JEE Main 2026 — Jan 21, Shift 2 — Mathematics solution

JEE Main2026Jan 21, Shift 2Mathematics

Let $f : R \to R$ be a twice differentiable function such that $f^{''} (x) > 0$ for all $x \in R$ and $f^{'} (a - 1) = 0$ , where a is real number. Let $g (x) = f (tan^{2} x - 2 tan x + a)$ , $0 < x < \frac{π}{2}$ .

Consider the following two statements :

(I) g is increasing in $(0, \frac{π}{4})$

(II) g is decreasing in $(\frac{π}{4}, \frac{π}{2})$

Then,

A
Neither (I) nor (II) is True
B
Only (II) is True
C
Only (I) is True
D
Both (I) and (II) are True

Solution

Write the inner argument by completing the square: $tan^{2} x - 2 tan x + a = (tan x - 1)^{2} + (a - 1)$ . So $g (x) = f ((tan x - 1)^{2} + a - 1)$ .

Differentiate: $g^{'} (x) = f^{'} ((tan x - 1)^{2} + a - 1) \cdot 2 (tan x - 1) sec^{2} x$ .

Since $f^{''} (x) > 0$ everywhere and $f^{'} (a - 1) = 0$ , $f^{'}$ is strictly increasing with a single zero at $a - 1$ . The argument $(tan x - 1)^{2} + (a - 1) \geq a - 1$ , with equality only at $x = π /4$ . So $f^{'} (\cdot) > 0$ for all $x \neq = π /4$ .

Therefore the sign of $g^{'} (x)$ matches the sign of $(tan x - 1)$ :

• On $(0, π /4)$ : $tan x < 1$ , so $g^{'} (x) < 0$ — g is decreasing. So (I) is false.

• On $(π /4, π /2)$ : $tan x > 1$ , so $g^{'} (x) > 0$ — g is increasing. So (II) is false.

Neither (I) nor (II) is true.

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