QJEEM21Maed26: JEE Main 2021 Mathematics - Class 12 JEEM21Maed26 Solution: If [ . ] represents the greatest integer function, then the…

$\int\limits_0^{\sqrt {{\pi \over 2}} } {\left[ {[{x^2}] - \cos x} \right]} dx$

$= \int\limits_0^1 {[ - \cos x]dx} + \int\limits_1^{\sqrt {{\pi \over 2}} } {[1 - \cos x]dx}$

= $\int\limits_0^1 {\left[ { - \cos x} \right]} dx + \int\limits_1^{\sqrt {{\pi \over 2}} } {1dx} + \int\limits_1^{\sqrt {{\pi \over 2}} } {\left[ { - \cos x} \right]} dx$

When 0 $\le$ x $\le$ 1 then -1 $\le$ -cosx $\le$ 0

$\therefore$ ${\left[ { - \cos x} \right]}$ = -1

When 1 $\le$ x $\le$ ${\sqrt {{\pi \over 2}} }$ = 1.24 then -0.33 $\le$ -cosx $\le$ 0

$\therefore$ ${\left[ { - \cos x} \right]}$ = -1 (Integer value present in the left side of -0.33)

$= - \int\limits_0^1 {dx + \int\limits_1^{\sqrt {{\pi \over 2}} } {dx - \int\limits_1^{\sqrt {{\pi \over 2}} } {dx} } }$

$= - (x)_0^1 = - 1$

$\therefore$ $\left| {\int\limits_0^{\sqrt {{\pi \over 2}} } {\left[ {[{x^2}] - \cos x} \right]dx} } \right|$ = 1