If ab > - 1, bc > - 1 and ca > - 1, then  is

• - 1

• ${\cot }^{-1}\left(\mathrm{abc}\right)$

• 0

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

• $\frac{1}{2}$

If xe^xy + ye^{- xy} = ${\sin }^2\left(\mathrm{x}\right)$, then $\frac{\mathrm{dy}}{\mathrm{dx}}$ at x = 0 is

• 2y² - 1

• 2y

• y² - y

• y² - 1

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

• $\frac{1}{2}$

• $\frac{2}{3}$

• 1

• $\frac{3}{2}$

If f(x) = ${\cos }^{-1}\left(\frac{2\cos \left(\mathrm{x} + 3\sin \left(\mathrm{x}\right)\right)}{\sqrt{13}}\right)$, then [f'(x)]² is equal to

• $\sqrt{1 + \mathrm{x}}$

• 1 + 2x

• 2

• 1

If u = ${\tan }^{-1}\left(\frac{\sqrt{1 - {\mathrm{x}}^2} - 1}{\mathrm{x}}\right)$ and v = ${\sin }^{-1}\left(\mathrm{x}\right)$, then $\frac{\mathrm{du}}{\mathrm{dv}}$ is equal to

• $\sqrt{1 - {\mathrm{x}}^2}$

• $- \frac{1}{2}$

• 1

• - x

17.

If y = $\frac{1}{1 + \mathrm{x} + {\mathrm{x}}^2}$, then $\frac{\mathrm{dy}}{\mathrm{dx}}$ is equal to

• y²(2 + 2x)

• $\frac{- \left(1 + 2\mathrm{x}\right)}{{\mathrm{y}}^2}$

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

• - y²(1 + 2x)

If g(x) is the inverse of f(x) and f'(x) = $\frac{1}{1 + {\mathrm{x}}^3}$, then g'(x) is equal to

• g(x)

• 1 + g(x)

• 1 + {g(x)}³

• $\frac{1}{1 + {\left\{\mathrm{g}\left(\mathrm{x}\right)\right\}}^3}$

The equation of the tangent to the curve $\sqrt{\frac{\mathrm{x}}{\mathrm{a}}} + \sqrt{\frac{\mathrm{y}}{\mathrm{b}}}$ = 1 at the point (x₁, y₁) is $\frac{\mathrm{x}}{\sqrt{{\mathrm{ax}}_1}} + \frac{\mathrm{y}}{\sqrt{{\mathrm{by}}_1}} = \mathrm{k}$. Then, the value of k is

• 2

• 1

• 3

• 3

The slope of the normal to the curve x = t² + 3t - 8 and y = 2t² - 2t - 5 at the point (2, - 1) is

• $\frac{6}{7}$

• $- \frac{6}{7}$

• $\frac{7}{6}$

• - $\frac{7}{6}$