Functions Question - None

Question

If

g(x) = x^{2} - x + 1

and

f(x) = \sqrt{\frac{1}{x} - x}

, then -

Accepted Answer

Step 1: Find the domain of f(x) = √(1/x - x)

For f(x) to be defined: 1/x - x ≥ 0 and x ≠ 0

This gives: 1/x ≥ x ⟹ 1 ≥ x² ⟹ -1 ≤ x ≤ 1

Since x ≠ 0, domain of f is [-1, 0) ∪ (0, 1]

Step 2: Find when g(x) lies in domain of f

We need: g(x) ∈ [-1, 0) ∪ (0, 1]

Since g(x) = x² - x + 1 = (x - 1/2)² + 3/4

The minimum value of g(x) is 3/4 (at x = 1/2)

For g(x) ∈ [-1, 0) ∪ (0, 1]: We need g(x) ∈ (0, 1]

So: 0 < x² - x + 1 ≤ 1

The inequality x² - x + 1 > 0 holds for all real x (discriminant = 1 - 4 = -3 < 0)

For x² - x + 1 ≤ 1: x² - x ≤ 0 ⟹ x(x - 1) ≤ 0 ⟹ x ∈ [0, 1]

Step 3: Verify domain

Domain of f(g(x)) = [0, 1] ✓

Step 4: Find range of f(g(x))

For x ∈ [0, 1], we have g(x) ∈ [3/4, 1]

f(g(x)) = √(1/g(x) - g(x))

When g(x) = 1: f(1) = √(1 - 1) = 0

When g(x) = 3/4: f(3/4) = √(4/3 - 3/4) = √(16/12 - 9/12) = √(7/12) = √7/(2√3)

Let h(t) = 1/t - t for t ∈ [3/4, 1]

h'(t) = -1/t² - 1 < 0, so h is decreasing

Thus f(g(x)) ranges from 0 to √(7/12) = √7/(2√3)

Range is [0, √7/(2√3)) (note: √7/(2√3) ≈ 0.54, not 7/(2√3))

Step 5: Check if f(g(x)) is many-one

g(x) = x² - x + 1 is not one-one on [0, 1] since it's a parabola with vertex at x = 1/2

For example: g(0) = 1 and g(1) = 1, so different x-values map to same g(x)

Therefore f(g(x)) is many-one ✓

Step 6: Check if f(g(x)) is bounded

f(g(x)) has range [0, √7/(2√3)), which is bounded ✗

∴ Answer: A,C

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