Question

Let the sum of the maximum and the minimum values of the function \( f(x) = \frac{2x^2 - 3x + 8}{2x^2 + 3x + 8} \) be \( \frac{m}{n} \), where \( \text{gcd}(m, n) = 1 \). Then \( m + n \) is equal to:

• A. 182
• B. 217
• C. 195
• D. 201

Answer 1

The Correct Option is D

Analyze the function \( f(x) = \frac{2x^2 - 3x + 8}{2x^2 + 3x + 8} \).

To find the maximum and minimum values, let \( y = \frac{2x^2 - 3x + 8}{2x^2 + 3x + 8} \).

Multiply both sides by the denominator to rewrite this as:
\[ y(2x^2 + 3x + 8) = 2x^2 - 3x + 8 \]

Expanding and rearranging terms, we get:
\[ 2x^2y + 3xy + 8y = 2x^2 - 3x + 8 \]
\[ 2x^2(y - 1) + x(3y + 3) + 8(y - 1) = 0 \]

This is a quadratic equation in \(x\). For real values of \(x\), the discriminant \(D\) must satisfy \(D \geq 0\).
Using the Discriminant Condition \(D \geq 0\):

For the quadratic equation \(2x^2(y - 1) + x(3y + 3) + 8(y - 1) = 0\), calculate the discriminant \(D\):
\[ D = (3y + 3)^2 - 4 \times 2(y - 1) \times 8(y - 1) \] 

Simplifying this inequality yields: 

\[ y \in \left[\frac{5}{11}, 1\right] \]

Therefore, the minimum value of \(y\) is \(\frac{5}{11}\), and the maximum value of \(y\) is \(1\).
Sum of Maximum and Minimum Values: \[ \frac{5}{11} + 1 = \frac{5}{11} + \frac{11}{11} = \frac{146}{55} \]
Thus, \(m = 146\) and \(n = 55\), so \(m + n = 146 + 55 = 201\).