Let f:(0, ∞) → R be a twice differentiable function. If for some a ≠ 0, ∫_0^1 f(λ x) d λ= a f(x), f(1)=1 and f(16)= 1 8, then 16-f^prime ( 1 16 ) is…

Given, $\displaystyle \int_0^1 f(\lambda x) d \lambda=a f(x)$∫01​f(λx)dλ=af(x) Let $\lambda x=u$λx=u $d \lambda=\frac{1}{x} d u$dλ=x1​du $\therefore$∴ From (1) $\displaystyle \frac{1}{x} \int_0^x f(u) d u=a f(x)$x1​∫0x​f(u)du=af(x) $\displaystyle \Rightarrow \int_0^x f(u) d u=\operatorname{axf}(x)$⇒∫0x​f(u)du=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. ($Integratebothsidew.r.t.(x$)$)\begin{aligned}
& \Rightarrow \quad \int \frac{f^{\prime}(x)}{f(x)} d x=\frac{(1-a)}{a} \int \frac{1}{x} d x \
& \Rightarrow \ln f(x)=\left(\frac{1-a}{a}\right) \ln x+c
\end{aligned}$Now at$Nowatx=1 f(1)=1$$\Rightarrow c=0$Also given$Alsogivenf(16)=\frac{1}{8}$$\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{aligned} & \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^{-4 x\left(\frac{-7}{4}\right)}=-96 \ & \therefore \quad 16-f^{\prime}\left(\frac{1}{16}\right) \Rightarrow 16-(-96)=112\end{aligned}$