Difference between revisions of "MIE 2016/Problem 2"

Latest revision as of 18:30, 2 October 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

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)_/¯

@@ Line 14: / Line 14: @@
 (e) <math>k\geq8</math>
+==Solution 1==
+===Objective:===
+We can solve this problem in two steps: First, we solve for the range of <math>\frac{x^2-2x-14}{x}>3</math>, then combine it with the range of <math>x\leq12</math> 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 <math>\frac{x^2-2x-14}{x}>3</math>.
+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.
+Case 1: <math>0<x\leq12</math>
+<cmath>\frac{x^2-2x-14}{x}>3</cmath>
+<cmath>x^2 - 2x - 14>3x</cmath>
+<cmath>x^2 - 5x - 14 > 0</cmath>
+Factoring, we get
+<cmath>(x-7)(x+2)>0</cmath>
+Now, <math>x</math> can be greater than <math>7</math> or less than <math>-2</math>.
+But in this case, <math>0<x\leq12</math>, and this further restricts our solutions.
+So, for the case where <math>0<x\leq12</math>, our solutions are <math>7<x\leq12</math>
+Case 2: <math>x<0</math>
+<cmath>\frac{x^2-2x-14}{x}>3</cmath>
+<cmath>x^2-2x-14<3x</cmath>
+<cmath>x^2-5x - 14<0</cmath>
+<cmath>(x-7)(x+2)<0</cmath>
+We have in this case that <math>-2<x<7</math>, but the case statement further restricts our solutions.
+For this case, the solutions are <math>-2<x<0</math>
+===Step 2===
+Now, we know the solutions for <math>x</math>: in the first case, where <math>7<x\leq12</math>, the integer solutions are <math>x = {8, 9, 10, 11, 12}</math>
+In the second case, where <math>-2<x<0</math>, the only integer solution is <math>x = {-1}</math>
+The union of these two cases gives <math>x = {-1, 8, 9, 10, 11, 12}</math>.
+There are <math>k=6</math> solutions and <math>6\leq k\leq8</math>, giving <math>\boxed{D)}</math>
+~Windigo
 ==Solution 2==
+Because <math>x \le 12</math> is the simpler condition, we can apply it to the solution of <math>\frac{x^2-2x-14}{x} < 3</math>. 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 <math>0</math> on one of the sides, and factor the other. To do this, we can subtract <math>\frac{3x}{x}</math> from both sides, going from
+<cmath>\frac{x^2-2x-14}{x} < 3</cmath> to <cmath>\frac{x^2-2x-14-3x}{x} < 0</cmath> and combining like terms to get <cmath>\frac{x^2-5x-14}{x} < 0</cmath> factoring to get the usable form of <cmath>\frac{(x+2)(x-7)}{x} < 0</cmath> Using the snake method, we can build a table.
+{| class="wikitable"
+|+Snake Table
+|-
+|<math>\space</math>
+|<math>x<-2</math>
+|<math>-2<x<0</math>
+|<math>0<x<7</math>
+|<math>7<x</math>
+|-
+|<math>x+2</math>
+|<math>-</math>
+|<math>+</math>
+|<math>+</math>
+|<math>+</math>
+|-
+|<math>x</math>
+|<math>-</math>
+|<math>-</math>
+|<math>+</math>
+|<math>+</math>
+|-
+|<math>x-7</math>
+|<math>-</math>
+|<math>-</math>
+|<math>-</math>
+|<math>+</math>
+|-
+|multiplied
+|<math>-</math>
+|<math>+</math>
+|<math>-</math>
+|<math>+</math>
+|}
+Using this table, we can find out that this inequality is true when <math>-2<x<0</math> or <math>7<x</math>. Applying the second inequality, we can modify these to <math>-2<x<0</math> or <math>7<x\le12</math> so they fit the whole set. Counting the number of integers that satisfy these conditions, we can see that the six numbers <math>-1, 8, 9, 10, 11, 12</math> fit these conditions. Therefore, the answer is <math>\boxed{D)}</math>. <cmath>\space</cmath>
+-Aurora64 ¯\_(ツ)_/¯
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