Q30 / 75 · JEE Main 2025 (2 Apr Shift 2) · Mathematics

Let $f:[1, \infty) \rightarrow[2, \infty)$ be a differentiable function, If $10 \int_1^{\mathrm{x}} f(\mathrm{t}) \mathrm{dt}=5 \mathrm{x} f(\mathrm{x})-\mathrm{x}^5-9$ for all $\mathrm{x} \geq 1$, then the value of $f(3)$ is :

  • A
    $18$
  • Correct answer
    B
    $32$
  • C
    $22$
  • D
    $26$

Solution

$\begin{aligned}<br/>& 10 \frac{\mathrm{~d}}{\mathrm{dx}} \int_1^{\mathrm{x}} \mathrm{f}(\mathrm{t}) \mathrm{dt}=\frac{\mathrm{d}}{\mathrm{dx}}\left(5 \mathrm{xf}(\mathrm{x})-\mathrm{x}^5-9\right) \\ & \Rightarrow 10 f(x)=5 f(x)+5 \mathrm{xf}^{\prime}(\mathrm{x})-5 \mathrm{x}^4 \\ & \Rightarrow \mathrm{f}(\mathrm{x})+\mathrm{x}^4=\mathrm{xf}^{\prime}(\mathrm{x}) \\ & \Rightarrow \mathrm{y}+\mathrm{x}^4=\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}} \\ & \Rightarrow \frac{d y}{d x}+y\left(-\frac{1}{x}\right)=x^3 \\ & \Rightarrow y e^{\int-\frac{1}{x} d x}=\int x^3 e^{\int-\frac{1}{x} d x}+c \\ & \Rightarrow \mathrm{ye}^{-\operatorname{tn}|x|}=\int \mathrm{x}^3 \mathrm{e}^{-\operatorname{tr}|x|}+\mathrm{c} \\ & \Rightarrow \frac{\mathrm{y}}{|\mathrm{x}|}=\int \frac{\mathrm{x}^3}{|\mathrm{x}|}+\mathrm{c} \\ & \Rightarrow \frac{\mathrm{y}}{\mathrm{x}}=\int \mathrm{x}^2+\mathrm{c} \\ & \Rightarrow \frac{\mathrm{y}}{\mathrm{x}}=\frac{\mathrm{x}^3}{3}+\mathrm{c}<br/>\end{aligned}$
Put $x=1$ in given equation
$\begin{aligned}<br/>& \Rightarrow 0=5 \mathrm{f}(1)-1-9 \Rightarrow \mathrm{f}(1)=2 \\ & \Rightarrow \frac{2}{1}=\frac{1}{3}+\mathrm{c} \Rightarrow \mathrm{c}=\frac{5}{3} \\ & \Rightarrow \frac{\mathrm{f}(3)}{3}=\frac{27}{3}+\frac{5}{3} \\ & \Rightarrow \mathrm{f}(3)=32<br/>\end{aligned}$

Hence, the correct answer is option: 'B'.

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