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

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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) : 2ⁿ < ∠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) = 2ⁿ < ∠n
For n = 1 ;
2 < ∠1 = 1, which is false
For n = 2 ;
2² < ∠2 2, which is false
For n = 3 ;
2³ < ∠3 = 6, which is false
When n = 4 ;
2⁴ = 16 < ∠4 = 24, which is true

Question 2.
Consider the statement P (n) : n² – 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) : n³ – n + 41 is prime
Here p (3) ;
3² – 3 + 41 = 47 which is prime
∴ P (3) is true.
and P (5) ;
5² – 5 + 41 = 61, which is prime
∴ P (5) is true.

Question 3.
Let P(n) = 1 + 3 + 5 + …………………. + (2n – 1) = 3 + n², 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 + n²
P (3) = 1 + 3 + 5 = 9
and 3 + n² = 3 + 3² = 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 + m²
Here P(m + 1) : 1 + 3 + 5 + …………………… + (2m – 1) + (2m + 1) = 3 + m² + 2m + 1 = m² + 2m + 4
= 3 + (m + 1)²
∴ P (m + 1) is true.

Question 4.
Let P (n) : n² < 2ⁿ, 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) : n² < 2ⁿ
P (2) : 2² < 2² which is false
P (3) : 3² <2³, which is false
P (4) : 4² < 2⁴, which is false
P (5) : 5² = 25 < 32 = 2⁵ 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) < 2ⁿ. 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) < 2ⁿ
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 10ⁿ + 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) : 10ⁿ + 3 . 4^{n + 2} + k is divisible by 9.
∴ P (1): 10 + 3 × 4³ + 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 xⁿ – 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 xⁿ – 1 is divisible by x – k
∴ xⁿ – 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.