1
$\begingroup$

I wanted to know, how can I determine all the values of the parameter $a$ for which the inequality $3 - |x-a| > x^2$ is satisfied by at least one negative $x$.

I tried for $x<a, |x-a|=-(x-a)$

and for $x>a, |x-a|=(x-a)$

substituting and solving I get a region but the answer given is $\frac{-13}{4}$$<a$$<3$.

I don't want a graph of the two equations saying that's the region.

I am stuck. Help(Hint) appreciated.

$\endgroup$
  • $\begingroup$ Hint: First try $a = -3 ~\text{and}~ 3$, then try $-2 \le a \le 2$. Can you now figure out why? $\endgroup$ – Amzoti Aug 6 '13 at 0:56
4
$\begingroup$

Written in the form $|x-a|<3-x^2$, we see that $x^2<3$ is a requirement for any solution, so that if there are negative $x$ as solutions they must satisfy $-\sqrt{3}<x<0.$ We may remove the absolute values, using that $|x-a|=|a-x|$, to obtain $$x^2-3<a-x<3-x^2,$$ $$ x^2+x-3<a<-x^2+x+3. \tag{1}$$ The minimum of the left side here is $-13/4=-3.25$ taken on at $x=-1/2=-0.5,$ and since $-\sqrt{3} \approx -1.732$ we see by computing the two sides of (1) at $x=-1/2$ that $a$ may take on any value in the interval $(-13/4,9/4)$ and there will be negative $x$ solutions. The right side is $3$ at $x=0$, and for $x<0$ the right side is less than $3$, so that $a<3$ is necessary for there to be any negative solution $x$. Now let $x=-t$ for $t>0$ where we will let $t \to 0^+.$ Then computing the two sides of (1) shows that $a$ may be anywhere in the interval $(-t-(3-t^2),-t+(3-t^2).$ As $t\to 0^+$ this variable interval approaches the interval $(-3,3)$, whose union with the previously established interval $(-13/4,9/4)$ gives the interval $(-13/4,3)$ as the set of possible $a$ for which the inequality has a negative $x$ as solution.

$\endgroup$

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.