SAT 8

224

7. D Secant is the reciprocal function of the cosine function. If sec θ = 2, then cos θ = 12 .

cos θ sec θ = 12 (2) = 1.

8.B

16x4 − 9 = 4.

16x4 = 13.

x4 = 1613 .

1

x= ± 13 4 = ±0.95.

16

9.D The probability of rolling a 6 is 61 , so the

	probability of NOT rolling a 6 is							1 −	1	or	5	.	The
									6		6
	probability of not rolling a 6 on both rolls is:
	5		5			25
					=		.
	6			6		36

10. D Graph f (x) = 4 − x2 on your graphing calculator to see that it is the graph of a semicircle centered at the origin with a radius of 2 units. Because f(x) equals the square root of an expression, it must be a positive value, and y ≥ 0 is part of the range. There is an upper limit on y, however. The maximum y-value occurs when x = 0.

f (0) = 4 − 02 = 2.

The range is between 0 and 2, inclusive.

11. E Because the two events are independent, the probability that Meghan wins and John loses is the product of the two probabilities. The probability that

John loses is: 1 − 45 = 15 .

P =	9		1		=	9	.
	11		5			55

12. C You know the length of the side opposite and the side adjacent to θ, so use arctangent to solve for the angle measure.

Tan−1		11		= 40.2º.
		13

PART III / EIGHT PRACTICE TESTS

13. E Remember that a logarithm is an exponent. You’re trying to determine to what exponent to raise

the base, 2, to equal 16 2 .

log2 16 2 = log2 24 (2)	1
	2 .

By the properties of exponents

24 (2)	2		= 24 + 2				= 2	2 .
	1					1		9
log2 2		9	=	9	.
		2
				2

14. C Complex roots occur in conjugate pairs. If i is a root of the polynomial, then −i is also a root. Use the four roots to determine the factors of the polynomial. Then multiply to get the polynomial.

x(x − 4)(x − i)(x + i) =

x(x − 4)(x2 + 1) =

x(x3 − 4x2 + x − 4) =

x4 − 4x3 + x2 − 4x.

15. B Recall that in the polar coordinate system x = r cos θ. Alternatively, use right triangle trigonometry to determine cosine θ.

cos θ = xr .

Now solve for x: x = r cos θ.

16. A In a right triangle, if sin θ = 34 , the leg oppo-

site θ measures 3 and the hypotenuse measures 4. Use the Pythagorean Theorem to solve for the side adjacent to θ.

	a2 + 32	= 42.
	a2	= 16 − 9 = 7.
	a =		7.
	cos θ =			7	.
			4

Recall that the double angle formula for cosine is:

cos	2θ = cos2		θ − sin2					θ.
		7	2		3			2	=	7		−	9		= −	2	= −	1
cos	2θ =			−															.
		4			4					17			16			16		8

PRACTICE TEST 1

17. B The inverse of the logarithmic function is the exponential function. The inverse of f (x) = log3 x is, therefore, f −1(x) = 3x.

You can also determine this by graphing f (x) = log3x and f−1(x) = 3x on your calculator and observing that their graphs are reflections of each other over the line y = x.

18. D In a rational function, asymptotes occur when the denominator approaches 0 because division by 0 is undefined.

4 − x2 = 0.

4= x2. x = ± 2.

Vertical asymptotes, therefore, occur at x = 2 and x = −2. Because the degree of the numerator equals the degree of the denominator, a horizontal asymptote occurs at the quotient of the coefficients of the x2 terms.

6 = −6.

−1

The line y = −6 is a horizontal asymptote. (To verify the three asymptotes, graph the function on your graphing calculator.)

19. E Because 81 = 34, both sides of the equation can be written in base 3.

35− x = 81x+1.

35− x = 34( x+1).

Now, set the exponents equal to each other and solve for x.

5 − x = 4x + 4.

1= 5x. x = 15 .

20.B Because both the x and y terms are raised to the 2nd power, the graph cannot be a parabola. Note that the x2 and y2 terms have different coefficients, and both are positive. The graph would, therefore, be an ellipse.

16x2 + 8y2 − 32x + 8y − 1 = 0.

					1	2
(x −	2		y +
					2
	1)	+					= 1.
9				9

84

225

21.C fg(12) is the product of f (12) and g(12). f (12) = 2(12) + 5 = 29.

	g(12) =		1				=		1	.
			+ 12					18
	6
	fg(12) = 29			1			=	29			.
				18				18

22. A Because the median is 9, the 4th term when arranged from least to greatest is 9. The mode is 7, so either 2 or 3 of the integers less than 9 equal 7. Note that 7 is the only mode. Because the problem asks for the least possible range, assume that three of the integers equal 7.

