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(D) The given differential equation is $y \sin 2x - \cos x + (1 + \sin^2 x) \frac{dy}{dx} = 0$,where $y = f(x)$.Rearranging the terms,we get $\frac{dy

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$\frac{2}{5}$

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(D) The given differential equation is $y \sin 2x - \cos x + (1 + \sin^2 x) \frac{dy}{dx} = 0$,where $y = f(x)$.Rearranging the terms,we get $\frac{dy}{dx} + \left(\frac{\sin 2x}{1 + \sin^2 x}\right) y = \frac{\cos x}{1 + \sin^2 x}$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{\sin 2x}{1 + \sin^2 x}$ and $Q(x) = \frac{\cos x}{1 + \sin^2 x}$.The Integrating Factor ($I$.$F$.) is $e^{\int P(x) dx} = e^{\int \frac{\sin 2x}{1 + \sin^2 x} dx}$.Let $u = 1 + \sin^2 x$,then $du = 2 \sin x \cos x dx = \sin 2x dx$. Thus,$\int P(x) dx = \int \frac{du}{u} = \ln(1 + \sin^2 x)$.So,$I$.$F$. $= e^{\ln(1 + \sin^2 x)} = 1 + \sin^2 x$.The general solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$.$y(1 + \sin^2 x) = \int \frac{\cos x}{1 + \sin^2 x} \cdot (1 + \sin^2 x) dx + C = \int \cos x dx + C = \sin x + C$.Given $f(0) = 0$,we have $0(1 + 0) = \sin(0) + C$,which implies $C = 0$.Thus,$f(x) = \frac{\sin x}{1 + \sin^2 x}$.For $x = \frac{\pi}{6}$,$f(\frac{\pi}{6}) = \frac{\sin(\pi/6)}{1 + \sin^2(\pi/6)} = \frac{1/2}{1 + (1/2)^2} = \frac{1/2}{1 + 1/4} = \frac{1/2}{5/4} = \frac{1}{2} \cdot \frac{4}{5} = \frac{2}{5}$.

$A$ function $y = f(x)$ satisfies the differential equation $f(x) \sin 2x - \cos x + (1 + \sin^2 x) f'(x) = 0$ where $f(0) = 0$. Then the value of $f(\frac{\pi}{6})$ is equal to

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