JEE Advanced 2026 — Paper 1 — MATH-II solution

JEE Advanced2026Paper 1MATH-II

Let $R$ denote the set of all real numbers. Let $f : R \to R$ be an arbitrary function and let $g : R \to R$ be the function defined by

$g (x) = x f (x)$ , for all $x \in R$ .

Then which of the following statements is (are) TRUE ?

A
The function $g$ is always continuous at $x = 0$
B
If $f$ is continuous at $x = 0$ , then $g$ is differentiable at $x = 0$
C
If $g$ is differentiable at $x = 0$ , then $f$ is continuous at $x = 0$
D
If $g$ is differentiable at $x = 0$ , then $x \to 0 lim f (x)$ exists

Solution

(A) Take $f (x) = \frac{1}{x} sin (\frac{1}{x})$ for $x \neq = 0$ and $f (0) = 0$ . Then $g (x) = x f (x) = sin (1/ x)$ for $x \neq = 0$ and $g (0) = 0$ . Since $sin (1/ x)$ has no limit as $x \to 0$ , $g$ is not continuous at $0$ . FALSE.

(B) Suppose $f$ is continuous at $0$ . Then $g (0) = 0$ and

$g^{'} (0) = lim_{h \to 0} \frac{g ( h ) - g ( 0 )}{h} = lim_{h \to 0} \frac{h f ( h )}{h} = lim_{h \to 0} f (h) = f (0)$

The limit exists (finite), so $g$ is differentiable at $0$ . TRUE.

(C) If $g$ is differentiable at $0$ , then $lim_{h \to 0} f (h)$ exists (as shown below), but this limit need not equal $f (0)$ . For example, redefining $f (0)$ to be any other value keeps $g (x) = x f (x)$ differentiable at $0$ while breaking continuity of $f$ . FALSE.

(D) Differentiability of $g$ at $0$ requires $lim_{h \to 0} \frac{g ( h ) - g ( 0 )}{h} = lim_{h \to 0} f (h)$ to exist (note $g (0) = 0 \cdot f (0) = 0$ ). TRUE.

$∴$ The correct options are (B) and (D).

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