JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Application of Derivatives

Rewrite $g(x)=f((\tan x-1{)}^2+a-1)$. Then $g^{\prime }(x)=f^{\prime }((\tan x-1{)}^2+a-1)\cdot 2(\tan x-1){\sec }^{2}x$. Since $f^{\prime }(a-1)=0$ and $f^{\prime \prime }(x)>0$, the function $f^{\prime }$ is strictly increasing. At the minimum point $(\tan x-1{)}^2=0$ (i.e., $\tan x=1$ or $x=\frac{\pi }{4}$), we have $f^{\prime }(a-1)=0$. For $(\tan x-1{)}^2>0$, we have $f^{\prime }((\tan x-1{)}^2+a-1)>0$. Therefore, $g^{\prime }(x)>0$ when $\tan x-1>0$ (i.e., $x\in (\frac{\pi }{4},\frac{\pi }{2})$), so g is increasing on $(\frac{\pi }{4},\frac{\pi }{2})$. Similarly, $g^{\prime }(x)<0$ when $\tan x-1<0$ (i.e., $x\in (0,\frac{\pi }{4})$), so g is decreasing on $(0,\frac{\pi }{4})$. Thus, statement (I) is false and statement (II) is false.