Let $f:[1,\infty)\to R$ be a monotonic and differentiable function and $f(1)=1.$If $N$ is the number of solutions of $f(f(x))=\frac{1}{x^2-2x+2}$. Let $f:[1,\infty)\to R$ be a monotonic and differentiable function and $f(1)=1.$If $N$ is the number of solutions of $f(f(x))=\frac{1}{x^2-2x+2}$.Find $N.$

$\frac{1}{x^2-2x+2}=\frac{1}{(x-1)^2+1}$.Its graph look like the graph of $\frac{1}{x^2+1}$ but peak at $(1,1)$ buti do not know how does the graph of $f(f(x))$ behave and hence i am not able to find the number of points of intersection of $f(f(x))$ and $\frac{1}{x^2-2x+2}$. 

Hint: If $f(x)$ is monotonic, then $f(f(x))$ is (possibly non-strictly) increasing, so…

Full solution: Clearly if $f$ is increasing, so is $f\circ f$. Furthermore,  if $f$ is decreasing, then $x \leq y$ implies $f(x) \geq f(y)$, which in turn implies $f(f(x)) \leq f(f(y))$. Either way,  $f \circ f$ is increasing.
Since $f$ is $1$ at $x=1$, it follows that $f \circ f$ must have minimum $1$ at $x=1$. But it is easy to see that $(x^2-2x+2)^{-1}$ has maximum $1$ at $x=1$, thus $N=1$.
In particular, note that you don't need the hypothesis that $f$ is differentiable.
