$\underset{\mathrm{x}\to 0}{\lim }\frac{\left(1 - \cos \left({\displaystyle \frac{{\mathrm{x}}^2}{2}}\right)\right)\left(1 - \cos \left({\displaystyle \frac{{\mathrm{x}}^2}{4}}\right)\right)}{{\mathrm{x}}^8} = 2^{ - \mathrm{k}}, \mathrm{find} \mathrm{k}$

$\mathrm{The} \mathrm{value} \mathrm{of} 0 . {16}^{{\log }_{2 . 5}\left(\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + ... \infty \right)} \mathrm{is}$

$\underset{\mathrm{x}\to \mathrm{a}}{\lim }\frac{{\left(\mathrm{a} + 2\mathrm{x}\right)}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}} - {\left(3\mathrm{x}\right)}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}{{\left(3\mathrm{a} + \mathrm{x}\right)}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}} - {\left(4\mathrm{x}\right)}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}} \left(\mathrm{a} \ne 0\right) = ?$

• $\left(\frac{2}{9}\right){\left(\frac{2}{3}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

• ${\left(\frac{2}{3}\right)}^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

• ${\left(\frac{2}{9}\right)}^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

• $\left(\frac{2}{3}\right){\left(\frac{2}{9}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

$$\begin{gathered} \mathrm{Let} \mathrm{f} : \left(0, \infty \right) \to \left(0, \infty \right) \mathrm{be} \mathrm{a} \mathrm{differentiable} \mathrm{function} \mathrm{such} \mathrm{that} \mathrm{f}\left(1\right) = \mathrm{e} \mathrm{and} \\ \underset{\mathrm{t}\to \mathrm{x}}{\lim } \frac{{\mathrm{t}}^2{\mathrm{f}}^2\left(\mathrm{x}\right) - {\mathrm{x}}^2{\mathrm{f}}^2\left(\mathrm{t}\right)}{\mathrm{t} - \mathrm{x}}. \mathrm{If} \mathrm{f}\left(\mathrm{x}\right) = 1, \mathrm{then} \mathrm{x} \mathrm{is} \mathrm{equal} \mathrm{to} \end{gathered}$$

• $\frac{1}{\mathrm{e}}$

• 2e

• $\frac{1}{2\mathrm{e}}$

• e

$\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{x}\left({\mathrm{e}}^{\left({}^{\sqrt{1 + {\mathrm{x}}^2 + {\mathrm{x}}^4} - 1}\right)/\mathrm{x}} - 1\right)}{\sqrt{1 + {\mathrm{x}}^2 + {\mathrm{x}}^4} - 1}$

• does not exist

• is equal to 1

• is equal to $\sqrt{\mathrm{e}}$

• is equal to 0

$\mathrm{A} \underset{\mathrm{x}\to 1}{\lim }\left(\frac{{\int }_0^{{\left(\mathrm{x} - 1\right)}^2}\mathrm{tcos}\left({\mathrm{t}}^2\right)d\mathrm{t}}{\left(\mathrm{x} - 1\right)\sin \left(\mathrm{x} - 1\right)}\right) = ?$

• 1

• does not exist

• $\frac{1}{2}$

• $- \frac{1}{2}$

137.

If [x] denotes the greatest integer not exceeding x and if the function f defined by

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right) = \left\{ \frac{\mathrm{a} + 2\cos \left(\mathrm{x}\right)}{{\mathrm{x}}^2}, \mathrm{x}< 0 \\ \mathrm{btan}\left(\frac{\mathrm{\pi }}{\left(\mathrm{x} + 4\right)}\right), \mathrm{x} \ge 0\right\} \end{gathered}$$

is continuous at x = 0, then the ordered pair(a, b) is equal to

• (- 2, 1)

• (- 2, - 1)

• $\left(- 1, \sqrt{3}\right)$

• $\left(- 2, \sqrt{3}\right)$

$$\begin{gathered} \mathrm{If} \mathrm{y} = \left(1 +\mathrm{x}\right)\left(1 + {\mathrm{x}}^2\right)\left(1 + {\mathrm{x}}^4\right) . . . \left(1 + {\mathrm{x}}^{2\mathrm{n}}\right), \\ \mathrm{then} {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\mathrm{x} = 0} = ? \end{gathered}$$

• 0

• $\frac{1}{2}$

• 1

• 2

$$\begin{gathered} \mathrm{The} \mathrm{quadratic} \mathrm{equation} \mathrm{whose} \mathrm{roots} \mathrm{are} \mathrm{l} \mathrm{and} \mathrm{m},\mathrm{where} \\ \mathrm{l} = \underset{\mathrm{\theta } \to 0}{\lim } \left(\frac{3\sin \left(\mathrm{\theta }\right) - 4{\sin }^2\left(\mathrm{\theta }\right)}{\mathrm{\theta }}\right), \\ \mathrm{m} = \underset{\mathrm{\theta } \to 0}{\lim } \left(\frac{2\tan \left(\mathrm{\theta }\right)}{\mathrm{\theta } \left(1 - {\tan }^2\left(\mathrm{\theta }\right) \right)}\right) \mathrm{is} \end{gathered}$$

• ${\mathrm{x}}^2 + 5\mathrm{x} + 6$

• ${\mathrm{x}}^2 - 5\mathrm{x} + 6$

• ${\mathrm{x}}^2 - 5\mathrm{x} - 6$

• ${\mathrm{x}}^2 + 5\mathrm{x} - 6$

$\underset{\mathrm{x}\to 0}{\lim }\frac{\sqrt{1 + \sqrt{1 + {\mathrm{x}}^2 }} - 2}{\mathrm{x} - 8} = ?$

• $\frac{3}{2}$

• $\frac{1}{4}$

• $\frac{1}{24}$

• $\frac{1}{12}$