Class 11 Maths Chapter 4 Principle of Mathematical Induction MCQs

Class 11 Maths Chapter 4 Principle of Mathematical Induction MCQs are available here for the students who are preparing for the exams. These multiple-choice questions are a useful resource for the students to improve their analytical skills and problem-solving skills. We also provided detailed explanations for each of these MCQs along with correct options.

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Class 11 Maths Chapter 4 Principle of Mathematical Induction MCQs – Download PDF

MCQs for Class 11 Maths Chapter 4 Principle of Mathematical Induction with Answers

1. If xn – 1 is divisible by x – k, then the least positive integral value of k is

(a) 1

(b) 2

(c) 3

(d) 4

Correct option: (a) 1

Solution:

Given,

P(n): xn – 1 is divisible by x – k

Let us substitute n = 1, 2, 3,..

⇒ P(1) : x – 1

⇒ P(2) : x2 – 1 = (x−1)(x+1)

⇒ P(3) : x3 – 1 = (x − 1)(x2 + x + 1)

⇒ P(4) : x4 − 1 = (x2 − 1)(x2 + 1) = (x − 1)(x + 1)(x2 + 1)

Therefore, the least positive integral value of k is 1.

2. For any natural number n, 22n – 1 is divisible by

(a) 2

(b) 3

(c) 4

(d) 5

Correct option: (b) 3

Solution:

Let P(n) = 22n – 1

Substituting n = 1, 2, 3,….

P(1) = 22(1) – 1 = 4 – 1 = 3

This is divisible by 3.

P(2) = 22(2) – 1 = 16 – 1 = 15

This is divisible by 3.

P(3) = 22(3) – 1 = 256 – 1 = 255

This is also divisible by 3.

Assume that P(n) is true for some natural number k, i.e., P(k): 22k – 1 is divisible by 3, i.e., 22k – 1 = 3q, where q ∈ N

Now,

P(k + 1) : 22(k+1) – 1

= 22k + 2 – 1

= 22k . 22 – 1

= 22k . 4 – 1

= 3.22k + (22k – 1)

= 3.22k + 3q

= 3 (22k + q) = 3m, where m ∈ N

Thus P(k + 1) is true, whenever P(k) is true.

Therefore, for any natural number n, 22n – 1 is divisible by 3.

3. A student was asked to prove a statement P(n) by induction. He proved that P(k + 1) is true whenever P(k) is true for all k > 5 ∈ N and also that P(5) is true. Based on this, he could conclude that P(n) is true

(a) for all n ∈ N

(b) for all n > 5

(c) for all n ≥ 5

(d) for all n < 5

Correct option: (c) for all n ≥ 5

Solution:

The student could be able to conclude that P(n) is true for all n ≥ 5 since P(5) is true for all k > 5 ∈ N as well as true for P(5) and P(k + 1) is true, whenever P(k) is true.

4. If P (n): “49n + 16n + k is divisible by 64 for n ∈ N” is true, then the least negative integral value of k is

(a) 1

(b) -2

(c) -1

(d) -3

Correct option: (c) -1

Solution:

Given that P(n) : 49n + 16n + k is divisible by 64 for n ∈ N

For n = 1,

P(1) : 49 + 16 + k = 65 + k is divisible by 64.

Thus k, should be -1 since, 65 – 1 = 64 is divisible by 64.

5. For all n ∈ N, 3.52n+1 + 23n+1 is divisible by

(a) 19

(b) 17

(c) 23

(d) 25

Correct option: (b) 17

Solution:

Let P(n) be the statement that 3.52n + 1 + 23n + 1 is divisible by 17

If n = 1, then given expression = 3 * 53 + 24 + 375 + 16 = 391 = 17 * 23, divisible by 17.

P(1) is true

Assume that P(k) is true.

3.52k + 1 + 23k + 1 is divisible by 17.

3.52k = 1 + 23k + 1 = 17m where m ∈ N

3.52(k + 1) + 1 + 23(k + 1) + 1

= 3.52k + 1 * 52 + 23k + 1 * 23

= 25(17m – 23k+1) + 8.23k + 1

= 425m – 25.23k + 1 + 8.23k + 1

= 425m – 17.23k + 1

= 17(25m – 23k + 1), divisible by 17

P(k + 1) is true by Principle of Mathematical Induction

P(n) is true for all n ∈ N. 3.52n + 1 + 23n + 1 is divisible by 17 for all n ∈ N

6. n(n + 1) (n + 5) is a multiple of

(a) 3

(b) 8

(c) 5

(d) 7

Correct option: (a) 3

Solution:

Let P(n) = n(n + 1)(n + 5)

Substituting n = 1, 2, 3,….

P(1) = 1(1 + 1)(1 + 5) = 2(6) = 12; multiple of 2, 3, 4, 6

P(2) = 2(2 + 1)(2 + 5) = 2(3)(7) = 42; multiple of 2, 3, 6, 7

P(3) = 3(3 + 1)(3 + 5) = 3(4)(8) = 96; multiple of 2, 3, 4, 6, 8, 12

..

Thus, from the above statements and verifying the options, we can say that n(n + 1)(n + 5) is a multiple of 3.

7. n2 < 2n for all natural numbers

(a) n ≥ 5

(b) n < 5

(c) n > 1

(d) n ≤ 3

Correct option: (a) n ≥ 5

Solution:

Consider, P(n) : n2 < 2n

Substituting n = 1, 2, 3,…

P(1): 12 < 21

1 < 2 (not true)

P(2): 22 < 22

4 < 4 (not true)

P(3): 32 < 23

9 < 8 (not true)

P(4): 42 < 24

16 < 16 (not true)

P(5): 52 < 25

25 < 32 (true)

P(6): 62 < 26

26 < 64 (true)

Thus, n2 < 2n for all natural numbers n ≥ 5.

8. If 10n + 3.4n+2 + k is divisible by 9 for all n ∈ N, then the least positive integral value of k is

(a) 5

(b) 3

(c) 7

(d) 1

Correct option: (a) 5

Solution:

Given that 10n + 3.4n+2 + k is exactly divisible by 9.

Consider: P(n) = 10n + 3.4n+2 + k

Substituting n = 1,

P(1) = 101 + 3.41+2 + k

= 10 + 3(64) + k

= 10 + 192 + k

= 202 + k is exactly divisible by 9, the value of k will be 5.

9. Let P(n) : “2n < (1 × 2 × 3 × … × n)”. Then the smallest positive integer for which P(n) is true is

(a) 1

(b) 2

(c) 3

(d) 4

Correct option: (d) 4

Solution:

P(1) : 21 < 1

2< 1 is false

P(2) : 22 < 1 × 2

4 < 2 is false

P(3) : 23 < 1 × 2 × 3

8 < 6 is false

P(4) : 24 < 1 × 2 × 3 × 4

16 < 24 is true

10. For every positive integer n, 7n – 3n is divisible by

(a) 3

(b) 4

(c) 7

(d) 5

Correct option: (b) 4

Solution:

Let P(n) = 7n – 3n

Substituting n = 1, 2, 3,…

P(1) = 71 – 31 = 7 – 3 = 4

P(2) = 72 – 32 = 49 – 9 = 40

P(3) = 73 – 33 = 343 – 27 = 316

Thus, for every positive integer n, 7n – 3n is divisible by 4.

Also, check:

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