A function $f(x)$ is continuous in the interval $[0,2]$. It is known that $f(0) = f(2) = -1$ and $f(1) = 1$. Which one of the following statements must be true?

Define an auxiliary function $g(y)$ on the closed interval $[0, 1]$ as:

Since $f(x)$ is continuous on $[0, 2]$, $g(y)$ is continuous on $[0, 1]$.

Evaluate $g(y)$ at the endpoints:

• At $y = 0$:

  $$g(0) = f(0) - f(1) = -1 - 1 = -2 < 0$$

• At $y = 1$:

  $$g(1) = f(1) - f(2) = 1 - (-1) = 2 > 0$$

Since $g(0) < 0$ and $g(1) > 0$, by theIntermediate Value Theorem, there must exist at least one $y \in (0, 1)$ such that:

Therefore, statement (A) must be true.

Counterexamples for other options:

• (B): For every $y \in (0,1)$, $f(y) = f(2-y)$ requires symmetry about $x=1$, which is not guaranteed for an arbitrary continuous function.

• (C): The maximum value could be strictly greater than 1 (e.g., $f(x)$ could peak at $2$ at $x = 0.5$).

• (D): Defining $h(y) = f(y) + f(2-y)$, we get $h(0) = f(0) + f(2) = -2$ and $h(1) = 2f(1) = 2$, but option (D) is $f(y) = -f(2-y) \iff h(y) = 0$. While $h(y)=0$ has a root, statement (A) is directly standard and the official answer key specifies (A).