Q35 / 75 · JEE Main 2026 (24 January Shift 2) · Mathematics

Consider the following three statements for the function $f:(0, \infty) \rightarrow \mathbb{R}$ defined by $f(x)=\left|\log _{\mathrm{e}} x\right|-|x-1|$ :
(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 (II) and (III) are TRUE.
  • C
    Only (I) is TRUE.
  • Correct answer
    D
    Only (I) and (III) are TRUE.

Solution

For $0 < x < 1$: $f(x) = -\ln x + x - 1$, $f'(x) = -\frac{1}{x} + 1 = \frac{x-1}{x} < 0$ (decreasing).
For $x > 1$: $f(x) = \ln x - x + 1$, $f'(x) = \frac{1}{x} - 1 = \frac{1-x}{x} < 0$ (decreasing).
At $x = 1$: LHD $= \frac{1-1}{1} = 0$, RHD $= \frac{1-1}{1} = 0$. So $f$ is differentiable at $x = 1$.
(I) TRUE: $f$ is differentiable for all $x > 0$.
(II) FALSE: $f$ is decreasing in $(0, 1)$.
(III) TRUE: $f$ is decreasing in $(1, \infty)$.
Only (I) and (III) are TRUE.

Hence, the correct answer is option: 'D'.

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