Question

If \(x\) is real, then the difference between the greatest and least values of \(\frac{x^2 - x + 1}{x^2 + x + 1}\) is

• A. \(\frac{10}{3}\)
• B. \(\frac{8}{3}\)
• C. \(\frac{5}{3}\)
• D. \(\frac{1}{3}\)

Hint:

For rational expressions, setting the expression equal to \(y\) and then using the discriminant condition is one of the fastest ways to find the range.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Let \(y = \frac{x^2 - x + 1}{x^2 + x + 1}\). For real \(x\), the discriminant of the resulting quadratic must be \(\ge 0\).
Step 2: Key Formula or Approach:
Rearrange to \(x^2(y-1) + x(y+1) + (y-1) = 0\).
Set \(D = B^2 - 4AC \ge 0\).
Step 3: Detailed Explanation:
\(D = (y+1)^2 - 4(y-1)(y-1) = (y+1)^2 - 4(y-1)^2 \ge 0\)
\((y+1 - 2y + 2)(y+1 + 2y - 2) \ge 0 \implies (3 - y)(3y - 1) \ge 0\)
\((y - 3)(3y - 1) \le 0 \implies \frac{1}{3} \le y \le 3\).
Greatest value \(M = 3\), Least value \(m = 1/3\).
Difference \(= 3 - 1/3 = 8/3\).
Step 4: Final Answer:
The difference is \(\frac{8}{3}\).