Let $f : R \to R$ is a function satisfying $f(10-x) = f(x)$ and $f(2-x) = f(2+x), \forall x \in R$. If $f(0) = 101$. Then, the minimum possible number of values of $x$ satisfying $f(x) = 101, x \in [0, 25]$ is_____.
Step-by-Step Solution
Key Concept: Combining the two functional equations reveals periodicity with period 6, and symmetry about $x = 2$ constrains the number of solutions per period.
From $f(10-x) = f(x) = f(4-x)$, we deduce $f(10-x) = f(4-x)$. Substituting $t = 4-x$ gives $f(6+t) = f(t)$, so $f$ has period 6. On $[0,25]$, the equation $f(x) = 101$ is satisfied at $x = 0, 6, 12, 18, 24$ (five solutions). By symmetry about $x = 2$, there is one solution at $x = 4$ in each period, giving solutions at $x = 4, 10, 16, 22$. The minimum total is $5 + 4 = 9$ solutions.
Correct Answer: 9