There are two possible scenarios for determining the numbers greater than the median:

7, 7, 7, 9, 10, 10, 11,

7, 7, 7, 9, 10, 11, 11.

Note that if 7 occurs three times, then one of the integers greater than 9 can occur twice. In either case, however, the range equals 11–7, or 4.

23. C To check for x-axis symmetry, replace x by −x.

(−x)2 − (−x)y = 4 is not equivalent to the original equation.

To check for y-axis symmetry, replace y by −y.

x2 − x(−y) = 4 is not equivalent to the original equation.

To check for origin symmetry, replace x by −x and y by −y.

(−x)2 − (−x)(−y) = 4 is equivalent to the original equation, so the graph is symmetric with respect to only the origin.

24. E Raising the imaginary number i to an exponent follows the pattern:

i1 = i.

i2 = −1.

i3 = −i.

i4 = 1.

21 ÷ 4 = 5 remainder 1, so i21 is equivalent to i1 = i. Raising i to any power that results in a remainder of 1 when the exponent is divided by 4 will also equal i1 or i. 31 ÷ 4 = 7 remainder 3, so i31 is equivalent to i3 = −i, not i.

226

25. B The minimum value of a quadratic equation ax2 + bx + c occurs when x = − 2ba . When graphed the

minimum occurs at the vertex of a parabola that is concave up.

C(n) = 0.01x2 − 90x + 25,000.

		n = −	b		= −		(−90)				= 4,500 units.
			2a				2(0.01)
26. D	( (3 (			)))			(( 2			3 )4 )
5	4										1		1
			n			=	(n	1	)	1			5	.

When raising a power to a power, multiply the exponents.

1	×	1		×	1	×		1	=		1			.
2		3			4			5		120
5	(4	(3		(			)))					1
					n				= n120 .

27. A The slope of the line with x-intercept (6, 0) and y-intercept (0, −15) is:

	m =	y2	− y1	=	−15 − 0	=	15		=	5	.
		x2	− x1		0 − 6			6		2

The y-intercept is given as −15. In slope-intercept form y = mx + b the equation is therefore:

y= 52 x + (−15).

y= 52 x − 15.

28.D The degree of the polynomial is odd and the leading coefficient is negative. (an = −3) By the Leading Coefficient Test, the graph rises left and falls right.

29.E If x + 2 is a factor of x4 + x3 + 3x2 + kx − 10, then f(−2) = 0.

f (−2) = (−2)4 + (−2)3 + 3(−2)2 + k(−2) − 10 = 0.

16 − 8 + 12 + k(−2) − 10 = 0.

−2k + 10 = 0.

−2k = −10.

k = 5.

PART III / EIGHT PRACTICE TESTS

Alternatively, divide x4 + x3 + 3x2 + kx − 10 by x + 2 using either synthetic or long division. The remainder, −2k + 10, must equal zero for x + 2 to be a factor of the polynomial.

−2k + 10 = 0.

k = 5.

30. D The Rational Root Test states that if a polynomial function has integer coefficients, every rational

zero of the function has the form qp (simplified to

lowest terms) where p = a factor of the constant term a0 and q = a factor of the leading coefficient an. Here a0 = −15 and an = 6.

The factors of −15 are 1, 15, 3, and 5, and the factors of 6 are 1, 6, 2, and 3.

qp = ±1, ± 12 , ± 13 , ± 61 , ± 3, ± 32 , ± 5,

± 52 , ± 53 , ± 65 , ± 15, ± 152 .

Remember not to duplicate terms and write each in simplest form. There are 24 possible roots.

31. C Draw the radius to the point of tangency T. The radius is perpendicular to the tangent segment. Then, draw the segment connecting P to the center of the circle. P is bisected and the new segment is also perpendicular to ST. The central angle measures 75° and the new figure is as follows:

S

75°	15°	P

	3	x
		15°

T

Let x = half the length of ST as shown. Use the radius, 3, as the hypotenuse of the small right triangle in the interior of the circle to solve for x.

cos 15º = 3x .

x = 3 cos 15º ≈ 2.898.

ST = 2(2.898) ≈ 5.796 ≈ 5.8.

PRACTICE TEST 1

32. B If the investment doubles, A = $5,000.

						0.045		12t
	5,000	= 2,500			1 +		.
						12
				0.045		12t
	2	= 1	+			.
					12
	2	= 1.0037512t.

Now, take either the log or natural log of both sides of the equation to solve for the variable, t, in the exponent.

log 2 = log (1.00375)12t .

log 2 = 12t log 1.00375.

log 2 = 0.0195t.

t≈ 15.4 years.

