Question : If $16-x^2>|x-a|$ is to be satisfied by at least one negative value of x, then the complete set of values of 'a' is ?

Attempt :

We can write the inequality as : $$x^2-16 < x-a < 16-x^2$$ $$\Rightarrow x^2-x+(a-16) < 0 \ldots (i)$$and$$ x^2+x-(a+16) < 0 \ldots (ii)$$

For the quadratic inequality (ii), the vertex of the parabola occurs at $x=-1/2$

So it suffices that discriminant $D>0$ for the inequality to be true for atleast one negative value of $x$

However for inequality (i) I'm not able to figure out the conditions required for it to be true. Any help is appreciated. Thanks



The Graph of $y=|x-a|$ is V-shaped. If $a$ is positive, $a$ must take the value such that the $y$-intercept of $y=|x-a|$ is less than the $y$-intercept of $y=16-x^2$, i.e. $a<16$.

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If $a$ is negative, $a$ must take the value such that $y=x-a$ cut $y=16-x^2$ at two distinct points, i.e. $a>-16.25$

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Therefore, $-16.25<a<16$.

  • $\begingroup$ Just confirming, you obtained the lower limit of $-16.25$ by considering the case of tangency of $y=x-a$ on $y=16-x^2$ right? Thanks a lot for your answer the method is quite simpler $\endgroup$ – Shreyas S May 1 '17 at 16:51
  • $\begingroup$ Yes. Using discriminant. $\endgroup$ – CY Aries May 1 '17 at 16:57
  • $\begingroup$ Alright thanks. $\endgroup$ – Shreyas S May 1 '17 at 17:07

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