【SAT数学】今日份的练习题（一）

北京sat培训,sat备考资料,sat网课,sat培训机构,sat保分班,sat真题

今日份的SAT数学题来啦，备考SAT的宝宝们快拿起笔练一练吧，开动小脑筋，找找做题手感。

1. If f(x) = x2 – 3, where x is an integer, which of the following could be a

value of f(x)?

I 6

II 0

III -6

A. I only

B. I and II only

C. II and III only

D. I and III only

E. I, II and III

Correct Answer: A

解析:

Choice I is correct because f(x) = 6 when x=3. Choice II is incorrect because

to make f(x) = 0, x2 would have to be 3. But 3 is not the square of an integer.

Choice III is incorrect because to make f(x) = 0, x2 would have to be –3 but

squares cannot be negative. (The minimum value for x2 is zero; hence, the

minimum value for f(x) = -3)

2. For how many integer values of n will the value of the expression 4n + 7

be an integer greater than 1 and less than 200?

A. 48

B. 49

C. 50

D. 51

E. 52

Correct Answer: C

解析:

1 4n + 7 200. n can be 0, or -1. n cannot be -2 or any other negative integer or the expression 4n + 7 will be less than1. The largest value for n will be an integer (200 - 7) /4. 193/4 = 48.25, hence 48. The number of integers between -1 and 48 inclusive is 50

这是今日份的数学练习题哦~备考SAT的宝宝们一定要找好刷题的感觉，易伯华会一直陪在大家身边，祝考试的宝宝们取得好成绩！