The solutions of the equation $\left|\begin{array}{ccc}\mathrm{x}& 2& - 1\\ 2& 5& \mathrm{x}\\ - 1& 2& \mathrm{x}\end{array}\right|$ = 0 are

• 3, - 1

• - 3, 1

• 3, 1

• - 3, - 1

32.

If A = $\left[\begin{array}{cc}3& 5\\ 2& 0\end{array}\right] \mathrm{and} \mathrm{B} = \left[\begin{array}{cc}1& 17\\ 0& - 10\end{array}\right]$, then $\left|\mathrm{AB}\right|$ is equal to

• 80

• 100

• - 110

• 92

The inverse of the matrix $\left[\begin{array}{cc}5& - 2\\ 3& 1\end{array}\right]$ is

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

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

• $\frac{1}{13}\left[\begin{array}{cc}- 2& 5\\ 1& 3\end{array}\right]$

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

The identity element in the group $\mathrm{M} = \left\{\left|\begin{array}{cc}\mathrm{x}& \mathrm{x}\\ \mathrm{x}& \mathrm{x}\end{array}\right|, \mathrm{x} \in \mathrm{R}; \mathrm{x} \ne 0\right\}$with respect to matrix multiplication is

• $\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]$

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

• $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

• $\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$

In the group G = {1, 3, 7, 9} under multiplication modulo 10, the inverse of 3 is

• 1

• 3

• 7

• 9

In the group (Q⁺, *) of positive rational numbers w.r.t. the binary operation * defined by a * b = $\frac{\mathrm{ab}}{3}, \forall \mathrm{a}, \mathrm{b} \in {\mathrm{Q}}^{+}$, the solution of the 3 equation 5 * 4 = 4^-1 in Q⁺ is

• $\frac{27}{20}$

• $\frac{20}{27}$

• $\frac{1}{20}$

• 20

If f(x) = $$\begin{gathered} \left\{\frac{\sin \left(5\mathrm{x}\right)}{{\mathrm{x}}^2 + 2\mathrm{x}}, \mathrm{x} \ne 0 \\ \mathrm{k} + \frac{1}{2}, \mathrm{x} = 0\right\} \end{gathered}$$ is continuous at x = 0, then the value of k is

• 1

• - 2

• 2

• 1/2

A population p(t) of 1000 bacteria introduced into nutrient medium grows according to the relation p(t) = 1000 + $\frac{1000\mathrm{t}}{100 +\hspace{0.17em}{\mathrm{t}}^2}$. The maximum 100+t size of this bacterial population is

• 1100

• 1250

• 1050

• 5250

$\mathrm{y} = {\tan }^{-1}\left(\frac{\sqrt{1 + {\mathrm{x}}^2} - \sqrt{1 - {\mathrm{x}}^2}}{\sqrt{1 + {\mathrm{x}}^2} + \sqrt{1 - {\mathrm{x}}^2}}\right), \mathrm{then} \frac{\mathrm{dy}}{\mathrm{dx}}$ is equal to

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

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

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

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

If ST and SN are the lengths of the subtangent and the subnormal at the point $\mathrm{\theta } = \frac{\mathrm{\pi }}{2}$ on the curve x = a$\left(\mathrm{\theta } + \sin \left(\mathrm{\theta }\right)\right)$, y = a$\left(1 - \cos \left(\mathrm{\theta }\right)\right)$, $\mathrm{a} \ne 1$, then

• ST = SN

• ST = 2SN

• ST² = aSN³

• ST³ = aSN