2017 AMC 10B 第 3 题

Question

实数

x

、

y

、

z

满足

0 < x < 1

、

-1 < y < 0

、

1 < z < 2

。

下面哪个数一定为正？

Real numbers

x,

y,

and

z

satisfy the inequalities

0 < x < 1,

-1 < y < 0,

and

1 < z < 2.

Which of the following numbers is necessarily positive?

y+x^2

y+xz

y+y^2

y+2y^2

y+z

选项

y+z

可以直接由给定范围判断。

The choice

y+z

can be tested directly from the given bounds

其余表达式各可用一个小反例排除。

Use one small counterexample to reject each other expression

因为

-1 < y

且

1 < z

，所以

0 < y+z

，因此选项 E 一定正确。

此外，用下面这组取值可以排除其余每个选项：

x=0.1\text{，}

y=-0.25\text{，}

z=1.25\text{。}

所以正确答案是 E。

Since

-1 < y

and

1 < z,

we can add the inequalities to see that

0 < y+z.

This naturally proves choice E correct.

Furthermore, we can eliminate every other choice with the following values:

x=0.1,

y=-0.25,

z=1.25.

Thus, the correct answer is E .

其他年份的第 3 题