Which of the following principles are strictly decreasing on 0, π/2 ?
(A) cos x   (B) cos 2x   (C) cos 3x   (D) tan x

Solution:

Increasing functions are those functions that increase monotonically within a particular domain,

and decreasing functions are those which decrease monotonically within a particular domain.

(A) Let f₁ (x) = cos x

Therefore, f' (x) = - sin x

In interval (0, π/2) f₁' (x) = - sin x < 0

Thus, cos x is strictly decreasing in (0, π/2).

(B) Let f₂ (x) = cos 2x

Therefore,

f'₂ (x) = - 2sin 2x

Now,

⇒ 0 < x < π / 2

⇒ 0 < 2x < π

⇒ sin 2x > 0

⇒ - 2 sin 2x < 0

Hence,

f'₂ (x) = - 2sin 2x < 0 in (0, π/2).

Thus, cos 2x is strictly decreasing in (0, π/2).

(C) Let f₃ (x) = cos 3x

Therefore, f'₃ (x) = - 3sin 3x

Now,

f'₃ (x) = 0

⇒ sin 3x = 0

⇒ 3x = π [∵ x ∈ (0, π/2)]

⇒ x = π/3

The point x = π/3, divides (0, π/2) into (0, π/3) and (π/3, π/2)

In interval (0, π/3)

f'₃ (x)

= - 3sin 3x < 0 [0 < x < π/3; 0 < 3x < π]

Hence,

f₃ is strictly decreasing in (0, π/3)

In interval (π/3, π/2)

f₃ (x) = - 3 sin 3x > 0 [π/3 < x < π/2 < π < 3x < 3π/2]

Hence,

f₃ is strictly increasing in (π/3, π/2)

Thus,

cos 3x is neither increasing nor decreasing in the interval (0, π/2).

(D) Let f₄ (x) = tan x

Therefore,

f'₄ (x) = sec² x

In interval (0, π/2),

f'₄ (x) = sec² x > 0

Thus, tan x is strictly increasing in (0, π/2)

Thus, the correct options are A and B

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NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.2 Question 12

Which of the following principles are strictly decreasing on (0, π/2)? (A) cos x (B) cos 2x (C) cos 3x (D) tan x

Summary:

Hence we have concluded that tan x is strictly increasing in (0, π/2). Thus, the correct options are A and B