Question

Let $f$ be $a$ differentiable function defined on $\left[0, \frac{\pi}{2}\right]$ such that $f(x)>0$ and $f(x)+\int\limits_0^x f(t) \sqrt{1-\left(\log _e f(t)\right)^2} d t=e, \forall x \in\left[0, \frac{\pi}{2}\right]$. Then $\left(6 \log _e f\left(\frac{\pi}{6}\right)\right)^2$ is equal to _______

Answer 1

$ f(x)+\int\limits_0^x f(t) \sqrt{1-\left(\log _e f(t)\right)^2} d t=e$
$ \Rightarrow f(0)=e$
$ f^{\prime}(x)+f(x) \sqrt{1-(\ln f(x))^2}=0 $
$ f(x)=y $
$ \frac{d y}{d x}=-y \sqrt{1-(\ln y)^2}$
$ \int \frac{d y}{y \sqrt{1-(\ln y)^2}}=-\int d x $
Put $\ln y=t $
$ \int \frac{d t}{\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 $\left(6 \times \frac{\sqrt{3}}{2}\right)^2=27$