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

Answer: (D)

201

Answer 2

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 $m + n = 146 + 55 = 201$.