The differential equation corresponding to the family of circles inthe plane touching the Y-axis at the origin, is

• $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{\mathrm{y}}^2 - {\mathrm{x}}^2}{2\mathrm{xy}}$

• $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2\mathrm{xy}}{{\mathrm{x}}^2 + {\mathrm{y}}^2}$

• $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{\mathrm{x}}^2 - {\mathrm{y}}^2}{2\mathrm{xy}}$

• $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{\mathrm{y}}^2 + {\mathrm{x}}^2}{2\mathrm{xy}}$

Let y = y(x) be the solution of the differential equation,

$$\begin{gathered} \frac{2 + \sin \left(\mathrm{x}\right)}{\mathrm{y} + 1} . \frac{\mathrm{dy}}{\mathrm{dx}} = - \cos \left(\mathrm{x}\right), \mathrm{y} > 0, \mathrm{y}\left(0\right) = 1. \mathrm{If} \mathrm{y}\left(\mathrm{\pi }\right) = \mathrm{a} \\ \mathrm{and} \frac{\mathrm{dy}}{\mathrm{dx}} \mathrm{at} \mathrm{x} = \mathrm{\pi } \mathrm{is} \mathrm{b}, \mathrm{then} \mathrm{the} \mathrm{ordered} \mathrm{pair} \left(\mathrm{a}, \mathrm{b}\right) = ? \end{gathered}$$

• $\left(2, \frac{3}{2}\right)$

• (1, - 1)

• (2, 1)

• (1, 1)

The equation of the normal to the curve y = (1 + x)^2y + cos²(sin ^{– 1}(x)) at x = 0 is :

• y = 4x + 2

• y + 4x = 2

• x + 4y = 8

• 2y + x = 4

If a curve y = f(x), passing through the point (1,2), is the solution of the differential equation, $2{\mathrm{x}}^2\mathrm{dy} = \left(2\mathrm{xy} + {\mathrm{y}}^2\right)\mathrm{dx}, \mathrm{then} \mathrm{f}\left(\frac{1}{2}\right)$ is equal to

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

• $1 + \log {\left(\mathrm{e}\right)}^2$

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

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

$\mathrm{If} \mathrm{y} =\sum _{\mathrm{k} = 1}^6 {\cos }^{-1}\left\{\frac{3}{5}\cos \left(\mathrm{kx}\right) - \frac{4}{5}\sin \left(\mathrm{kx}\right)\right\} \mathrm{then} \frac{\mathrm{dy}}{\mathrm{dx}} \mathrm{at} \mathrm{x} = 0 \mathrm{is}$

The equation of curve satisfying differential equation $\left(1 + {\mathrm{y}}^2\right)\left({\mathrm{e}}^{\mathrm{x}} + 1\right)\mathrm{dy} = {\mathrm{e}}^{\mathrm{x}}{\mathrm{y}}^2\mathrm{dx} \mathrm{and} \mathrm{also} \mathrm{passes} \mathrm{through} \mathrm{the} \mathrm{point} \left(0, 1\right) \mathrm{is}$

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

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

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

• $2{\mathrm{y}}^2 +\hspace{0.17em}1 = \mathrm{ylog}\left(\frac{1 + {\mathrm{e}}^{\mathrm{x}}}{2}\right)$

$\mathrm{If} {\mathrm{y}}^2 + \log \left({\cos }^2\left(\mathrm{x}\right)\right) = \mathrm{y} \mathrm{then}$

• $\left|\mathrm{y}\text{'}\text{'}\left(0\right)\right| = 2$

• $\left|\mathrm{y}\text{'}\left(0\right)\right| +\left| \mathrm{y}\text{'}\text{'}\left(0\right)\right| = 1$

• $\left|\mathrm{y}\text{'}\left(0\right) \right| + \left|\mathrm{y}\text{'}\text{'}\left(0\right)\right| = 3$

• None of these

$\mathrm{If} {\mathrm{x}}^3\mathrm{dy} + \mathrm{xydx} = {\mathrm{x}}^2\mathrm{dy} + 2\mathrm{ydx}; \mathrm{y}\left(2\right) = \mathrm{e} \mathrm{and} \mathrm{x} > 1, \mathrm{then} \mathrm{y}\left(4\right) = ?$

• $\frac{\sqrt{\mathrm{e}}}{2}$

• $\frac{1}{2} + \sqrt{\mathrm{e}}$

• $\frac{3}{2}\sqrt{\mathrm{e}}$

• $\frac{3}{2} + \sqrt{\mathrm{e}}$

$$\begin{gathered} \mathrm{If} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{series} 20 + 19\frac{3}{5} + 19\frac{1}{5} + 18\frac{4}{5} + ... \mathrm{upto} {\mathrm{n}}^{\mathrm{th}} \mathrm{term} \mathrm{is} 488 \mathrm{and} \\ \mathrm{the} {\mathrm{n}}^{\mathrm{th}} \mathrm{term} \mathrm{is} \mathrm{negative} \end{gathered}$$

• ${\mathrm{n}}^{\mathrm{th}} \mathrm{term} \mathrm{is} - 4\frac{2}{5}$

• n = 41

• n^th term is - 4

• n = 60

If the surface area of a cube is increasing at a rate of 3.6 cm²/sec, retaining its shape; then the rate of change of its volume (in cm³/sec.), when the length of a side of the cube is 10 cm, is

• 20

• 10

• 18

• 9