JEE Main 2026 — 24 January, Evening Shift — Question 15

For $0 < x < 1$: $\ln x < 0$, $x-1 < 0$, so $f(x) = -\ln x - (1-x) = -\ln x + x - 1$. For $x \ge 1$: $\ln x \ge 0$, $x-1 \ge 0$, so $f(x) = \ln x - (x-1) = \ln x - x + 1$. Differentiate: $f'(x) = -\frac{1}{x} + 1$ for $0 < x < 1$, and $f'(x) = \frac{1}{x} - 1$ for $x > 1$. At $x=1$, left derivative = $-1+1=0$, right derivative = $1-1=0$, so $f$ is differentiable at $x=1$.

Hence $f$ is differentiable for all $x>0$. Statement (I) is true. For $0 < x < 1$: $f'(x) = 1 - \frac{1}{x} < 0$? Actually $1 - \frac{1}{x} < 0$ for $0 1$: $f'(x) = \frac{1}{x} - 1 < 0$, so $f$ is decreasing in (1,∞). Statement (III) is true. Thus only (I) and (III) are true. Correct option: D.