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QJEEM23Ma0638: JEE Main 2023 Mathematics - Class 12 JEEM23Ma0638 Solution: Let be differentiable function defined on such that and . Then…

Let ff be aa differentiable function defined on [0,π2]\left[ {0,{\pi \over 2}} \right] such that f(x) > 0 and f(x)+∫0xf(t)1−(log⁡ef(t))2dt=e,∀x∈[0,π2]f(x) + \int_0^x {f(t)\sqrt {1 - {{({{\log }_e}f(t))}^2}} dt = e,\forall x \in \left[ {0,{\pi \over 2}} \right]} . Then (6log⁡ef(π6))2\left( {6{{\log }_e}f\left( {{\pi \over 6}} \right)} \right)^2 is equal to __________.

Step-by-Step Verified Solution

f(x)+∫0xf(t)1−(log⁡ef(t))2dt=e...(1)f(x)+\int_{0}^{x} f(t) \sqrt{1-\left(\log _{e} f(t)\right)^{2}} d t=e\quad...(1)

So, f(0)=ef(0)=e

Now differentiate w.r. to xx

f′(x)+f(x)1−(log⁡ef(x)2=0f′(x)f(x)1−(log⁡ef(x))2=−1\begin{gathered}f^{\prime}(x)+f(x) \sqrt{1-\left(\log _{e} f(x)^{2}\right.}=0 \\\\\frac{f^{\prime}(x)}{f(x) \sqrt{1-\left(\log _{e} f(x)\right)^{2}}}=-1\end{gathered}

Let log⁡ef(x)=t\log _{e} f(x)=t

∴∫dt1−t2=−x+c\therefore \int \frac{d t}{\sqrt{1-t^{2}}}=-x+c

⇒sin⁡−1t=−x+c\Rightarrow \sin ^{-1} t=-x+c

Now f(0)=e⇒t=1f(0)=e \Rightarrow t=1 So, c=π2c=\frac{\pi}{2}

∴t=sin⁡(π2−x)=cos⁡x(∵x∈[0,π2])\therefore t=\sin \left(\frac{\pi}{2}-x\right)=\cos x \quad\left(\because x \in\left[0, \frac{\pi}{2}\right]\right)

∴(6log⁡ef(π6))2=27\therefore\left(6 \log _{e} f\left(\frac{\pi}{6}\right)\right)^{2}=27

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