QJEEM23Maa98c: JEE Main 2023 Mathematics - Class 12 JEEM23Maa98c Solution: Let denote the greatest integer . Consider the function . Then…

Combining both the graph we get,

Both graph cuts each other at $y=2$ and $y=1$

$\therefore x^2=2$

$\Rightarrow x=\sqrt2$

Two graphs cuts each other at $x=1$ and $x=\sqrt2$

In 0 to 1, $f(x) = \max ({x^2},1 + [x]) = 1$

In 1 to $\sqrt2$ , $f(x) = \max ({x^2},1 + [x]) = 2$

In $\sqrt2$ to 2, $f(x) = \max ({x^2},1 + [x]) = {x^2}$

$\therefore$ $\int\limits_0^2 {f(x)dx}$

$= \int\limits_0^1 {1\,dx + \int\limits_1^{\sqrt 2 } {2\,dx + \int\limits_{\sqrt 2 }^2 {{x^2}\,dx} } }$

$= \left[ x \right]_0^1 + 2\left[ x \right]_1^{\sqrt 2 } + \left[ {{{{x^3}} \over 3}} \right]_{\sqrt 2 }^2$

$= 1 + 2(\sqrt 2 - 1) + {8 \over 3} - {{2\sqrt 2 } \over 3}$

$= {8 \over 3} - 1 + 2\sqrt 2 - {{2\sqrt 2 } \over 3}$

$= {5 \over 3} + {{6\sqrt 2 - 2\sqrt 2 } \over 3}$

$= {5 \over 3} + {{4\sqrt 2 } \over 3}$

$= {{5 + 4\sqrt 2 } \over 3}$