The set of all a R for which the equation x|x-1|+|x+2|+a=0 has exactly one real root, is :

Question

The set of all $a\in R$ for which the equation $x|x-1|+|x+2|+a=0$ has exactly one real root, is :

Accepted Answer

$x|x-1|+|x+2|+a=0$

Case I : If $x<-2$ then

$-x^2+x-x-2+a=0$

$a=x^2+2$

$y=x^2+2$ is decreasing $\forall x \in(-\infty,-2)$

Case II : If $-2 \leq x<1$ then

$$
\begin{aligned}
& -x^2+x+x+2+a=0 \\
& a=x^2-2 x-2
\end{aligned}
$$

$y=x^2-2 x-2$ is decreasing $\forall x \in[-2,1)$.

Case III: If $x \geq 1$ then

$$
\begin{aligned}
& x^2-x+x+2+a=0 \\
& a=-\left(x^2+2\right)
\end{aligned}
$$

$y=-\left(x^2+2\right)$ is decreasing $\forall x \in[1, \infty)$

$\therefore$ Exactly one real root $\forall x \in R$

Solutions · 1