33.C The nth term of an arithmetic sequence is given by the equation: an = al + (n − 1)d, where d = the common difference between consecutive terms and a1 = the first term of the sequence.

The difference between 23 and 7 represents 4d because the sequence increases by 6 − 2 = 4 terms.

4d = 23 − 7.

4d = 16d

d = 4

Knowing that a2 = 7, you can solve for al:

a2 = al + (2 − 1)4 = 7.

al + 4 = 7.

al = 3.

The 90th term is, therefore:

a90 = 3 + (90 − 1)4 = 359.

34. D Multiplying 45,454,545,454,545 by the units digit of 1,234, 4, results in a 14-digit product. Multiplying 45,454,545,454,545 by the tens digit of 1,234, 3, results in a 15-digit product because it is necessary to use a zero placeholder for the units digit. Similarly, multiplying by the hundreds digit requires 2 placeholders, and multiplying by the thousands digit requires 3 placeholders. The product will contain:

14 + 3 = 17 digits.

On your calculator, the product may be displayed as 5.609 . . . E16, which represents 5.609 . . . × 1016. In decimal form, this results in a 5 followed by 16 digits.

227

35. E The figure is the graph of the sine function with

amplitude 1 and period π. The equation is, therefore, 2

y = 12 sin 2x.

36. A Use the Law of Sines to determine the length of BC.

	sin A	=		sin B	=		sin C		.
	a
				b			c
sin 29º		=		sin 43º		.
3.4
				BC
	BC =			3.4(sin 43º )				.
				sin 29º

BC ≈ 4.8 cm.

37.D

0.75(84) + 0.25(92) = 86.

Her overall grade will be 86%.

38.B The distance between ordered triples (x1, y1, z1) and (x2, y2, z2) is given by the formula:

Distance = (x2 − x1)2 + ( y2 − y1)2 + (z2 − z1)2 .

Distance = (7 − −4)2 + (5 − 2)2 + (−3 − 1)2 .

=112 + 32 + (−4)2 .

=146 ≈ 12.1.

39. B

To find the slope of the line, write the parametric equations as a single equation in terms of x and y. Because x = 8 − t, t = 8 − x. Substitute this value of t into the equation for y:

y = 10 + 2(8 − x).

y= 10 + 16 − 2x.

y= −2x + 26.

The equation of the line is in slope-intercept form, so the slope of the line is the coefficient of the x term, −2.

228

40.C Because a Φ b = a Φ b = a−b − 3b, n Φ − 2 = n− (−2) − 3(−2).

n2 + 6 = 70. n2 = 64.

n = ±8.

Answer C, −8, is one possible solution for n.

41. C AD is the hypotenuse of a right triangle with legs measuring s and 2s . Use the Pythagorean Theorem to solve for AD.

	s		2
AD2 = s2 +	2	.

AD2 = s2 + s2 . 4

AD2 = 52 . 4

AD = s 5 2

AD = BC, so the perimeter of ABCD is:

	s +	s 5	+ s +	s 5	= 2s + s 5.
		2		2

42. C The standard form of the equation of an ellipse is:

	(x − h)2	+	( y − k)2		= 1 for an ellipse whose major axis
	a2		b2
	is horizontal, and			(x − h)2		+	( y − k)2	= 1 for an ellipse
							a2
					b2

whose major axis is vertical. In both cases, 2a = the length of the major axis.

First, divide both sides of the equation by 48 to write it in standard form.

12x2 + 8y2 = 48.

x2 + y2 = 1. 4 6

This ellipse has a vertical major axis because a must

be greater than b. a = 6. The length of the major axis

is 2 6 ≈ 4.90.

PART III / EIGHT PRACTICE TESTS

43. D The line tangent to the circle x2 + y2 = 1 at

		3	,	4		is perpendicular to the radius drawn from the
		5		5

point of tangency.

Note that the equation x2 + y2 = 1 represents a circle centered at the origin with a radius of one unit.

									3		4
	The line containing the radius from point									,			has
									5		5
			3
						4
	a slope of m =		5		=		.	Any line perpendicular to the
			4			3

5

radius will have a slope equivalent to the opposite

reciprocal of 4 , which is − 3 . 3 4

Now, find the equation of the line containing the point

		3	,	4		with a slope of	−	3	. In point-slope form, the
		5		5				4

equation is:

y − 45 = − 34 x − 35 .

y = − 34 x + 209 + 45 .

y = − 34 x + 2025 = − 34 x + 54 .

