61.

A point moves in the xy-plane such that the sum of its distance from two mutually perpendicular lines is always equal to 5 units. The area(in sq units) enclosed by the locus of the point,is

• $\frac{25}{4}$

• 25

• 50

• 100

If the pair of lines given by $\left({\mathrm{x}}^2 + {\mathrm{y}}^2\right){\cos }^2\left(\mathrm{\theta }\right) = {\left(\mathrm{xcos}\left(\mathrm{\theta }\right) + \mathrm{ysin}\left(\mathrm{\theta }\right)\right)}^2$ are perpendicular to each other, then $\mathrm{\theta }$ is equal to

• 0

• $\frac{\mathrm{\pi }}{4}$

• $\frac{\mathrm{\pi }}{3}$

• $\frac{3\mathrm{\pi }}{4}$

If the foot of the perpendicular from (0, 0, 0) to a plane is (1, 2, 3), then the equation of the plane is

• 2x + y + 3z = 14

• x + 2y + 3z = 14

• x + 2y + 3z + 14 = 0

• x + 2y - 3z = 14

$$\begin{gathered} \mathrm{If} \mathrm{u} = \mathrm{f}\left(\mathrm{r}\right), \mathrm{where} {\mathrm{r}}^2 = {\mathrm{x}}^2 + {\mathrm{y}}^2, \mathrm{then} \\ \left(\frac{\partial \mathrm{u}}{\partial {\mathrm{x}}^2} + \frac{{\partial }^2\mathrm{u}}{\partial {\mathrm{y}}^2}\right) = ? \end{gathered}$$

• f''(r)

• f''(r) +f'(r)

• f''(r) + $\frac{1}{\mathrm{r}}$f'(r)

• f''(r) + rf'(r)

$\int \frac{\mathrm{dx}}{{\mathrm{x}}^2\sqrt{4 + {\mathrm{x}}^2}} =$

• $\frac{1}{4}\sqrt{4 + {\mathrm{x}}^2} + \mathrm{C}$

• $- \frac{1}{4}\sqrt{4 + {\mathrm{x}}^2} + \mathrm{C}$

• $- \frac{1}{4\mathrm{x}}\sqrt{4 + {\mathrm{x}}^2} + \mathrm{C}$

• $\frac{9}{4\mathrm{x}}\sqrt{4 + {\mathrm{x}}^2} + \mathrm{C}$

$\int {\sec }^2\left(\mathrm{x}\right){\csc }^4\left(\mathrm{x}\right)\mathrm{dx} = ?$

• 4

• 3

• 2

• 1

$\int \frac{\mathrm{dx}}{\sqrt{\mathrm{x} - {\mathrm{x}}^2}} = ?$

• $2{\sin }^{-1}\left(\sqrt{\mathrm{x}}\right) + \mathrm{C}$

• $2{\sin }^{-1}\left(\mathrm{x}\right) + \mathrm{C}$

• $2x{\sin }^{-1}\left(\mathrm{x}\right) + \mathrm{C}$

• ${\sin }^{-1}\left(\sqrt{\mathrm{x}}\right) + \mathrm{C}$

$\mathrm{If} \mathrm{a} > 0, \mathrm{then} {\int }_{- \mathrm{\pi }}^{\mathrm{\pi }}\frac{{\sin }^2\left(\mathrm{x}\right)}{1 + {\mathrm{a}}^{\mathrm{x}}}\mathrm{dx} = ?$

• $\frac{\mathrm{\pi }}{2}$

• $\mathrm{\pi }$

• $\frac{2\mathrm{\pi }}{2}$

• a$\mathrm{\pi }$

The area (in sq units) bounded by the curves y² = 4x and x² = 4y is

• $\frac{64}{3}$

• $\frac{16}{3}$

• $\frac{8}{3}$

• $\frac{2}{3}$

$$\begin{gathered} \mathrm{The} \mathrm{value} \mathrm{of} \mathrm{the} \mathrm{integral} {\int }_0^4\frac{\mathrm{dx}}{1 + {\mathrm{x}}^2} \mathrm{obtained} \mathrm{by} \mathrm{using} \mathrm{trapezoidal} \\ \mathrm{rule} \mathrm{with} \mathrm{h} = 1 \mathrm{is} \end{gathered}$$

• $\frac{63}{85}$

• ${\tan }^{-1}\left(4\right)$

• $\frac{108}{85}$

• $\frac{113}{85}$