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

The correct answer is 27
f(x)+0∫x​f(t)1−(loge​f(t))2​dt=e
⇒f(0)=e
f′(x)+f(x)1−(lnf(x))2​=0
f(x)=y
dxdy​=−y1−(lny)2​
∫y1−(lny)2​dy​=−∫dx
Put lny=t
∫1−t2​dt​=−x+C
sin−1t=−x+C⇒sin−1(lny)=−x+C
sin−1(lnf(x))=−x+C
f(0)=e
⇒2π​=C
⇒sin−1(lnf(x))=−x+2π​
⇒sin−1(lnf(6π​))=6−π​+2π​
⇒sin−1(lnf(6π​))=3π​
⇒lnf(6π​)=23​​, we need (6×23​​)2=27