JEE Main 2023 — Mathematics Calculus

Given:

$$f(x)+{\int }_{0}^{x}f(t)\sqrt{1-{({\mathrm{log}}_{e}(f(t)))}^{2}}dt=e...(1)$$f(x)+∫0x​f(t)1−(loge​(f(t)))2​dt=e...(1)

Put $x=0$x=0, then

$f(0)=e$f(0)=e.

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

$${f}^{'}(x)+f(x)\sqrt{1-{({\mathrm{log}}_{e}(f(x)))}^{2}}=0$$f′(x)+f(x)1−(loge​(f(x)))2​=0

Put $y=f(x)$y=f(x).

$$\frac{dy}{dx}+y\sqrt{1-{({\mathrm{log}}_{e}y)}^{2}}=0$$dxdy​+y1−(loge​y)2​=0

$$\Rightarrow \int \frac{dy}{y\sqrt{1-{({\mathrm{log}}_{e}y)}^{2}}}=-\int dx$$⇒∫y1−(loge​y)2​dy​=−∫dx

$$\Rightarrow \int \frac{d({\mathrm{log}}_{e}y)}{\sqrt{1-{({\mathrm{log}}_{e}y)}^{2}}}=-\int dx$$⇒∫1−(loge​y)2​d(loge​y)​=−∫dx

$$\Rightarrow {\mathrm{sin}}^{-1}({\mathrm{log}}_{e}y)=-x+C$$⇒sin−1(loge​y)=−x+C

Put $x=0,y=e$x=0,y=e

$${\mathrm{sin}}^{-1}({\mathrm{log}}_{e}e)=-0+C$$sin−1(loge​e)=−0+C

$$\Rightarrow C=\frac{\pi }{2}$$⇒C=2π​

So,

$${\mathrm{sin}}^{-1}({\mathrm{log}}_{e}y)=-x+\frac{\pi }{2}$$sin−1(loge​y)=−x+2π​

$$\Rightarrow {\mathrm{log}}_{e}y=\mathrm{sin}(-x+\frac{\pi }{2})$$⇒loge​y=sin(−x+2π​)

$$\Rightarrow {\mathrm{log}}_{e}f(x)=\mathrm{sin}(-x+\frac{\pi }{2})$$⇒loge​f(x)=sin(−x+2π​)

Put $x=\frac{\pi }{6}$x=6π​

$$\Rightarrow {\mathrm{log}}_{e}f(\frac{\pi }{6})=\mathrm{sin}(-\frac{\pi }{6}+\frac{\pi }{2})=\mathrm{sin}(\frac{\pi }{3})=\frac{\sqrt{3}}{2}$$⇒loge​f(6π​)=sin(−6π​+2π​)=sin(3π​)=23​​

$$\Rightarrow 6{\mathrm{log}}_{e}f(\frac{\pi }{6})=3\sqrt{3}$$⇒6loge​f(6π​)=33​

$$\Rightarrow {{6{\mathrm{log}}_{e}f(\frac{\pi }{6})}}^{2}=27$$⇒6loge​f(6π​)2=27