MCQMediumJEE Main 2023 · 1 February, Shift 1Linear Differential Equations

Mathematics Question from JEE Main 2023 · 1 February, Shift 1

If y=y(x)y = y(x) is the solution curve of the differential equation dydx+ytanx=xsecx,0xπ3\frac{dy}{dx} + y \tan x = x \sec x, \quad 0 \leq x \leq \frac{\pi}{3} and y(0)=1y(0) = 1, then y(π6)y \left( \frac{\pi}{6} \right) is equal to:

  • A

    π12+32loge(2e3)\frac{\pi}{12} + \frac{\sqrt{3}}{2} \log_e \left( \frac{2}{e\sqrt{3}} \right)

  • B

    π12+32loge(23e)\frac{\pi}{12} + \frac{\sqrt{3}}{2} \log_e \left( \frac{2\sqrt{3}}{e} \right)

  • C

    π1232loge(23e)\frac{\pi}{12} - \frac{\sqrt{3}}{2} \log_e \left( \frac{2\sqrt{3}}{e} \right)

  • D

    π1232loge(2e3)\frac{\pi}{12} - \frac{\sqrt{3}}{2} \log_e \left( \frac{2}{e\sqrt{3}} \right)

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