JEE Main 2026 — Jan 24, Shift 2 — Mathematics solution

JEE Main2026Jan 24, Shift 2Mathematics

Consider the following three statements for the function $f : (0, \infty) \to R$ defined by $f (x) = ∣ lo g_{e} x ∣ - ∣ x - 1∣$ :

(I) $f$ is differentiable at all $x > 0$ .

(II) $f$ is increasing in $(0, 1)$ .

(III) $f$ is decreasing in $(1, \infty)$ .

Then

A
All (I), (II) and (III) are TRUE.
B
Only (I) is TRUE.
C
Only (II) and (III) are TRUE.
D
Only (I) and (III) are TRUE.

Solution

Split based on $x \geq 1$ or $0 < x < 1$ :

For $x \geq 1$ : $∣ ln x ∣ = ln x$ , $∣ x - 1∣ = x - 1$ , so $f (x) = ln x - x + 1$ . $f^{'} (x) = \frac{1}{x} - 1 \leq 0$ for $x \geq 1$ , with equality only at $x = 1$ . So $f$ is decreasing on $(1, \infty)$ . ✓ (III true)

For $0 < x < 1$ : $∣ ln x ∣ = - ln x$ , $∣ x - 1∣ = 1 - x$ , so $f (x) = - ln x + x - 1$ . $f^{'} (x) = - \frac{1}{x} + 1 < 0$ on $(0, 1)$ . So $f$ is decreasing (not increasing) on $(0, 1)$ . ✗ (II false)

At $x = 1$ : $f^{'} (1^{+}) = 0$ and $f^{'} (1^{-}) = 0$ , so $f$ is differentiable at $x = 1$ . Elsewhere both pieces are differentiable. ✓ (I true)

$∴$ Only (I) and (III) are TRUE.

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