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): 2^2k – 1 is divisible by 3, i.e., 2^2k – 1 = 3q, where q ∈ N

Now,

P(k + 1) : 2^2(k+1) – 1

= 2^{2k + 2} – 1

= 2^2k . 2² – 1

= 2^2k . 4 – 1

= 3.2^2k + (2^2k – 1)

= 3.2^2k + 3q

= 3 (2^2k + 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 3.5^{2n + 1} + 2^{3n + 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.

3.5^{2k + 1} + 2^{3k + 1} is divisible by 17.

3.5^2k = 1 + 2^{3k + 1} = 17m where m ∈ N

3.5^{2(k + 1) + 1} + 23^{(k + 1) + 1}

= 3.5^{2k + 1} * 5² + 2^{3k + 1} * 2³

= 25^{(17m – 23k+1)} + 8.2^{3k + 1}

= 425m – 25.2^{3k + 1} + 8.2^{3k + 1}

= 425m – 17.2^{3k + 1}

= 17(25m – 2^{3k + 1}), divisible by 17

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

P(n) is true for all n ∈ N. 3.5^{2n + 1} + 2^{3n + 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:

• Principle of Mathematical Induction Class 11 Notes
• Important Questions for Class 11 Maths Chapter 4