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 g c d ( m , n ) = 1 . Then m + n is equal to [2024]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
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 $g c d (m, n) = 1 .$ Then $m + n$ is equal to [2024]

1 182

2 217

3 201

4 195

Ans.
(3)

Let $f (x) = \frac{2 x^{2} - 3 x + 8}{2 x^{2} + 3 x + 8} = y,$ $2 x^{2} + 3 x + 8 > 0 \forall x \in R$

$\Rightarrow x^{2} (2 y - 2) + x (3 y + 3) + 8 y - 8 = 0$

For real roots, $D \geq 0$

$\Rightarrow {(3 y + 3)}^{2} - 4 (2 y - 2) (8 y - 8) \geq 0$

$\Rightarrow {(3 y + 3)}^{2} - {(8 y - 8)}^{2} \geq 0$

$\Rightarrow (11 y - 5) (- 5 y + 11) \geq 0$

$\Rightarrow (y - \frac{5}{11}) (y - \frac{11}{5}) \leq 0 \Rightarrow y \in [\frac{5}{11}, \frac{11}{5}]$

Sum of maximum and minimum values

$y_{max} + y_{min} = \frac{5}{11} + \frac{11}{5} = \frac{146}{55} = \frac{m}{n} \Rightarrow m + n = 201$

Want Us to Email you About Special Offers & Updates?