71.

$\mathrm{If} \int {\mathrm{x}}^3{\mathrm{e}}^{5\mathrm{x}}\mathrm{dx} = \frac{{\mathrm{e}}^{5\mathrm{x}}}{5^4}\left[\mathrm{f}\left(\mathrm{x}\right)\right] + \mathrm{C},\mathrm{then} \mathrm{f}\left(\mathrm{x}\right) = ?$

• $\frac{{\mathrm{x}}^3}{5} - \frac{3{\mathrm{x}}^2}{5^2} + \frac{6\mathrm{x}}{5^3} - \frac{6}{5^4}$

• $5{\mathrm{x}}^3 - 5^2{\mathrm{x}}^2 + 5^3\mathrm{x} - 6$

• $5^3{\mathrm{x}}^3 - 15{\mathrm{x}}^2 + 30\mathrm{x} - 6$

• $5^3{\mathrm{x}}^3 - 75{\mathrm{x}}^2 + 30\mathrm{x} - 6$

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

• $\frac{{\mathrm{x}}^2 + 2}{{\mathrm{x}}^2 +\hspace{0.17em}2\mathrm{x} + 2} - \frac{1}{2}{\tan }^{-1}\left(\mathrm{x} - 1\right) + \mathrm{C}$

• $\frac{{\mathrm{x}}^2 - 2}{4\left({\mathrm{x}}^2 + 2\mathrm{x} +\hspace{0.17em}2\right)} - \frac{1}{2}{\tan }^{-1}\left(\mathrm{x} +\hspace{0.17em}1\right) +\hspace{0.17em}\mathrm{C}$

• $\frac{{\mathrm{x}}^2 + 2}{2\left({\mathrm{x}}^2 + 2\mathrm{x} +\hspace{0.17em}2\right)} - \frac{1}{2}{\tan }^{-1}\left(\mathrm{x} +\hspace{0.17em}1\right) +\hspace{0.17em}\mathrm{C}$

• $\frac{2\left(\mathrm{x} - 1\right)}{{\mathrm{x}}^2 +\hspace{0.17em}2\mathrm{x} +\hspace{0.17em}2} + \frac{1}{2}{\tan }^{-1}\left(\mathrm{x} +\hspace{0.17em}1\right) +\hspace{0.17em}\mathrm{C}$

$\mathrm{If} \int \log \left({\mathrm{a}}^2 + {\mathrm{x}}^2\right)\mathrm{dx} = \mathrm{h}\left(\mathrm{x}\right) + \mathrm{C}, \mathrm{then} \mathrm{h}\left(\mathrm{x}\right) = ?$

• $\mathrm{xlog}\left({\mathrm{a}}^2 +\hspace{0.17em}{\mathrm{x}}^2\right) + 2{\tan }^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)$

• ${\mathrm{x}}^2\log \left({\mathrm{a}}^2 +\hspace{0.17em}{\mathrm{x}}^2\right) +\hspace{0.17em}\mathrm{x} +\hspace{0.17em}{\mathrm{atan}}^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)$

• $\mathrm{xlog}\left({\mathrm{a}}^2 + {\mathrm{x}}^2\right) - 2\mathrm{x} + 2{\mathrm{atan}}^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)$

• ${\mathrm{x}}^2\log \left({\mathrm{a}}^2 +\hspace{0.17em}{\mathrm{x}}^2\right) + 2\mathrm{x} - {\mathrm{a}}^2{\tan }^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)$

$$\begin{gathered} \mathrm{For} \mathrm{x} > 0, \mathrm{if} \int \left(\log {\left(\mathrm{x}\right)}^5\right)\mathrm{dx} = ? \\ \mathrm{x}\left[\mathrm{A}\left(\log {\left(\mathrm{x}\right)}^5\right) + \mathrm{B}{\left(\log \left(\mathrm{x}\right)\right)}^4+\hspace{0.17em}\mathrm{C}{\left(\log \left(\mathrm{x}\right)\right)}^3 + \mathrm{D}{\left(\log \left(\mathrm{x}\right)\right)}^2 +\hspace{0.17em}\mathrm{E}\left(\log \left(\mathrm{x}\right)\right) + \mathrm{F}\right] + \mathrm{Constant}, \mathrm{then} \\ \mathrm{A} +\hspace{0.17em}\mathrm{B} +\hspace{0.17em}\mathrm{C} +\hspace{0.17em}\mathrm{D}\hspace{0.17em}+\hspace{0.17em}\mathrm{E}\hspace{0.17em}+ \mathrm{F} = ? \end{gathered}$$

