If $f(x)$ and $g(x)$ are functions satisfying $f(g(x)) = x^3 + 3x^2 + 3x + 4$ and $f(x) = (\ln x)^3 + 3$,then the slope of the tangent to the curve $y = g(x)$ at $x = -1$ is:

Let $Q$ be the set of all rational numbers in $[0,1]$ and $f:[0,1] \rightarrow [0,1]$ be defined by $f(x) = \begin{cases} x & \text{for } x \in Q \\ 1-x & \text{for } x \notin Q \end{cases}$. Then,the set $S = \{x \in [0,1] : (f \circ f)(x) = x\}$ is equal to

Let $R$ be a relation '$ < $' from $A$ to $B$,where $A = \{1, 2, 3, 4\}$ and $B = \{1, 3, 5\}$ such that $(a, b) \in R \iff a < b$. Then $R \circ R^{-1}$ is:

The composite mapping $fog$ of the maps $f:R \to R$,$f(x) = \sin x$,and $g:R \to R$,$g(x) = x^2$ is:

If $g(f(x)) = |\sin x|$ and $f(g(x)) = (\sin \sqrt{x})^2$,then

If $f(x) = 3x + 10$ and $g(x) = x^2 - 1$,then $(fog)^{-1}(x) = $