3
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ My intuition tells me p=0 $\endgroup$ – YuiTo Cheng Jan 30 at 15:07
2
$\begingroup$

Hint:

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 $p=0$

Edit:

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.