• - 44

• - 42

• - 40

• - 36

By the definition of the definite integral, the value of $\underset{\mathrm{n}\to \infty }{\lim }\left(\frac{1}{\sqrt{{\mathrm{n}}^2 - 1}} + \frac{{\displaystyle 1}}{{\displaystyle \sqrt{{\mathrm{n}}^2 - 2^2}}} + ... \frac{{\displaystyle 1}}{{\displaystyle \sqrt{{\mathrm{n}}^2 - {\left(\mathrm{n} - 1\right)}^2}}}\right)$ is equal to

• $\mathrm{\pi }$

• $\frac{\mathrm{\pi }}{2}$

• $\frac{\mathrm{\pi }}{4}$

• $\frac{\mathrm{\pi }}{6}$

${\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\left(\frac{\mathrm{x} + \frac{\mathrm{\pi }}{4}}{2 - \cos \left(2\mathrm{x}\right)}\right)\mathrm{dx}$ is equal to

• $\frac{8\mathrm{\pi }\sqrt{3}}{5}$

• $\frac{2\mathrm{\pi }\sqrt{3}}{9}$

• $\frac{4{\mathrm{\pi }}^2\sqrt{3}}{9}$

• $\frac{{\mathrm{\pi }}^2}{6\sqrt{3}}$

The solution of the differential equation $\left(1 + {\mathrm{y}}^2\right) + \left(\mathrm{x} - {\mathrm{e}}^{{\tan }^{-1}\left(\mathrm{y}\right)}\right)\frac{\mathrm{dy}}{\mathrm{dx}} = 0$, is

• ${\mathrm{xe}}^{{\tan }^{-1}\left(\mathrm{y}\right)} = {\tan }^{-1}\left(\mathrm{y}\right) + \mathrm{C}$

• ${\mathrm{xe}}^{2{\tan }^{-1}\left(\mathrm{y}\right)} = {\tan }^{-1}\left(\mathrm{y}\right) + \mathrm{C}$

• $2{\mathrm{xe}}^{{\tan }^{-1}\left(\mathrm{y}\right)} = e^{2{\tan }^{-1}\left(\mathrm{y}\right)} + \mathrm{C}$

• ${\mathrm{x}}^2{\mathrm{e}}^{{\tan }^{-1}\left(\mathrm{y}\right)} = 4{\mathrm{e}}^{2{\tan }^{-1}\left(\mathrm{y}\right)} + \mathrm{C}$

The solution of the differential equation $\left(2\mathrm{x} - 4\mathrm{y} + 3\right)\frac{\mathrm{dy}}{\mathrm{dx}} + \left(\mathrm{x} - 2\mathrm{y} + 1\right) = 0$ is

• $\left[\log \left(2\mathrm{x} - 4\mathrm{y}\right) + 3\right] = \mathrm{x} - 2\mathrm{y} + \mathrm{C}$

• $\left[\log \left[2\left(2\mathrm{x} - 4\mathrm{y}\right) + 3\right]\right] = 2\left(\mathrm{x} - 2\mathrm{y}\right)+ \mathrm{C}$

• $\log \left[2\left(\mathrm{x} - 2\mathrm{y}\right) + 5\right] = 2\left(\mathrm{x} + \mathrm{y}\right) + \mathrm{C}$

• $\log \left[4\left(\mathrm{x} - 2\mathrm{y}\right) + 5\right] = 4\left(\mathrm{x} + 2\mathrm{y}\right) + \mathrm{C}$

The mid-point of the line segment joining the centroid and the orthocentre of the triangle whose vertices are (a, b),(a, c) and (d, c), is

• $\left(\frac{5\mathrm{a} +\hspace{0.17em}\mathrm{d}}{6}, \frac{\mathrm{b} + 5\mathrm{c}}{6}\right)$

• `$\left(\frac{\mathrm{a} + 5\mathrm{d}}{6}, \frac{5\mathrm{b} + \mathrm{c}}{6}\right)$

• (a, 0)

• (0, 0)

The orthocentre of the triangle formed by the lines x + y = 1 and 2y² - xy - 6x² = 0

• $\left(\frac{4}{3}, \frac{4}{3}\right)$

• $\left(\frac{2}{3}, \frac{2}{3}\right)$

• $\left(\frac{2}{3}, \frac{- 2}{3}\right)$

• $\left(\frac{4}{3}, \frac{ - 4}{3}\right)$