﻿ DIAGNOSTIC TEST - SAT SUBJECT TEST MATH LEVEL 2 ﻿

## DIAGNOSTIC TEST

DIAGNOSTIC TEST

### Diagnostic Test

 The following directions are for the print book only. Since this is an e-Book, record all answers and self-evaluations separately.

The diagnostic test is designed to help you pinpoint your weaknesses and target areas for improvement. The answer explanations that follow the test are keyed to sections of the book.

To make the best use of this diagnostic test, set aside between 1 and 2 hours so you will be able to do the whole test at one sitting. Tear out the preceding answer sheet and indicate your answers in the appropriate spaces. Do the problems as if this were a regular testing session.

When finished, check your answers against the Answer Key at the end of the test. For those that you got wrong, note the sections containing the material that you must review. If you do not fully understand how to get a correct answer, you should review those sections also.

The Diagnostic Test questions contain a hyperlink to their Answer Explanations. Simply click on the question numbers to move back and forth between questions and answers.

Finally, fill out the self-evaluation on a separate sheet of paper in order to pinpoint the topics that gave you the most difficulty.

 50 questions: 1 hour Directions: Decide which answer choice is best. If the exact numerical value is not one of the answer choices, select the closest approximation. Fill in the oval on the answer sheet that corresponds to your choice. Notes: (1)   You will need to use a scientific or graphing calculator to answer some of the questions. (2)   You will have to decide whether to put your calculator in degree or radian mode for some problems. (3)   All figures that accompany problems are plane figures unless otherwise stated. Figures are drawn as accurately as possible to provide useful information for solving the problem, except when it is stated in a particular problem that the figure is not drawn to scale. (4)   Unless otherwise indicated, the domain of a function is the set of all real numbers for which the functional value is also a real number.

TIP

For the Diagnostic Test, practice exercises, and sample tests, an asterisk in the Answers and Explanations section indicates that a graphing calculator is necessary.

 Reference Information. The following formulas are provided for your information. Volume of a right circular cone with radius r and height h:  Lateral area of a right circular cone if the base has circumference C and slant height is l:  Volume of a sphere of radius r:  Surface area of a sphere of radius r: S = 4πr2 Volume of a pyramid of base area B and height h:

1.      A linear function, f, has a slope of –2. f(1) = 2 and f(2) = q. Find q.

(A) 0

(B)

(C)

(D) 3

(E) 4

2.      A function is said to be even if f(x) = f(–x). Which of the following is not an even function?

(A) = | |

(B) = sec x

(C) = log x2

(D) x2 + sin x

(E) = 3x4 – 2x2 + 17

3.      What is the radius of a sphere, with center at the origin, that passes through point (2,3,4)?

(A) 3

(B) 3.31

(C) 3.32

(D) 5.38

(E) 5.39

4.      If a point (x,y) is in the second quadrant, which of the following must be true?

I. y

II. > 0

III.

(A) only I

(B) only II

(C) only III

(D) only I and II

(E) only I and III

5.      If f(x) = x2 – ax, then f(a) =

(A) a

(B) a2 – a

(C) 0

(D) 1

(E) – 1

(A) 80

(B) 82

(C) 84

(D) 86

(E) 88

7.      log7 9 =

(A) 0.89

(B) 0.95

(C) 1.13

(D) 1.21

(E) 7.61

8.      If log2m = x and log2n = y, then mn =

(A) 2x+y

(B) 2xy

(C) 4xy

(D) 4x+y

(E) cannot be determined

9.      How many integers are there in the solution set of | – 2 | ≤ 5?

(A) 0

(B) 7

(C) 9

(D) 11

(E) an infinite number

10.       If , then f(x) can also be expressed as

(A) x

(B) –x

(C) ± x

(D) | |

(E) f (x) cannot be determined because is unknown.

11.       The graph of (x2 – 1)x2 – 4 has

(A) one horizontal and one vertical asymptote

(B) two vertical but no horizontal asymptotes

(C) one horizontal and two vertical asymptotes

(D) two horizontal and two vertical asymptotes

(E) neither a horizontal nor a vertical asymptote

12.

(A) –5

(B)

(C)

(D)  1

(E)  This expression is undefined.

