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(D) The given differential equation is $\frac{dy}{dx} \sin x - y \cos x = -\frac{\sin^2 x}{x^2}$.Dividing by $\sin^2 x$,we get $\frac{1}{\sin x} \frac

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All of the above

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(D) The given differential equation is $\frac{dy}{dx} \sin x - y \cos x = -\frac{\sin^2 x}{x^2}$.Dividing by $\sin^2 x$,we get $\frac{1}{\sin x} \frac{dy}{dx} - y \frac{\cos x}{\sin^2 x} = -\frac{1}{x^2}$.This is equivalent to $\frac{d}{dx} \left( \frac{y}{\sin x} ight) = -\frac{1}{x^2}$.Integrating both sides,we get $\frac{y}{\sin x} = \int -\frac{1}{x^2} dx = \frac{1}{x} + C$.So,$y = \sin x \left( \frac{1}{x} + C ight)$.Given $y ightarrow 0$ as $x ightarrow \infty$,we must have $C = 0$,so $f(x) = \frac{\sin x}{x}$.$1$. $\mathop {Lim}\limits_{x 	o 0} f(x) = \mathop {Lim}\limits_{x 	o 0} \frac{\sin x}{x} = 1$.$2$. Since $\sin x  0$,$f(x) = \frac{\sin x}{x} $3$. Using the series expansion $\frac{\sin x}{x} = 1 - \frac{x^2}{6} + \frac{x^4}{120} - \dots$,the integral $\int_0^{\pi/2} (1 - \frac{x^2}{6}) dx = [x - \frac{x^3}{18}]_0^{\pi/2} = \frac{\pi}{2} - \frac{\pi^3}{144} \approx 1.57 - 0.21 = 1.36 > 1$.Thus,all statements are correct.

$A$ function $y = f(x)$ satisfying the differential equation $\frac{dy}{dx} \sin x - y \cos x + \frac{\sin^2 x}{x^2} = 0$ is such that $y \rightarrow 0$ as $x \rightarrow \infty$. Which of the following statements is correct?

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