<s>[INST] 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 $\frac{1}{x^2-2x+2}$. 
 [/INST] Let $g(x)=f(f(x))$, then $g'(x)=f'(x)f'(f(x))\geq0$ since $f(x)$ is monotonic function. 
Note that $x=1$ is the solution of $g(x)=f(f(x))=\frac{1}{x^2-2x+2}=\frac{1}{(x-1)^2+1}$
Now, since $\frac{1}{(x-1)^2+1}$ is strictly decresing function on interval $[1,\infty)$, and $g(x)$ is nondecreasing on that interval, the solution $x=1$ is unique. Hence, $N=1$.
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