13.       A linear function has an x-intercept of  and a y-intercept of . The graph of the function has a slope of

(A) –1.29

(B) –0.77

(C) 0.77

(D) 1.29

(E) 2.24

14.       If f(x) = 2– 1, find the value of that makes f(f(x)) = 9.

(A) 2

(B) 3

(C) 4

(D) 5

(E) 6

15.       The plane 2+ 3– 4= 5 intersects the x-axis at (a,0,0), the y-axis at (0,b,0), and the z-axis at (0,0,c). The value of a + b + c is

(A) 1

(B)

(C) 5

(D)

(E) 9

16.       Given the set of data 1, 1, 2, 2, 2, 3, 3, 4, which one of the following statements is true?

(A) mean ≤ median ≤ mode

(B) median ≤ mean ≤ mode

(C) median ≤ mode ≤ mean

(D) mode ≤ mean ≤ median

(E) The relationship cannot be determined because the median cannot be calculated.

17.       If , what is the value of  ?

(A)

(B) – 2

(C)

(D)

(E) 2

18.       Find all values of that make .

(A) 0

(B) ±1.43

(C) ±3

(D) ±4.47

(E) 5.34

19.       Suppose  for –4 ≤ ≤ 4, then the maximum value of the graph of | f (x) | is

(A) –8

(B) 0

(C) 2

(D) 4

(E) 8

20.       If tan , then sin  =

(A) ±0.55

(B) ±0.4

(C) 0.55

(D) 0.83

(E) 0.89

21.       If and are the domain of a function and f(b) < f(a), which of the following must be true?

(A) b

(B) a

(C) b

(D) b

(E) = 0 or = 0

22.       Which of the following is perpendicular to the line y = – 3x + 7 ?

(A)

(B) y = 7x – 3

(C)

(D)

(E) y = 3x – 7

23.       The statistics below provide a summary of IQ scores of 100 children.

Mean: 100
Median: 102
Standard Deviation: 10
First Quartile: 84
Third Quartile: 110
About 50 of the children in this sample have IQ scores that are

(A) less than 84

(B) less than 110

(C) between 84 and 110

(D) between 64 and 130

(E) more than 100

24.       If , then

(A) f(x) = f(−x)

(B)

(C) f(−x) = −f(x)

(D)

(E)

25.       The polar coordinates of a point P are (2,240°). The Cartesian (rectangular) coordinates of P are

(A)

(B)

(C)

(D)

(E)

26.       The height of a cone is equal to the radius of its base. The radius of a sphere is equal to the radius of the base of the cone. The ratio of the volume of the cone to the volume of the sphere is

(A)

(B)

(C)

(D)

(E)

27.       In how many distinguishable ways can the seven letters in the word MINIMUM be arranged, if all the letters are used each time?

(A) 7

(B) 42

(C) 420

(D) 840

(E) 5040

28.       Which of the following lines are asymptotes of the graph of ?

I. = 1

II. = –1

III. = 1

(A) I only

(B) II only

(C) III only

(D) I and II

(E) II and III

29.       What is the probability of getting at least three heads when flipping four coins?

(A)

(B)

(C)

(D)

(E)

30.       The positive zero of = 3x2 – 4– 5 is, to the nearest tenth, equal to

(A) 0.8

(B) 0.7 + 1.1i

(C) 0.7

(D) 2.1

(E) 2.2

31.       In the figure above, is the set of all points in the shaded region. Which of the following represents the set consisting of all points (2x,y), where (x,y) is a point in S?

(A)

(B)

(C)

(D)

(E)

32.       If a square prism is inscribed in a right circular cylinder of radius 3 and height 6, the volume inside the cylinder but outside the prism is

(A) 2.14

(B) 3.14

(C) 61.6

(D) 115.6

(E) 169.6

33.       What is the length of the major axis of the ellipse whose equation is 10x2 + 20y2 = 200?

(A) 3.16

(B) 4.47

(C) 6.32

(D) 8.94

(E) 14.14

34.       The fifth term of an arithmetic sequence is 26, and the eighth term is 41. What is the first term?

(A) 3

(B) 4

(C) 5

(D) 6

(E) 7

35.       What is the measure of one of the larger angles of the parallelogram that has vertices at (−2,−2), (0,1), (5,1), and (3,−2)?

(A) 117.2°

(B) 123.7°

(C) 124.9°

(D) 125.3°

(E) 131.0°

36.       If  for all nonzero real numbers, for what value of does f(f(x)) = x ?

(A) only 1

(B) only 0

(C) all real numbers

(D) all real numbers except 0

(E) no real numbers

37.        For what value(s) of is a continuous function?

(A) 1

(B) 2

(C) 3

(D) 6

(E) no value of k

38.       If f(x) = 2x2 – 4 and g(x) = 2x, the value of g(f(1)) is

(A) –4

(B) 0

(C)

(D) 1

(E) 4

39.       If , what is the value of –1(15)?

(A) 0.65

(B) 0.90

(C) 5.00

(D) 7.5

(E) 25.98

40.       Which of the following could be the equation of one cycle of the graph in the figure above?

I. = sin 4x

II.

