SAT 8

366

41.The graph in Figure 1 could be a portion of the graph of which of the following functions?

I.f (x) = x3 + ax2 + bx1 + c

II.g(x) = −x3 + ax2 + bx1 + c

III.h(x) = −x5 + ax 4 + bx3 + cx2 + dx + e

(A)I only

(B)II only

(C)I and II only

(D)II and III only

(E)I, II, and III

42.If f (x) = ln e2x, then what is the smallest possible integer x such that f (x) > 10,000?

(A)2,501

(B)4,999

(C)5,000

(D)5,001

(E)10,001

43.If log2 (x − 1) = log4 (x − 9), then x =

(A)−2

(B)2

(C)5

(D)−2 or 5

(E)No solution

44.If [n] represents the greatest integer less than or equal

to n, then which of the following is the solution to 2[n] − 16 = 6?

(A)n = 11

(B)11 ≤ n < 12

(C)n = 12

(D)11 < n < 12

(E)11 < n ≤ 12

45.If a rectangular prism has faces with areas of 10, 12 and 30 units2, then what is its volume?

(A)30

(B)60

(C)90

(D)120

(E)150

PART III / EIGHT PRACTICE TESTS

USE THIS SPACE AS SCRATCH PAPER

y

4

2

x

–2

–4

Figure 1

GO ON TO THE NEXT PAGE

1. D Take either the log or natural log of both sides of the equation to solve for x.

2. A

1

(k + 12)3 = −3.

Cube each side of the equation to solve for k.

=n2.

4.B (− 4, 5) is the only given point that is both in

the second quadrant and at a distance of 41 from the origin.

(−4 − 0)2 + (5 − 0)2 =

42 + 52 =

16 + 25 =

41.

370

5. D

tan θ(sin θ) + cos θ =

sin θ (sin θ) + cos θ = cos θ

6. C The radicand must be greater than or equal to zero.

2 − x2 ≥ 0.

2 ≥ x2.

x ≤ − 2 or x ≥ 2.

7. D

6n3 = 48n2.

6n3 − 48n2 = 0.

6n2 (n − 8) = 0n = 0 or 8.

8. A Because you know the composition of f and g

results in 2 x , you need to determine what input

value of f will result in 2 x.

4x = 2 x.

Therefore, g(x) = 4x. Test your answer by checking the composition.

g(x) = 4x, so f ( g(x)) = f (4x) = 4x = 2 x.

9. C

10! = 10! . 3!(10 − 3)! 3!7!

=8 × 9 × 10

1 × 2 × 3 .

=120.

PRACTICE TEST 7

16. C The largest angle of a triangle is opposite its longest side. Let θ = the largest angle of the triangle. θ is opposite the side measuring 7 inches. Using the Law of Cosines:

72 = 32 + 52 − 2(3)(5) cos θ. 49 = 9 + 25 − 30 cos θ.

15 = −30 cos θ. cos θ = − 12 .

θ= cos−1− 12 ≈ 120°.

17.E Because the center of the circle must lie in the 2nd quadrant, its coordinates are (−3, 3). The radius of the circle is 32 = 9.

Therefore, the standard form of the equation of the circle is (x + 3)2 + (y − 3)2 = 9.

18. 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) = 0.02n2 − 42n + 5,000.

19. D For the three terms to be part of a geometric sequence there must be a common ratio between consecutive terms. Add n to each of the three terms to get the progression −2 + n, 10 + n, 94 + n. Now, set the common ratios equal to each other and solve for n.

20. C Think of A and B as vertices of a right triangle with one leg measuring 8 units (the diameter of the cylinder’s base) and one leg measuring 15 units (the height).

AB = 82 + 152 .

AB = 64 + 225.

AB = 289 = 17.

371

21. D Because the median is 12, the 3rd term when arranged from least to greatest is 12. The mode is 5, so the 2 integers less than 12 equal 5. Note that 5 is the only mode. Because the problem asks for the least possible range, assume the five integers are:

5, 5,12,13,14

The range equals 14 − 5, or 9.

22. D An equation with roots of 4 and − 12 has factors x − 4 and x + 12.

(x − 4) x + 12 = 0. (x − 4)(2x + 1) = 0.

2x2 + x − 8x − 4 = 0.

2x2 − 7x − 4 = 0.

23. E The maximum value of f(x) = 7 − (x − 6)2 is the y-coordinate of its vertex.

f (x) = 7 − (x − 6)2

y − 7 = −(x − 6)2

The vertex is (6, 7), so the maximum value is 7.

24.C

If sec θ < 0, then cos θ < 0. If cot θ > 0, then tan θ > 0.

