JEE Main 2025 (Online) 2nd April Evening ShiftQuestion 1
Let $f:[1,\infty)\to[2,\infty)$ be a differentiable function. If $10\int_{1}^{x}f(t)\,dt=5xf(x)-x^{5}-9$ for all $x\ge1$, then the value of $f(3)$ is :
Differential Equation
Solution
Concept. Put $x=1$ for $f(1)$, then differentiate: $\left(\frac fx\right)'=x^2$. Formulas used. (1) $\frac{d}{dx}\int_1^xf(t)\,dt=f(x)$ and $\left(\dfrac fx\right)'=\dfrac{xf'-f}{x^2}$Step 1 - $f(1)$. At $x=1$: $0=5f(1)-1-9$, so $f(1)=2$. Step 2 - Differentiate. $10f=5f+5xf'-5x^4$, so $xf'-f=x^4$ and by (1), $\left(\frac fx\right)'=x^2$. Step 3 - Integrate. $\frac fx=\frac{x^3}{3}+C$, so $f=\frac{x^4}{3}+Cx$. $f(1)=2$ gives $C=\frac53$. Step 4 - Value. $f(3)=27+5=32$. Answer: (D)