9. Given: $c>0$, let $p$ : the function $y=c^{x}$ is monotonically decreasing on $\mathbf{R}$, $q$ : the solution set of the inequality $x+|x-2 c|>1$ is
R. If $p$ and $q$ are such that exactly one of them is true, find the range of values for $c$.
9. The function $y=c^{x}$ is monotonically decreasing on $\mathbf{R}$ if and only if $0<c<1$. The solution set of $01$ is $\mathbf{R} \Leftrightarrow y=x+|x-2 c|$ is always greater than 1 on $\mathbf{R}$.
Because $x+|x-2 c|=\left\{\begin{array}{ll}2 x-2 c, & x \geqslant 2 c \\ 2 c, & x<2 c\end{array}\right.$, the minimum value of $x+|x-2 c|$ is $2c$. For $x+|x-2 c|>1$ to hold, we need $2c>1$ which means $c>\frac{1}{2}$. If $p$ is correct and $q$ is not, then $0<c \leqslant \frac{1}{2}$; if $p$ is not correct and $q$ is, then $c \geqslant 1$. Therefore, the range of $c$ is $\left(0, \frac{1}{2}\right] \cup[1,+\infty)$.