SAT SUBJECT TEST MATH LEVEL 2

PART 2

REVIEW OF MAJOR TOPICS

CHAPTER 1
Functions

 
1.1 Overview

Answers and Explanations

Definitions

1.     (A) Either (3,2) or (3,1), which is not an answer choice, must be removed so that 3 will be paired with only one number.

2.     (E) For each value of there is only one value 
for in each case. Therefore, f, g, and are all functions.

3.     (C) Since division by zero is forbidden, cannot equal 2.

Combining Functions

1.     (E) f(–2) = 3(–2)– 2(–2) + 4 = 20.

2.     (D) g(2) = 32 = 9. f(g(2)) = f(9) = 31.

3.     (C) To get from f(x) to f(g(x)), x2 must become 4x2. Therefore, the answer must contain 2since (2x)= 4x2.

4.     (D) g(x) cannot equal 0. Therefore,  .

5.     (A) Since f(2) implies that = 2, g(f(2)) = 2. Therefore, g(f(2)) = 3(f(2)) + 2 = 2. Therefore, f(2) = 0.

6.     (C) p(a) = 0 implies 4– 6 = 0, so  .

* 7.      (E) 

Inverses

1.     (E) If = 2– 3, the inverse is = 2– 3, which is equivalent to .

2.     (A) By definition.

3.     (B) The inverse is {(2,1),(3,2),(4,3),(1,4),(2,5)}, which is not a function because of (2,1) and (2,5). Therefore, the domain of the original function must lose either 1 or 5.

4.     (E) If this line were reflected about the line y = x to get its inverse, the slope would be less than 1 and the y-intercept would be less than zero. The only possibilities are Choices D and E. Choice D can be excluded because since the x-intercept of f(x) is greater than –1, the y-intercept of its inverse must be greater than –1.

Odd and Even Functions

1.     (D) Use the appropriate test for determining whether a relation is even.

                I. The graph of = 2 is a horizontal line, which is symmetric about the y-axis, so = 2 is even.

              II.   Since f(–x) = – f(x) unless = 0, this function is not even.

            III.   Since (–x)y2 = 1 whenever x2 + y2 = 1, this relation is even.

2.     (D) Use the appropriate test for determining whether a relation is odd.

                I. The graph of = 2 is a horizontal line, which is not symmetric about the origin, so = 2 is not odd.

              II.   Since f(–x) = –= –f(x), this function is odd.

            III.   Since (–x)2 + (–y)2 = 1 whenever x2 + y2 = 1, this relation is odd.

3.     (B) The analysis of relation III in the above examples indicates that I and II are both even and odd. Since – 0 when = 0 unless = 0, III is not even, and is therefore not both even and odd.

4.     (C) A is even, B is odd, D is even, and E is odd. C is not even because (–x)– 1 = –x3 – 1, which is neither x3 – 1 nor –x3 + 1.