JEE Main 2026 (Online) 21st January Evening ShiftQuestion 1
Let $f:\mathbf{R}\to\mathbf{R}$ be a twice differentiable function such that $f''(x)>0$ for all $x\in\mathbf{R}$ and $f'(a-1)=0$, where $a$ is a real number. Let $g(x)=f(\tan^2x-2\tan x+a)$, $0<x<\frac\pi2$. Consider the following two statements: (I) $g$ is increasing in $\left(0,\frac\pi4\right)$ (II) $g$ is decreasing in $\left(\frac\pi4,\frac\pi2\right)$. Then,
Differentiation and its Application
Solution
Concept. Write $g'=f'(u)\,u'$. Since $u\ge a-1$ and $f'$ is increasing with $f'(a-1)=0$, we get $f'(u)\ge0$; so the sign of $g'$ is the sign of $u'$. Formulas used. (1) $f''>0$ $\Rightarrow$ $f'$ is increasing(2) $u=\tan^2x-2\tan x+a=(\tan x-1)^2+a-1$Step 1 - Sign of $f'(u)$. By (2), $u\ge a-1$. By (1), $f'(u)\ge f'(a-1)=0$ (equality only at $x=\frac\pi4$). Step 2 - $g'$. $$g'(x)=f'(u)\cdot2(\tan x-1)\sec^2x.$$ Step 3 - Signs. On $\left(0,\frac\pi4\right)$, $\tan x<1$, so $g'<0$: $g$ decreases. On $\left(\frac\pi4,\frac\pi2\right)$, $g'>0$: $g$ increases. Step 4 - Conclude. (I) says increasing on $\left(0,\frac\pi4\right)$: false. (II) says decreasing on $\left(\frac\pi4,\frac\pi2\right)$: false. Answer: (A)