JEE Main 2026 — Application of Derivatives Question with Solution

Given $f^{\prime \prime }(x)>0$ for all $x\in \mathbb{R}$, which means $f^{\prime }(x)$ is a strictly increasing function.
We are given $f^{\prime }(a-1)=0$. Since $f^{\prime }(x)$ is strictly increasing, $f^{\prime }(x)<0$ for $x<a-1$ and $f^{\prime }(x)>0$ for $x>a-1$.
The function $g(x)$ is defined as $g(x)=f({\tan }^{2}x-2\tan x+a)$ for $x\in (0,\pi /2)$.
Let $u(x)={\tan }^{2}x-2\tan x+a=(\tan x-1{)}^2+a-1$.
Differentiating $g(x)$ with respect to $x$:
$g^{\prime }(x)=f^{\prime }(u(x))\cdot u^{\prime }(x)=f^{\prime }((\tan x-1{)}^2+a-1)\cdot (2\tan x{\sec }^{2}x-2{\sec }^{2}x)$
$g^{\prime }(x)=f^{\prime }((\tan x-1{)}^2+a-1)\cdot 2{\sec }^{2}x(\tan x-1)$.

Case 1: $x\in (0,\pi /4)$.
In this interval, $0<\tan x<1$, so $(\tan x-1)<0$.
Also, $(\tan x-1{)}^2>0$, so $u(x)=(\tan x-1{)}^2+a-1>a-1$.
Since $f^{\prime }(x)$ is increasing and $f^{\prime }(a-1)=0$, for $u(x)>a-1$, we have $f^{\prime }(u(x))>0$.
Thus, $g^{\prime }(x)=(+)(+)(-)<0$. So $g(x)$ is decreasing in $(0,\pi /4)$.
Statement (I) is False.

Case 2: $x\in (\pi /4,\pi /2)$.
In this interval, $\tan x>1$, so $(\tan x-1)>0$.
Also, $(\tan x-1{)}^2>0$, so $u(x)=(\tan x-1{)}^2+a-1>a-1$.
Thus, $f^{\prime }(u(x))>0$.
Then $g^{\prime }(x)=(+)(+)(+)>0$. So $g(x)$ is increasing in $(\pi /4,\pi /2)$.
Statement (II) is False.

Therefore, neither (I) nor (II) is true.