Let $f$ and $g$ be functions defined by $f(x) = \frac{x}{x + 1}$ and $g(x) = \frac{x}{1 - x}$. Then $(fog)(x)$ is

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x)=x^3-x$ and $g(x)=\sin 2x$,then the values of $x \in (0, 2\pi)$ that satisfy $f(g(x)) > 0$ lie in the interval

For a suitably chosen real constant $a$,let a function $f: R-\{-a\} \rightarrow R$ be defined by $f(x)=\frac{a-x}{a+x}$. Further suppose that for any real number $x \neq-a$ and $f(x) \neq-a$,$(f \circ f)(x)=x$. Then $f\left(-\frac{1}{5}\right)$ is equal to

For a suitably chosen real constant $a$,let the function $f: R-\{-a\} \rightarrow R$ be defined by $f(x)=\frac{a-x}{a+x}$. Further,suppose that for any real number $x \neq-a$ and $f(x) \neq-a$,$(f \circ f)(x)=x$. Then,$f\left(-\frac{1}{2}\right)$ is equal to

If $f(x)$ and $g(x)$ are two real valued functions such that $f(x)=3x-2$ and $g(x)=x^2+2$, then $[(g \circ f)+(f \circ g)](x) = $

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) = $