Question

If \( f(x) > 0 \; \forall x \in \mathbb{R} \), \( f'(3) = 0 \) and \( g(x) = f(\tan^2 x - 2\tan x + 4) \), \( 0 < x < \frac{\pi}{2} \), then \( g(x) \) is increasing in

• A. \(\left(0,\frac{\pi}{4}\right)\)
• B. \(\left(\frac{\pi}{6},\frac{\pi}{3}\right)\)
• C. \(\left(0,\frac{\pi}{3}\right)\)
• D. \(\left(\frac{\pi}{4},\frac{\pi}{2}\right)\)

Hint:

For composite functions, analyze inner function first and use critical point of outer derivative.

Answer 1

The Correct Option is D

Concept:
Use chain rule and monotonicity of inner function.
Step 1: Rewrite function.
\[ g(x) = f((\tan x - 1)^2 + 3) \]
Step 2: Let \( u = (\tan x - 1)^2 + 3 \).
Then: \[ g'(x) = f'(u)\cdot u' \] Given \( f'(3) = 0 \), so critical point at \( u = 3 \).
Step 3: Find when \( u = 3 \).
\[ (\tan x - 1)^2 + 3 = 3 \Rightarrow (\tan x - 1)^2 = 0 \Rightarrow \tan x = 1 \Rightarrow x = \frac{\pi}{4} \]
Step 4: Check monotonicity.
For \( x>\frac{\pi}{4} \), \( \tan x>1 \Rightarrow u \) increases. Thus \( g(x) \) increases in: \[ \left(\frac{\pi}{4}, \frac{\pi}{2}\right) \]