JEE Main 2023 (Online) 13th April Morning ShiftQuestion 1
The set of all $a\in\mathbb{R}$ for which the equation $x|x-1|+|x+2|+a=0$ has exactly one real root, is
Quadratic Equation · Equations with Modulus and Greatest Integer
Solution
Concept. $f(x)=x|x-1|+|x+2|$ is continuous and strictly increasing on all of $\mathbb R$, so $f(x)=-a$ always has exactly one root. Formulas used. (1) A continuous strictly increasing function from $-\infty$ to $\infty$ takes every value exactly onceStep 1 - $x\ge1$. $f=x(x-1)+x+2=x^2+2$, increasing. Step 2 - $-2\le x<1$. $f=x(1-x)+x+2=-x^2+2x+2$, $f'=2-2x>0$. Step 3 - $x<-2$. $f=x(1-x)-x-2=-x^2-2$, $f'=-2x>0$. Step 4 - Conclude. $f$ is continuous, increases on each piece, and $f\to\mp\infty$ as $x\to\mp\infty$. By (1), $f(x)=-a$ has exactly one root for every real $a$. Answer: (C)