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

Concept:
To find the range of a rational function, set \[ y=\frac{x^2-x+1}{x^2+x+1} \] and then solve for \(x\) in terms of \(y\). For real \(x\), the resulting quadratic in \(x\) must have real roots, so its discriminant must be non-negative. ip

Step 1: Set the expression equal to \(y\).
Let \[ y=\frac{x^2-x+1}{x^2+x+1} \] Cross multiply: \[ y(x^2+x+1)=x^2-x+1 \] ip

Step 2: Form a quadratic in \(x\).
\[ yx^2+yx+y=x^2-x+1 \] \[ (y-1)x^2+(y+1)x+(y-1)=0 \] ip

Step 3: Use the discriminant condition.
For real \(x\), \[ (y+1)^2 - 4(y-1)^2 \ge 0 \] \[ y^2+2y+1 - 4(y^2-2y+1)\ge 0 \] \[ y^2+2y+1 -4y^2+8y-4\ge 0 \] \[ -3y^2+10y-3\ge 0 \] \[ 3y^2-10y+3\le 0 \] ip

Step 4: Find the range of \(y\).
Solve: \[ 3y^2-10y+3=0 \] \[ y=\frac{10\pm\sqrt{100-36}}{6} =\frac{10\pm 8}{6} \] So, \[ y=3 \quad \text{or} \quad y=\frac{1}{3} \] Thus the range is: \[ \frac{1}{3}\le y\le 3 \] ip

Step 5: Find the required difference.
Greatest value: \[ 3 \] Least value: \[ \frac{1}{3} \] Difference: \[ 3-\frac{1}{3}=\frac{8}{3} \] ip Hence, the correct answer is:
\[ \boxed{(B)\ \frac{8}{3}} \]