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

Correct answer: B. 201 aligned & y=(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 & use D ≥ 0 & (3 y+3)^2-4(2 y-2)(8 y-8) ≥ 0 & (11 y-5)(5 y-11) ≤ 0 & y ∈[(5)/(11), (11)/(5)] & y=1 is also included aligned

JEE Main 20244 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 4 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Functions

JEE Main 2024 · 4 April · Morning Shift · Mathematics · Functions

Let the sum of the maximum and the minimum values of the function...

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}

\frac{\mathrm{m}}{\mathrm{n}}

, where

\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1

. Then

\mathrm{m}+\mathrm{n}

is equal to :

Options

A195
B201
C217
D182

Correct answer

B. 201

Step-by-step solution

\begin{aligned} & 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 \text{ is also included }\end{aligned}

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