Let the sum of the maximum and the minimum values of the function f ( x ) = 2 x 2 - 3 x + 8 2 x 2 + 3 x + 8 ? be m n , where g c d ( m , n ) = 1 . Then m + n is equal to [2024]

Question

Let the sum of the maximum and the minimum values of the function $f (x) = \frac{2 x^{2} - 3 x + 8}{2 x^{2} + 3 x + 8}$ be $\frac{m}{n}$ , where $g c d (m, n) = 1 .$ Then $m + n$ is equal to [2024]

Smart Achievers · Accepted Answer

(3)

Let $f (x) = \frac{2 x^{2} - 3 x + 8}{2 x^{2} + 3 x + 8} = y,$ $g c d (m, n) = 1 .$

$\Rightarrow x^{2} (2 y - 2) + x (3 y + 3) + 8 y - 8 = 0$

For real roots, $D \geq 0$

$\Rightarrow {(3 y + 3)}^{2} - 4 (2 y - 2) (8 y - 8) \geq 0$

$\Rightarrow {(3 y + 3)}^{2} - {(8 y - 8)}^{2} \geq 0$

$\Rightarrow (11 y - 5) (- 5 y + 11) \geq 0$

$\Rightarrow (y - \frac{5}{11}) (y - \frac{11}{5}) \leq 0 \Rightarrow y \in [\frac{5}{11}, \frac{11}{5}]$

Sum of maximum and minimum values

$y_{max} + y_{min} = \frac{5}{11} + \frac{11}{5} = \frac{146}{55} = \frac{m}{n} \Rightarrow m + n = 201$

Solution

Q.
Let the sum of the maximum and the minimum values of the function $f (x) = \frac{2 x^{2} - 3 x + 8}{2 x^{2} + 3 x + 8}$ be $\frac{m}{n}$ , where $g c d (m, n) = 1 .$ Then $m + n$ is equal to [2024]

1 182

2 217

3 201

4 195

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