SAT 8

394

6. A Think of sine either in terms of the opposite leg and hypotenuse of a right triangle or in terms of the

point (x, y) and r of a unit circle. Because π < θ < 32π ,

θ lies in quadrant III and its cosine is negative.

sin θ = − 419 = − ry .

Because r = x2 + y2 ,41 = x2 + (−9)2 .

x = 40.

cos θ = − xr = − 4041

7. C The line passes through the points (5, 0) and (0, −1). The slope of the line is m = 15 .

Because the y-intercept is given, you can easily write the equation in slope-intercept form.

y = 15 x − 1. x − 5y = 5.

8. C Recall that a quadratic equation can be thought of as: a[x2 − (sum of the roots)x + (product of the roots)] = 0. Substitute the sum = −4, and the product = −5 to get:

a(x2 − −4x + −5) = 0.

a(x2 + 4x − 5) = 0.

When a = 1, the result is the equation given in Answer C: x2 + 4x − 5 = 0.

9. C
		1				−1	)			1
f	−			=	9−( 4				= 9	4
		4
							1			2				1
				=	(32 )				= 3				= 3
								4			4			2
				=	3.

10.D

log4 x + 3log4 x = 9 4 log4 x = 9

log4 x = 94

4 9 = x

4

x = 22.6.

PART III / EIGHT PRACTICE TESTS

11. D Graph f(x) = x2 − 5 to determine its range on the specified interval. Because the domain is specified as −1 ≤ x ≤ 4, the curve has a beginning and an ending point.

When x = −1, y = 4, and when x = 4, y = 11. The range is the set of all possible y values, so realize that the y values decrease between 4 and 11. The range is 0 ≤ y ≤ 11.

12. B Recall that sin2 θ + cos2 θ = 1, so cos2 θ = 1 − sin2 θ.

(1 + sin θ)(1 − sin θ) =

1 + sin θ + sin θ − sin2 θ =

1 − sin2 θ =

cos2 θ.

13.E The minimum value of the function is the

y-coordinate of the parabola’s vertex. For the function f (x) = (x − 1)2 + 18, the vertex is (1, 18). (You can check this by graphing the parabola on your graphing calculator.) The minimum value is, therefore, 18.

14.C

6 C3 =	6!
	3!(6 − 3)!

=4 × 5 × 6

1 × 2 × 3

=20.

15.C The mean is the sum of the data divided by the number of terms.

(0 + 2 + 2 + 1 + 0 + 2 + 2 + 0 + 1 + 0 + 2 + 0 + 0 + 2) = 14

1414 = 1

16. C There are 4 possible points of intersection as shown:

PRACTICE TEST 8
		π
17. A	The graph of y = −2cos x +		is the graph of
		4

y = cos x with a phase shift of π units left, an amplitude

4

of 2, and reflected over the x-axis. The maximum value occurs at the point where x = π − π4 = 34π . The y-coordinate at that point is 2.

18. B Let s = the sum of the scores of Matt’s first four tests.

4s = 84. s = 336.

Matt’s new average is 336 + 94 = 86%. 5

19. E Use the formula s = r θ, where s = the arc length and r = the radius of the circle. Convert 80° to radian measure first.

	π		=	4π	radians.
80	180			9

Now, solve for the arc length:

	4π			16π
s = 4			=			cm.
				9
	9

20. E The denominator cannot equal zero.

x2 + 2 ≠ 0 x2 ≠ −2.

Recall that on the SAT Subject Test, unless otherwise stated, the domain of a function f is assumed to be the set of real numbers x for which f (x) is a real number. Because x is a squared term, it will, therefore, never equal a negative number. The domain is the set of all real numbers.

21. C This problem can be done quickly and with little work if you recall that the composition of a function and its inverse function, f −1[ f (x)] and f [ f −1(x)], equal x.

f −1[ f (2)] = 2.

395

22. B

(x2 + y2 ) = (2cos θ)2 + (2sin θ)2

=[4(cos2 θ + sin2 θ)].

