JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Basic Mathematics

Let $f(x)=x|x-1|+|x+2|$; the equation is $f(x)=-a$. For $x\ge 1$, $f(x)=x^2+2$; for $-2\le x<1$, $f(x)=-x^2+2x+2$; for $x<-2$, $f(x)=-x^2-2$. Each piece is increasing, and the pieces join continuously at $x=-2$ with value $-6$ and at $x=1$ with value $3$, so $f$ is strictly increasing on all of $\mathbb{R}$ with range $(-\infty ,\infty )$. A strictly monotonic function attains every value exactly once, so $f(x)=-a$ has exactly one root for each real $a$. The required set of values of $a$ is $(-\infty ,\infty )$.