Functions Question - None

Question

The range of the function

f(x) = \sqrt{4 - x^{2}} + \sqrt{x^{2} - 1}

is

Accepted Answer

Step 1: Find the Domain

For f(x) = √(4 - x²) + √(x² - 1) to be defined, we need:

• 4 - x² ≥ 0 ⟹ x² ≤ 4 ⟹ -2 ≤ x ≤ 2

• x² - 1 ≥ 0 ⟹ x² ≥ 1 ⟹ x ≤ -1 or x ≥ 1

Combined domain: x ∈ [-2, -1] ∪ [1, 2]

Step 2: Use Substitution

Let t = x², where t ∈ [1, 4] (from domain constraints).

Then f(x) = g(t) = √(4 - t) + √(t - 1)

Step 3: Find Critical Points

g'(t) = -1/(2√(4-t)) + 1/(2√(t-1))

Setting g'(t) = 0:

1/(2√(t-1)) = 1/(2√(4-t))

√(4-t) = √(t-1)

4 - t = t - 1

5 = 2t ⟹ t = 5/2

Step 4: Evaluate at Critical Point and Endpoints

• At t = 1: g(1) = √(4-1) + √(1-1) = √3 + 0 = √3

• At t = 5/2: g(5/2) = √(4 - 5/2) + √(5/2 - 1) = √(3/2) + √(3/2) = 2√(3/2) = √6

• At t = 4: g(4) = √(4-4) + √(4-1) = 0 + √3 = √3

Step 5: Verify Maximum and Minimum

g''(t) = -1/(4(4-t)^(3/2)) - 1/(4(t-1)^(3/2)) < 0 for all t ∈ (1,4)

So t = 5/2 gives a maximum. The minimum value √3 occurs at both endpoints.

∴ Range = [√3, √6]

∴ Answer: D

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