Recognize that you can use one of the Pythagorean Identities, cos2 θ + sin2 θ = 1, to simplify the expression.

= [4(cos2 θ + sin2 θ)] = 4(1) = 2.

23. D The sum of the roots is: 7 + i + 7 − i = 14.

The product of the roots is: (7 + i)(7 − i) = 49 − i2 = 50.

The quadratic equation is, therefore, given by the equation:

x2	− (sum of the roots)x	= 0.
a
	+(product of the roots)
	a(x2 − 14x + 50)	= 0.

Setting a equal to 1 results in one possible answer:

x2 − 14x + 50 = 0.

24. B Because the circle is a unit circle, the coordinates of P are (cos 45°, sin 45°). This can be simplified

to		2	,	2
		2		2	.

If you don’t know what the cosine and sine of 45° equal, let (x, y) be the coordinates of P, and draw a right triangle with legs of length x and y. The triangle is a 45°− 45°−90° triangle, so use the ratios of the sides of this special right triangle to determine that the coordinates

of point P are		2	,	2
		2		2	.

25. D Use either synthetic or long division to divide x4 − 2x3 − 8x + 5 by x − 3. Remember to include a zero placeholder for the x2 term.

3 1 − 2 0 − 8 5

3 3 9 3

The remainder is 8.

396

26. A Use right triangle trigonometry to determine values for the three trigonometric functions.

sin A =

opposite

=

a

.

hypotenuse

c

tan B =

opposite

=

b

.

adjacent

a

sec A =

1

=

hypotenuse

=

c

.

cos A

adjacent

b

a b

sin A tan B

=

c a

.

sec A

c

b

b

b2

.

=

c

=

c

c2

b

27. C Recognize that the heights of the bouncing ball form a geometric sequence with a common ratio of

2 and an initial term of 8. After hitting the ground 3

for the first time, the ball will reach a height of

		2
	(8)			= 5.33. After the second bounce, the ball will
		3
							2
					2
	reach a height of (8)						= 3.56. After the third bounce,
					3

	2		3
the ball will reach a height of (8)			= 2.37 feet.
	3

28. C Solve the inequality for y to get y > 72 x. Then, graph the linear equation y = 52 x. The solution to the

inequality is the shaded area above the line, and that region falls in quadrants I, II, and III.

29. E Because θ is an acute angle, think of the right triangle that contains one angle of measure θ.

	cot θ = 5 =		adjacent		.
			opposite
	Solve for the hypotenuse: x =							(52 + 12 ) = 26.
	sin θ =	opposite		=		1	=	26	.
		hypotenuse				26		26

30.C

G = 0.03m + 0.2.

G = 0.03(100) + 0.2.

G = 3.2.

				PART III / EIGHT PRACTICE TESTS
31.	B
			n!	= (n − 1)!
	2
		n!		= 2.
		(n − 1)!
			n = 2.
32.	D The Law of Cosines states: c2 = a2 + b2 − 2ab
cos C.

42 = 52 + 82 − 2(5)(8) cos C, where C is the angle opposite the shortest side.

	16		= 89 − 80 cos C.
	−73		= −80 cos C.
	73		= 24.1º
cos−1
	80

33. C The length of the major axis equals 2a. In this

problem, a =	9.
2a =	6.

34. A Take the log of both sides of the equation to solve for k.

log(4k )			= log(5k + 3 ).
	k log 4		= (k + 3)log 5.
log 4			= k + 3.
k
log 5
k(0.86135)			= k + 3.
		k	= −21.6.

35. D

317n = (3 17 )(3 n ) = (3 17 )(7.128)

=18.3.

36.B Filling the cone-shaped cup with water creates

a cone similar to the cup itself. The radii and heights of the two cones are proportional. Let r = the radius of the surface of the water.

106 = 3r .

18 = 10r.

1.8 = r.

