Question

If x is real and y = $\frac{x^{2} - x + 3}{x + 2}$, then

• A. y ≥ 10
• B. y ≥ 11
• C. y ≤ –11 or y ≥ 1
• D. –11 < y < 1

Answer 1

Answer: (C)

y ≤ –11 or y ≥ 1

Answer 2

y = $\frac{x^{2} - x + 3}{x + 2}$

xy + 2y = x² – x + 3

x² – x(1 + y) + 3 – 2y = 0

Q x ∈ R

∴ D ≥ 0

(1 + y)² – 4 (3 – 2y) ≥ 0

y² + 2y + 1 – 12 + 8y ≥ 0

y² + 10y – 11 ≥ 0

(y + 11)(y – 1) ≥ 0

⇒ y ∈ (– ∞, –11] ∪ [1, ∞)