JEE Main 2025 — Definite Integration Question with Solution

Given, ${\int }_0^{1}f(\lambda x)d\lambda =af(x)$ Let $\lambda x=u$ $d\lambda =\frac{1}{x}du$ $\therefore$ From (1) $\frac{1}{x}{\int }_0^{x}f(u)du=af(x)$ $\Rightarrow {\int }_0^{x}f(u)du=\mathrm{axf}(x)$ Differentiate both sides $\begin{aligned} & f(x)=a\left(x f^{\prime}(x)+f(x)\right) \\ & \Rightarrow \quad f(x)=a x f^{\prime}(x)+a f(x) \\ & \Rightarrow \quad(1-a) f(x)=a x f^{\prime}(x) \\ & \Rightarrow \quad \frac{f^{\prime}(x)}{f(x)}=\frac{(1-a)}{a} \cdot \frac{1}{x} \end{aligned}$ Integrate both side w.r.t. ( $x$ ) $\begin{aligned} & \Rightarrow \quad \int \frac{f^{\prime}(x)}{f(x)} d x=\frac{(1-a)}{a} \int \frac{1}{x} d x \\ & \Rightarrow \ln f(x)=\left(\frac{1-a}{a}\right) \ln x+c \end{aligned}$ Now at $x=1f(1)=1$ $\Rightarrow c=0$ Also given $f(16)=\frac{1}{8}$ $\begin{aligned} & \therefore \quad \frac{1}{8}=(16)^{\frac{1-a}{a}} \\ & \Rightarrow \quad 2^{-3}=2^{\frac{4-4 a}{a}} \\ & \Rightarrow \quad-3=\frac{4-4 a}{a} \\ & \Rightarrow-3 a=4-4 a \\ & \Rightarrow a=4 \\ & \therefore \quad f(x)=x^{-3 / 4} \\ & f(x)=\frac{-3}{4} x^{-\frac{7}{4}} \end{aligned}$ $\begin{array}{rl}& \text{ Put }x=\frac{1}{16}\\ & f^{\prime }\left(\frac{1}{16}\right)=\frac{-3}{4}{\left(\frac{1}{16}\right)}^{-7/4}=\frac{-3}{4}\cdot 2^{-4x\left(\frac{-7}{4}\right)}=-96\\ & \therefore 16-f^{\prime }\left(\frac{1}{16}\right)\Rightarrow 16-(-96)=112\end{array}$