Verified Solution • 2025–26 Session moderate

QJEEM24Maf06e: JEE Main 2024 Mathematics - Class 12 JEEM24Maf06e Solution: A function satisfies with condition . Then, is equal to

A function y=f(x)y=f(x) satisfies f(x)sin⁡2x+sin⁡x−(1+cos⁡2x)f′(x)=0f(x) \sin 2 x+\sin x-\left(1+\cos ^2 x\right) f^{\prime}(x)=0 with condition f(0)=0f(0)=0 . Then, f(π2)f\left(\frac{\pi}{2}\right) is equal to

Multiple Choice Options

  • A 2
  • B 1
  • C −- 1
  • D 0

Step-by-Step Verified Solution

amp;dydx−(sin⁡2x1+cos⁡2x)y=sin⁡xamp; I.F. =1+cos⁡2xamp;y⋅(1+cos⁡2x)=∫(sin⁡x)dxamp;=−cos⁡x+Camp;x=0,C=1amp;y(π2)=1\begin{aligned}& \frac{d y}{d x}-\left(\frac{\sin 2 x}{1+\cos ^2 x}\right) y=\sin x \\& \text { I.F. }=1+\cos ^2 x \\& y \cdot\left(1+\cos ^2 x\right)=\int(\sin x) d x \\& =-\cos x+C \\& x=0, C=1 \\& y\left(\frac{\pi}{2}\right)=1\end{aligned}

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