Put x=2 in the power of e, we get e^(1/(2-2)) which is e^infinity which is infinity.. Now 1/(1+infinity) = 1/ infinity which is equal to zero. Hence answer is zero.

Let $f(x)$ be a continuous function from $[0,1]$ to $[0,1]$ satisfying the following properties. $f(0)=0$, $f(1)=1$, and $f(x_1)<f(x_2)$ for $x_1 < x_2$ with $0 < x_1, \: x_2<1$. Then the number of such functions is $0$ $1$ $2$ $\infty$