Question

Let $f(x) = 3\sqrt{x - 2} + \sqrt{4 - x}$ be a real-valued function. If $\alpha$ and $\beta$ are respectively the minimum and the maximum values of $f$, then $\alpha^2 + 2\beta^2$ is equal to:

• A. 44
• B. 42
• C. 24
• D. 38

Answer 1

Answer: (B)

42

Answer 2

To determine the range of $f(x) = 3\sqrt{x-2} + \sqrt{4-x}$, we analyze the domain of $f(x)$ by finding values of $x$ for which both square roots are real.

1. For $\sqrt{x-2}$ to be real, we need $x \geq 2$.
2. For $\sqrt{4-x}$ to be real, we need $x \leq 4$.

Therefore, $x$ is restricted to the interval $[2, 4]$.

Next, we evaluate $f(x)$ at the endpoints to determine the minimum and maximum values.

• At $x = 2$:
  $f(2) = 3\sqrt{2-2} + \sqrt{4-2} = 0 + \sqrt{2} = \sqrt{2}$.
• At $x = 4$:
  $f(4) = 3\sqrt{4-2} + \sqrt{4-4} = 3 \cdot \sqrt{2} + 0 = 3\sqrt{2}$.

Thus, the minimum value $\alpha$ is $\sqrt{2}$, and the maximum value $\beta$ is $3\sqrt{2}$.

Now, we calculate $\alpha^2 + 2\beta^2$:

$\alpha^2 + 2\beta^2 = (\sqrt{2})^2 + 2(3\sqrt{2})^2 = 2 + 2 \times 18 = 2 + 36 = 42.$

Therefore, $\alpha^2 + 2\beta^2$ is equal to 42.