The relationship between the correlation coefficient r and the regression coefficients b_xy and b_yx, is

• $\mathrm{r} = \sqrt{{\mathrm{b}}_{\mathrm{xy}}{\mathrm{b}}_{\mathrm{yx}}}$

• $\mathrm{r} = \frac{2{\mathrm{b}}_{\mathrm{xy}}{\mathrm{b}}_{\mathrm{yx}}}{{\mathrm{b}}_{\mathrm{xy}} + {\mathrm{b}}_{\mathrm{yx}}}$

• $\mathrm{r} = \frac{1}{2}\left({\mathrm{b}}_{\mathrm{xy} }+ {\mathrm{b}}_{\mathrm{yx}}\right)$

• $\mathrm{r} = {\mathrm{b}}_{\mathrm{xy}} . {\mathrm{b}}_{\mathrm{yx}}$

If $\overline{\mathrm{x}} = \overline{\mathrm{y}} = 0, \sum \mathrm{xy} = 12, {\mathrm{\sigma }}_{\mathrm{x}} = 2, {\mathrm{\sigma }}_{\mathrm{y}} = 3, \mathrm{n} = 10$ then the coefficient of correlation is

• 0.4

• 0.3

• 0.2

• 0.1

43.

Performing 3 iterations of bisection method the smallest positive approximate root of the equation x³ - 5x + 1 = 0, is

• 0.25

• 0.125

• 0.52

• 0.1875

If $\left[\begin{array}{ccc}1& 3& \mathrm{\lambda } + 2\\ 2& 4& 8\\ 3& 5& 10\end{array}\right]$ is a singular matrix, then the value of $\mathrm{\lambda }$ is

• - 2

• 4

• 2

• - 4

$\left|\begin{array}{ccc}1 + \mathrm{x}& 1& 1\\ 1& 1 + \mathrm{y}& 1\\ 1& 1& 1 +\hspace{0.17em}\mathrm{z}\end{array}\right|$ is equal to

• $\mathrm{xyz}\left(1 + \frac{1}{\mathrm{x}} + \frac{1}{\mathrm{y}} + \frac{1}{\mathrm{z}}\right)$

• xyz

• $1 + \frac{1}{\mathrm{x}} + \frac{1}{\mathrm{y}} + \frac{1}{\mathrm{z}}$

• $\frac{1}{\mathrm{x}} + \frac{1}{\mathrm{y}} + \frac{1}{\mathrm{z}}$

$\sin \left[{\cot }^{-1}\left\{{\tan }^{-1}\left(\mathrm{x}\right)\right\}\right]$ is equal to

• 1

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

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

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

Inverse of the function f(x) = $\frac{{\mathrm{e}}^{\mathrm{x}} - {\mathrm{e}}^{- \mathrm{x}}}{{\mathrm{e}}^{\mathrm{x}} + {\mathrm{e}}^{- \mathrm{x}}} + 2$ is

• ${\log }_{\mathrm{e}}{\left(\frac{\mathrm{x} - 2}{\mathrm{x} - 1}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

• ${\log }_{\mathrm{e}}{\left(\frac{\mathrm{x} - 1}{3 - \mathrm{x}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

• ${\log }_{\mathrm{e}}{\left(\frac{\mathrm{x}}{2 - \mathrm{x}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

• ${\log }_{\mathrm{e}}{\left(\frac{\mathrm{x} - 1}{\mathrm{x} + 1}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

If f(x) = $\frac{\mathrm{x} - 1}{\mathrm{x}}, \mathrm{x} \ne 0, 0 \in \mathrm{R}$ and g(u) = u² + 1, $\mathrm{u} \in \mathrm{R}$ then g[f(1)] and f[g(- 1)] is equal to

• 1, 1/2

• - 1, 1/2

• 0, - 1

• None of these

If f(x) = $$\begin{gathered} \left\{\mathrm{b} + \mathrm{ax}, \mathrm{x} < 0 \\ \mathrm{b}{\left(\mathrm{x} - 1\right)}^2, \mathrm{x} \ge 0\right\} \end{gathered}$$, is differentiable at x = 0, then (a, b) will be

• (- 3, - 1)

• (- 3, 1)

• (3, 1)

• (2, - 1)

If f(x) = $\frac{2\mathrm{x} - 1}{\mathrm{x} +\hspace{0.17em}5}; \left(\mathrm{x} \ne - 5\right)$, then f^-1(x) is equal to

• $\frac{\mathrm{x} + 5}{2\mathrm{x} - 1}, \mathrm{x} \ne \frac{1}{2}$

• $\frac{5\mathrm{x} + 1}{2 - \mathrm{x}}, \mathrm{x} \ne 2$

• $\frac{\mathrm{x} - 5}{2\mathrm{x} + 1}, \mathrm{x} \ne \frac{1}{2}$

• $\frac{5\mathrm{x} - 1}{\mathrm{x} - 2}, \mathrm{x} \ne 2$