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

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:

For $x \in R - \{0, 1\}$,let ${f_1}(x) = \frac{1}{x}$,${f_2}(x) = 1 - x$,and ${f_3}(x) = \frac{1}{1 - x}$ be three given functions. If a function $J(x)$ satisfies $(f_2 \circ J \circ f_1)(x) = f_3(x)$,then $J(x)$ is equal to:

Let $f(x)=\sqrt{x^{2}-3x+2}$ and $g(x)=\sqrt{x}$ be two given functions. If $S$ is the domain of $f \circ g$ and $T$ is the domain of $g \circ f$,then:

For $x \in R$,two real-valued functions $f(x)$ and $g(x)$ are such that $g(x) = \sqrt{x} + 1$ and $(f \circ g)(x) = x + 3 - \sqrt{x}$. Then $f(0)$ is equal to:

If $f(x) = \sin^2 x + \sin^2(x + \frac{\pi}{3}) + \cos x \cos(x + \frac{\pi}{3})$ and $g(\frac{5}{4}) = 1$,then $(g \circ f)(x) = $