Question

If \( x \) is real and \( \alpha, \beta \) are maximum and minimum values of \( \frac{x^2 - x + 1}{x^2 + x + 1} \) respectively, then \( \alpha + \beta = \):

• A. \( \frac{10}{3} \)
• B. \( \frac{8}{3} \)
• C. \( \frac{4}{3} \)
• D. \( \frac{-2}{3} \)

Hint:

For rational functions, sometimes analyzing the behavior as \( x \to \infty \) or solving the derivative can provide insights into the maximum and minimum values.

Answer 1

The Correct Option is A

Step 1: Expression for the function. We are given the function: \[ f(x) = \frac{x^2 - x + 1}{x^2 + x + 1}. \]

Step 2: Differentiating the function. We differentiate the function with respect to \(x\) using the quotient rule: \[ f'(x) = \frac{(2x - 1)(x^2 + x + 1) - (x^2 - x + 1)(2x + 1)}{(x^2 + x + 1)^2}. \]

Step 3: Solving for the critical points. We solve for the critical points by setting the numerator of \(f'(x)\) equal to zero.

Step 4: Evaluating the maximum and minimum values. After evaluating the function at the critical points, we find the maximum and minimum values of \(f(x)\) to be \( \frac{10}{3} \) .

Step 5: Sum of the maximum and minimum values. The sum is: \[ \frac{10}{3} \]

Answer 2

Given: For real \( x \), let \[ f(x) = \frac{x^2 - x + 1}{x^2 + x + 1}. \] Let \( \alpha \) and \( \beta \) be the maximum and minimum values of \( f(x) \) respectively. Find the value of \[ \alpha + \beta. \]

Step 1: Simplify the function using substitution Let \[ y = x + \frac{1}{2}, \] to rewrite the function in a simpler form. Express numerator and denominator in terms of \( y \): \[ x = y - \frac{1}{2}, \] Calculate numerator: \[ x^2 - x + 1 = \left(y - \frac{1}{2}\right)^2 - \left(y - \frac{1}{2}\right) + 1 = y^2 - y + \frac{1}{4} + \frac{1}{2} + 1 = y^2 - y + \frac{7}{4}. \] Calculate denominator: \[ x^2 + x + 1 = \left(y - \frac{1}{2}\right)^2 + \left(y - \frac{1}{2}\right) + 1 = y^2 - y + \frac{1}{4} + y - \frac{1}{2} + 1 = y^2 + \frac{3}{4}. \] Therefore, \[ f(x) = \frac{y^2 - y + \frac{7}{4}}{y^2 + \frac{3}{4}}. \]

Step 2: Express \( f(x) \) as: \[ f(y) = \frac{y^2 - y + \frac{7}{4}}{y^2 + \frac{3}{4}}. \] Rewrite numerator: \[ y^2 - y + \frac{7}{4} = (y^2 + \frac{3}{4}) - y + 1. \] Hence, \[ f(y) = \frac{(y^2 + \frac{3}{4}) - y + 1}{y^2 + \frac{3}{4}} = 1 + \frac{-y + 1}{y^2 + \frac{3}{4}}. \]

Step 3: Find extrema of \( f(y) \) Set derivative \( f'(y) = 0 \). Let \[ g(y) = \frac{-y + 1}{y^2 + \frac{3}{4}}. \] Compute derivative using quotient rule: \[ g'(y) = \frac{(-1)(y^2 + \frac{3}{4}) - (-y + 1)(2y)}{(y^2 + \frac{3}{4})^2} = 0. \] Numerator: \[ -(y^2 + \frac{3}{4}) + 2y(y - 1) = -y^2 - \frac{3}{4} + 2y^2 - 2y = y^2 - 2y - \frac{3}{4} = 0. \] Multiply both sides by 4: \[ 4y^2 - 8y - 3 = 0. \] Solve quadratic: \[ y = \frac{8 \pm \sqrt{64 + 48}}{8} = \frac{8 \pm \sqrt{112}}{8} = \frac{8 \pm 4\sqrt{7}}{8} = 1 \pm \frac{\sqrt{7}}{2}. \]

Step 4: Calculate \( f(y) \) at these points For \( y = 1 + \frac{\sqrt{7}}{2} \): Calculate numerator: \[ -y + 1 = -\left(1 + \frac{\sqrt{7}}{2}\right) + 1 = -\frac{\sqrt{7}}{2}. \] Calculate denominator: \[ y^2 + \frac{3}{4} = \left(1 + \frac{\sqrt{7}}{2}\right)^2 + \frac{3}{4} = 1 + \sqrt{7} + \frac{7}{4} + \frac{3}{4} = 1 + \sqrt{7} + \frac{10}{4} = 1 + \sqrt{7} + 2.5 = 3.5 + \sqrt{7}. \] Then, \[ f(y) = 1 + \frac{-\frac{\sqrt{7}}{2}}{3.5 + \sqrt{7}} = 1 - \frac{\sqrt{7}/2}{3.5 + \sqrt{7}}. \] Similarly for \( y = 1 - \frac{\sqrt{7}}{2} \), \[ f(y) = 1 + \frac{\frac{\sqrt{7}}{2}}{3.5 - \sqrt{7}}. \]

Step 5: Sum of maximum and minimum values \[ \alpha + \beta = \left(1 - \frac{\sqrt{7}/2}{3.5 + \sqrt{7}}\right) + \left(1 + \frac{\sqrt{7}/2}{3.5 - \sqrt{7}}\right) = 2 + \frac{\sqrt{7}/2}{3.5 - \sqrt{7}} - \frac{\sqrt{7}/2}{3.5 + \sqrt{7}}. \] Calculate difference: \[ \frac{\sqrt{7}}{2} \left(\frac{1}{3.5 - \sqrt{7}} - \frac{1}{3.5 + \sqrt{7}}\right) = \frac{\sqrt{7}}{2} \times \frac{(3.5 + \sqrt{7}) - (3.5 - \sqrt{7})}{(3.5)^2 - (\sqrt{7})^2} = \frac{\sqrt{7}}{2} \times \frac{2 \sqrt{7}}{12.25 - 7} = \frac{\sqrt{7}}{2} \times \frac{2 \sqrt{7}}{5.25}. \] Simplify: \[ \frac{\sqrt{7}}{2} \times \frac{2 \sqrt{7}}{5.25} = \frac{7}{5.25} = \frac{28}{21} = \frac{4}{3}. \] Thus, \[ \alpha + \beta = 2 + \frac{4}{3} = \frac{6}{3} + \frac{4}{3} = \frac{10}{3}. \]

Final answer: \[ \boxed{\frac{10}{3}}. \]