Let f(x) = $\frac{{\left({\mathrm{e}}^{\mathrm{x}} - 1\right)}^2}{\sin \left({\displaystyle \frac{\mathrm{x}}{\mathrm{a}}}\right)\log \left(1 + {\displaystyle \frac{\mathrm{x}}{4}}\right)}$ for $\mathrm{x} \ne 0$ and f(0) = 12.  If f is continuous at x = 0, then the value of a is equal to

• 1

• - 1

• 2

• 3

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

• $\frac{1}{\sqrt{1 - \mathrm{x}}}$

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

• $\frac{1}{\sqrt{\mathrm{x}}}$

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

The derivative of ${\sin }^{-1}\left(2\mathrm{x}\sqrt{1 - {\mathrm{x}}^2}\right)$ with respect to ${\sin }^{-1}\left(3\mathrm{x} - 4\mathrm{x}\right)$ is

• $\frac{2}{3}$

• $\frac{3}{2}$

• $\frac{1}{2}$

• 1

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

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

• $\mathrm{\pi }$

• 0

• 1

If $\mathrm{f}\left(\mathrm{x}\right) = \left|\mathrm{x} - 2\right| + \left|\mathrm{x} + 1\right| - \mathrm{x}$, then $\mathrm{f}\text{'}\left(- 10\right)$ is equal to

• - 3

• - 2

• - 1

• 0

If $\mathrm{x} = \mathrm{a}\left(1 + \cos \left(\mathrm{\theta }\right)\right), \mathrm{y} = \mathrm{a}\left(\mathrm{\theta } + \sin \left(\mathrm{\theta }\right)\right)$, then $\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}$ at $\mathrm{\theta } = \frac{\mathrm{\pi }}{2}$ is

• - $\frac{1}{\mathrm{a}}$

• $\frac{1}{\mathrm{a}}$

• - 1

• - 2

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

• $\frac{1}{2}$

• 2

• - 2

• - $\frac{1}{2}$

The distance between the origin and the normal to the curve y = e^2x + x² at x = 0 is

• 2

• $\frac{2}{\sqrt{3}}$

• $\frac{2}{\sqrt{5}}$

• $\frac{1}{2}$

The value of c in (0, 2) satisfying the mean value theorem for the function f(x) = x(x - 1)², x $\in$ [0, 2] is equal to

• $\frac{3}{4}$

• $\frac{4}{3}$

• $\frac{1}{3}$

• $\frac{2}{3}$

The point on the curve x² + y² = a², y $\ge$ 0 at which the tangent is parallel to x-axis is

• (a, 0)

• (- a, 0)

• $\left(\frac{\mathrm{a}}{2}, \frac{\sqrt{3}}{2}\mathrm{a}\right)$

• (0, a)