Regular engagement with Understanding ISC Mathematics Class 11 Solutions Chapter 4 Principle of Mathematical Induction MCQs can boost students’ confidence in the subject.

## ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction MCQs

Choose the correct answer from the given four options in questions (1 to 11) :

Question 1.

Let P (n) : 2^{n} < ∠n then the smallest positive integer for which P (n) is true is

(a) 1

(b) 2

(c) 3

(d) 4

Answer:

(d) 4

Given P (n) = 2^{n} < ∠n

For n = 1 ;

2 < ∠1 = 1, which is false

For n = 2 ;

2^{2} < ∠2 2, which is false

For n = 3 ;

2^{3} < ∠3 = 6, which is false

When n = 4 ;

2^{4} = 16 < ∠4 = 24, which is true

Question 2.

Consider the statement P (n) : n^{2} – n + 41 is prime. Then which of the following is true ?

(a) Both P (3) and P (5) are true.

(b) P (3) is true but P (5) is false.

(c) Both P (3) and P (5) are false.

(d) P (3) is false but P (5) is true.

Answer:

(a) Both P (3) and P (5) are true.

Given P(n) : n^{3} – n + 41 is prime

Here p (3) ;

3^{2} – 3 + 41 = 47 which is prime

∴ P (3) is true.

and P (5) ;

5^{2} – 5 + 41 = 61, which is prime

∴ P (5) is true.

Question 3.

Let P(n) = 1 + 3 + 5 + …………………. + (2n – 1) = 3 + n^{2}, then which of the following is true?

(a) P (3) is correct

(b) P (2) is correct

(c) P (m) ⇒ P (m + 1)

(d) P (m) ⇏ P (m + 1)

Answer:

(c) P (m) ⇒ P (m + 1)

Given P (n) = 1 + 3 + 5 + ………………….. + (2n – 1) = 3 + n^{2}

P (3) = 1 + 3 + 5 = 9

and 3 + n^{2} = 3 + 3^{2} = 12

∴ P (3) is not correct.

Here P (2) = 1 + 3 ≠ 3 + 2

∴ P (2) is not correct.

Let P (m) is true

i.e. 1 + 3 + 5 + ……………………. + 2m – 1 = 3 + m^{2}

Here P(m + 1) : 1 + 3 + 5 + …………………… + (2m – 1) + (2m + 1) = 3 + m^{2} + 2m + 1 = m^{2} + 2m + 4

= 3 + (m + 1)^{2}

∴ P (m + 1) is true.

Question 4.

Let P (n) : n^{2} < 2^{n}, n > 1, then the smallest positive integer for which P (n) is true ?

(a) 2

(b) 3

(c) 4

(d) 5

Solution:

(d) 5

Given P (n) : n^{2} < 2^{n}

P (2) : 2^{2} < 2^{2} which is false

P (3) : 3^{2} <2^{3}, which is false

P (4) : 4^{2} < 2^{4}, which is false

P (5) : 5^{2} = 25 < 32 = 2^{5} which is true.

Question 5.

Consider the statement P (n): iOn + 3 is prime, then which of the following is not true?

(a) P (1)

(b) P (2)

(c) P (3)

(d) P (4)

Answer:

(c) P (3)

Given P (n) : 10n + 3 is prime

Hence P (1) = 10 + 3 = 13 is prime

P (2) : 10 × 2 + 3 = 13 is prime

P (3) : 10 × 3 + 3 = 33 which is not prime.

Question 6.

Let P (n) be the statement n (n + 1) (n + 2) is an integral multiple of 12, then which of the following is not true?

(a) P (3)

(b) P (4)

(c) P (5)

(d) P (6)

Answer:

(c) P (5)

Let P (n) : n (n + 1) (n + 2) is an integral multiple of 12.

Here P (3) : 3 (3 + 1) (3 + 2) = 60 which is an integral multiple of 12.

P (4) : 4 (4 + 1) (4 + 2) = 120, which is an integral multiple of 12.

P (5) : 5 (5 + 1) (5 + 2) = 5 × 6 × 7 = 210, which is not an integral multiple of 12.

Question 7.

Let P (n) : (2n + 1) < 2^{n}. then the smallest positive integer for which P (n) is true ?

(a) 2

(b) 3

(c) 4

(d) 5

Answer:

(b) 3

Given P (n) : (2n + 1) < 2^{n}

P (2) : 2 × 2 1 = 5 > 22 = 4

∴ P(2) is false.

P (3) : 2 × 3 + 1 = 7 < 2 = 8

∴ P (3) is true.

Question 8.

If 10^{n} + 3 . 4^{n + 2} + k is divisible by 9, ∀ n ∈ N. then the least positive integral value of k is

(a) 1

(b) 3

(c) 5

(d) 7

Answer:

(c) 5

Let P (n) : 10^{n} + 3 . 4^{n + 2} + k is divisible by 9.

∴ P (1): 10 + 3 × 4^{3} + k i.e. 202 + k is divisible by 9

∴ k = 5

∵ 207 is divisible by 9. .

Question 9.

For all n ∈ N, 3 . 5^{2n + 1} + 2^{3n + 1} is divisible by

(a) 17

(b) 19

(c) 23

(d) 25

Answer:

(a) 17

Let P (n) : 3 . 5^{2n + 1} + 2^{3n + 1} is divisible by same number.

P (1) ; 3 × 5^{2 + 1} + 2^{3 + 1} = 375 + 16 = 391 which is divisibel by 17

Question 10.

If x^{n} – 1 is divisible by x – k, then the least positive integral value of k is

(a) 1

(b) 2

(c) 3

(d) 4

Answer:

(a) 1

Since x^{n} – 1 is divisible by x – k

∴ x^{n} – 1 = f (x) (x – k)

⇒ f (x) = \(\frac{x^n-1}{x-k}\)

Since f (x) is a polynomial in x

∴ k must be equal to 1.

Thus,

\(\frac{x^n-1}{x-1}=\frac{(x-1)\left(x^{n-1}+x^{n-2}+\ldots .+1\right)}{x-1}\)

= x^{n – 1} + x^{n – 2} + ………………………… + 1

Question 11.

A student was asked to prove a statement P (n) by induction. 11e proved that P (k + 1) is true whenever P (k) ¡s true for all k ≥ 5 ∈ N and also that P (5) is true on the basis of this he conclude that P (n) is true

(a) ∀ n ∈ W

(b) ∀ n > 5

(c) ∀ n ≥ 5

(d) ∀ n < 5

Solution:

(c) ∀ n ≥ 5

Since P (5) is true.

∴ P (n) is true for n 5 and P (k) is given to be true for all k ≥ 5 ∈ N and student proved that P (k + 1) is true.

Thus by Mathematical induction, P (n) is true for all n ≥ 5.