Question

The range of the function \( f(x) = \frac{x^2 + x + 1}{x^2 - x + 1} \) is:

• A. \( \left[\frac{1}{3}, 3\right] \)
• B. \( \left[\frac{1}{2}, 2\right] \)
• C. \( \left[-\frac{1}{2}, -\frac{1}{4}\right] \)
• D. \( \left[-\frac{1}{2}, 2\right] \)

Hint:

Finding Range of Rational Functions}
Express the function as \( y = f(x) \) and cross-multiply
Use discriminant \( D \geq 0 \) to ensure real solutions exist
Solve resulting inequality for range

Answer 1

The Correct Option is A

Let \( y = \frac{x^2 + x + 1}{x^2 - x + 1} \). Cross-multiplying: \[ y(x^2 - x + 1) = x^2 + x + 1 \Rightarrow yx^2 - yx + y = x^2 + x + 1. \] Rewriting: \[ (y - 1)x^2 - (y + 1)x + (y - 1) = 0. \] This is a quadratic in \( x \), for \( x \in \mathbb{R} \), discriminant \( D \geq 0 \): \[ D = (y + 1)^2 - 4(y - 1)^2 \geq 0. \] Simplify: \[ D = y^2 + 2y + 1 - 4(y^2 - 2y + 1) = -3y^2 + 10y - 3 \geq 0. \] Solve the inequality: \[ -3y^2 + 10y - 3 \geq 0 \Rightarrow y \in \left[\frac{1}{3}, 3\right]. \]