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 ?
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