I'm having a problem with this question:
For which values of the real parameter a the equation:


has exactly 4 solutions?
The solution is this: $$0 < a < 1$$

What I tried was this:

1) Make the first condition, $a>0$,
Split the equation in these two: $(1)~|x|-1=a;~(2)~|x|-1=-a$.
2) Make the second conditions, $(1)~a+1>0,~a>-1; (2)~-a+1>0,~a<1$
If the conditions are true, the equation will have 4 solutions, therefore, a is contained in the interval $<0,1>$, which is $0 < a < 1$.

Was the described procedure correct?

Thanks in advance.

  • 1
    $\begingroup$ You should draw the graph of the function $f(x)=\vert \vert x\vert -1\vert$. $\endgroup$ – Etienne Mar 10 '14 at 19:36

Notice that $$|x|=\alpha\iff x=\pm\alpha$$ so we have two values of $x$ if $\alpha>0$ hence

$$||x|-1|=a\iff |x|-1=\pm a\iff|x|=1+\pm a\iff x=\pm(1+\pm a)$$ and we have $4$ values of $x$ if $a>0$ and $1+\pm a>0$ hence $a>-1$ and $a<1$ so $a\in(0,1)$.

  • $\begingroup$ So I did it the right way... $\endgroup$ – user134480 Mar 10 '14 at 19:47
  • $\begingroup$ Yes you did it. $\endgroup$ – user63181 Mar 10 '14 at 19:49

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.