Difference between revisions of "MIE 2016/Problem 2"

Revision as of 15:56, 12 September 2020

Problem 2

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

$\begin{cases}\frac{x^2-2x-14}{x}>3\\\\x\leq12\end{cases}$

(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

Objective:

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

@@ Line 22: / Line 22: @@
 We now simplify the inequality:
 Case 0: <math>x=0</math>
 This has no solutions since <math>x=0</math> will make the function undefined.

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