≡ 0 (mod 17)
Hence divisible by 17.

Que-5: If 49ⁿ + 16n + λ is divisible by 64 for all n ∈ N, then the least negative integral value of λ is

(a) −1         (b) −2          (c) −3         (d) −4

Sol: (a) -1
49 ≡ -15 (mod 64)
Using expansion and simplifying, expression ≡ 1 + 16n + λ (mod 64)
For divisibility ⇒ λ = -1

Que-6: For positive integer n, n³ + 2n is always divisible by

(a) 3         (b) 7         (c) 5         (d) 6

Sol: (a) 3
n³ + 2n = n(n² + 2)
Check modulo 3 cases (n ≡ 0,1,2)
Always divisible by 3.

Que-7: For n ∈ N, n(n + 1)(n + 5) is a multiple of

(a) 4        (b) 12          (c) 3          (d) none of these

Sol: (c) 3
Among any three consecutive integers n, (n+1), (n+5), one is always divisible by 3.

Que-8: (2³ⁿ − 1) will be divisible by (∀ n ∈ N)

(a) 25        (b) 8          (c) 7         (d) 3

Sol: (c) 7
Since 2³ = 8 ≡ 1 (mod 7),
therefore 2³ⁿ − 1 is divisible by 7.

Que-9: For a natural number n, which one is the correct statement?

(a) 1³ + 2³ + …. + n³ = (1 + 2 + …. + n)²
(b) 1³ + 2³ + …. + n³ > (1 + 2 + …. + n)²
(c) 1³ + 2³ + …. + n³ < (1 + 2 + …. + n)²
(d) 1³ + 2³ + …. + n³ ≠ (1 + 2 + …. + n)²

Sol:  (a) 1³ + 2³ + …. + n³ = (1 + 2 + …. + n)²
Formula: 1³ + 2³ + … + n³
= [n(n+1)/2]² = (1+2+…+n)²

Que-10: Which of the following results is valid?

(a) (1 + x)ⁿ > (1 + nx) ∀ x ∈ N
(b) (1 + x)ⁿ ≥ (1 + nx) ∀ x ∈ N, where x > −1
(c) (1 + x)ⁿ ≤ (1 + nx) ∀ x ∈ N
(d) (1 + x)ⁿ < (1 + nx) ∀ x ∈ N

Sol: (b) (1 + x)ⁿ ≥ (1 + nx) ∀ x ∈ N, where x > −1
By Binomial Theorem: (1+x)ⁿ ≥ 1+nx for x > −1.

Que-11: If n is a natural number, then

(a) 1² + 2² + …. + n² < n³/3
(b) 1² + 2² + …. + n² = n³/3
(c) 1² + 2² + …. + n² > n³
(d) 1² + 2² + …. + n² > n³/3

 Sol: (d) 1² + 2² + …. + n² > n³/3
Sum formula: 1²+2²+…+n²
= n(n+1)(2n+1)/6 > n³/3

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