Q36 / 75 · JEE Main 2026 (22 January Shift 1) · Mathematics
Let $f:[1, \infty) \rightarrow \mathbb{R}$ 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
- Correct answerB3
- C4
- D-3
Solution
Differentiating $6\int_1^x f(t)dt = 3xf(x) + x^3 - 4$:
$6f(x) = 3f(x) + 3xf'(x) + 3x^2 \Rightarrow f(x) - xf'(x) = x^2$.
$\frac{d}{dx}\left(\frac{f(x)}{x}\right) = \frac{xf'(x)-f(x)}{x^2} = -1$.
Integrating: $\frac{f(x)}{x} = -x + C \Rightarrow f(x) = -x^2 + Cx$.
At $x = 1$: $0 = 3f(1) - 3 \Rightarrow f(1) = 1$. So $-1 + C = 1 \Rightarrow C = 2$.
$f(x) = -x^2 + 2x$. $f(2) = 0$, $f(3) = -3$.
$f(2) - f(3) = 3$.
$6f(x) = 3f(x) + 3xf'(x) + 3x^2 \Rightarrow f(x) - xf'(x) = x^2$.
$\frac{d}{dx}\left(\frac{f(x)}{x}\right) = \frac{xf'(x)-f(x)}{x^2} = -1$.
Integrating: $\frac{f(x)}{x} = -x + C \Rightarrow f(x) = -x^2 + Cx$.
At $x = 1$: $0 = 3f(1) - 3 \Rightarrow f(1) = 1$. So $-1 + C = 1 \Rightarrow C = 2$.
$f(x) = -x^2 + 2x$. $f(2) = 0$, $f(3) = -3$.
$f(2) - f(3) = 3$.
Hence, the correct answer is option: 'B'.