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NVAMediumJEE 2023Separation of Variables

JEE Mathematics 2023 Question with Solution

Let ff be a differentiable function defined on (0,π2)\left(0, \frac{\pi}{2}\right) such that f(x)>0f(x) > 0 and

f(x)+∫0xf(t)1−(log⁡ef(t))2 dt=e,∀x∈[0,π2].f(x) + \int_0^x f(t)\sqrt{1 - (\log_e f(t))^2} \, dt = e, \quad \forall x \in \left[0, \frac{\pi}{2}\right].

Then (6log⁡ef(π6))2\left(6 \log_e f\left(\frac{\pi}{6}\right)\right)^2 is equal to:

Answer

Correct answer:27

Step-by-step solution

Standard Method

Given:

f(x)+∫0xf(t)1−(log⁡ef(t))2 dt=ef(x) + \int_0^x f(t)\sqrt{1 - (\log_e f(t))^2} \, dt = e

with f(x)>0f(x) > 0.

Find: (6log⁡ef(π6))2\left(6 \log_e f\left(\frac{\pi}{6}\right)\right)^2.

From the given equation, at x=0x=0,

f(0)=ef(0) = e

Detailed Working from Extracted Solution

Differentiate the given relation with respect to xx:

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

So,

f′(x)=−f(x)1−(ln⁡f(x))2f'(x) = -f(x)\sqrt{1-(\ln f(x))^2}

Let

f(x)=yf(x)=y

Then the extracted working gives

dxdy=−1y1−(ln⁡y)2\frac{dx}{dy} = -\frac{1}{y\sqrt{1-(\ln y)^2}}

which is written as

∫1y1−(ln⁡y)2 dy=−∫dx\int \frac{1}{y\sqrt{1-(\ln y)^2}} \, dy = -\int dx

Put

ln⁡y=t\ln y = t

Then

∫11−t2 dt=−x+C\int \frac{1}{\sqrt{1-t^2}} \, dt = -x + C

Hence,

sin⁡−1t=−x+C\sin^{-1} t = -x + C

that is,

sin⁡−1(ln⁡y)=−x+C\sin^{-1}(\ln y) = -x + C

So,

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

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

ln⁡f(0)=1\ln f(0)=1

Therefore, from the extracted solution,

π2=C\frac{\pi}{2} = C

Hence,

sin⁡−1(ln⁡f(x))=−x+π2\sin^{-1}(\ln f(x)) = -x + \frac{\pi}{2}

Now substitute x=π6x=\frac{\pi}{6}:

sin⁡−1(ln⁡f(π6))=−π6+π2=π3\sin^{-1}\left(\ln f\left(\frac{\pi}{6}\right)\right) = -\frac{\pi}{6} + \frac{\pi}{2} = \frac{\pi}{3}

Thus,

ln⁡f(π6)=sin⁡π3=32\ln f\left(\frac{\pi}{6}\right) = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}

Therefore,

(6log⁡ef(π6))2=(6⋅32)2=(33)2=27\left(6\log_e f\left(\frac{\pi}{6}\right)\right)^2 = \left(6 \cdot \frac{\sqrt{3}}{2}\right)^2 = (3\sqrt{3})^2 = 27

So the required numerical value is 27.

The second provided approach contains an inconsistent intermediate form involving sin⁡(1)\sin(1), but the first extracted working and the final stated answer consistently give 27.

Common mistakes

  • Differentiating the integral term incorrectly. By the Fundamental Theorem of Calculus, ddx∫0xg(t) dt=g(x)\frac{d}{dx}\int_0^x g(t)\,dt = g(x), so the term becomes f(x)1−(log⁡ef(x))2f(x)\sqrt{1-(\log_e f(x))^2}. Do not differentiate the integrand with respect to tt again.

  • Forgetting to use the initial condition at x=0x=0. Substituting x=0x=0 in the original equation gives f(0)=ef(0)=e, which implies ln⁡f(0)=1\ln f(0)=1. This is essential to determine the constant of integration.

  • Using the substitution incorrectly. If u=ln⁡f(x)u=\ln f(x), then u′=f′(x)f(x)u' = \frac{f'(x)}{f(x)}, not f′(x)f'(x) itself. Hence dividing the differential equation by f(x)f(x) is necessary before integrating.

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