Question

Which of the following functions are strictly decreasing on (0,π/2)?

• A. cos x
• B. cos 2x
• C. cos 3x
• D. tan x

Answer 1

The Correct Option is A, B

(A) Let f₁(x) = cos x.

=f₁(x) = -sin x

In interval (0,\(\frac \pi2\)), f^'₁(x) = -sin x<0.

\(\implies\)f₁(x) = cos x is strictly decreasing in interval (0,\(\frac \pi2\)).

(B) Let f₂(x) = cos2x.

 f^'₂(x) = -2sin 2x

Now 0<x<\(\frac \pi2\) \(\implies\) 0<2x<π \(\implies\) sin 2x>0 \(\implies\) -2 sin 2x<0

f'₁(x) = -2sin 2x < 0 on (0,\(\frac \pi2\))

\(\implies\)f'2(x) = cos 2x is strictly decreasing in interval (0,\(\frac \pi2\)).

(C) Let f₃(x) = cos3x.

f'₃(x) = -sin3x

Now, f'₃(x) = 0

\(\implies\)sin3x=0\(\implies\)3x=π, as x ε (0,\(\frac \pi2\))

\(\implies\)x = \(\frac \pi2\)

The point x=\(\frac \pi3\) divides the interval (0,\(\frac \pi2\)) into two disjoint intervals

i.e., 0 (0,\(\frac \pi3\)) and (\(\frac \pi3\),\(\frac \pi2\)).

Now, in interval(0,\(\frac \pi3\)), f3(x) =-3 sin3x<0 [as 0<x<\(\frac \pi3\)=0<3x<\(\pi\)].

∴ f₃ is strictly decreasing in interval (0,\(\frac \pi3\))

However, in interval (\(\frac \pi3\),\(\frac \pi2\)), f₃(x)=-3sin 3x>0 [as \(\frac \pi3\)<x<\(\frac \pi2\)\(\implies\)π<3x<\(\frac {3\pi}{2}\)]

∴ f₃ is strictly increasing in interval (\(\frac \pi3\),\(\frac \pi2\)).

Hence, f₃ is neither increasing nor decreasing in interval (0,\(\frac \pi2\))

(D) Let f₄(x) = tan x.

\(\implies\)f₄(x) = sec² x

In interval (0,\(\frac \pi2\)), f'₄(x) = sec²x>0.

f₄ is strictly increasing in interval (0,\(\frac \pi2\)).

Therefore, functions cos x and cos 2x are strictly decreasing in (0,\(\frac \pi2\)).

Hence, the correct answers are (A) and (B).