71.

Solution of the differential equation $\frac{\mathrm{dy}}{\mathrm{dx}} + \frac{\mathrm{y}}{\mathrm{x}} = \sin \left(\mathrm{x}\right)$ is :

• $\mathrm{x}\left(\mathrm{y} + \cos \left(\mathrm{x}\right)\right) = \sin \left(\mathrm{x}\right) + \mathrm{c}$

• $\mathrm{x}\left(\mathrm{y} - \cos \left(\mathrm{x}\right)\right) = \sin \left(\mathrm{x}\right) + \mathrm{c}$

• $\mathrm{x}\left(\mathrm{ycos}\left(\mathrm{x}\right)\right) = \sin \left(\mathrm{x}\right) + \mathrm{c}$

• $\mathrm{x}\left(\mathrm{y} - \cos \left(\mathrm{x}\right)\right) = \cos \left(\mathrm{x}\right) + \mathrm{c}$

The projection of the vector $2\hat{\mathrm{i}} + \hat{\mathrm{j}} - 3\hat{\mathrm{k}}$ on the vector $\hat{\mathrm{i}} - 2\hat{\mathrm{j}} + \hat{\mathrm{k}}$ is

• $- \frac{3}{\sqrt{14}}$

• $\frac{3}{\sqrt{14}}$

• $- \sqrt{\frac{3}{2}}$

• $\frac{3}{\sqrt{2}}$

An unit vector perpendicular to the plane containing the vectors $\hat{\mathrm{i}} - \hat{\mathrm{j}} + \hat{\mathrm{k}}$ and $- \hat{\mathrm{i}} + \hat{\mathrm{j}} + \hat{\mathrm{k}}$ is

• $\pm \frac{\hat{\mathrm{i}} - \hat{\mathrm{j}}}{\sqrt{2}}$

• $\frac{\hat{\mathrm{i}} + \hat{\mathrm{k}}}{\sqrt{2}}$

• $\pm \frac{\hat{\mathrm{j}} - \hat{\mathrm{k}}}{\sqrt{2}}$

• $\mp \frac{\hat{\mathrm{i}} + \hat{\mathrm{j}}}{\sqrt{2}}$

If a, b and c are mutually perpendicular unit vectors, then $\left|\mathrm{a} + \mathrm{b} + \mathrm{c}\right|$ is equal to

• 3

• $\sqrt{3}$

• $\left(\sqrt{{\mathrm{a}}^2 + {\mathrm{b}}^2 + {\mathrm{c}}^2}\right)/3$

• 1

$\int \frac{\sin \left(2\mathrm{x}\right)}{1 + {\cos }^2\left(\mathrm{x}\right)}\mathrm{dx}$ =

• $- \frac{1}{2}\log \left(1 + {\cos }^2\left(\mathrm{x}\right)\right) + \mathrm{c}$

• $2\log \left(1 + {\cos }^2\left(\mathrm{x}\right)\right) + \mathrm{c}$

• $\frac{1}{2}\log \left(1 + \cos \left(2\mathrm{x}\right)\right) + \mathrm{c}$

• $- \log \left(1 + {\cos }^2\left(\mathrm{x}\right)\right) + \mathrm{c}$

$\int \frac{{\mathrm{e}}^{\mathrm{x}}\left(1 + \sin \left(\mathrm{x}\right)\right)}{1 + \cos \left(\mathrm{x}\right)}\mathrm{dx}$ =

• ${\mathrm{e}}^{\mathrm{x}}\tan \left(\frac{\mathrm{x}}{2}\right) +\hspace{0.17em}\mathrm{c}$

• ${\mathrm{e}}^{\mathrm{x}}\tan \left(\mathrm{x}\right) +\hspace{0.17em}\mathrm{c}$

• ${\mathrm{e}}^{\mathrm{x}}\left(\frac{1 + \sin \left(\mathrm{x}\right)}{1 - \cos \left(\mathrm{x}\right)}\right) + \mathrm{c}$

• ${\mathrm{e}}^{\mathrm{x}}\tan \left(\mathrm{x}\right) + \mathrm{c}$

$\int \frac{1 + \tan \left(\mathrm{x}\right)}{{\mathrm{e}}^{- \mathrm{x}}\cos \left(\mathrm{x}\right)}\mathrm{dx}$ is equal to

• ${\mathrm{e}}^{- \mathrm{x}}\tan \left(\mathrm{x}\right) +\hspace{0.17em}\mathrm{c}$

• ${\mathrm{e}}^{- \mathrm{x}}\sec \left(\mathrm{x}\right) + \mathrm{c}$

• ${\mathrm{e}}^{\mathrm{x}}\sec \left(\mathrm{x}\right) + \mathrm{c}$

• ${\mathrm{e}}^{\mathrm{x}}\tan \left(\mathrm{x}\right) + \mathrm{c}$

${\int }_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\csc }^2\left(\mathrm{x}\right)\mathrm{dx}$ is equal to

• - 1

• 1

• 0

• $\frac{1}{2}$

${\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\log \left(1 + \tan \left(\mathrm{x}\right)\right)\mathrm{dx}$ is equal to

• $\frac{\mathrm{\pi }}{8}{\log }_{\mathrm{e}}\left(2\right)$

• $\frac{\mathrm{\pi }}{4}{\log }_2\left(\mathrm{e}\right)$

• $\frac{\mathrm{\pi }}{4}{\log }_{\mathrm{e}}\left(2\right)$

• $\frac{\mathrm{\pi }}{8}{\log }_{\mathrm{e}}\left(\frac{1}{2}\right)$

The area bounded by the parabola y² = 4ax and the line x = a and x = 4a is

• $\frac{35{\mathrm{a}}^2}{3}$

• $\frac{4{\mathrm{a}}^2}{3}$

• $\frac{7{\mathrm{a}}^2}{3}$

• $\frac{56{\mathrm{a}}^2}{3}$