Mathematics · Definite Integration

JEE Main 2025 — 2 April, Evening Shift — Question 32

Let $f:[1, \infty) \rightarrow[2, \infty)$ be a differentiable function. If $10 \int_{1}^{x} f(t) d t=5 x f(x)-x^{5}-9$ for all

$x \geq 1$ , then the value of $f(3)$ is :

Answer: B

$10 \int_{1}^{x} f(t) d t=5 x f(x)-x^{5}-9$

differentiate w.r.t. $x$ both sides

\begin{aligned} & 10 f(x)=5 f(x)+5 x f^{\prime}(x)-5 x^{4} \\& \Rightarrow 5 f(x)=5 x f^{\prime}(x)-5 x^{4} \\& \text { Let } y=f(x) \Rightarrow \frac{d y}{d x}=f^{\prime}(x) \\& 5 y=5 x \frac{d y}{d x}-5 x^{4} \Rightarrow \frac{d y}{d x}-\frac{y}{x}=x^{3} \end{aligned}

I.F. of this linear differential equation

\begin{aligned} & \Rightarrow \text { I.F. }=e^{\int \frac{-1}{x} d x}=e^{-\ln x}=e^{\ln \left(\frac{1}{x}\right)}=\frac{1}{x} \\& \Rightarrow y\left(\frac{1}{x}\right)=\int x^{3} \cdot \frac{1}{x} d x=\frac{x^{3}}{3}+C \\& \Rightarrow y=\frac{x^{4}}{3}+C x, \text { at } x=1 \\& 10 \int_{1}^{1} f(t) d t=5 f(1)-1-9 \Rightarrow f(1)=2 \\& \Rightarrow f(1)=\frac{1}{3}+C=2 \quad \Rightarrow C=\frac{5}{3} \\& \Rightarrow f(3)=\frac{3^{4}}{3}+\frac{5}{3} \cdot 3=3^{3}+5=32 \end{aligned}

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