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(D) Let $y = \frac{2x^2 - 3x + 8}{2x^2 + 3x + 8}$.Rearranging the terms,we get $y(2x^2 + 3x + 8) = 2x^2 - 3x + 8$.$x^2(2y - 2) + x(3y + 3) + 8y - 8 =

Vedclass · Accepted Answer

$201$

Vedclass · Answer

(D) Let $y = \frac{2x^2 - 3x + 8}{2x^2 + 3x + 8}$.Rearranging the terms,we get $y(2x^2 + 3x + 8) = 2x^2 - 3x + 8$.$x^2(2y - 2) + x(3y + 3) + 8y - 8 = 0$.For $x$ to be real,the discriminant $D \geq 0$.$D = (3y + 3)^2 - 4(2y - 2)(8y - 8) \geq 0$.$9(y + 1)^2 - 4 \cdot 2(y - 1) \cdot 8(y - 1) \geq 0$.$9(y^2 + 2y + 1) - 64(y^2 - 2y + 1) \geq 0$.$9y^2 + 18y + 9 - 64y^2 + 128y - 64 \geq 0$.$-55y^2 + 146y - 55 \geq 0$.$55y^2 - 146y + 55 \leq 0$.Solving the quadratic equation $55y^2 - 146y + 55 = 0$ using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$y = \frac{146 \pm \sqrt{21316 - 12100}}{110} = \frac{146 \pm \sqrt{9216}}{110} = \frac{146 \pm 96}{110}$.$y_1 = \frac{242}{110} = \frac{11}{5}$ and $y_2 = \frac{50}{110} = \frac{5}{11}$.Thus,the range is $[\frac{5}{11}, \frac{11}{5}]$.The maximum value is $\frac{11}{5}$ and the minimum value is $\frac{5}{11}$.Sum $= \frac{11}{5} + \frac{5}{11} = \frac{121 + 25}{55} = \frac{146}{55}$.Here $m = 146$ and $n = 55$. Since $\gcd(146, 55) = 1$,$m + n = 146 + 55 = 201$.

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 $\gcd(m, n) = 1$. Then $m + n$ is equal to :

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