The number of integral values of $a$ for which the function $f: R \to R, f(x) = 2x^3 - 3(a + 2)x^2 + 12ax - 7$ where $a \in [-4, 6]$ is invertible,is

Let $f: R - \{\frac{\alpha}{6}\} \rightarrow R$ be defined by $f(x) = \frac{5x + 3}{6x - \alpha}$. Then the value of $\alpha$ for which $(f \circ f)(x) = x$,for all $x \in R - \{\frac{\alpha}{6}\}$,is:

If $f(x) = x + \tan x$ and $g(x)$ is the inverse of $f(x)$,then $g'(x)$ is equal to

If $f: R \rightarrow R$ is defined by $f(x)=2x+3$,then $f^{-1}(x)$

Consider $f: R_{+} \rightarrow [4, \infty)$ given by $f(x) = x^{2} + 4$. Show that $f$ is invertible with the inverse $f^{-1}$ of $f$ given by $f^{-1}(y) = \sqrt{y - 4}$,where $R_{+}$ is the set of all non-negative real numbers.

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x) = 5x - 3$ and $g(x) = x^2 + 3$,then $g \circ f^{-1}(3)$ is equal to