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 :

JEE Main 2024 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 4 April 2024, Shift 1, Maths Q12

Question 12 of 90 in this shift, Maths question 12 of 30, Section A.

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

f(x)=\frac{2x^2-3x+8}{2x^2+3x+8}

\frac{m}{n}

, where

\gcd(m, n)=1

. Then

m+n

is equal to :

(1) $182$
(2) $195$
(3) $201$ Official answer
(4) $217$

Official answer

Option 3

NTA final key.

Chapter: Application of Derivatives
Topic: Maxima and Minima
Idea tested: Extrema of Quadratic over Quadratic

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Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.