Difference between revisions of "MIE 2016/Problem 2"

(Problem 2)
(Solution 2)
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(e) <math>k\geq8</math>
 
(e) <math>k\geq8</math>
  
===Solution 2===
+
==Solution 2==
 
 
  
 
===See Also===
 
===See Also===

Revision as of 20:20, 7 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

See Also