MIE 2016/Problem 2: Difference between revisions

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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 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.