MIE 2016/Day 1/Problem 2

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Problem 2

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


(a) $0\leq k\leq 2$

(b) $2\leq k\leq 4$

(c) $4\leq k\leq6$

(d) $6\leq k\leq8$

(e) $k\geq8$





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

Adding the other interval we get


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

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