Functions Question - None

Question

Range of function

f(x) = \log_2 \left( \frac{4}{\sqrt{x+2}+\sqrt{2-x}} \right)

is given by

Accepted Answer

Step 1: Find the Domain

For f(x) to be defined, we need:

• x + 2 ≥ 0 ⟹ x ≥ -2

• 2 - x ≥ 0 ⟹ x ≤ 2

Therefore, Domain = [-2, 2]

Step 2: Find the Range of the Denominator

Let g(x) = √(x+2) + √(2-x) where x ∈ [-2, 2]

Using Cauchy-Schwarz inequality: [√(x+2) + √(2-x)]² ≤ (1² + 1²)[(x+2) + (2-x)] = 2 × 4 = 8

So √(x+2) + √(2-x) ≤ 2√2

To find the minimum, let u = √(x+2) and v = √(2-x).

Then u² + v² = (x+2) + (2-x) = 4 (constant)

g(x) = u + v, where u² + v² = 4

At x = -2: g(-2) = 0 + 2 = 2

At x = 2: g(2) = 2 + 0 = 2

At x = 0: g(0) = √2 + √2 = 2√2

By calculus: dg/dx = 1/(2√(x+2)) - 1/(2√(2-x))

Setting dg/dx = 0: √(2-x) = √(x+2) ⟹ x = 0

Therefore, max[g(x)] = 2√2 and min[g(x)] = 2

Step 3: Find the Range of f(x)

Since 2 ≤ g(x) ≤ 2√2, we have:

1/(2√2) ≤ 1/g(x) ≤ 1/2

Therefore: 4/(2√2) ≤ 4/g(x) ≤ 4/2

Simplifying: 2/√2 ≤ 4/g(x) ≤ 2

Which gives: √2 ≤ 4/g(x) ≤ 2

Step 4: Apply Logarithm

f(x) = log₂(4/g(x))

When g(x) = 2√2: f(x) = log₂(4/(2√2)) = log₂(√2) = log₂(2^(1/2)) = 1/2

When g(x) = 2: f(x) = log₂(4/2) = log₂(2) = 1

Since g(x) is continuous on [-2, 2] and varies from 2 to 2√2 to 2, f(x) varies continuously from 1 to 1/2 to 1.

Therefore, Range of f(x) = [1/2, 1]

∴ Answer: B

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