10.1. For the function $f(x)=2013-8 x^{3}+12 x^{2}-14 x-a-\sin 2 \pi x$ find the number of integer values of $a$, for each of which the equation

$$
\underbrace{f(f(\ldots f}_{2013 \text { times }}(x) \ldots))=2 x-1
$$

on the interval $[50 ; 51]$ has a unique solution.
Solution. Since

$$
2013-8 x^{3}+12 x^{2}-14 x-a-\sin 2 \pi x=2008-(2 x-1)^{3}-4(2 x-1)-a+\sin \pi(2 x-1)
$$

then after the substitution of the variable $t=2 x-1$, we get a new problem: "For the function $F(t)=2008-t^{3}-4 t- a+\sin \pi t$, find the number of integer values of $a$, for each of which the equation

$$
\underbrace{F(F(\ldots F}_{2013 \text { times }}(t) \ldots))=t
$$

on the interval $[99 ; 101]$ has a unique solution."

The function $F(t)=2008-t^{3}-4 t-a+\sin \pi t$ is monotonically decreasing over the entire number line (which can be verified either by calculating $F^{\prime}(t)$ or by directly using the definition of a monotonically decreasing function). Therefore,

$$
\underbrace{F(F(\ldots F}_{2013 \text { times }}(t) \ldots))=t \Longleftrightarrow F(t)=t .
$$

The function $g(t)=t-F(t)$ is monotonically increasing. Consequently, the equation $g(t)=0$ has a unique solution on $[99 ; 101]$ if and only if

$$
\left\{\begin{array} { l } 
{ g ( 1 0 1 ) \geqslant 0 , } \\
{ g ( 9 9 ) \leqslant 0 , }
\end{array} \Longleftrightarrow \left\{\begin{array}{l}
101-2008+101^{3}+404+a \geqslant 0 \\
99-2008+99^{3}+396+a \leqslant 0
\end{array}\right.\right.
$$

From the obtained two-sided estimates for $a$, it follows that the number of integer values of $a$ is

$$
g(101)-g(99)+1=2+101^{3}-99^{3}+404-396+1=60013
$$

## Answer 60013 .