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.

  • $\begingroup$ 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

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.


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