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JEE Main 29-Jan-2025 Paper - Shift 1 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 29-Jan-2025 Paper - Shift 14 Marks
Question
Let $\mathrm{f}:(0, \infty) \rightarrow \mathrm{R}$ be a twice differentiable function. If for some $a \neq 0, \int_{0}^{1} \mathrm{f}(\lambda \mathrm{x}) \mathrm{d} \lambda=\operatorname{af}(\mathrm{x})$, $f(1)=1$ and $f(16)=\frac{1}{8}$, then $16-f^{\prime}\left(\frac{1}{16}\right)$ is equal to __________ .
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Answer

(112)
Sol. $\int_{0}^{1} f(\lambda x) d \lambda=a f(x)$
$\lambda \mathrm{x}=\mathrm{t}$
$\mathrm{d} \lambda=\frac{1}{\mathrm{x}} \mathrm{dt}$
$\frac{1}{x} \int_{0}^{x} f(t) d t=a f(x)$
$\int_{0}^{x} f(t) d t=\operatorname{axf}(x)$
$\mathrm{f}(\mathrm{x})=\mathrm{a}\left(\mathrm{x} \mathrm{f}^{\prime}(\mathrm{x})+\mathrm{f}(\mathrm{x})\right)$
$(1-a) f(x)=a . x f^{\prime}(x)$
$\frac{f^{\prime}(x)}{f(x)}=\frac{(1-a)}{a} \frac{1}{x}$
$\operatorname{lnf}(\mathrm{x})=\frac{1-\mathrm{a}}{\mathrm{a}} \operatorname{\ell n} \mathrm{x}+\mathrm{c}$
$\mathrm{x}=\mathrm{l}, \mathrm{f}(1)=\mathrm{l} \Rightarrow \mathrm{c}=0$
$\mathrm{x}=16, \mathrm{f}(16)=\frac{1}{8}$
$\frac{1}{8}=(16)^{\frac{1-a}{a}} \Rightarrow-3=\frac{4-4 a}{a} \Rightarrow a=4$
$f(x)=x^{-\frac{3}{4}}$
$f^{\prime}(x)=-\frac{3}{4} x^{-\frac{7}{4}}$
$\therefore 16-\mathrm{f}^{\prime}\left(\frac{1}{16}\right)$
$=16-\left(-\frac{3}{4}\left(2^{-4}\right)^{-7 / 4}\right)$
$=16+96=112$

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