Relations and F...

If $$f(x)=\displaystyle \dfrac{x}{\sqrt{1-x^{2}}},g(x)=\displaystyle \dfrac{x}{\sqrt{1+x^{2}}} $$, then $$(fog)(x)=$$       

If $$f(x)=1+x+x^{2}+x^{3}+\ldots\ldots $$ for $$\left | x \right |<1$$  then $$f^{-1}(x)=$$

If the function is $$f:R\rightarrow R,  g:R\rightarrow R$$ are defined as $$f(x)=2x+3, g(x)=x^{2}+7$$  and  $$f[g(x)]=25$$  then  $$x=$$    

If $$f(x)=\displaystyle \frac{2^{x}+2^{-x}}{2^{x}-2^{-x}}$$,  then  $$f^{-1}(x)=$$

I: If $$f:A\rightarrow B$$ is a bijection only then does $$f$$ have an inverse function
II: The inverse function $$f:R^{+}\rightarrow R^{+}$$ defined by $$f(x)=x^{2}$$ is $$f^{-1}(x)=\sqrt{x}$$

If $$F(n)=(-1)^{k-1}(n-1), G(n)=n-F(n)$$ then $$ (GoG)(n)=$$ (where $$k$$ is odd)

If $$f(x)=\dfrac{x}{\sqrt{1-x^{2}}}$$, then $$ (fof)(x)=$$

If $$f(x)=\sin^{-1}\left \{ 3-(x-6)^{4} \right \}^{1/3}$$  then $$ f^{-1}(x)=$$

Which of the following functions defined from $$(-\infty ,\infty )$$ to $$ (-\infty ,\infty )$$ is invertible ?

If $$f(x)=\dfrac{x}{\sqrt{1+x^{2}}}$$ then $$fofof(x)=$$