Let f be a differentiable function defined on [ 0,\pi \over 2 ] such that f(x) \gt 0 and f(x) + \int_0^x f(t)\sqrt 1 - (\log _ef(t))^2 dt = e,\forall x \in [ 0,\pi \over 2 ] . Then ( 6\log _ef( \pi \over 6 ) )^2 is equal to \underline\hspace2cm .

Definite Integration PYQ — JEE Main 2023 · 24 Jan · Shift 2 · Q42 with Solution

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Correct answer: 27

Given equation

We are given, for all $x \in \left[0,\frac{\pi}{2}\right]$ ,

f(x)+\int_0^x f(t)\sqrt{1-(\ln f(t))^2}\,dt=e,

with $f(x)>0$ and $f$ differentiable.

We need to find

\left(6\ln f\left(\frac{\pi}{6}\right)\right)^2.

Differentiate the given relation

Differentiate both sides with respect to $x$ :

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

So,

f'(x)=-f(x)\sqrt{1-(\ln f(x))^2}.

Since $f(x)>0$ , let

y(x)=\ln f(x).

Then

f(x)=e^{y(x)},\qquad f'(x)=e^{y(x)}y'(x)=f(x)y'(x).

Substitute into the differential equation:

f(x)y'(x)=-f(x)\sqrt{1-y(x)^2}.

Because $f(x)>0$ , divide by $f(x)$ :

y'(x)=-\sqrt{1-y(x)^2}.

Find the initial condition

Put $x=0$ in the original equation:

f(0)+\int_0^0 \cdots \,dt=e \implies f(0)=e.

Hence,

y(0)=\ln f(0)=\ln e=1.

So $y$ satisfies

y'(x)=-\sqrt{1-y(x)^2},\qquad y(0)=1.

Solve the differential equation

We know that $y=\cos x$ satisfies:

y'(x)=-\sin x,

and on $\left[0,\frac{\pi}{2}\right]$ ,

\sqrt{1-y^2}=\sqrt{1-\cos^2 x}=\sin x,

so indeed

y'(x)=-\sqrt{1-y(x)^2}.

Also,

y(0)=\cos 0=1.

Thus,

y(x)=\cos x.

Therefore,

\ln f(x)=\cos x.

So,

f(x)=e^{\cos x}.

Evaluate at $x=\frac{\pi}{6}$

\ln f\left(\frac{\pi}{6}\right)=\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}.

Hence,

6\ln f\left(\frac{\pi}{6}\right)=6\cdot \frac{\sqrt{3}}{2}=3\sqrt{3}.

Therefore,

\left(6\ln f\left(\frac{\pi}{6}\right)\right)^2=(3\sqrt{3})^2=27.

Comparison with stored answer

Derived answer = $27$ .

This matches the stored correct answer.

More from Definite Integration

View written solutionFree

Correct answer: 27

Given equation

We are given, for all $x \in \left[0,\frac{\pi}{2}\right]$ ,

f(x)+\int_0^x f(t)\sqrt{1-(\ln f(t))^2}\,dt=e,

with $f(x)>0$ and $f$ differentiable.

We need to find

\left(6\ln f\left(\frac{\pi}{6}\right)\right)^2.

Differentiate the given relation

Differentiate both sides with respect to $x$ :

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

So,

f'(x)=-f(x)\sqrt{1-(\ln f(x))^2}.

Since $f(x)>0$ , let

y(x)=\ln f(x).

Then

f(x)=e^{y(x)},\qquad f'(x)=e^{y(x)}y'(x)=f(x)y'(x).

Substitute into the differential equation:

f(x)y'(x)=-f(x)\sqrt{1-y(x)^2}.

Because $f(x)>0$ , divide by $f(x)$ :

y'(x)=-\sqrt{1-y(x)^2}.

Find the initial condition

Put $x=0$ in the original equation:

f(0)+\int_0^0 \cdots \,dt=e \implies f(0)=e.

Hence,

y(0)=\ln f(0)=\ln e=1.

So $y$ satisfies

y'(x)=-\sqrt{1-y(x)^2},\qquad y(0)=1.

Solve the differential equation

We know that $y=\cos x$ satisfies:

y'(x)=-\sin x,

and on $\left[0,\frac{\pi}{2}\right]$ ,

\sqrt{1-y^2}=\sqrt{1-\cos^2 x}=\sin x,

so indeed

y'(x)=-\sqrt{1-y(x)^2}.

Also,

y(0)=\cos 0=1.

Thus,

y(x)=\cos x.

Therefore,

\ln f(x)=\cos x.

So,

f(x)=e^{\cos x}.

Evaluate at $x=\frac{\pi}{6}$

\ln f\left(\frac{\pi}{6}\right)=\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}.

Hence,

6\ln f\left(\frac{\pi}{6}\right)=6\cdot \frac{\sqrt{3}}{2}=3\sqrt{3}.

Therefore,

\left(6\ln f\left(\frac{\pi}{6}\right)\right)^2=(3\sqrt{3})^2=27.

Comparison with stored answer

Derived answer = $27$ .

This matches the stored correct answer.

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