4-2.Quadratic Equations and Inequations
hard

જો વિધેય $f(x)=\frac{2 x^2-3 x+8}{2 x^2+3 x+8}$ ની મહતમ અને ન્યૂનતમ કિમંતો નો સરવાળો $\frac{m}{n}$ છે કે જ્યાં $\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$. તો  $\mathrm{m}+\mathrm{n}$ ની કિમંત મેળવો. 

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

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