If the function f(x) = $$\begin{gathered} \left\{\mathrm{x}, \mathrm{if} \mathrm{x} \le 1 \\ \mathrm{cx} + \mathrm{k}, \mathrm{if} 1 < \mathrm{x} < 4 \\ - 2\mathrm{x}, \mathrm{if} \mathrm{x} \ge 4\right\} \end{gathered}$$ is continuous everywhere, then the values of c and k are respectively.

• - 3, - 5

• - 3, 5

• - 3, - 4

• - 3, 4

If y = $5^{\tan \left(\mathrm{x}\right)}$, then $\frac{\mathrm{dy}}{\mathrm{dx}}$ at $\mathrm{x} = \frac{\mathrm{\pi }}{4}$ is equal to

• $5\log \left(5\right)$

• $10\log \left(5\right)$

• 0

• ${\left(\log \left(5\right)\right)}^2$

If $\mathrm{y} = {\sin }^{-1}\left(\mathrm{x}\right) \mathrm{and} \mathrm{z} = {\cos }^{-1}\left(\sqrt{1 - {\mathrm{x}}^2}\right)$, then $\frac{\mathrm{dy}}{\mathrm{dz}}$ is equal to

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

• $\frac{1}{2}$

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

• 1

If u = 2(t - sin(t)) and v = 2 (1 - cos(t)), then $\frac{\mathrm{dv}}{\mathrm{du}}$ at t = $\frac{2\mathrm{\pi }}{3}$ is equal to

• $\sqrt{3}$

• $- \sqrt{3}$

• $2\sqrt{3}$

• $\frac{1}{\sqrt{3}}$

If $\mathrm{f}\left(\mathrm{x}\right) = \log \left[{\mathrm{e}}^{\mathrm{x}}{\left(\frac{3 - \mathrm{x}}{3 + \mathrm{x}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right]$, then f'(1) is equal to

• $\frac{3}{4}$

• $\frac{2}{3}$

• $\frac{1}{3}$

• $\frac{1}{2}$

If y = ${\left(\log \left(\mathrm{x}\right)\right)}^2$, then $\frac{\mathrm{dy}}{\mathrm{dx}}$ at x = e is equal to

• 2

• $\frac{\mathrm{e}}{2}$

• e

• $\frac{2}{\mathrm{e}}$

747.

If y^x = 2^x, then $\frac{\mathrm{dy}}{\mathrm{dx}}$ is equal to

• $\frac{\mathrm{y}}{\mathrm{x}}\log \left(\frac{2}{\mathrm{y}}\right)$

• $\frac{\mathrm{x}}{\mathrm{y}}\log \left(\frac{2}{\mathrm{y}}\right)$

• $\frac{\mathrm{y}}{\mathrm{x}}\log \left(\frac{\mathrm{y}}{2}\right)$

• $\frac{\mathrm{x}}{\mathrm{y}}\log \left(\frac{\mathrm{y}}{2}\right)$

If x² + 2xy + 2y² = 1, then $\frac{\mathrm{dy}}{\mathrm{dx}}$  at the point where y = 1 is equal to

• 1

• 2

• - 1

• 0

Let f(x) = x³ - x + p ($0 \le \mathrm{x} \le 2$) where p is a constant. The value c of mean value theorem is

• $\frac{\sqrt{3}}{2}$

• $\frac{\sqrt{6}}{3}$

• $\frac{\sqrt{3}}{3}$

• $\frac{2\sqrt{3}}{3}$

If the function $$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right) = \left\{\frac{{\mathrm{x}}^2 - \left(\mathrm{k} + 2\right)\mathrm{x} +\hspace{0.17em}2\mathrm{k}}{\mathrm{x} - 2} \mathrm{for} \mathrm{x} \ne 2 \\ 2 \mathrm{for} \mathrm{x} = 2\right\} \end{gathered}$$ is continuous at x = 2, then k is equal to

• $- \frac{1}{2}$

• - 1

• 0

• $\frac{1}{2}$