Let f be a differentiable function defined on [0, 2 ] such that f… - Vidyadip

MCQ

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 $.............$

A
$25$
B
$26$
C
$23$
✓
$27$

✓

Answer

Correct option: D.

$27$

d
$f ( x )+\int_0^{ x } f ( t ) \sqrt{1-\left(\log _{ e } f ( t )\right)^2} dt = e$

$\Rightarrow f (0)= e$

$f ^{\prime}( x )+ f ( x ) \sqrt{1-(\ln f ( x ))^2}=0$

$f ( x )= y$

$\frac{ dy }{ dx }=- y \sqrt{1-(\ln y )^2}$

$\int \frac{ dy }{ y \sqrt{1-(\ln y )^2}}=-\int dx$

$\text { Put } \ln y = t$

$\int \frac{ dt }{\sqrt{1-t^2}}=- x + C$

$\sin ^{-1} t=-x+C \Rightarrow \sin ^{-1}(\ln y )=- x + C$

$\sin ^{-1}(\ln f ( x ))=- x + C$

$f(0)= e$

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

$\Rightarrow \sin ^{-1}(\ln f ( x ))=- x +\frac{\pi}{2}$

$\Rightarrow \sin ^{-1}\left(\ln f \left(\frac{\pi}{6}\right)\right)=\frac{-\pi}{6}+\frac{\pi}{2}$

$\Rightarrow \sin ^{-1}\left(\ln f \left(\frac{\pi}{6}\right)\right)=\frac{\pi}{3}$ $\Rightarrow \ln f \left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}, \text { we need }\left(6 \times \frac{\sqrt{3}}{2}\right)^2=27$

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