Number of answers : $f(x)=f^{-1}(x)$ 
let $f(x)= 1+\sqrt{x+k+1}-\sqrt{x+k} \ \ k \in \mathbb{R}$
Number of answers :
$$f(x)=f^{-1}(x) \ \   \ \ :f^{-1}(f(x))=x$$

MY Try :
$$y=1+\sqrt{x+k+1}-\sqrt{x+k} \\( y-1)^2=x+k+1-x-k-2\sqrt{(x+k+1)(x+k)}\\(y-1)^2+k-1=-2\sqrt{(x+k+1)(x+k)}\\ ((y-1)^2+k-1)^2=4(x^2+x(2k+1)+k^2+k)$$
now what do i do ?

Hint:
Point of intersection of $f(x)$ and $f^{-1}(x)$ while same as that  of $f(x)$ and the line $y=x$.
