If $y = \frac{\tan x \cos^{-1} x}{\sqrt{1-x^2}}$, then the value of $\frac{dy}{dx}$ when $x = 0$ is:

Let $f: R \to R$ be a differentiable function such that $f(2) = 6$ and $f'(2) = \frac{1}{48}.$ Then $\lim_{x \to 2} \int_{6}^{f(x)} \frac{4t^3}{x - 2} dt$ equals

Let $y = f(x) = \begin{cases} e^{-\frac{1}{x^2}}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}$. Then which of the following can best represent the graph of $y = f(x)$?

If $f(x) = \begin{cases} \frac{8}{x^3} - 6x, & x \le 1 \\ \sqrt{x} + 1, & x > 1 \end{cases}$, then at $x = 1$, $f$ is:

The set of values of $p$ for which the equation $|\ln x| - px = 0$ possesses three distinct roots is

Observe the following statements:
$I. f(x) = a x^{41} + b x^{-40} \Rightarrow \frac{f^{\prime \prime}(x)}{f(x)} = 1640 x^{-2}$
$II. \frac{d}{d x} \tan ^{-1}\left(\frac{2 x}{1-x^2}\right) = \frac{1}{1+x^2}$
Which of the following is correct?