MIE 2016/Problem 2

Problem 2

The following system has $k$ integer solutions. We can say that:


(a) $0\leq k\leq 2$

(b) $2\leq k\leq 4$

(c) $4\leq k\leq6$

(d) $6\leq k\leq8$

(e) $k\geq8$

Solution 1


We can solve this problem in two steps: First, we solve for the range of $\frac{x^2-2x-14}{x}>3$, then combine it with the range of $x\leq12$ to get a compound inequality which we can use to find all possible integer solutions.

Step 1

We first find the range of the inequality $\frac{x^2-2x-14}{x}>3$.

We now simplify the inequality:

Case 0: $x=0$

This has no solutions since $x=0$ will make the function undefined.

Case 1: $0<x\leq12$

\[\frac{x^2-2x-14}{x}>3\] \[x^2 - 2x - 14>3x\] \[x^2 - 5x - 14 > 0\] Factoring, we get \[(x-7)(x+2)>0\] Now, $x$ can be greater than $7$ or less than $-2$. But in this case, $0<x\leq12$, and this further restricts our solutions. So, for the case where $0<x\leq12$, our solutions are $7<x\leq12$

Case 2: $x<0$ \[\frac{x^2-2x-14}{x}>3\] \[x^2-2x-14<3x\] \[x^2-5x - 14<0\] \[(x-7)(x+2)<0\] We have in this case that $-2<x<7$, but the case statement further restricts our solutions.

For this case, the solutions are $-2<x<0$

Step 2

Now, we know the solutions for $x$: in the first case, where $7<x\leq12$, the integer solutions are $x = {8, 9, 10, 11, 12}$

In the second case, where $-2<x<0$, the only integer solution is $x = {-1}$

The union of these two cases gives $x = {-1, 8, 9, 10, 11, 12}$.

There are $k=6$ solutions and $6\leq k\leq8$, giving $\boxed{D)}$ ~Windigo

Solution 2

Because $x \le 12$ is the simpler condition, we can apply it to the solution of $\frac{x^2-2x-14}{x} < 3$. We can find the solution of the first inequality given in the problem by simplifying and using the Snake Method. To use the Snake Method, we need to have $0$ on one of the sides, and factor the other. To do this, we can subtract $\frac{3x}{x}$ from both sides, going from \[\frac{x^2-2x-14}{x} < 3\] to \[\frac{x^2-2x-14-3x}{x} < 0\] and combining like terms to get \[\frac{x^2-5x-14}{x} < 0\] factoring to get the usable form of \[\frac{(x+2)(x-7)}{x} < 0\] Using the snake method, we can build a table.

Snake Table
$\space$ $x<-2$ $-2<x<0$ $0<x<7$ $7<x$
$x+2$ $-$ $+$ $+$ $+$
$x$ $-$ $-$ $+$ $+$
$x-7$ $-$ $-$ $-$ $+$
multiplied $-$ $+$ $-$ $+$

Using this table, we can find out that this inequality is true when $-2<x<0$ or $7<x$. Applying the second inequality, we can modify these to $-2<x<0$ or $7<x\le12$ so they fit the whole set. Counting the number of integers that satisfy these conditions, we can see that the six numbers $-1, 8, 9, 10, 11, 12$ fit these conditions. Therefore, the answer is $\boxed{D)}$. \[\space\] -Aurora64 ¯\_(ツ)_/¯

See Also

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