jee-advanced 2014 Q41

jee-advanced · India · paper2 First order differential equations (integrating factor)
The function $y = f(x)$ is the solution of the differential equation
$$\frac{dy}{dx} + \frac{xy}{x^2 - 1} = \frac{x^4 + 2x}{\sqrt{1 - x^2}}$$
in $(-1,1)$ satisfying $f(0) = 0$. Then
$$\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x)\, dx$$
is
(A) $\frac{\pi}{3} - \frac{\sqrt{3}}{2}$
(B) $\frac{\pi}{3} - \frac{\sqrt{3}}{4}$
(C) $\frac{\pi}{6} - \frac{\sqrt{3}}{4}$
(D) $\frac{\pi}{6} - \frac{\sqrt{3}}{2}$
The function $y = f(x)$ is the solution of the differential equation

$$\frac{dy}{dx} + \frac{xy}{x^2 - 1} = \frac{x^4 + 2x}{\sqrt{1 - x^2}}$$

in $(-1,1)$ satisfying $f(0) = 0$. Then

$$\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x)\, dx$$

is\\
(A) $\frac{\pi}{3} - \frac{\sqrt{3}}{2}$\\
(B) $\frac{\pi}{3} - \frac{\sqrt{3}}{4}$\\
(C) $\frac{\pi}{6} - \frac{\sqrt{3}}{4}$\\
(D) $\frac{\pi}{6} - \frac{\sqrt{3}}{2}$