JEE Main 2026 — 21 January, Evening Shift — Question 2

Given $f''(x) > 0$, so $f'(x)$ is strictly increasing. $f'(a-1) = 0$. $g(x) = f(\tan^2 x - 2\tan x + a) = f((\tan x - 1)^2 + a - 1)$. $g'(x) = f'((\tan x - 1)^2 + a - 1) \cdot 2(\tan x - 1) \sec^2 x$. Since $(\tan x - 1)^2 + a - 1 \ge a - 1$ and $f'$ is increasing,

$f'((\tan x - 1)^2 + a - 1) \ge f'(a-1) = 0$. Equality only when $\tan x = 1$, otherwise $f' > 0$. Thus sign of $g'(x)$ is same as sign of $\tan x - 1$. For $0 < x < \frac{\pi}{4}$, $\tan x < 1$, so $g'(x) < 0$ → $g$ decreasing. For $\frac{\pi}{4} < x < \frac{\pi}{2}$, $\tan x > 1$, so $g'(x) > 0$ → $g$ increasing. Hence (I) false, (II) false.

Neither (I) nor (II) is True.