The y-intercept is 5 or 1.25. 4

44. C
	3x2 − 7x − 6	=	3x + 2, x − 3
	x2 − 9		x − 3, x + 3

=3x + 2 x + 3 .

The values of the function are equal to 3x + 2 for all x + 3

x except x = ±3. As x approaches 3, the value of the function approaches:

3(3) + 2	=	11	≈ 1.8.
3 + 3		6

PRACTICE TEST 1

45. A A function is odd if replacing x with −x results in the opposite of the original function. Its graph is symmetric with respect to the origin. Instead of algebraically determining which of the given functions satisfy the equation f (−x) = − f (x), graph them on your graphing calculator. You can see that the sine function is symmetric to the origin.

f (x) = cos x is symmetric with respect to the y-axis. f (x) = x2 − 10 is also symmetric with respect to the y-axis. f (x) = 4x and f (x) = log2 x are inverse functions and reflections of each other over the line y = x. They are not symmetric with respect to the origin.

Answer A, f (x) = sin x, is the correct answer choice.

46.E Factor the equation to get: (4sin x − 1)(3sin x + 1) = 0.

	sin x =	1	or sin x = −					1	.
		4						3
	x = sin−1				1		= 14.478º.
					4

x = sin−1 − 13 = 19.471º.

Considering 0° ≤ x ≤ 360°, the possible solutions are:

x = 14.48º , 165.52º , 199.47º , or 340.53º.

The problem specifies the interval 180° ≤ x ≤ 360°, however. 199.5° or 340.5° are the correct answers.

47. B The problem is asking for the square root of 15 − 8i. Square the answer choices to determine which one results in 15 − 8i. Remember that i2 = −1.

(4 − i)(4 − i) = 16 − 8i + i2.

= 16 − 8i − 1 = 15 − 8i.

229

48. B

tan θ cot θ						=
	tan θ
	sin θ		+	cos θ
				sin θ .
	cos θ
	sin θ
	cos			θ

Multiply the numerator and denominator by the LCD, cos θ sin θ:

	sin θ				+		cos θ				cos θ sin θ
	cos θ						sin θ			×
			sin θ								cos θ sin θ
			cos θ
=		sin2			θ + cos2					θ
					sin2 θ
=		1
		sin2			θ

=csc2 θ.

49.A The first term in the series, a1, is 1. Divide the second term by the first term to determine the common

	ratio, r = −							1		. Because this is an infinite geometric
							5
	sequence, find its sum by using the formula:
	S =			a1							.
		1 − r
	S =	1												.
		1 −							−1
									5
	S =		1							=		5			.
				6
							6
		5

50. D The middle term of

3x −

1

8

y

has the coeffi-

3

cient 8C4.

8!

=

5 × 6 × 7 × 8

=

70.

4!4!

1 × 2 × 3 × 4

70(3x)4

−

1

y 4

=

3

1 4

70(34 )(x4 ) −

( y4 ) =

3

70x4 y4.

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PRACTICE TEST 1

231

DIAGNOSE YOUR STRENGTHS AND WEAKNESSES

Check the number of each question answered correctly and “X” the number of each question answered incorrectly.

Algebra	1	2	6	8	13	19	26	32	47		50		Total Number Correct
10 questions
Solid Geometry	38	41	Total Number Correct
2 questions
Coordinate	3	15	20	27	42	43	Total Number Correct
Geometry
6 questions
Trigonometry	7	12	16	31	36	46	48	Total Number Correct
7 questions
Functions	4	5	10	14	17	18	21	23	25		28		29	30	35			39		45	Total Number Correct
15 questions
Data Analysis,	9	11	22	37	Total Number Correct
Statistics,
and Probability
4 questions
Numbers	24	33	34	40	44	49	Total Number Correct
and Operations
6 questions

Number of correct answers − 14 (Number of incorrect answers) = Your raw score

___________________________ − 14 (_____________________________) = ________________

232	PART III / EIGHT PRACTICE TESTS

Compare your raw score with the approximate SAT Subject Test score below:

		SAT Subject Test
	Raw Score	Approximate Score
Excellent	43–50	770–800
Very Good	33–43	670–770
Good	27–33	620–670
Above Average	21–27	570–620
Average	11–21	500–570
Below Average	< 11	< 500

PRACTICE TEST 2

233

PRACTICE TEST 2

Treat this practice test as the actual test and complete it in one 60-minute sitting. Use the following answer sheet to fill in your multiple-choice answers. Once you have completed the practice test:

1.Check your answers using the Answer Key.

2.Review the Answers and Solutions.

3.Fill in the “Diagnose Your Strengths and Weaknesses” sheet and determine areas that require further preparation.