QJEEM22Ma88b0: JEE Main 2022 Mathematics - Class 12 JEEM22Ma88b0 Solution: The sum of the maximum and minimum values of the function in…

$f(x) = |5x - 7| + [{x^2} + 2x]$

$= |5x - 7| + [{(x + 1)^2}] - 1$

Critical points of

$f(x) = {7 \over 5},\sqrt 5 - 1,\,\sqrt 6 - 1,\,\sqrt 7 - 1,\,\sqrt 8 - 1,\,2$

$\therefore$ Maximum or minimum value of $f(x)$ occur at critical points or boundary points

$\therefore$ $f\left( {{5 \over 4}} \right) = {3 \over 4} + 4 = {{19} \over 4}$

$f\left( {{7 \over 5}} \right) = 0 + 4 = 4$

as both $|5x - 7|$ and ${x^2} + 2x$ are increasing in nature after $x = {7 \over 5}$

$\therefore$ $f(2) = 3 + 8 = 11$

$\therefore$ $f{\left( {{7 \over 5}} \right)_{\min }} = 4$ and $f{(2)_{\max }} = 11$

Sum is $4 + 11 = 15$