The volume of sphere is increasing at the rate of 1200 cu cm/s. The rate of increase in its surface area when the radius is 10 cm is

• 120 sqcm/s

• 240 sqcm/s

• 200 sqcm/s

• 100 sqcm/s

The slope of the tangent to the curve

$\mathrm{y} = {\int }_0^{\mathrm{x}}\frac{1}{1 + {\mathrm{t}}^3}d\mathrm{t} \mathrm{at} \mathrm{the} \mathrm{point}, \mathrm{where} \mathrm{x} = 1 \mathrm{is}$

• $\frac{1}{4}$

• $\frac{1}{3}$

• $\frac{1}{2}$

• 1

$\mathrm{The} \mathrm{solution} \mathrm{of} \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{1}{\mathrm{x}} = \frac{{\mathrm{e}}^{\mathrm{y}}}{{\mathrm{x}}^2} \mathrm{is}$

• $2\mathrm{x} = \left(1 +{\mathrm{Cx}}^2 \right){\mathrm{e}}^{\mathrm{y}}$

• $\mathrm{x} = \left(1 +{\mathrm{Cx}}^2 \right){\mathrm{e}}^{\mathrm{y}}$

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

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

The differential equation $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\mathrm{ax} +\hspace{0.17em}\mathrm{by} +\hspace{0.17em}\mathrm{c}}$ ,where a, b, c are all non-zero real numbers, is

• linear in y

• linear in x

• linear in both x and y

• homogeneous equation

$$\begin{gathered} \mathrm{Let} \mathrm{D} \mathrm{be} \mathrm{the} \mathrm{domain} \mathrm{of} \mathrm{a} \mathrm{twice} \mathrm{differentiable} \mathrm{function} \mathrm{f}. \\ \mathrm{For} \mathrm{all} \mathrm{x} \in \mathrm{D}, \mathrm{f}"\left(\mathrm{x}\right) + \mathrm{f}\left(\mathrm{x}\right) = 0 \mathrm{and} \mathrm{f}\left(\mathrm{x}\right)= \int \mathrm{g}\left(\mathrm{x}\right)\mathrm{dx} + \mathrm{constant}. \\ \mathrm{If} \mathrm{h}\left(\mathrm{x}\right) = {\left\{\mathrm{f}\left(\mathrm{x}\right)\right\}}^2 + {\left\{\mathrm{g}\left(\mathrm{x}\right)\right\}}^2 \mathrm{and} \mathrm{h}\left(0\right) = 5, \mathrm{then} \mathrm{h}\left(2015\right) - \mathrm{h}\left(2014\right) \mathrm{is} \mathrm{equal} \mathrm{to} \end{gathered}$$

• 5

• 3

• 0

• 1

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

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

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

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

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

$$\begin{gathered} \mathrm{For} {\mathrm{x}}^2 - 4 \ne 0, \mathrm{the} \mathrm{value} \mathrm{of} \\ \frac{\mathrm{d}}{\mathrm{dx}}\left[\log \left({\mathrm{e}}^{\mathrm{x}}{\left(\frac{\mathrm{x} - 2}{\mathrm{x} + 2}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\right)\right] \mathrm{at} \mathrm{x} = 3 \mathrm{is} \end{gathered}$$

• $\frac{8}{5}$

• 2

• 1

• $\frac{8{\mathrm{e}}^3}{5}$

$\mathrm{If} \mathrm{y} = \frac{{\sinh }^{-1}\left(\mathrm{x}\right)}{\sqrt{1 + {\mathrm{x}}^2}}, \mathrm{then} \left(1 + {\mathrm{x}}^2\right){\mathrm{y}}^2 + 3{\mathrm{xy}}_1 + \mathrm{y} = ?$

• 2

• 1

•  - 1 

• 0

390.

The solution of the equation

$\left(\mathrm{x} - 4{\mathrm{y}}^3\right)\frac{\mathrm{dy}}{\mathrm{dx}} - \mathrm{y} = 0, \left(\mathrm{y} > 0\right) \mathrm{is}$

• x = y³ + cy

• x + 2y³ = cy

• y = x³ + cx

• y + 2x³ = cx