Let f be a differentiable function defined on [0, (π)/(2)] such...

Given:

$f\left(x\right) + {∫}_{0}^{x}f$\left(t\right)\sqrt{1 - {\left(\log_{e}\left(f\left(t\right)\right)\right)}^{2}}dt = e . . . \left(1\right)$</p><p>Put$x = 0$, then</p><p>$f\left(0\right) = e$.</p><p>Differentiating \left(1\right)$ w.r.t. $x$, we get

${f}^{'}\left(x\right) + f\left(x\right)\sqrt{1 - {\left(\log_{e}\left(f\left(x\right)\right)\right)}^{2}} = 0$

Put $y = f\left(x\right)$.

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

$⇒ ∫ \dfrac{dy}{y\sqrt{1 - {\left(\log_{e}y\right)}^{2}}} = - ∫ dx$

$⇒ ∫ \dfrac{d\left(\log_{e}y\right)}{\sqrt{1 - {\left(\log_{e}y\right)}^{2}}} = - ∫ dx$

$⇒ \sin^{- 1}\left(\log_{e}y\right) = - x + C$

Put $x = 0, y = e$

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

$⇒ C = \dfrac{\pi}{2}$

So,

$\sin^{- 1}\left(\log_{e}y\right) = - x + \dfrac{\pi}{2}$

$⇒ \log_{e}y = \sin \left(- x + \dfrac{\pi}{2}\right)$

$⇒ \log_{e}f\left(x\right) = \sin \left(- x + \dfrac{\pi}{2}\right)$

Put $x = \dfrac{\pi}{6}$

$⇒ \log_{e}f\left(\dfrac{\pi}{6}\right) = \sin \left(- \dfrac{\pi}{6} + \dfrac{\pi}{2}\right) = \sin \left(\dfrac{\pi}{3}\right) = \dfrac{\sqrt{3}}{2}$

$⇒ 6\log_{e}f\left(\dfrac{\pi}{6}\right) = 3\sqrt{3}$

$⇒ {\left\{6\log_{e}f\left(\dfrac{\pi}{6}\right)\right\}}^{2} = 27$