$\mathrm{A} = \left[\begin{array}{cc}1& 2\\ 3& 2\\ - 1& 0\end{array}\right], \mathrm{B} = \left[\begin{array}{ccc}1& 3& 2\\ 4& - 1& 3\end{array}\right]$, then order of AB is

• 2 x 2

• 3 x 3

• 1 x 3

• 3 x 2

$\frac{\mathrm{d}}{\mathrm{dx}}\left[\sec \left\{{\cos }^{-1}\left(\frac{\mathrm{x}}{8}\right)\right\}\right]$ is equal to

• $\frac{1}{8}$

• $- \frac{1}{8}$

• $\frac{8}{{\mathrm{x}}^2}$

• - $\frac{8}{{\mathrm{x}}^2}$

If f(x) = $\sqrt{1 + {\cos }^2\left({\mathrm{x}}^2\right)}$, then $\mathrm{f}\text{'}\left(\frac{\sqrt{\mathrm{\pi }}}{2}\right)$ is

• $\frac{\sqrt{\mathrm{\pi }}}{6}$

• $- \sqrt{\frac{\mathrm{\pi }}{6}}$

• $\frac{1}{\sqrt{6}}$

• $\frac{\mathrm{\pi }}{\sqrt{6}}$

If A + I = $\left[\begin{array}{cc}3& - 2\\ 4& 1\end{array}\right]$, then (A + I)(A - I) is equal to

• $\left[\begin{array}{cc}- 5& - 4\\ 8& - 9\end{array}\right]$

• $\left[\begin{array}{cc}- 5& 4\\ - 8& 9\end{array}\right]$

• $\left[\begin{array}{cc}5& 4\\ 8& 9\end{array}\right]$

• $\left[\begin{array}{cc}- 5& - 4\\ - 8& - 9\end{array}\right]$

If y = ${\mathrm{asin}}^3\left(\mathrm{\theta }\right)$ and x = ${\mathrm{acos}}^3\left(\mathrm{\theta }\right)$, then at $\mathrm{\theta } = \frac{\mathrm{\pi }}{3}, \frac{\mathrm{dy}}{\mathrm{dx}}$ is equal to

• $\frac{\sqrt{\mathrm{\pi }}}{6}$

• $- \sqrt{3}$

• $\frac{- 1}{\sqrt{3}}$

• $\sqrt{3}$

If x + y = 8, then maximum value of x²y is

• $\frac{2048}{9}$

• $\frac{2048}{81}$

• $\frac{2048}{3}$

• $\frac{2048}{27}$

If A = $\left[\begin{array}{cc}1& - 1\\ 2& - 1\end{array}\right]$ and B = $\left[\begin{array}{cc}1& \mathrm{a}\\ 4& \mathrm{b}\end{array}\right]$ and (A + B)² = A² + B². Then, a and b are respectively

• 1, - 1

• 2, - 3

• - 1, 1

• 3, - 2

If A = $\left[\begin{array}{ccc}2& - 1& 1\\ - 2& 3& - 2\\ - 4& 4& - 3\end{array}\right]$, then A² is equal to

• null matrix

• itself A

• unit matrix

• Scalar matrix

If f(1) = 10, f(2) = 14, then using Newton's forward formula f(1.3) is equal to 

• 12.2

• 11.2

• 10.2

• 15.2

10.

If f(x) = x³ - 9x + 1 and x = 3, then using Newton Raphson method, first iteration is

• $\frac{53}{18}$

• $\frac{53}{9}$

• $\frac{35}{18}$

• $\frac{35}{9}$