The solution of the differential equation$\frac{\mathrm{dy}}{\mathrm{dx}} + \sin \left(\frac{\mathrm{y} + \mathrm{x}}{2}\right) + \sin \left(\frac{\mathrm{y} - \mathrm{x}}{2}\right) = 0$ is

• $\log \left(\tan \left(\frac{\mathrm{y}}{2}\right)\right) = \mathrm{C} - 2\sin \left(\mathrm{x}\right)$

• $\log \left(\tan \left(\frac{\mathrm{y}}{4}\right)\right) = \mathrm{C} - 2\sin \left(\frac{\mathrm{x}}{2}\right)$

• $\log \left(\tan \left(\frac{\mathrm{y}}{2} + \frac{\mathrm{\pi }}{4}\right)\right) = \mathrm{C} - 2\sin \left(\mathrm{x}\right)$

• $\log \left(\tan \left(\frac{\mathrm{y}}{2} + \frac{\mathrm{\pi }}{4}\right)\right) = \mathrm{C} - 2\sin \left(\frac{\mathrm{x}}{2}\right)$

The part of straight line y = x between x = 1 and x = 2 is revolved about X-axis, then the curved surface area of the solid thus generated is

• $2\sqrt{2}\mathrm{\pi }$

• $3\sqrt{2}$

• $3\sqrt{2}\mathrm{\pi }$

• None of these

If ${\int }_0^{\mathrm{\pi }}\mathrm{xf}\left({\sin }^2\left(\mathrm{x}\right) + {\sec }^2\left(\mathrm{x}\right)\right)\mathrm{dx} = \mathrm{k}{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\mathrm{f}\left({\sin }^2\left(\mathrm{x}\right) + {\sec }^2\left(\mathrm{x}\right)\right)\mathrm{dx}$, then the value of k is

• $\frac{\mathrm{\pi }}{2}$

• $\mathrm{\pi }$

• $- \frac{\mathrm{\pi }}{2}$

• None of these

24.

A body of 6 kg rests in limiting equilibrium on an inclined plane whose slope is 30°. If the plane is raised to slope of 60°, then force (in kg-wt) along the plane required to support it is

• 3

• $2\sqrt{3}$

• $\sqrt{3}$

• $3\sqrt{3}$