Difference between revisions of "MIE 2016/Day 1/Problem 2"

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===Problem 2===
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==Problem 2==
 
The following system has <math>k</math> integer solutions. We can say that:
 
The following system has <math>k</math> integer solutions. We can say that:
  
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===Solution 2===
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==Solution 2==
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<math>\frac{x^2-2x-14}{x}>3</math>
  
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<math>\frac{x^2-2x-14}{x}-\frac{3x}{x}>0</math>
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<math>\frac{x^2-5x-14}{x}>0\begin{cases}x^2-5x-14>0\\x>0\end{cases}</math>
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<math>\left.\begin{array}{l}x > 0\\x > 7\\x < -2\end{array}\right\}x\in(-2,0)\cup(7,+\infty)</math>
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Adding the other interval we get
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<math>x\in(-2,0)\cup(7,12]</math>
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If <math>k</math> is the number of integer solutions, then <math>k=7</math>. <math>\boxed{D}</math>
  
  
 
===See Also===
 
===See Also===

Revision as of 18:29, 8 January 2018

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 2

$\frac{x^2-2x-14}{x}>3$

$\frac{x^2-2x-14}{x}-\frac{3x}{x}>0$

$\frac{x^2-5x-14}{x}>0\begin{cases}x^2-5x-14>0\\x>0\end{cases}$

$\left.\begin{array}{l}x > 0\\x > 7\\x < -2\end{array}\right\}x\in(-2,0)\cup(7,+\infty)$


Adding the other interval we get


$x\in(-2,0)\cup(7,12]$


If $k$ is the number of integer solutions, then $k=7$. $\boxed{D}$


See Also

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