Find the minimum integral value of $\alpha$ for which the graph of $f(x) = ||x - 2| - \alpha| - 5$ has exactly four $x-$intercepts.

Assertion $(A)$: $\coth x = \frac{1-k}{1+k}$ where $0 < k < 2$.
Reason $(R)$: The graph of $y = \tanh x$ always lies between the lines $y = -1$ and $y = 1$.
Choose the correct option:

Let the function $g : (-\infty, \infty) \to \left( - \frac{\pi}{2}, \frac{\pi}{2} \right)$ be given by $g(u) = 2 \tan^{-1}(e^u) - \frac{\pi}{2}$. Then,$g$ is -

Let the function $g: (-\infty, \infty) \to \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ be defined by $g(u) = 2 \tan^{-1}(e^u) - \frac{\pi}{2}$. Then $g(u)$ is:

If $f(x) = \frac{1}{\sqrt{x+2 \sqrt{2x-4}}} + \frac{1}{\sqrt{x-2 \sqrt{2x-4}}}$ for $x > 2$,then $f(11)$ is equal to

The minimum value of $|x| + |x + \frac{1}{2}| + |x - 3| + |x - \frac{5}{2}|$ is