If $f : R \to R$ and $g : R \to R$ are defined as $f(x) = 2x - |x|$ and $g(x) = 2x + |x|$, then

If $f(x)=-|x|$, then $(f \circ f \circ f)(x) + (f \circ f \circ f)(-x) =$

Consider the function $f: R \rightarrow R$ defined by $f(x)=\frac{2x}{\sqrt{1+9x^2}}$. If the composition of $f$,$\underbrace{(f \circ f \circ \ldots \circ f)}_{10 \text{ times }}(x) = \frac{2^{10}x}{\sqrt{1+9\alpha x^2}}$,then the value of $\sqrt{3\alpha+1}$ is equal to:

If $f: R \to R$ is defined by $f(x) = (x + 1)^2$ and $g: R \to R$ is defined by $g(x) = x^2 + 1$,then $(fog)(-3)$ equals:

Let $R = \{(1, 3), (2, 2), (3, 2)\}$ and $S = \{(2, 1), (3, 2), (2, 3)\}$ be two relations on the set $A = \{1, 2, 3\}$. Then $R \circ S = $

If $f(x) = \log_a x$ and $F(x) = a^x$,then $F[f(x)]$ is