JEE Main 2023 — Definite Integration Question with Solution

Given:

$f\left(x\right)+{\int }_0^{x}f\left(t\right)\sqrt{1-{\left({\log }_e\left(f\left(t\right)\right)\right)}^2}dt=e ...\left(1\right)$

Put $x=0$, then

$f\left(0\right)=e$.

Differentiating $\left(1\right)$ w.r.t. $x$, we get

$f^{\text{'}}\left(x\right)+f\left(x\right)\sqrt{1-{\left({\log }_e\left(f\left(x\right)\right)\right)}^2}=0$

Put $y=f\left(x\right)$.

$\frac{dy}{dx}+y\sqrt{1-{\left({\log }_{e}y\right)}^2}=0$

$\Rightarrow \int \frac{dy}{y\sqrt{1-{\left({\log }_{e}y\right)}^2}}=-\int dx$

$\Rightarrow \int \frac{d\left({\log }_{e}y\right)}{\sqrt{1-{\left({\log }_{e}y\right)}^2}}=-\int dx$

$\Rightarrow {\sin }^{-1}\left({\log }_{e}y\right)=-x+C$

Put $x=0, y=e$

${\sin }^{-1}\left({\log }_{e}e\right)=-0+C$

$\Rightarrow C=\frac{\mathrm{\pi }}{2}$

So,

${\sin }^{-1}\left({\log }_{e}y\right)=-x+\frac{\mathrm{\pi }}{2}$

$\Rightarrow {\log }_{e}y=\sin \left(-x+\frac{\mathrm{\pi }}{2}\right)$

$\Rightarrow {\log }_{e}f\left(x\right)=\sin \left(-x+\frac{\mathrm{\pi }}{2}\right)$

Put $x=\frac{\mathrm{\pi }}{6}$

$\Rightarrow {\log }_{e}f\left(\frac{\mathrm{\pi }}{6}\right)=\sin \left(-\frac{\mathrm{\pi }}{6}+\frac{\mathrm{\pi }}{2}\right)=\sin \left(\frac{\mathrm{\pi }}{3}\right)=\frac{\sqrt{3}}{2}$

$\Rightarrow 6{\log }_{e}f\left(\frac{\mathrm{\pi }}{6}\right)=3\sqrt{3}$

$\Rightarrow {\left\{6{\log }_{e}f\left(\frac{\mathrm{\pi }}{6}\right)\right\}}^2=27$