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

The Correct Option is B

To establish the range of \( f(x) = 3\sqrt{x-2} + \sqrt{4-x} \), we first define the domain of \( f(x) \) by identifying \( x \) values that yield real results for both square roots.

1. For \( \sqrt{x-2} \) to be real, \( x \geq 2 \) is required.
2. For \( \sqrt{4-x} \) to be real, \( x \leq 4 \) is required.

Consequently, the domain for \( x \) is the interval \([2, 4]\).

Next, we evaluate \( f(x) \) at the domain's endpoints to ascertain the minimum and maximum values.

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

Therefore, the minimum value, denoted as \( \alpha \), is \( \sqrt{2} \), and the maximum value, denoted as \( \beta \), is \( 3\sqrt{2} \).

Finally, we compute \( \alpha^2 + 2\beta^2 \):

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

The result of \( \alpha^2 + 2\beta^2 \) is 42.