III.

(A) only I

(B) only I and II

(C) only II and III

(D) only II

(E) I, II, and III

41.       If 2 sin2– 3 = 3 cos and 90° < < 270°, the number of values that satisfy the equation is

(A) 0

(B) 1

(C) 2

(D) 3

(E) 4

42.       If = tan–1 and = 315°, then =

(A) 278.13°

(B) 351.87°

(C) –8.13°

(D) 171.87°

(E) 233.13°

43.       Observers at locations due north and due south of a rocket launchpad sight a rocket at a height of 10 kilometers. Assume that the curvature of Earth is negligible and that the rocket’s trajectory at that time is perpendicular to the ground. How far apart are the two observers if their angles of elevation to the rocket are 80.5° and 68.0°?

(A) 0.85 km

(B) 4.27 km

(C) 5.71 km

(D) 20.92 km

(E) 84.50 km

44.       The vertex angle of an isosceles triangle is 35°. The length of the base is 10 centimeters. How many centimeters are in the perimeter?

(A) 16.6

(B) 17.4

(C) 20.2

(D) 43.3

(E) 44.9

45.       If the graph below represents the function f(x), which of the following could represent the equation of the inverse of ?

(A) y2 – 8– 1

(B) y2 + 11

(C) = (– 4)– 3

(D) = (+ 4)– 3

(E) = (+ 4)+ 3

46.       If > 4 is a constant, how would you translate the graph of x2 to get the graph of x2 + 4k?

(A) left 2 units and up units

(B) right 2 units and up (k– 4) units

(C) left 2 units and up (k– 4) units

(D) right 2 units and down (k– 4) units

(E) left 2 units and down (k– 4) units

47.       If f(x) = logb and f(2) = 0.231, the value of is

(A) 0.3

(B) 1.3

(C) 13.2

(D) 20.1

(E) 32.5

48.       If fn+1 = fn-1 + 2 fn for n = 2, 3, 4, . . . , and f1 = 1 and f2 = 1, then f5 =

(A) 7

(B) 11

(C) 17

(D) 21

(E) 41

49.       Suppose cos  = in . Then tan  =

(A) 1

(B)

(C)

(D)

(E)

50.       A certain component of an electronic device has a probability of 0.1 of failing. If there are 6 such components in a circuit, what is the probability that at least one fails?

(A) 0.60

(B) 0.47

(C) 0.167

(D) 0.000006

(E) 0.000001

DIAGNOSTIC TEST

 1. A                    2. D                    3. E                    4. E                    5. C                    6. D                    7. C                    8. A                    9. D                    10. D                    11. C                    12. C                    13. A                    14. B                    15. B                    16. C                    17. C 18. D                    19. E                    20. A                    21. D                    22. C                    23. C                    24. A                    25. A                    26. B                    27. C                    28. E                    29. C                    30. D                    31. C                    32. C                    33. D                    34. D 35. B                    36. D                    37. D                    38. C                    39. C                    40. E                    41. D                    42. B                    43. C                    44. D                    45. C                    46. C                    47. D                    48. C                    49. E                    50. B

The following explanations are keyed to the review portions of this book. The number in brackets after each explanation indicates the appropriate section in the Review of Major Topics (Part 2). If a problem can be solved using algebraic techniques alone, [algebra] appears after the explanation, and no reference is given for that problem in the Self-Evaluation Chart at the end of the test.

An asterisk appears next to those solutions for which a graphing calculator is necessary.

1. (A) f (1) = 2 means that the line goes through point (1,2). f(2) = means that the line goes through point (2,q). Slope  implies , so = 0. [1.2]

2. * (D) Even functions are symmetric about the y-axis. Graph each answer choice to see that Choice D is not symmetric about the y-axis.

An alternative solution is to use the fact that sin  sin(–x), from which you deduce the correct answer choice. [1.1]

 TIP  Properties of even and odd functions: Even + even is always an even function. Odd + odd is always an even function. Odd x even is always an odd function.

