Consider the equation $||x-1|-|x+2||=p$

Find the value of $p$ for which the above equation has one solution.

  • $\begingroup$ My intuition tells me p=0 $\endgroup$ – YuiTo Cheng Jan 30 at 15:07


Notice the range of $||x-1|-|x+2||=[0,3]$, so $p\in [0,3]$

for $x\leqslant-2$ or $x \geqslant1$ , $p=3$

if $p\neq0$, there are $2$ distinct solutions for $x$ (why?)

so $p=0$


The graph of $||x-1|-|x+2||$ enter image description here


You need $|x-1|=|x+2|$ and you need that this has only one solution, which is the case for $x=-0.5$. So $p=0$is indeed the answer.


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