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) = $

Let $f(x)$ and $g(x)$ be two continuous functions defined from $R \rightarrow R$,such that $f(x_1) > f(x_2)$ and $g(x_1) < g(x_2)$ for all $x_1 > x_2$. Then the solution set of $f(g(\alpha^2 - 2\alpha)) > f(g(3\alpha - 4))$ is

Let $f(x)=x^2+2x+2$,$g(x)=-x^2+2x-1$,and $a, b$ be the extreme values of $f(x)$ and $g(x)$ respectively. If $c$ is the extreme value of $\frac{f}{g}(x)$ (for $x \neq 1$),then $a+2b+5c+4=$

Let $g(x) = ||x + 2| - 3|$. If $a$ denotes the number of relative minima,$b$ denotes the number of relative maxima,and $c$ denotes the product of the zeroes of $g(x)$,then the value of $(a + 2b - c)$ is:

If $f(x)=2\{x\}+5x$,where $\{x\}$ is the fractional part function,then $f(-1.4)$ is

The function $y = \frac{2x - 1}{x - 2}$ $(x \neq 2)$: