Mathematics - Definite Integration Question with Solution | TestHub

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 $\begin{array}{rl}& f(x)=a\left(x{f}^{\prime }(x)+f(x)\right)\\ & \Rightarrow f(x)=ax{f}^{\prime }(x)+af(x)\\ & \Rightarrow (1-a)f(x)=ax{f}^{\prime }(x)\\ & \Rightarrow \frac{f^{\prime }(x)}{f(x)}=\frac{(1-a)}{a}\cdot \frac{1}{x}\end{array}$ Integrate both side w.r.t. ( $x$ ) $\begin{array}{rl}& \Rightarrow \int \frac{f^{\prime }(x)}{f(x)}dx=\frac{(1-a)}{a}\int \frac{1}{x}dx\\ & \Rightarrow \ln f(x)=\left(\frac{1-a}{a}\right)\ln x+c\end{array}$ Now at $x=1f(1)=1$ $\Rightarrow c=0$ Also given $f(16)=\frac{1}{8}$ $\begin{array}{rl}& \therefore \frac{1}{8}=(16{)}^{\frac{1-a}{a}}\\ & \Rightarrow 2^{-3}=2^{\frac{4-4a}{a}}\\ & \Rightarrow -3=\frac{4-4a}{a}\\ & \Rightarrow -3a=4-4a\\ & \Rightarrow a=4\\ & \therefore f(x)=x^{-3/4}\\ & f(x)=\frac{-3}{4}{x}^{-\frac{7}{4}}\end{array}$ $\begin{array}{rl}& \text{ Put }x=\frac{1}{16}\\ & f^{\prime }\left(\frac{1}{16}\right)=\frac{-3}{4}{\left(\frac{1}{16}\right)}^{-7/4}=\frac{-3}{4}\cdot 2^{-4x\left(\frac{-7}{4}\right)}=-96\\ & \therefore 16-f^{\prime }\left(\frac{1}{16}\right)\Rightarrow 16-(-96)=112\end{array}$