Differentiation of $(x^2-5x+8) \times (x^3+7x+9)$ can be done by

Let $f(x) = (\sin(\tan^{-1} x) + \sin(\cot^{-1} x))^2 - 1$ for $|x| > 1$. If $\frac{dy}{dx} = \frac{1}{2} \frac{d}{dx}(\sin^{-1}(f(x)))$ and $y(\sqrt{3}) = \frac{\pi}{6}$,then $y(-\sqrt{3})$ is equal to

If $y = \log(\tan(x/2)) + \sin^{-1}(\cos x)$,then $dy/dx$ is

$\frac{1}{e^{3x}}(e^x + e^{5x}) = a_0 + a_1x + a_2x^2 + \ldots$
$\Rightarrow 2a_1 + 2^3a_3 + 2^5a_5 + \ldots$ is equal to

$(i)$ $f(x)$ is continuous and defined for all real numbers.
$(ii)$ $f'(-5) = 0$; $f'(2)$ is not defined and $f'(4) = 0$.
$(iii)$ $(-5, 12)$ is a point which lies on the graph of $f(x)$.
$(iv)$ $f''(2)$ is undefined,but $f''(x)$ is negative everywhere else.
$(v)$ The signs of $f'(x)$ are given below:
| $x$ | $(-\infty, -5)$ | $-5$ | $(-5, 2)$ | $2$ | $(2, 4)$ | $4$ | $(4, \infty)$ |
|---|---|---|---|---|---|---|---|
| $f'(x)$ | $+$ | $0$ | $-$ | Undefined | $+$ | $0$ | $-$ |
Possible graph of $y = f(x)$ is:

Let $f(x) = \begin{cases} \frac{5 e^{1/x} + 2}{3 - e^{1/x}}, & x \neq 0 \\ 0, & x = 0 \end{cases}$. Then at $x = 0$,$x f(x)$ and $f(x)$ are respectively: