QJEEM22Ma5bf2: JEE Main 2022 Mathematics - Class 12 JEEM22Ma5bf2 Solution: Let f(x) = max {

For $\left| {x + 1} \right|$ critical point, x + 1 = 0 $\Rightarrow$ x = $-$ 1

For $\left| {x + 2} \right|$ critical point, x + 2 = 0 $\Rightarrow$ x = $-$ 2

For $\left| {x + 3} \right|$ critical point, x + 3 = 0 $\Rightarrow$ x = $-$ 3

For $\left| {x + 4} \right|$ critical point, x + 4 = 0 $\Rightarrow$ x = $-$ 4

For $\left| {x + 5} \right|$ critical point, x + 5 = 0 $\Rightarrow$ x = $-$ 5

Here maximum function is represent by the dotted line.

$\therefore$ Point of intersection A of line y = $-$ x $-$ 1 and y = x + 5 :

$- x - 1 = x + 5$

$\Rightarrow 2x = - 6$

$\Rightarrow x = - 3$

$\therefore$ $y = - 3 + 5 = 2$

$\therefore$ Point $A = ( - 3,2)$

$\therefore$ $\int_{ - 6}^0 {f(x)dx}$

$= \int_{ - 6}^{ - 3} {( - x - 1)dx + \int_{ - 3}^0 {(x + 5)dx} }$

$= \left( { - {{{x^2}} \over 2} - x} \right)_{ - 6}^{ - 3} + \left[ {{{{x^2}} \over 2} + 5x} \right]_{ - 3}^0$

$= \left[ {\left( { - {9 \over 2} + 3} \right) - \left( { - {{36} \over 2} + 6} \right)} \right] + \left[ {0 - \left( {{9 \over 2} - 15} \right)} \right]$

$= \left( { - {3 \over 2} + 12} \right) + {{21} \over 2}$

$= + {{21} \over 2} + {{21} \over 2}$

$= 21$