Consider the following three statements for the function \(\mathrm{f}:(0, \infty) \rightarrow \mathbb{R}\) defined by \(\mathrm{f}(\mathrm{x})=\left|\log _{\mathrm{e}} \mathrm{x}\right|-|\mathrm{x}-1|:\) (I) f is differentiable at all \(\mathrm{x}>0\). (II) f is increasing in \((0,1)\). (III) f is decreasing in ( \(1, \infty\) ). Then.

(D) Only (I) and (III) are TRUE.

(A) All (I), (II) and (III) are TRUE.

(C) Only (II) and (III) are TRUE.

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Continuity and Differentiability

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.

Answer: (D)

For

x \geq 1

f (x) = ln x - (x - 1)

with

f^{'} (x) = \frac{1}{x} - 1

; for

0 < x < 1

f (x) = - ln x + (x - 1)

with

f^{'} (x) = 1 - \frac{1}{x}

. At

x = 1

both one-sided derivatives equal

0

and

f

is continuous, so

f

is differentiable throughout

(0, \infty)

, and statement (I) holds. On

(0, 1)

1 - \frac{1}{x} < 0

, so

f

is decreasing and statement (II) is false. On

(1, \infty)

\frac{1}{x} - 1 < 0

, so

f

is decreasing and statement (III) holds. Thus only (I) and (III) are true.

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Single correct option