If $f$ is an exponential function and $g$ is a logarithmic function,then $fog(1)$ will be

If $f(x) = \frac{2x + 1}{3x - 2}$,then $(fof)(2)$ is equal to

Let $f: R - \{-\frac{1}{2}\} \rightarrow R$ and $g: R - \{-\frac{5}{2}\} \rightarrow R$ be defined as $f(x) = \frac{2x+3}{2x+1}$ and $g(x) = \frac{|x|+1}{2x+5}$. Then the domain of the function $f \circ g$ is:

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x)=x-[x]$ and $g(x)=[x]$ for $x \in R$, where $[x]$ is the greatest integer not exceeding $x$, then for every $x \in R, f(g(x))$ is equal to

If $f: R - \{\frac{3}{7}\} \rightarrow R - \{\frac{3}{7}\}$ is given by $f(x) = \frac{3x+5}{7x-3}$,then the statement which is not true is:

If $f(x)=3x+6$,$g(x)=4x+k$ and $f \circ g(x)=g \circ f(x)$,then $k =$