Finding the value of $6b-3a$ 
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:
$$6b-3a=1-(-1)=2$$