3. * (E) Since the radius of a sphere is the distance between the center, (0,0,0), and a point on the surface, (2,3,4), use the distance formula in three dimensions to get

Use your calculator to find . [2.2]

4. (E) A point in the second quadrant has a negative x-coordinate and a positive y-coordinate. Therefore, y, and  must be true, but can be less than or equal to zero. The correct answer is E. [1.1]

5. (C) f(a) means to replace in the formula with an a. Therefore, f(a) = a2· = 0. [1.1]

6. * (D) Since the average of your first three test grades is 78, each test grade could have been a 78. If represents your final test grade, the average of the four test grades is , which is to be equal to 80. Therefore, .

234 + = 320. So = 86. [4.1]

7. * (C) Use the change-of-base theorem and your calculator to get:
. [1.4]

8. (A) Add the two equations: log+ logn = x + y, which becomes logmn = x+y (basic property of logs). 2x + y = mn. [1.4]

9. * (D) Plot the graph of = abs(x – 2) – 5 in the standard window that includes both x-intercepts. You can count 11 integers between –3 and 7 if you include both endpoints.

The inequality |–2| ≤ 5 means that is less than or equal to 5 units away from 2. Therefore, –3 ≤ ≤ 7, and there are 11 integers in this interval. [1.6]

10. (D)  indicates the need for the positive square root of x2. Therefore,  if  and  if . This is just the definition of absolute value, and so  is the only answer for all values of x. [1.6]

11. * (C) Solve for y, and plot the graph of  in the standard window to observe two vertical and one horizontal asymptotes.

An alternative solution is to use the facts that  has vertical asymptotes when the denominator is zero, i.e., when = ±1, and a horizontal asymptote of = 1 as . [1.5]

12. * (C) Enter the given expression into Yand key in TBLSET with TblStart = 0 and  Observe Yapproach 0.5 as gets larger.

Divide the numerator and denominator of the expression by x2 and observe that the expression approaches  as . [1.5]

13. (A) y = mx + b. Use the x-intercept to get  and the y-intercept to get  . Therefore,  and . [1.2]

14. (B) f(f(x)) = 2(2x–1)–1 = 4–3. Solve 4–3 = 9 to get = 3. [1.1]

15. (B) Substituting the points into the equation gives , and . [2.2]

16. (C) Mode = 2, median  , mean . Thus, median ≤ mode ≤ mean. [4.1]

17. (C) Multiply  through by to get – 3= 7x. Subtract from both sides to get –3= 6x. Divide through by 6so that  will be on one side of the equals sign.

This gives . [algebra]

18. * (D) Enter the 3 by 3 matrix into the graphing calculator and evaluate its determinant as 0. The 2 by 2 matrix on the right side of the equation has the determinant x2– 20. Solve this for to get ±4.47. [3.3]

19. * (E) Plot the graph of = abs((1/2)x2–8) in a [–4,4] by [–10,10] window and observe that the maximum value is 8 (at = 0).

An alternative solution is to recognize that the graph of is a parabola that is symmetric about the y-axis and opens up. The minimum value = –8 occurs when = 0, so 8 is the maximum value of |f(x)|. [1.2]

20. * (A) Tan is positive in the first and third quadrants, so  is a first or third quadrant angle. The sine of a first quadrant angle is positive, but the sine of a third quadrant angle is negative. Press sin(tan–1(2/3))  0.55. The correct answer choice is therefore ±0.55. [1.3]

21. (D) If b, then f(a) = f(b). Since f(a f(b), it follows that b. [1.1]

22. (C) The slope of the given line is –3. Therefore, the slope of a perpendicular line is the negative reciprocal, or . [1.2]

23. (C) Fifty children is half of 100 children, and half of the data points lie between the first and third quartiles. [4.1]

24. * (A) Plot the graph of  and observe that it is symmetric about the y-axis. Hence f(x) = sec is an even function and f(x) = f(–x). An alternative solution is to recall that cos is even, so its reciprocal is also even. [1.1]

25. (A) From the figure . [2.1]

26. (B) Since r = h in the cone,

27. * (C) The word MINIMUM contains 7 letters, which can be permuted 7! ways. The 3 M’s can be permuted 3! ways, and the 2 I’s can be permuted in 2! ways, so only  permutations look different from each other. Therefore, there are  distinguishable ways the letters can be arranged. [3.1]

28. (E) Vertical asymptotes occur when the denominator is zero, so = –1 is the only vertical asymptote. Since  the asymptote is = 1. [1.5]

