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?
[Math] 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
inequalityquadratics
Related Solutions
Firstly we require $c>0$ for the second log to be real.
Now considering your inequality $$(4-c)x^2+4x+(7-c)\geq0$$
If $c<4$ there will always be at least one solution to the inequality aince the curve is a positive quadratic.
If $c>4$, we require that the discriminant of this negative quadratic to be positive or equal to zero so that the curve does not lie entirely below the $x$ axis.
Therefore $$16-4(4-c)(7-c)\geq0$$ $$\Rightarrow (c-3)(c-8)\leq 0$$ $$\Rightarrow 3\leq c\leq8$$ Hence the requirement for $c$, when we combine these results, is $$0<c\leq8$$
https://www.desmos.com/calculator/eupzet7eav
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. $a<16$.
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. $a>-16.25$
Therefore, $-16.25<a<16$.
Best Answer
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 $x^2$