A) \[(2y-1)\frac{dy}{dx}-\sin x=0\]
B) \[(2y-1)\cos x+\frac{dy}{dx}=0\]
C) \[(2y-1)\cos x-\frac{dy}{dx}=0\]
D) \[(2y-1)\cos x+\frac{dy}{dx}=0\]
Correct Answer: D
Solution :
\[y=\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+.....\infty }}}\] \[\Rightarrow y=\sqrt{\sin x+y}\Rightarrow {{y}^{2}}=\sin x+y\] On differentiating both sides, we get \[2y\frac{dy}{dx}=\cos x+\frac{dy}{dx}\Rightarrow \frac{dy}{dx}(2y-1)=\cos x\].
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