Mathematics · Functions

JEE Main 2024 — 4 April, Shift 1 — Question 5

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

Option A:
182
Option B:
217
Option C:
195
Option D:
201
Correct

Answer: D

Step-by-step solution

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

Rewrite:

f(x)=1-\frac{6x}{2x^2+3x+8}

Let

g(x)=\frac{6x}{2x^2+3x+8}

Differentiate:

g'(x)=\frac{6(2x^2+3x+8)-6x(4x+3)}{(2x^2+3x+8)^2}

Set numerator $=0$ :

12x^2+18x+48-24x^2-18x=0

-12x^2+48=0 \Rightarrow x^2=4 \Rightarrow x=\pm2

Evaluate:

f(2)=\frac{5}{11}, \qquad f(-2)=\frac{11}{5}

Sum of maximum and minimum:

\frac{11}{5}+\frac{5}{11} =\frac{146}{55}

m+n=146+55=\boxed{201}

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 4 April, Shift 1
Subject: Mathematics
Chapter: Functions
Topic: Domain & range of functions

If the domain of the function sin ^-1 (3 x-22/2 x-19 )+log e (frac3 x^2-8 x+5x^2-3 x-10 ) is (alpha, beta] , then 3 alpha+10 beta is equal…

Answer: D

Step-by-step solution

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