Relations and F...

Let $$f(x) = \frac{ax}{x + 1}, x \neq  -1$$. Then for what value of $$\alpha$$ is $$f(f(x)) = x$$?                                                       (IIT-JEE, 2001)

If $$f(x + 1) + f(x - 1) = 2f(x) $$ and $$f(0) = 0$$, then $$f(n+1), n \space \epsilon \space N$$, is

If $$ f : R^{+} \rightarrow R, f(x) + 3xf \left (\frac {1} {x} \right ) = 2 \left ( x + 1 \right )$$, then $$f(99)$$ is equal to 

Let $$f:R\to R$$ be the function defined by $$f(x)=\left\{ \begin{matrix} 2x:x>3 \\ { x }^{ 2 }:1<x\le 3 \\ 3x:x\le 1 \end{matrix} \right.$$ Then $$f(-1)+f(2)+f(4)$$ is 

If A  = { a , b , c , d} and B = { p , q ,r ,s} then relation from A and B is

Rule from of relation {(1 ,2) , ( 2 , 5) , (3 , 10) , ( 4 , 17), ......} in N

A relation R in N is defined such that $$ xRy \Leftrightarrow  x +  4y  =  16 ,$$ then the range of R is

If A  = { a ,b , c} , then numbers of possible non zero relations in A is

A real valued function $$f(x)$$ satisfies the functional equation $$f(x-y)=f(x)f(y)-f(a-x)f(a+y)$$, where $$a$$ is a given constant and $$f(0)=1$$, then

If $$f(x) + 2f(1 - x) = x^2 + 2 $$; $$\forall x \in R$$, then $$f(x)$$ is given by