The number of integral values of $a$ for which the inequality $3- |x-a |>x^2$ is satisfied by at least one negative $x$, must be equal to 6.

I don't know how to solve this. Can you help?

  • $\begingroup$ First, remove the $|...|$ If $x>a$, then $x-a$ is positive, so the mod can be removed without changing. If not, then add a negative sign when removing the mod. Make cases and solve :) $\endgroup$ – mikhailcazi Jul 21 '13 at 13:35

enter image description herethis is the graph of x^2 and 3-|x-a|

as you can see the graph of $x^2$ and $3-|x-a|$ will intersect on negative x- axis if a<3 and a>-3.25 so that right part of inverted V shaped curve is tangent to $x^2$


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