JEE Mathematics 2026 Question with Solution

Given: $f''(x) > 0$ for all $x \in \mathbb{R}$, so $f'(x)$ is strictly increasing. Also, $f'(a-1) = 0$.

Find: Whether statements (I) and (II) about monotonicity of $g(x) = f(\tan^2 x - 2\tan x + a)$ are true.

Let

Then,

Using the chain rule,

Now differentiate $\nu(x)$:

Hence,

Since

for all admissible $x$, and $f'(x)$ is strictly increasing with $f'(a-1)=0$, we get

Now check the sign on each interval.

For $x \in \left(0, \frac{\pi}{4}\right)$, we have $0 < \tan x < 1$, so

Also, $\sec^2 x > 0$ and $f'(\nu(x)) \ge 0$. Therefore,

So $g$ is decreasing on $\left(0, \frac{\pi}{4}\right)$. Hence statement (I) is false.

For $x \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)$, we have $\tan x > 1$, so

Again, $\sec^2 x > 0$ and $f'(\nu(x)) \ge 0$. Therefore,

So $g$ is increasing on $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$. Hence statement (II) is false.

Therefore, both statements (I) and (II) are false. The correct option is D.

Given: $f''(x) > 0$ and $f'(a-1)=0$.

Find: The monotonic behaviour of $g(x)=f\big((\tan x-1)^2+a-1\big)$.

Because $f''(x)>0$, the function $f'(x)$ is strictly increasing. Therefore:

• if $y > a-1$, then $f'(y) > 0$,
• if $y = a-1$, then $f'(y) = 0$.

Now,

so

Thus the argument of $f'$ in $g'(x)$ is never less than $a-1$.

Differentiate:

The factor $f'\big((\tan x-1)^2+a-1\big)$ is non-negative, and $\sec^2 x$ is always positive. So the sign of $g'(x)$ is determined by the sign of $\tan x-1$.

On $\left(0,\frac{\pi}{4}\right)$, $\tan x<1$, hence $\tan x-1<0$ and so $g'(x)\le 0$.

On $\left(\frac{\pi}{4},\frac{\pi}{2}\right)$, $\tan x>1$, hence $\tan x-1>0$ and so $g'(x)\ge 0$.

Therefore, $g$ is not increasing on $\left(0,\frac{\pi}{4}\right)$ and not decreasing on $\left(\frac{\pi}{4},\frac{\pi}{2}\right)$. The correct option is D.