JEE 2022 — Mathematics PYQ

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Question

JEE 2022 — Mathematics PYQ

JEE | Mathematics | 2022

If $y = y(x)$ is the solution of the differential equation $(1 + e^{2x}) \frac{dy}{dx} + 2(1 + y^2) e^x = 0$ and $y(0) = 0$ , then $6 \left( y'(0) + (y(\log_e \sqrt{3})) \right)^2$ is equal to:

Choose the correct answer:

Explanation

Step 1: Variable Separation aur Integration

Terms ko separate karne par:

\Rightarrow \frac{dy}{1 + y^2} = -\frac{2e^x}{1 + e^{2x}} dx

Dono taraf integration karne par:

\Rightarrow \int \frac{dy}{1 + y^2} = -2 \int \frac{e^x}{1 + e^{2x}} dx

\Rightarrow \tan^{-1} (y) = -2 \tan^{-1} e^x + C

Step 2: Constant $C$ ki value nikalna

Hame pata hai $y(0) = 0$ , isliye $x = 0$ aur $y = 0$ rakhne par:

\therefore \tan^{-1} (0) = -2 \tan^{-1} (1) + C

\Rightarrow C = 2 \left( \frac{\pi}{4} \right) = \frac{\pi}{2}

Ab hum solution ko aise likh sakte hain:

\Rightarrow \tan^{-1} y + 2 \tan^{-1} e^x = \frac{\pi}{2} \dots (i)

Step 3: $y'(0)$ ki value nikalna

Equation se $\frac{dy}{dx}$ ki value $x=0$ par:

\text{Now, } \left( \frac{dy}{dx} \right)_{x=0} = \left( -\frac{2(1 + y^2) e^x}{1 + e^{2x}} \right)_{x=0}

\Rightarrow \left( \frac{dy}{dx} \right)_{x=0} = -1

Step 4: $y(\log_e \sqrt{3})$ ki value nikalna

Equation $(i)$ mein $x = \log_e \sqrt{3}$ rakhne par:

\tan^{-1} y + 2 \tan^{-1} e^{\log_e \sqrt{3}} = \frac{\pi}{2}

\Rightarrow \tan^{-1} y + 2 \tan^{-1} \sqrt{3} = \frac{\pi}{2}

\Rightarrow \tan^{-1} y + \frac{2\pi}{3} = \frac{\pi}{2}

\Rightarrow \tan^{-1} y = -\frac{\pi}{6}

\Rightarrow y(\log_e \sqrt{3}) = -\frac{1}{\sqrt{3}}

Step 5: Final Calculation

Hame iski value nikalni hai:

\therefore 6 [y'(0) + y(\log_e \sqrt{3})^2] = 6 \left[ -1 + \frac{1}{3} \right] = -4

Final Answer: -4

Choose the correct answer:

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