We are asked to determine whether the function $y = \sqrt{4-x^2} - 2x^2$ is an even function, an odd function, both, or neither.

1. Problem Description

We are asked to determine whether the function $y = \sqrt{4-x^2} - 2x^2$ is an even function, an odd function, both, or neither.

2. Solution Steps

First, we need to determine the domain of the function. Since we have a square root, the expression inside the square root must be non-negative:

$4 - x^2 \ge 0$

$x^2 \le 4$

$-2 \le x \le 2$

So, the domain of the function is $[-2, 2]$.

Next, we need to check if the function is even, i.e., if $f(-x) = f(x)$ for all $x$ in the domain.

$f(-x) = \sqrt{4 - (-x)^2} - 2(-x)^2 = \sqrt{4 - x^2} - 2x^2 = f(x)$

Since $f(-x) = f(x)$, the function is an even function.

We also need to check if the function is odd, i.e., if $f(-x) = -f(x)$ for all $x$ in the domain.

$-f(x) = -(\sqrt{4-x^2} - 2x^2) = -\sqrt{4-x^2} + 2x^2$

Since $f(-x) = \sqrt{4-x^2} - 2x^2$ and $-f(x) = -\sqrt{4-x^2} + 2x^2$, it is clear that $f(-x) \ne -f(x)$ in general.

For example, let $x = 1$. Then $f(1) = \sqrt{4-1} - 2(1)^2 = \sqrt{3} - 2$. Then $-f(1) = -\sqrt{3} + 2$.

Now, $f(-1) = \sqrt{4-(-1)^2} - 2(-1)^2 = \sqrt{4-1} - 2(1) = \sqrt{3} - 2$.

So, $f(-1) = f(1)$, but $f(-1) \ne -f(1)$. Therefore, the function is not odd.

Since the function is even but not odd, the answer is that it is an even function.

3. Final Answer

B. 偶函数

The function is an even function.

Final Answer: The final answer is $\boxed{B}$