Let $f:(0, \infty) \rightarrow \mathbf{R}$ be a twice differentiable function. If for some $\mathrm{a} \neq 0, \int_0^1 f(\lambda x) \mathrm{d} \lambda=\mathrm{a} f(x), f(1)=1$ and $f(16)=\frac{1}{8}$, then $16-f^{\prime}\left(\frac{1}{16}\right)$ is equal to _______.

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Aicharya · Accepted Answer

Given, $\int_0^1 f(\lambda x) d \lambda=a f(x)$
Let $\lambda x=u$
$d \lambda=\frac{1}{x} d u$
$\therefore$ From (1) $\frac{1}{x} \int_0^x f(u) d u=a f(x)$
$\Rightarrow \int_0^x f(u) d u=\operatorname{axf}(x)$
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 $x=1 f(1)=1$
$\Rightarrow c=0$
Also given $f(16)=\frac{1}{8}$
$\begin{aligned}
& \therefore \quad \frac{1}{8}=(16)^{\frac{1-a}{a}} \\ & \Rightarrow \quad 2^{-3}=2^{\frac{4-4 a}{a}} \\ & \Rightarrow \quad-3=\frac{4-4 a}{a} \\ & \Rightarrow-3 a=4-4 a \\ & \Rightarrow a=4 \\ & \therefore \quad f(x)=x^{-3 / 4} \\ & f(x)=\frac{-3}{4} x^{-\frac{7}{4}}
\end{aligned}$
$\begin{aligned} & \text { Put } x=\frac{1}{16} \\ & f^{\prime}\left(\frac{1}{16}\right)=\frac{-3}{4}\left(\frac{1}{16}\right)^{-7 / 4}=\frac{-3}{4} \cdot 2^{-4 x\left(\frac{-7}{4}\right)}=-96 \\ & \therefore \quad 16-f^{\prime}\left(\frac{1}{16}\right) \Rightarrow 16-(-96)=112\end{aligned}$

Let \(f:(0, \infty) \rightarrow \mathbf{R}\) be a twice differentiable function. If for some \(\mathrm{a}…

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