Consider $f(x)= [x]^2 - [x+6]$ and $g(x)= 3kx^2+ 2x + 4(1-3k)$ where $[x]$ denotes the floor function.
Let $A= \{x ~|~ f(x)= 0\}$ and $k \in [a,b]$ for which every element of set A satisfies the inequality $g(x)\ge 0$.
1) The set $A$ is equal to ?
2) The value of $6b- 3a =$?
The first part is easy. It's just the solution of the quadratic; $[x]^2- [x]- 6 = 0 $
$\therefore \mathbb A = [-2, -1) \cup [3,4)$ which is the right answer.
But I am stuck on the second part of the question. I am unable to use the constraint on the values of $x$ to find k.
using the condition I have got:
$k(12- 3x^2) \ge -{2(x+2)} $
For $x=2$ this is true, For $x\ne 2$:
$\implies3k(2-x) \ge -2 $
What do I do next?
The answer is :
$2$
You need for $k$ to satify: $$k (3x^2-12) =3k(x-2)(x+2)\geq -2(x+2)$$ for all $x \in \Bbb A= [-2, -1) \cup [3,4)$.
You can then consider the three cases:
$x=-2$ then the inequality is verified.
For all $x \in (-2, -1)$, as $x-2 <0$ and $x+2>0$ the inequality is equivalent to: $$k \leq \frac{2}{3(2-x)}$$ and as it must be true for any $x$ and $\inf_{x \in(-2,-1)} \frac{2}{3(2-x)}=\frac{2}{3(2-(-2))}=\frac{1}{6}$ it is the same as: $$k \leq \frac{1}{6}$$
For all $x \in (3, 4)$, as $x-2 >0$ and $2+x>0$ the inequality is equivalent to: $$k \geq -\frac{2}{3(x-2)}$$ and once again as $\sup_{x \in(3,4)} \frac{-2}{3(x-2)}= \frac{-2}{3(4-2)}=\frac{-1}{3}$ it is true iff: $$k \geq -\frac{1}{3}$$
Thus $g(x) \geq 0$ for all $x \in \{-2\} \cup (-2, -1) \cup [3,4)$ iff: $$-\frac{1}{3} \leq k \leq \frac{1}{6}$$ i.e:
$$a=-\frac{1}{3}$$
$$b=\frac{1}{6}$$ and: $$6b-3a=1-(-1)=2$$