A.

$\frac{\mathrm{y}}{\mathrm{x}}$

$$\begin{gathered} \mathrm{Given}, {\cos }^{-1}\left(\frac{{\mathrm{x}}^2 - {\mathrm{y}}^2}{{\mathrm{x}}^2 + {\mathrm{y}}^2}\right) = \mathrm{k} \\ \frac{{\mathrm{x}}^2 - {\mathrm{y}}^2}{{\mathrm{x}}^2 + {\mathrm{y}}^2} = \cos \left(\mathrm{k}\right) \\ \mathrm{On} \mathrm{differentiating} \mathrm{w}.\mathrm{r}.\mathrm{t}. \mathrm{x}, \mathrm{we} \mathrm{get} \\ \frac{\left({\mathrm{x}}^2 + {\mathrm{y}}^2\right)\left(2\mathrm{x} - 2\mathrm{y}{\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}}\right) - \left({\mathrm{x}}^2 - {\mathrm{y}}^2\right)\left(2\mathrm{x} + 2\mathrm{y}{\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}}\right) }{{\left({\mathrm{x}}^2 + {\mathrm{y}}^2\right)}^2} = 0 \\ \Rightarrow - 4{\mathrm{x}}^2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}} + 4{\mathrm{xy}}^2 = 0 \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{y}}{\mathrm{x}} \end{gathered}$$

C.

$\frac{1}{24}$

$$\begin{gathered} \underset{\mathrm{x}\to 0}{\lim }\frac{\sqrt{1 + \sqrt{1 + {\mathrm{x}}^2}} - 2}{\mathrm{x} - 8} \times \frac{\sqrt{1 + \sqrt{1 + {\mathrm{x}}^2}} + 2}{\sqrt{1 + \sqrt{1 + {\mathrm{x}}^2}} + 2} \\ = \underset{\mathrm{x}\to 0}{\lim }\frac{1 + \sqrt{1 + \mathrm{x}} - 4}{\sqrt{1 + \sqrt{1 + \mathrm{x}} + 2}\left(\mathrm{x} - 8\right)} \\ = \underset{\mathrm{x}\to 0}{\lim }\frac{1 + \sqrt{1 + \mathrm{x}} - 4}{\sqrt{1 + \sqrt{1 + \mathrm{x}} + 2}\left(\mathrm{x} - 8\right)} \times \frac{\sqrt{1 + \mathrm{x}} + 3}{\left(\sqrt{1 + \mathrm{x}} + 3\right)} \\ = \underset{\mathrm{x}\to 0}{\lim }\frac{1 + \mathrm{x} - 9}{\left(\sqrt{1 + \mathrm{x}} + 3\right)\left(\sqrt{1 + \sqrt{1 + \mathrm{x}} + 2}\right)\left(\mathrm{x} - 8\right)} \\ = \underset{\mathrm{x}\to 0}{\lim }\frac{1}{\left(\sqrt{1 + \mathrm{x}} + 3\right)\left(\sqrt{1 + \sqrt{1 + \mathrm{x}} + 2}\right)} \\ = \frac{1}{\left(\sqrt{1 + 8} + 3\right)\left(\sqrt{1 + \sqrt{1 + 8} + 2}\right)} \\ = \frac{1}{\left(3 + 3\right)\left(2 + 2\right)} = \frac{1}{24} \end{gathered}$$

B.

${\mathrm{x}}^2 - 5\mathrm{x} + 6$

$$\begin{gathered} \mathrm{We} \mathrm{have}, \\ \mathrm{l} = \underset{\mathrm{\theta }\to 0}{\lim }\left[\frac{3\sin \left(\mathrm{\theta }\right) - 4{\sin }^2\left(\mathrm{\theta }\right)}{\mathrm{\theta }}\right] \\ \mathrm{Using} \mathrm{L}-\mathrm{hospital} \mathrm{rule} \\ = \underset{\mathrm{\theta }\to 0}{\lim } \frac{3\cos \left(\mathrm{\theta }\right) - 8\sin \left(\mathrm{\theta }\right)\cos \left(\mathrm{\theta }\right)}{1} = 3 \\ \mathrm{and} \mathrm{m} = \underset{\mathrm{\theta }\to 0}{\lim }\frac{2\tan \left(\mathrm{\theta }\right)}{1 -{\tan }^2\left(\mathrm{\theta }\right)} \\ = \underset{\mathrm{\theta }\to 0}{\lim } \frac{\tan \left(2\mathrm{\theta }\right)}{\mathrm{\theta }} \\ = \underset{\mathrm{\theta }\to 0}{\lim } \frac{\tan \left(2\mathrm{\theta }\right)}{2\mathrm{\theta }} = 2 \\ \mathrm{The} \mathrm{required} \mathrm{quadratic} \mathrm{equation} \mathrm{is} \\ \Rightarrow {\mathrm{x}}^2 - \left(\mathrm{l} + \mathrm{m}\right)\mathrm{x} +\mathrm{lm} = 0 \\ \Rightarrow {\mathrm{x}}^2 - \left(3 + 2\right)\mathrm{x} + 3 . 2 = 0 \\ {\mathrm{x}}^2 - 5\mathrm{x} + 6 = 0 \end{gathered}$$