Applications of Derivatives
Increasing/Decreasing — Composite Function with f'' > 0
nta_pyq_2026_jan
Grade 12
Question:
Let $f:\mathbb{R}\to\mathbb{R}$ be a twice differentiable function such that $f''(x)>0$ for all $x\in\mathbb{R}$ and $f'(a-1)=0$, where $a$ is a real number. Let $g(x)=f(\tan^2x-2\tan x+a)$, $0<x<\dfrac{\pi}{2}$.
Consider the following two statements:
(I) $g$ is increasing in $\left(0,\dfrac{\pi}{4}\right)$
(II) $g$ is decreasing in $\left(\dfrac{\pi}{4},\dfrac{\pi}{2}\right)$. Then,
Neither (I) nor (II) is True
Only (I) is True
Both (I) and (II) are True
Only (II) is True
Step-by-Step Solution
Key Concept: Let $u(x)=(\tan x-1)^2+a-1$. Since $f''(x)>0$, $f'$ is strictly increasing with $f'(a-1)=0$, so $f'(t)<0$ for $t<a-1$ and $f'(t)>0$ for $t>a-1$. $g'(x)=f'(u(x))\cdot2\sec^2x(\tan x-1)$.
Both statements are false. Neither (I) nor (II) is true.
Correct Answer: 1