JEE 2023 — Mathematics PYQ

To solve ${\int }_0^2\max \{x^2,1+[x]\}dx$, we must break the limit $[0,2]$ into intervals where $[x]$ is constant.

1. Break down by intervals

• Case 1: 0 \leq x < 1

  • $[x]=0$

  • $f(x)=\max \{x^2,1+0\}=\max \{x^2,1\}$

  • Since x^2 < 1 in this range, $f(x)=1$.

• Case 2: 1 \leq x < 2

  • $[x]=1$

  • $f(x)=\max \{x^2,1+1\}=\max \{x^2,2\}$

  • We need to find where $x^2=2⟹x=\sqrt{2}$.

  • For 1 \leq x < \sqrt{2}, x^2 < 2, so $f(x)=2$.

  • For \sqrt{2} \leq x < 2, $x^2\ge 2$, so $f(x)=x^2$.

2. Set up the Definite Integral

The integral splits into three parts based on the points $x=1$ and $x=\sqrt{2}$:

3. Calculation

• First part: ${\int }_0^{1}1dx=[x{]}_0^1=1$

• Second part: ${\int }_1^{\sqrt{2}}2dx=[2x{]}_1^{\sqrt{2}}=2\sqrt{2}-2$

• Third part: ${\int }_{\sqrt{2}}^2{x}^{2}dx={\left[\frac{x^3}{3}\right]}_{\sqrt{2}}^2=\frac{8}{3}-\frac{(\sqrt{2}{)}^3}{3}=\frac{8-2\sqrt{2}}{3}$

4. Final Addition

Summing all the parts:

Correct Option: (1)