Question

If $f:R\to R^{+}$ is a function defined as $\left[f\right(x){]}^2=\underset{0}{\overset{x}{\int }}[\{f\left(t\right){\}}^2+\left\{f^{\prime }\right(t){\}}^2]dt+100,$ then $f\left(\ln 2\right)=$

Answer 1

$\left[f\right(x){]}^2=\underset{0}{\overset{x}{\int }}[\{f\left(t\right){\}}^2+\left\{f^{\prime }\right(t){\}}^2]dt+100$
Differentiating w.r.t $x$
$2f\left(x\right).f^{\prime }\left(x\right)=\left[f\right(x){]}^2+[f^{\prime }\left(x\right){]}^2[\because \frac{d}{dx}\underset{a}{\overset{x}{\int }}f(t)dt=f(x\left)\right]$
$\Rightarrow \left[f\right(x)-f^{\prime }(x){]}^2=0\Rightarrow f(x)=f^{\prime }(x)\Rightarrow \int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}dx=\int dx+c\Rightarrow \ln |f(x)|=x+\ln C\Rightarrow f(x)=C{e}^x$
Now, for $x=0$
$\left[f\right(0){]}^2=\underset{0}{\overset{0}{\int }}[\{f\left(0\right){\}}^2+\left\{f^{\prime }\right(0){\}}^2]dt+100\Rightarrow [f\left(0\right){]}^2=100\Rightarrow f\left(0\right)=10(\because f:R\to R^{+})$
$\therefore f\left(x\right)=10e^x$