JEE Main 2026 — Mathematics Calculus

Given $f''(x) > 0$f′′(x)>0 for all $x \in \mathbb{R}$x∈R, which means $f'(x)$f′(x) is a strictly increasing function.

We are given $f'(a-1) = 0$f′(a−1)=0. Since $f'(x)$f′(x) is strictly increasing, $f'(x) < 0$f′(x)<0 for $x < a-1$x<a−1 and $f'(x) > 0$f′(x)>0 for $x > a-1$x>a−1.

The function $g(x)$g(x) is defined as $g(x) = f(\tan^2 x - 2\tan x + a)$g(x)=f(tan2x−2tanx+a) for $x \in (0, \pi/2)$x∈(0,π/2).

Let $u(x) = \tan^2 x - 2\tan x + a = (\tan x - 1)^2 + a - 1$u(x)=tan2x−2tanx+a=(tanx−1)2+a−1.

Differentiating $g(x)$g(x) with respect to $x$x:

$$g'(x) = f'(u(x)) \cdot u'(x) = f'((\tan x - 1)^2 + a - 1) \cdot (2\tan x \sec^2 x - 2\sec^2 x)$$g′(x)=f′(u(x))⋅u′(x)=f′((tanx−1)2+a−1)⋅(2tanxsec2x−2sec2x)

$g'(x) = f'((\tan x - 1)^2 + a - 1) \cdot 2\sec^2 x (\tan x - 1)$g′(x)=f′((tanx−1)2+a−1)⋅2sec2x(tanx−1).

Case 1: $x \in (0, \pi/4)$x∈(0,π/4).

In this interval, $0 < \tan x < 1$0<tanx<1, so $(\tan x - 1) < 0$(tanx−1)<0.

Also, $(\tan x - 1)^2 > 0$(tanx−1)2>0, so $u(x) = (\tan x - 1)^2 + a - 1 > a - 1$u(x)=(tanx−1)2+a−1>a−1.

Since $f'(x)$f′(x) is increasing and $f'(a-1) = 0$f′(a−1)=0, for $u(x) > a - 1$u(x)>a−1, we have $f'(u(x)) > 0$f′(u(x))>0.

Thus, $g'(x) = (+)(+)(-) < 0$g′(x)=(+)(+)(−)<0. So $g(x)$g(x) is decreasing in $(0, \pi/4)$(0,π/4).

Statement (I) is False.

Case 2: $x \in (\pi/4, \pi/2)$x∈(π/4,π/2).

In this interval, $\tan x > 1$tanx>1, so $(\tan x - 1) > 0$(tanx−1)>0.

Also, $(\tan x - 1)^2 > 0$(tanx−1)2>0, so $u(x) = (\tan x - 1)^2 + a - 1 > a - 1$u(x)=(tanx−1)2+a−1>a−1.

Thus, $f'(u(x)) > 0$f′(u(x))>0.

Then $g'(x) = (+)(+)(+) > 0$g′(x)=(+)(+)(+)>0. So $g(x)$g(x) is increasing in $(\pi/4, \pi/2)$(π/4,π/2).

Statement (II) is False.

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