JEE Main 2025 — 29 January, Morning Shift — Question 65

$\int_{0}^{1} f(\lambda x) d \lambda=a f(x)$

$\lambda \mathrm{x}=\mathrm{t}$

$\mathrm{d} \lambda=\frac{1}{\mathrm{x}} \mathrm{dt}$

$\frac{1}{x} \int_{0}^{x} f(t) d t=a f(x)$

$\int_{0}^{x} f(t) d t=\operatorname{axf}(x)$

$f(x)=a\left(x f^{\prime}(x)+f(x)\right)$

$(1-a) f(x)=a . x f^{\prime}(x)$

$\frac{f^{\prime}(x)}{f(x)}=\frac{(1-a)}{a} \frac{1}{x}$

$\operatorname{lnf}(x)=\frac{1-\mathrm{a}}{\mathrm{a}} \operatorname{\ell n} \mathrm{x}+\mathrm{c}$

$\mathrm{x}=1, \mathrm{f}(1)=1 \Rightarrow \mathrm{c}=0$

$\mathrm{x}=16, \mathrm{f}(16)=\frac{1}{8}$

$\frac{1}{8}=(16)^{\frac{1-a}{a}} \Rightarrow-3=\frac{4-4 a}{a} \Rightarrow a=4$

$f(x)=x^{-\frac{3}{4}}$

$f^{\prime}(x)=-\frac{3}{4} x^{-\frac{7}{4}}$

$\therefore 16-\mathrm{f}^{\prime}\left(\frac{1}{16}\right)$

$=16-\left(-\frac{3}{4}\left(2^{-4}\right)^{-7 / 4}\right)$

$=16+96=112$