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:

If $f: R \rightarrow R$ is defined by $f(x)= \begin{cases} |[x-5]|, & \text{for } x < 5 \\ [|x-5|], & \text{for } x \geq 5 \end{cases}$ Then,$(f \circ f)\left(-\frac{7}{2}\right) = ?$ (here,$[x]$ is the greatest integer function)

Define $f(x) = \begin{cases} 1 + x, & 0 \leq x \leq 2 \\ 3 - x, & 2 < x \leq 3 \end{cases}$. If $f \circ f(x)$ is discontinuous at $a$ and $b$ in $[0, 3]$ and $a < b$, then $2 a + 3 b = $

If $R$ is a relation $  < $ from $A = \{1, 2, 3, 4\}$ to $B = \{1, 3, 5\}$ i.e.,$(a, b) \in R \iff a < b$,then $R \circ R^{-1}$ is

Let $f(x)=3+2x$ and $g_n(x)=(f \circ f \circ f \circ \dots \text{n times})(x)$. For all $n \in N$,if all the lines $y=g_n(x)$ pass through a fixed point $(\alpha, \beta)$,then $\alpha+\beta=$

If $f(x) = \begin{cases} 2+2x, & -1 \leq x < 0 \\ 1-\frac{x}{3}, & 0 \leq x \leq 3 \end{cases}$ and $g(x) = \begin{cases} -x, & -3 \leq x \leq 0 \\ x, & 0 < x \leq 1 \end{cases}$,then the range of $(f \circ g)(x)$ is: