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4-2.Quadratic Equations and Inequations
hard
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 $\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$. Then $\mathrm{m}+\mathrm{n}$ is equal to :
A
$182$
B
$217$
C
$195$
D
$201$
(JEE MAIN-2024)
Solution
$ y=\frac{2 x^2-3 x+8}{2 x^2+3 x+8} $
$ x^2(2 y-2)+x(3 y+3)+8 y-8=0 $
$ \text { use } D \geq 0 $
$ (3 y+3)^2-4(2 y-2)(8 y-8) \geq 0 $
$ (11 y-5)(5 y-11) \leq 0 $
$ \Rightarrow y \in\left[\frac{5}{11}, \frac{11}{5}\right] $
$ y=1$ is also included
Standard 11
Mathematics
