AMC10 2004 A

AMC10 2004 A · Q15

AMC10 2004 A · Q15. It mainly tests Linear inequalities, Absolute value.

Given that $-4 \leq x \leq -2$ and $2 \leq y \leq 4$, what is the largest possible value of $\frac{x + y}{x}$?

已知$-4\leq x\leq -2$且$2\leq y\leq 4$，求$\frac{x+y}{x}$的最大可能值？

(A) -1 -1

(B) -1/2 -$\frac{1}{2}$

(D) 1/2 $\frac{1}{2}$

(E) 1 1

Answer

Correct choice: (D)

正确答案：（D）

Solution

Because $\frac{x + y}{x} = 1 + \frac{y}{x}$ and $\frac{y}{x} < 0$, the value is maximized when $\left|\frac{y}{x}\right|$ is minimized, that is, when $|y|$ is minimized and $|x|$ is maximized. So $y = 2$ and $x = -4$ gives the largest value, which is $1 + (-1/2) = 1/2$.

因为$\frac{x+y}{x}=1+\frac{y}{x}$且$\frac{y}{x}<0$，该值在$\left|\frac{y}{x}\right|$最小时最大化，即$|y|$最小且$|x|$最大时。所以$y=2$且$x=-4$给出最大值，即$1+(-1/2)=1/2$。

Topics