The total number matrices $\mathrm{A} = \left[\begin{array}{ccc}0& 2\mathrm{y}& 1\\ 2\mathrm{x}& \mathrm{y}& - 1\\ 2\mathrm{x}& - \mathrm{y}& 1\end{array}\right]$, $\left(\mathrm{x}, \mathrm{y} \in \mathrm{R}, \mathrm{x} \ne \mathrm{y}\right)$ for which A^TA = 3I₃ is

• 2

• 6

• 4

• 3

If the function f(x) = $$\begin{gathered} \left\{\mathrm{a}|\mathrm{\pi } - \mathrm{x}| + 1, \mathrm{x} \le 5 \\ \mathrm{b}\left|\mathrm{x} - \mathrm{\pi }\right| + 3, \mathrm{x} > 5\right\} \end{gathered}$$ is continuous at x = 5, then the value of a - b is

• $\frac{- 2}{\mathrm{\pi } + 5}$

• $\frac{2}{\mathrm{\pi } + 5}$

• $\frac{2}{\mathrm{\pi } - 5}$

• $\frac{2}{5 - \mathrm{\pi }}$

3.

If $\mathrm{f}\left(\mathrm{x}\right) = \left[\mathrm{x}\right] - \left[\frac{\mathrm{x}}{4}\right], \mathrm{x} \in \mathrm{R}$ , where [x] denotes the greatest integer function, then:

• Both $\underset{\mathrm{x}\to 4^{-}}{\lim }\mathrm{f}\left(\mathrm{x}\right)$ and $\underset{\mathrm{x}\to 4^{+}}{\lim }\mathrm{f}\left(\mathrm{x}\right)$ exist but noot equal

• f is continuous at x = 4

• $\underset{\mathrm{x}\to 4^{-}}{\lim }\mathrm{f}\left(\mathrm{x}\right)$ exist but $\underset{\mathrm{x}\to 4^{+}}{\lim }\mathrm{f}\left(\mathrm{x}\right)$ does not exist

• $\underset{\mathrm{x}\to 4^{+}}{\lim }\mathrm{f}\left(\mathrm{x}\right)$ exist but $\underset{\mathrm{x}\to 4^{-}}{\lim }\mathrm{f}\left(\mathrm{x}\right)$ does not exist

If the system of equation 2x + 3y - z = 0, x + ky - 2z = 0 and 2x - y + z = 0 has a non – trivial solution (x, y, z) , then $\frac{\mathrm{x}}{\mathrm{y}} +\hspace{0.17em}\frac{\mathrm{y}}{\mathrm{z}} + \frac{\mathrm{z}}{\mathrm{x}} + \mathrm{k}$ is equal to

• $\frac{3}{4}$

• $\frac{1}{2}$

• - 4

• - $\frac{1}{4}$

A water tank has the shape of an inverted right circular cone, whose semi-vertical angle is ${\tan }^{-1}\left(\frac{1}{2}\right)$. Water is poured into it at a constant rate of 5 cubic meter per minute. Then the rate (in m/min), at which the level of water is rising at the instant when the depth of water in the tank is 10m; is :

• $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$10\mathrm{\pi }$}\right.$

• $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$15\mathrm{\pi }$}\right.$

• $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5\mathrm{\pi }$}\right.$

• $\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$\mathrm{\pi }$}\right.$

The vertices B and C of a $∆$ABC lie on line, $\frac{\mathrm{x} + 2}{3} = \frac{\mathrm{y} - 1}{0} = \frac{\mathrm{z}}{4}$ such that BC = 5 units. Then the area (in sq.units) of this triangle, given that the point A(1, - 1, 2)

• $\sqrt{34}$

• $5\sqrt{17}$

• $2\sqrt{34}$

• 6

The area (in sq.units) of the region A = $\left\{\left(\mathrm{x}, \mathrm{y}\right) : \frac{{\mathrm{y}}^2}{2} \le \mathrm{x} \le \mathrm{y} +\hspace{0.17em}4\right\}$ is

• 18

• 16

• $\frac{53}{3}$

• 30

If f : $\mathrm{R} \to \mathrm{R}$ is a differentiable function and f(2) = 6, then $\underset{\mathrm{x}\to 2}{\lim }{\int }_6^{\mathrm{f}\left(\mathrm{x}\right)}\frac{2\mathrm{t} \mathrm{dt}}{\left(\mathrm{x} - 2\right)}$ is

• 0

• 24f'(2)

• 12f'(2)

• 2f'(2)

If unit vector $\overrightarrow{\mathrm{a}}$ makes angles $\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$ with $\hat{\mathrm{i}}, \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$, with $\hat{\mathrm{j}}$ and $\mathrm{\theta } \in \left(0, \mathrm{\pi }\right)$ with $\hat{\mathrm{k}}$, then a value of $\mathrm{\theta }$ is

• $\frac{2\mathrm{\pi }}{3}$

• $\frac{5\mathrm{\pi }}{6}$

• $\frac{5\mathrm{\pi }}{12}$

• $\frac{\mathrm{\pi }}{4}$

The value of the integral ${\int }_0^1{\mathrm{xcot}}^{-1}\left(1 - {\mathrm{x}}^2 +\hspace{0.17em}{\mathrm{x}}^4\right)\mathrm{dx}$ is

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

• $\frac{\mathrm{\pi }}{4} - {\log }_{\mathrm{e}}\left(2\right)$

• $\frac{\mathrm{\pi }}{4} - \frac{1}{2}{\log }_{\mathrm{e}}\left(2\right)$

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