JEE Main 2022 · 29 Jul · Shift 2, Mathematics question 5

Inside. $\lfloor h+1\rfloor=\lfloor h\rfloor+1$ with $h=2x-|3x^2-5x+2|$. On $[0,\frac23]$, $h=-3x^2+7x-2$ rises from $-2$ to $\frac43$; on $[\frac23,1]$, $h=3x^2-3x+2$ rises from $\frac43$ to 2.

Break points. $h=-1$ at $x_1=\frac{7-\sqrt{37}}6$, $h=0$ at $\frac13$, $h=1$ at $x_2=\frac{7-\sqrt{13}}6$.

Integral of $\lfloor h\rfloor$: $-2x_1-\left(\frac13-x_1\right)+(1-x_2)=\frac{\sqrt{37}+\sqrt{13}-10}6$; adding 1 gives $\dfrac{\sqrt{37}+\sqrt{13}-4}6$.

(A)