29. * (C) Since each of the 4 flips has 2 possible outcomes (heads or tails), there are 24 = 16 outcomes in the sample space. At least 3 heads means 3 or 4 heads.

ways to get 3 heads.  way to get 4 heads and .[4.2]

30. * (D) Use the quadratic formula program on your graphing calculator to get both zeros and choose the positive one. [1.2]

31. (C) Since the values remain the same but the values are doubled, the circle is stretched along the x-axis. [2.1]

32. * (C) Volume of cylinder . Volume of square prism = Bh, where is the area of the square base, which is  on a side. Thus, . Therefore, the desired volume is 54 – 108, which (using your calculator) is approximately 61.6. [2.2]

33. * (D) Divide both sides of the equation by 200 to write the equation in standard form . The length of the major axis is  [2.1]

34. (D) There are three constant differences between the fifth and eighth terms. Since 41 – 26 = 15, the constant difference is 5. The fifth term, 26, is four constant differences (20) more than the first term. Therefore, the first term is 26–20 = 6. [3.4]

35. * (B) The figure displayed for this question shows a parallelogram. The tangent of the acute angle with vertex (−2,−2) is , so that angle has the measure . The larger angle is the supplement, or 123.7°. [1.3]

36. (D). Since k is in the denominator, it cannot equal 0. [1.1]

37. * (D) Enter (3x2–3)/(x–1) into Yand key TBLSET. Set Indpnt to Ask and key TABLE. Enter values of progressively closer to 1 (e. a. .9, .99, .999, etc.) and observe that Ygets progressively closer to 6, so choose = 6.

An alternative solution is to factor the numerator to 3(x+1)(x–1), divide out x–1 from the numerator and denominator. As . [1.5]

38. (C) , and . [1.1]

39. (C) –1(15) is the value of that makes  equal to 15. Set , divide both sides by 3 to get . [1.1]

40. * (E) Plot the graphs of all three functions in a  by [–2,2] window and observe that they coincide.

An alternative solution is to deduce facts about the graphs from the equations. All three equations indicate graphs that have period  . The graph of equation I is a normal sine curve. The graph of equation II is a cosine curve with a phase shift right of , one-fourth of the period. Therefore, it fits a normal sine curve. The graph of equation III is a sine curve that has a phase shift left of , one-half the period, and reflected through the x-axis. This also fits a normal sine curve. [1.3]

41. * (D) With your calculator in degree mode, plot the graphs of 2(sin(x))– 3 and y = 3 cosx in a [90,270] by [–3,0] window and observe that the graphs intersect in 3 places.

An alternative solution is to distribute and transform the equation to read: 2 cos2x + 3 cos x + 1 = 0. This factors to (2 cos x + 1)(cos x + 1) = 0, so  or cos = –1. For , there are three solutions: x = 120°, 240°, 180°. [1.3]

42. * (B) With your calculator in degree mode, evaluate 315° – tan–1(–3/4) to get the correct answer choice. [1.3]

 TIP  Always look for places throughout your solution where division by zero might occur.

43. (C) The problem information is illustrated in the figure below.

Points A and B represent the two observers. Point C is the base of the altitude from the rocket to the ground. We know that  and . Therefore,  [1.3]

44. (D) Drop the altitude from the vertex to the base. The altitude bisects both the vertex angle and the base, cutting the triangle into two congruent right triangles. Since sin  and the perimeter = 43.3 cm. [1.3]

45. (C) The given graph looks like the left half of a parabola with vertex (4,–3) (using the values given in the answer choices as guides) that opens up. The equation of such a parabola is = (–4)– 3. The vertex of the inverse is (–3,4), so its equation is = (–4)– 3. [1.1]

46. (C) Complete the square of x2 + 4by adding and subtracting 4 to get the translated function of = (+ 2)+ (k – 4). Translate x2 left 2 units and up (–4) units. [2.1]

47. (D) f(2) = log2 = 0.231. Therefore, b0.231 = 2, and so = 21/0.231 , which (using your calculator) is approximately 20.1. [1.4]

48. (C) Let = 2, f= 3; then let = 3, f= 7; and finally let = 4, f= 17. [3.4]

49. (E) Since , the figure below shows  in Quadrant I with cos  = u.

From the figure, . [1.3]

50. * (B) The probability that at least one component fails is 1 minus the probability that all succeed. Since the probability of one component succeeding is 1 minus 0.1, or 0.9, the probability that all succeed is (0.9)= 0.53, and 1–0.53 = 0.47.

Self-Evaluation Chart for Diagnostic Test

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