## SAT SUBJECT TEST MATH LEVEL 2

## PART 2

## REVIEW OF MAJOR TOPICS

## CHAPTER 1

Functions

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1.6 Miscellaneous Functions

### Answers and Explanations

Parametric Equations

* 1. **(B)** Graph these parametric equations for values of *t *between –5 and 5 and for *x *and *y *between –2.5 and 2.5.

Apparently the *x *values are always greater than some value. Use the TRACE function to move the cursor as far left on the graph as it will go. This leads to a (correct) guess of . This can be verified by completing the square on the *x *equation:

.

This represents a parabola that opens up with vertex at . Therefore, .

2. **(D)** D is the only reasonable answer choice. To verify this, note that . So . Adding this to *x *= sin^{2}*t *gives . Since 0 *x * 1 because 0 sin^{2} *t * 1, this can only be a portion of the parabola given by the equation *y*^{2} + 4*x *= 4.

* 3. **(A)** You could graph all three parametric pairs to discover that only I gives a circle. (II and III give semicircles). You can also see this by a simple analysis of the equations. Removing the parameter in I by squaring and adding gives *x*^{2} + *y*^{2} = 1, which is a circle of radius 1. Substituting*x *for *t *in the *y *equation of II and squaring gives *x*^{2} + *y*^{2} = 1, but *y *≥ 0 so this is only a semicircle. Squaring and substituting *x*^{2} for *s *in the *y *equation of III gives *x*^{2} + *y*^{2} = 1, but *x *≥ 0 and so this is only a semicircle.

Piecewise Functions

* 1. **(B)** Enter abs(2*x *– 1) into *Y*_{1} and 4*x *+ 5 into *Y*_{2}. It is clear from the standard window that the two graphs intersect only at one point.

* 2. **(E)** Enter abs(*x *– 2) into *Y*_{1}, 1 into *Y*_{2}, and 4 into *Y*_{3}. An inspection of the graphs shows that the values of *x *for which the graph of *Y*_{1} is between the other two graphs are in two intervals. E is the only answer choice having this configuration.

* 3. **(A)** Subtract |*x*| from both sides of the equation. Since |*y*| cannot be negative, graph the piecewise function

In the first command, the word “and” is in TEST/LOGIC. The result is a square that is on a side. Therefore, the area is 8.

4. **(A)** Since *f(x*) must = *f*(|*x*|), the graph must be symmetric about the *y*-axis. The only graph meeting this requirement is Choice A.

* 5. **(B)** Since the point where a major change takes place is at (1,1), the expression in the absolute value should equal zero when *x *= 1. This occurs only in Choice B. Check your answer by graphing the function in B on your graphing calculator.

6. **(E)** Choice A fails if *N *= 0.5. Choice B subtracts cents from ounces. Choice C fails if *N *= 1. Choice D adds cents to ounces.

7. **(C)** Since *f(x*) = an integer by definition, the answer is Choice C.

* 8. **(E)** Enter int(4*x*) – 2*x *into *Y*_{1}. The graph is shown in the figure below.

The breaks in the graph indicate that it cannot be the graph of any of the first four answer choices.