Question

If the range of the function \( f(x) = \sqrt{3 - x} + \sqrt{5 + x} \) is \([\alpha, \beta]\), then \( \alpha^2 + \beta^2 \) is equal to:

• A. \(18\)
• B. \(20\)
• C. \(24\)
• D. \(25\)

Hint:

For expressions like \( \sqrt{a-x} + \sqrt{b+x} \), the maximum often occurs when the two square roots are equal.

Answer 1

The Correct Option is D

Concept: For functions involving square roots:
The expression inside each square root must be non-negative.
To find the range, determine the minimum and maximum values of the function over its domain.
For sums of square roots, symmetry and critical point analysis are useful.
Step 1: Finding the domain For \( f(x) = \sqrt{3 - x} + \sqrt{5 + x} \), \[ 3 - x \ge 0 \quad \text{and} \quad 5 + x \ge 0 \] \[ \Rightarrow -5 \le x \le 3 \]
Step 2: Finding the maximum value Differentiate: \[ f'(x) = -\frac{1}{2\sqrt{3-x}} + \frac{1}{2\sqrt{5+x}} \] Set \( f'(x) = 0 \): \[ \frac{1}{\sqrt{5+x}} = \frac{1}{\sqrt{3-x}} \Rightarrow 5 + x = 3 - x \Rightarrow x = -1 \] \[ f(-1) = \sqrt{4} + \sqrt{4} = 4 \] Thus, \( \beta = 4 \).
Step 3: Finding the minimum value Check endpoints: \[ f(-5) = \sqrt{8} + 0 = 2\sqrt{2} \] \[ f(3) = 0 + \sqrt{8} = 2\sqrt{2} \] Thus, \( \alpha = 2\sqrt{2} \).
Step 4: Required value \[ \alpha^2 + \beta^2 = (2\sqrt{2})^2 + 4^2 = 8 + 16 = 25 \]