JEE Advanced 2026 — Paper 1 — MATH-I solution

JEE Advanced2026Paper 1MATH-I

Consider the function $f : (0, \infty) \to (- \infty, \infty)$ given by

$f (x) = x lo g_{e} (x) - x + 1$

Then which one of the following statements is TRUE ?

A
The derivative of the function $f$ is decreasing in the interval $(0, 1)$
B
The function $f$ has a local maximum at some point $a \in (0, \infty)$
C
The function $f$ has a local minimum at some point $b \in (0, \infty)$
D
The function $f$ has NEITHER a point of local maximum NOR a point of local minimum in the interval $(0, \infty)$

Solution

Differentiating $f (x) = x ln x - x + 1$ :

$f^{'} (x) = \frac{x}{x} + \frac{ln x}{2 x} - 1 = \frac{ln x + 2 - 2 x}{2 x}$

Define $g (x) = ln x + 2 - 2 x$ . Then $g (1) = 0$ and

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

So $g^{'} (x) > 0$ for $x \in (0, 1)$ and $g^{'} (x) < 0$ for $x \in (1, \infty)$ , which makes $x = 1$ the global maximum of $g$ . Therefore $g (x) < g (1) = 0$ for every $x \in (0, \infty) ∖ {1}$ , and $g (1) = 0$ .

Hence $f^{'} (x) \leq 0$ for all $x \in (0, \infty)$ , with equality only at $x = 1$ . So $f$ is strictly decreasing on $(0, \infty)$ and has no local maximum or local minimum.

$∴$ The correct option is (D).

← Back to full paper

Practice more MATH-I

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →