x = 0, the function f(x) =  is

• continuous but not differentiable

• discontinuous and differentiable

• discontinuous and not differentiable

• continuous and differentiable

, then A^-1 is equal to

• $\left[\begin{array}{cc}- 5& - 2\\ - 3& 1\end{array}\right]$

• $\left[\begin{array}{cc}5/11& 2/11\\ 3/11& - 1/11\end{array}\right]$

• $\left[\begin{array}{cc}- 5/11& - 2/11\\ - 3/11& - 1/11\end{array}\right]$

• $\left[\begin{array}{cc}5& 2\\ 3& - 1\end{array}\right]$

Let f(x) = $$\begin{gathered} \left\{\frac{\sin \left(\mathrm{\pi x}\right)}{5\mathrm{x}}, \mathrm{x} \ne 0 \\ \mathrm{k}, \mathrm{x} = 0\right\} \end{gathered}$$if f(x) is continuous at x = 0, then k is equal to

• $\frac{\mathrm{\pi }}{5}$

• $\frac{5}{\mathrm{\pi }}$

• 1

• 0

If $\overrightarrow{\mathrm{a}}$ is perpendicular to $\overrightarrow{\mathrm{b}} \mathrm{and} \overrightarrow{\mathrm{c}}$, $\left|\overrightarrow{\mathrm{a}}\right| = 2, \left|\overrightarrow{\mathrm{b}}\right| = 3, \left|\overrightarrow{\mathrm{c}}\right| = 4$ and the angle between $\overrightarrow{\mathrm{b}} \mathrm{and} \overrightarrow{\mathrm{c}}$ is $\frac{2\mathrm{\pi }}{3}$, then $\left[\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}\right]$ is equal to

• $4\sqrt{3}$

• $6\sqrt{3}$

• $12\sqrt{3}$

• $18\sqrt{3}$

The general solution of y²dx + (x² - xy + y²)dy = 0 is

• ${\tan }^{-1}\left(\frac{\mathrm{x}}{\mathrm{y}}\right) + \log \left(\mathrm{y}\right) + \mathrm{c} = 0$

• $2{\tan }^{-1}\left(\frac{\mathrm{x}}{\mathrm{y}}\right) + \log \left(\mathrm{x}\right) + \mathrm{c} = 0$

• $\log \left(\mathrm{y} + \sqrt{{\mathrm{x}}^2 + {\mathrm{y}}^2}\right) + \log \left(\mathrm{y}\right) + \mathrm{c} = 0$

• ${\sinh }^{-1}\left(\frac{\mathrm{x}}{\mathrm{y}}\right) + \log \left(\mathrm{y}\right) + \mathrm{c} = 0$

The vector $\hat{\mathrm{i}} + \mathrm{x}\hat{\mathrm{j}} + 3\hat{\mathrm{k}}$ is rotated through an angle $\mathrm{\theta }$ and doubled in magnitude, then it becomes $4\hat{\mathrm{i}} + \left(4\mathrm{x} - 2\right)\hat{\mathrm{j}} + 2\hat{\mathrm{k}}$. The value of x is

• $\left\{- \frac{2}{3}, 2\right\}$

• $\left\{\frac{1}{3}, 2\right\}$

• $\left\{\frac{2}{3}, 0\right\}$

• {2, 7}

A ball of mass 3 kg moving with a velocity of 3 m/s collides with another ball of mass 1 kg moving with velocity u in the opposite direction. If the first ball comes to rest after the impact and e  = $\frac{2}{7}$, then u is m/s, is

• $\frac{13}{3}$

• $\frac{17}{3}$

• $\frac{19}{3}$

• $\frac{23}{3}$

Three forces ofmagnitude 30, 60 and P acting at a point are in equilibrium. If the angle between the first two is 60°, the value of P is

• $30\sqrt{7}$

• $30\sqrt{3}$

• $20\sqrt{6}$

• $25\sqrt{2}$

Let a and b be two equal vectors inclined at an angle $\mathrm{\theta }$, then $\mathrm{a} \sin \left(\frac{\mathrm{\theta }}{2}\right)$ is equal to

• $\frac{\left|\overrightarrow{\mathrm{a}} - \overrightarrow{\mathrm{b}}\right|}{2}$

• $\frac{\left|\overrightarrow{\mathrm{a}} + \overrightarrow{\mathrm{b}}\right|}{2}$

• $\left|\overrightarrow{\mathrm{a}} - \overrightarrow{\mathrm{b}}\right|$

• $\left|\overrightarrow{\mathrm{a}} + \overrightarrow{\mathrm{b}}\right|$

20.

The solution of the equation $\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2} = {\mathrm{e}}^{- 2\mathrm{x}}$ is

• $\mathrm{y} = \frac{1}{4}{\mathrm{e}}^{- 2\mathrm{x}} + \frac{\mathrm{cx}}{2} +\hspace{0.17em}\mathrm{d}$

• $\mathrm{y} = \frac{1}{4}{\mathrm{e}}^{- 2\mathrm{x}} + \mathrm{cx} +\hspace{0.17em}\mathrm{d}$

• $\mathrm{y} = \frac{1}{4}{\mathrm{e}}^{- 2\mathrm{x}} + {\mathrm{cx}}^2 +\hspace{0.17em}\mathrm{d}$

• $\mathrm{y} = \frac{1}{4}{\mathrm{e}}^{- 2\mathrm{x}} + {\mathrm{cx}}^3 +\hspace{0.17em}\mathrm{d}$