Aspire Faculty ID #17943 · Topic: JEE Main 2026 (22 January Morning Shift) · Just now
JEE Main 2026 (22 January Morning Shift)

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

Solution

$6\int_{1}^{x} f(t),dt = 3x f(x) + x^3 - 4$

Differentiate both side

$6f(x) = 3x f'(x) + 3f(x) + 3x^2$

$\Rightarrow 3f(x) = 3x f'(x) + 3x^2$

$\Rightarrow x f'(x) - f(x) = -x^2$

$\Rightarrow x\frac{dy}{dx} - y = -x^2$

$\Rightarrow \frac{d}{dx}\left(\frac{y}{x}\right) = -1$

$\Rightarrow \frac{y}{x} = -x + C$

$\Rightarrow f(x) = -x^2 + Cx$

At $x = 1,; y = 1 \Rightarrow C = 2$

$\Rightarrow f(x) = -x^2 + 2x$

$f(2) - f(3) = ( -4 + 4 ) - ( -9 + 6 ) = 3$

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