If the area of the parallelogram with $\overrightarrow{\mathrm{a}} \mathrm{and} \overrightarrow{\mathrm{b}}$ as two adjacent sides is 15 sq unit, then the area of the parallelogram having $3\overrightarrow{\mathrm{a}} + 2\overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{a}} + 3\overrightarrow{\mathrm{b}}$ as two adjacent sides in sq unit is :

• 120

• 105

• 75

• 45

22.

$\int \frac{\left({\mathrm{x}}^3 +\hspace{0.17em}3{\mathrm{x}}^2 + 3\mathrm{x} + 1\right)}{{\left(\mathrm{x} + 1\right)}^5}\mathrm{dx}$ is equal to

• $- \frac{1}{\left(\mathrm{x} + 1\right)} +\hspace{0.17em}\mathrm{c}$

• $\frac{1}{5}\log \left(\mathrm{x} +\hspace{0.17em}1\right) + \mathrm{c}$

• $\log \left(\mathrm{x} + 1\right) +\hspace{0.17em}\mathrm{c}$

• ${\tan }^{-1}\left(\mathrm{x}\right) +\hspace{0.17em}\mathrm{c}$

$\int \frac{\csc \left(\mathrm{x}\right)}{{\cos }^2\left(1 + \log \left(\tan \left({\displaystyle \frac{\mathrm{x}}{2}}\right)\right)\right)}\mathrm{dx}$ is equal to :

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

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

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

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

$\int \frac{\mathrm{dx}}{\mathrm{x}\sqrt{{\mathrm{x}}^6 - 16}}$ is equa to :

• $\frac{1}{3}{\sec }^{-1}\left(\frac{{\mathrm{x}}^3}{4}\right) +\hspace{0.17em}\mathrm{c}$

• ${\cosh }^{-1}\left(\frac{{\mathrm{x}}^3}{4}\right) +\hspace{0.17em}\mathrm{c}$

• $\frac{1}{12}{\sec }^{-1}\left(\frac{{\mathrm{x}}^3}{4}\right) +\hspace{0.17em}\mathrm{c}$

• ${\sec }^{-1}\left(\frac{{\mathrm{x}}^3}{4}\right) +\hspace{0.17em}\mathrm{c}$

If I₁ = ${\mathrm{I}}_1 = {\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sin \left(\mathrm{x}\right)\mathrm{dx} \mathrm{and} {\mathrm{I}}_2 = {\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\mathrm{xcos}\left(\mathrm{x}\right)\mathrm{dx}$, then which one of the following is true ?

• ${\mathrm{I}}_1 + {\mathrm{I}}_{2 }= \frac{\mathrm{\pi }}{2}$

• ${\mathrm{I}}_2 - {\mathrm{I}}_1= \frac{\mathrm{\pi }}{2}$

• ${\mathrm{I}}_1 + {\mathrm{I}}_{2 }= 0$

• ${\mathrm{I}}_1 = {\mathrm{I}}_{2 }$

If f(x) is defined [- 2, 2] by f(x) = 4 - 3x + 1 and g(x) = $\frac{\mathrm{f}\left(- \mathrm{x}\right) - \left(\mathrm{f}\left(\mathrm{x}\right)\right)}{{\mathrm{x}}^2 + 3}$, then ${\int }_{- 2}^2\mathrm{g}\left(\mathrm{x}\right)\mathrm{dx}$ is equal to :

• 64

• - 48

• 0

• 24

The area enclosed between the parabola y = x² - x + 2 and the line y = x + 2 in sq unit equals :

• $\frac{8}{3}$

• $\frac{1}{3}$

• $\frac{2}{3}$

• $\frac{4}{3}$

The solution of the differential equation e^-x(y + 1)dy + (cos²(x) + sin(2x))y dx = 0 subjected to the condition that y = 1 when x = 0 is

• y + log(y) + e^xcos²(x) = 2

• log(y + 1) + e^xcos²(x) = 1

• y + log(y) = e^xcos²(x)

• (y + 1) + e^xcos²(x) = 2