JEE Main 2025 — 3 April, Morning Shift — Question 44

Define $f(x) = \max\{|x|, x|x-2|\}$. Determine intervals where each component dominates. For $x < 0$: $|x| = -x$ and $x|x-2| = x(2-x) = 2x - x^2$.

Since $|x| > x|x-2|$, $f(x) = -x$. For $0 \le x < 2$: $|x| = x$, $x|x-2| = x(2-x) = 2x - x^2$.

On $[0,1]$, $f(x) = 2x - x^2$; on $[1,2)$, $f(x) = x$. For $x \ge 2$: $|x| = x$, $x|x-2| = x(x-2) = x^2 - 2x$.

On $[2,3]$, $f(x) = x$; on $[3,4]$, $f(x) = x^2 - 2x$. Set up integrals for each subinterval: $A = \int_{-2}^0 (-x)\,dx + \int_0^1 (2x - x^2)\,dx + \int_1^2 x\,dx + \int_2^3 x\,dx + \int_3^4 (x^2 - 2x)\,dx$. Evaluate each: $\int_{-2}^0 (-x)\,dx = 2$, $\int_0^1 (2x - x^2)\,dx = \frac{2}{3}$, $\int_1^2 x\,dx = \frac{3}{2}$, $\int_2^3 x\,dx = \frac{5}{2}$, $\int_3^4 (x^2 - 2x)\,dx = \frac{16}{3}$. Sum: $2 + \frac{2}{3} + \frac{3}{2} + \frac{5}{2} + \frac{16}{3} = \frac{72}{6} = 12$. Therefore, the required area is $12$ square units.