Question

Let $f$ be $a$ differentiable function defined on $\left[0,\frac{\pi }{2}\right]$ such that $f(x)>0$ and $f(x)+\underset{0}{\overset{x}{\int }}f(t)\sqrt{1-{\left({\log }_{e}f(t)\right)}^2}dt=e,\forall x\in \left[0,\frac{\pi }{2}\right]$. Then ${\left(6{\log }_{e}f\left(\frac{\pi }{6}\right)\right)}^2$ is equal to _______

Answer 1

Answer: 27

$f(x)+\underset{0}{\overset{x}{\int }}f(t)\sqrt{1-{\left({\log }_{e}f(t)\right)}^2}dt=e$
$\Rightarrow f(0)=e$
$f^{\prime }(x)+f(x)\sqrt{1-(\ln f(x){)}^2}=0$
$f(x)=y$
$\frac{dy}{dx}=-y\sqrt{1-(\ln y{)}^2}$
$\int \frac{dy}{y\sqrt{1-(\ln y{)}^2}}=-\int dx$
Put $\ln y=t$
$\int \frac{dt}{\sqrt{1-t^2}}=-x+C$
${\sin }^{-1}t=-x+C\Rightarrow {\sin }^{-1}(\ln y)=-x+C$
${\sin }^{-1}(\ln f(x))=-x+C$
$f(0)=e$
$\Rightarrow \frac{\pi }{2}=C$
$\Rightarrow {\sin }^{-1}(\ln f(x))=-x+\frac{\pi }{2}$
$\Rightarrow {\sin }^{-1}\left(\ln f\left(\frac{\pi }{6}\right)\right)=\frac{-\pi }{6}+\frac{\pi }{2}$
$\Rightarrow {\sin }^{-1}\left(\ln f\left(\frac{\pi }{6}\right)\right)=\frac{\pi }{3}$
$\Rightarrow \ln f\left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2},$ we need ${\left(6\times \frac{\sqrt{3}}{2}\right)}^2=27$