JEE Main 2026 — Jan 22, Shift 1 — Mathematics solution

JEE Main2026Jan 22, Shift 1Mathematics

Let $f : [1, \infty) \to R$ be a differentiable function. If $6 \int_{1}^{x} f (t) d t = 3 x f (x) + x^{3} - 4$ for all $x \geq 1$ , then the value of $f (2) - f (3)$ is

A
−4
B
−3
C
4
D
3

Solution

Differentiate both sides w.r.t. $x$ (Leibniz on the LHS): $6 f (x) = 3 f (x) + 3 x f^{'} (x) + 3 x^{2}$ , so $3 f (x) - 3 x f^{'} (x) = 3 x^{2}$ , i.e. $x f^{'} (x) - f (x) = - x^{2}$ .

Divide by $x^{2}$ : $\frac{x f ^{'} ( x ) - f ( x )}{x ^{2}} = - 1$ , which is $\frac{d}{d x} (\frac{f ( x )}{x}) = - 1$ .

Integrate: $\frac{f ( x )}{x} = - x + C$ , so $f (x) = - x^{2} + C x$ .

Use the original equation at $x = 1$ : $0 = 3 f (1) + 1 - 4 \Rightarrow f (1) = 1$ . Then $1 = - 1 + C \Rightarrow C = 2$ , giving $f (x) = - x^{2} + 2 x$ .

$f (2) = 0$ , $f (3) = - 3$ . $f (2) - f (3) = 3$ .

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