If the curves x² + py² = l and qx² + y² = l are orthogonal to each other, then

• p - q = 2

• $\frac{1}{\mathrm{p}} - \frac{1}{\mathrm{q}} = 2$

• $\frac{1}{\mathrm{p}} + \frac{1}{\mathrm{q}} = - 2$

• $\frac{1}{\mathrm{p}} + \frac{1}{\mathrm{q}} = 2$

The focal length of a mirror is given by $\frac{2}{\mathrm{f}} = \frac{1}{\mathrm{v}} - \frac{1}{\mathrm{u}}$. In finding the values of u and v, the errors are equal to ' 'p'. Then, the relative error in f is

• $\frac{\mathrm{p}}{2}\left(\frac{1}{\mathrm{u}} + \frac{1}{\mathrm{v}}\right)$

• $\mathrm{p}\left(\frac{1}{\mathrm{u}} + \frac{1}{\mathrm{v}}\right)$

• $\frac{\mathrm{p}}{2}\left(\frac{{\displaystyle 1}}{{\displaystyle \mathrm{u}}} - \frac{{\displaystyle 1}}{{\displaystyle \mathrm{v}}}\right)$

• $\mathrm{p}\left(\frac{{\displaystyle 1}}{{\displaystyle \mathrm{u}}} - \frac{{\displaystyle 1}}{{\displaystyle \mathrm{v}}}\right)$

$\mathrm{If} \mathrm{u} = \log \left({\mathrm{x}}^3 +\hspace{0.17em}{\mathrm{y}}^3 + {\mathrm{z}}^3 - 3\mathrm{xyz}\right), \mathrm{then} \left(\mathrm{x} + \mathrm{y} + \mathrm{z}\right)\left({\mathrm{u}}_{\mathrm{x}} + {\mathrm{u}}_{\mathrm{y}} \hspace{0.17em}+{\mathrm{u}}_{\mathrm{z}}\right) = ?$

• 0

• x - y + z

• 2

• 3

An integrating factor of the equation $\left(1 + \mathrm{y} +\hspace{0.17em}{\mathrm{x}}^2\mathrm{y}\right)\mathrm{dx} + \left(\mathrm{x} + {\mathrm{x}}^3\right)\mathrm{dy} = 0$ is

• e^x

• x²

• $\frac{1}{\mathrm{x}}$

• x

375.

$$\begin{gathered} \mathrm{The} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{differential} \mathrm{equation} \\ \frac{\mathrm{dy}}{\mathrm{dx}} - 2\mathrm{ytan}\left(2\mathrm{x}\right) = {\mathrm{e}}^{\mathrm{x}}\sec \left(2\mathrm{x}\right) \mathrm{is} \end{gathered}$$

• $\mathrm{ysin}\left(2\mathrm{x}\right) = {\mathrm{e}}^{\mathrm{x}} + \mathrm{C}$

• $\mathrm{ycos}\left(2\mathrm{x}\right) = \int {\mathrm{e}}^{\mathrm{x}} . 1\mathrm{dx} + \mathrm{C}$

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

• $\mathrm{ycos}\left(2\mathrm{x}\right) + e^x = \mathrm{C}$

$\mathrm{If} \mathrm{f}\left(\mathrm{x}\right) = \frac{\mathrm{x}}{1 + \mathrm{x}} \mathrm{and} \mathrm{g}\left(\mathrm{x}\right) = \mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right), \mathrm{then} \mathrm{g}\text{'}\left(\mathrm{x}\right) =$

• $\frac{1}{{\left(2\mathrm{x} + 3\right)}^2}$

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

• $\frac{1}{{\mathrm{x}}^2}$

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

The differential equation of the family of parabolas with vertex at (0, - 1) and having axis along the Y-axis is

• yy' + 2xy + 1 = 0

• xy' + y + 1 = 0

• xy' - 2y - 2

• xy' - y - 1 = 0

$\mathrm{The} \mathrm{solution} \mathrm{of} \mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}} = \mathrm{y} + {\mathrm{e}}^{\raisebox{1ex}{$\mathrm{y}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{x}$}\right.} \mathrm{with} \mathrm{y}\left(1\right) = 0 \mathrm{is}$

• $1 = \log \left(\mathrm{x}\right) + {\mathrm{e}}^{ \raisebox{1ex}{$\mathrm{y}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{x}$}\right.}$

• $\log \left(\mathrm{x}\right) = {\mathrm{e}}^{ - \raisebox{1ex}{$\mathrm{y}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{x}$}\right.}$

• $1 = 2\log \left(\mathrm{x}\right) + {\mathrm{e}}^{ - \raisebox{1ex}{$\mathrm{y}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{x}$}\right.}$

• $\log \left(\mathrm{x}\right) + {\mathrm{e}}^{ - \raisebox{1ex}{$\mathrm{y}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{x}$}\right.} = 1$

The solution of cos(y) + (xsin(y) - 1)dy/dx = 0 is, 

• $\mathrm{xsec}\left(\mathrm{y}\right) = \tan \left(\mathrm{y}\right) + \mathrm{C}$

• $\tan \left(\mathrm{y}\right) - \sec \left(\mathrm{y}\right) = \mathrm{Cx}$

• $\tan \left(\mathrm{y}\right) + \sec \left(\mathrm{y}\right) = \mathrm{Cx}$

• $\mathrm{xsec}\left(\mathrm{y}\right) + \tan \left(\mathrm{y}\right) = \mathrm{C}$

If x is real, then the minimum value of y = $\frac{{\mathrm{x}}^2 - \mathrm{x} + 1}{{\mathrm{x}}^2 + \mathrm{x} + 1} \mathrm{is}$

• 3

• $\frac{1}{3}$

• $\frac{1}{2}$

• 2