PRACTICE TEST 8

37.E The critical points of the inequality x(x − 4) (x − 2) > 0 are x = 0, 4, and 2. Evaluate the 4 intervals created by these points by determining if the inequality is satisfied on each interval. 0 < x < 2 or x > 4 is the correct answer choice.

38.C The volume a rectangular prism is given by the formula V = × w × h, so you need to find three integers whose product is 18. There are four possibilities:

1 × 1 × 18

1 × 2 × 9

1 × 3 × 6

2 × 3 × 3

				r nt
	39. D	A = P 1	+				where n is the number of times
				n
	the investment is compounded per year.
				0.06			4(4)
	A = 2,200 1 +						.
				4

A= 2,200(1.015)16.

A≈ 2,792.

40.E Answer A equals one and Answer B is less

than one, so both can be eliminated. Because C and E have the same denominator, and a < a + 1, C will always be less than E. It can also be eliminated as a possible answer choice. Substitute a few values of a into answers D and E to compare the expressions.

If a = 7, 65 < 86 .

If a = 10, 98 < 119 .

Answer E will always result in a greater value.

41. D Because the expressions represent the terms of an arithmetic sequence, there must be a common difference between consecutive terms.

6n − (4n + 1) = 7n + 2 − 6n.

2n − 1 = n + 2.

n = 3.

The first three terms are, therefore, 13, 18, and 23, making the common difference between terms, d, equal 5. The first term of the sequence is a1 = 13, and the 20th term is a20 = a1 + (n − 1)d = 13 + (20 − 1)(5) = 108.

397

The sum of a finite arithmetic sequence is:

Sn = n2 (a1 + an ), where n = the number of terms.

Sn = 202 (13 + 108) = 1,210.

42. B Because f (x) is a third-degree function, it can have, at most, three zeroes.

x3 − 2x2 − 8x = 0. x(x2 − 2x − 8) = 0. x(x − 4)(x + 2) = 0.

x = 0, x = 4, and x = −2.

43.A The function f (x) = −3 sin(4x + π) + 1 has an amplitude of a = −3 = 3 and a vertical shift of 1 unit up. The range spans from y = 1 − 3 = −2 to y = 1 + 3 = 4, so −2 ≤ y ≤ 6 is the correct answer choice.

44.E Factor the numerator and denominator. Then, simplify the expression and evaluate it when x = −2.

	f (x) =	x2 + 7x + 10				=		(x + 5)(x + 2)
		2x2 + 3x − 2					(2x − 1)(x + 2)
	=	(x + 5)
		(2x − 1)
	When x = −2,		(x + 5)			= −		3	.
			(2x − 1)					5

45. B The figure shows the graph of y = tan x shifted 2 units down with a period of 2π. The correct equation

is		x	− 2.
	y = tan
		2

46. A Substitute k = 0, 1, 2, . . . 7 into the summation to get:

1 − 2 + 4 − 8 + 16 − 32 + 64 − 128 = −85.

47. C Because x = t − 12, t = x + 12. Substitute this value into the second equation to get:

y= 4(x + 12) − 1.

y= 4x + 48 − 1.

y= 4x + 47.

The y-intercept of the resulting line is (0, 47).

398	PART III / EIGHT PRACTICE TESTS

48.C

Area = 12 (base height)

A =

1

(6a).

2

Use trigonometry to determine a:

tan 21º =

a

.

6

a ≈ 2.303.

A =

1

(6)(2.303) ≈ 6.9.

2

49.

D Find the number of permutations of six letters

taken six at a time, 2 of which are repeated.

6!

=

6 5 4 3 2 1

= 360.

2!

2 1

50.

B

9 C2 (11C2 )

=

20 C4

9! 11!

2!7! 2!9!

=

20!

4!16!

1,980

=

132

.

4,845

323

PRACTICE TEST 8

399

DIAGNOSE YOUR STRENGTHS AND WEAKNESSES

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

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

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

___________________________ − 14 (_____________________________) = ________________

400	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