The cosine is negative in both quadrants II and III. For the tangent to be positive, however, sin θ must also be a negative value. This occurs in quadrant III only.

25. C The integers from 1 to 100 form an arithmetic sequence having 100 terms. n = 100, a1 = 1, and an = 100. Substitute these values into the formula for the sum of a finite arithmetic sequence to get:

Sn = 1 + 2 + 3 + 4 + 5 + ... + 100.

Sn = 2n (a1 + an ).

Sn = 1002 (1 + 100).

Sn = 50(101) = 5,050.

372

26. A Complete the square to write the equation of the ellipse in standard form.

9x2 + 4 y2 − 36x − 12y + 18 = 0.

(9x2 − 36x) + (4 y2 − 12y) = −18.

9(x2 − 4x + 4) + 4 y2 − 3y + 94 = −18 + 36 + 9.

asymptotes at the zeroes of the denominator.

x2 − 1 = 0.

x = ±1.

Because the degree of the numerator is less than the degree of the denominator, a horizontal asymptote exists at y = 0.

Statements II and III are, therefore, true.

28. A The probability that all 4 tools are in error of 2% or more is:

0.08(0.08) (0.08)(0.08) ≈ 04.10 × 10−5.

29.B The double angle formula for cosine is: cos 2θ = cos2 θ − sin2 θ.

Since cos 2θ = 27 , cos2 θ − sin2 θ also equals 27.

PART III / EIGHT PRACTICE TESTS

30. B Factor the equation and solve for θ.

31. E Graph the function to see that it has a horizontal asymptote at y = 3. (Because the degree of the numerator equals the degree of the denominator, a

horizontal asymptote occurs at 3 , the ratio of the 1

coefficients of the x terms.)

Alternately, you can evaluate the function at a large value of x. For example, let x = 10,000:

The function approaches 3 as x gets infinitely large.

32. B Take the logarithm of both sides of the equation to solve for a .

33. D Because the rectangular coordinates (x, y) are (2, 5):

PRACTICE TEST 7

34. A Translating the graph of f (x) = x2 2 units down and 4 units left results in:

g(x) = (x + 4)2 − 2

Now, evaluate the function for x = −5.

g(−5) = (−5 + 4)2 − 2 = 1 − 2 = −1.

35. C

(12 sin x)(2 sin x) − (8 cos x)(−3 cos x) =

24 sin2 x + 24 cos2 x =

24(sin2 x + cos2 x) =

24(1) = 24.

36. E There are 36 possible outcomes when two dice are rolled. The following rolls result in a sum of 4 or 5:

{(2,2), (3,1), (1,3), (4,1), (1,4), (2,3), (3,2)}

Seven out of the 36 have a sum of 4 or 5, so the probability is 367 .

37. B Reflecting the graph of f (x) = 12x over the line

y = x results in the original function. The inverse function of f (x) = 12x is, therefore, f −1(x) = 12x .

f −1(−8) = 12−8 = −1.5.

38. A Recall that a function is even if it is symmetric with respect to the y-axis and f(−x) = f(x).

Because the cosine function is an even function, its reciprocal function, the secant function, is also even. Answer A is the correct answer choice.

39. C All of the functions have a period of 2π or π 2

with the exception of y = 12 cos 2π x. It has a period of

2π = 1. Answer C is the correct answer choice. 2π

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40. D There are 10 possible people that could serve as president. Once the president is chosen, there are 9 possible people that could serve as secretary, and once that person is chosen, there are 8 remaining people that could serve as treasurer. The total number of ways of selecting the three officers is:

10 × 9 × 8 = 720.

41.D The graph has three zeroes. Because it rises to the left and falls to the right, it represents an odddegree polynomial with a negative leading coefficient. Either statement II or III are possible answers.

42.D Because f (x) = ln e2x must be greater than 10,000:

2x > 10,000

x > 5,000

The smallest possible integer greater than 5,000 is 5,001.

43. E Use the change of base formula to rewrite the right side of the equation.

log2 (x − 1) = log4 (x − 9).

log2 (x − 1) = log2 (x − 9) . log2 4

log2 (x − 1) = log2 (x − 9) . 2

Now, solve for x.

2 log2 (x − 1) = log4 (x − 9).

(x − 1)2 = x − 9.

x2 − 2x + 1 = x − 9.

x2 − 3x + 10 = 0.

Solving for x using the quadratic formula results in no real solution.

44. B

2[n] − 16 = 6.

2[n] = 22.

[n] = 11.

11 is the greatest integer less than or equal to n, so n must be on the interval 11 ≤ n < 12.