SAT SUBJECT TEST MATH LEVEL 1
CHAPTER 21 Model Test 2
50 Questions / 60 MINUTES
Directions: For each question, determine which of the answer choices is correct and fill in the oval on the answer sheet that corresponds to your choice.
1. You will need to use a scientific or graphing calculator to answer some of the questions.
2. Be sure your calculator is in degree mode.
3. Each figure on this test is drawn as accurately as possible unless it is specifically indicated that the figure has not been drawn to scale.
4. The domain of any function f is the set of all real numbers x for which f (x) is also a real number, unless the question indicates that the domain has been restricted in some way.
5. The box below contains five formulas that you may need to answer one or more of the questions.
Reference Information. This box contains formulas for the volumes of three solids and the areas of two of them.
For a sphere with radius r :
• A = 4πr2
For a right circular cone with radius r, circumference c, height h, and slant height l:
For a pyramid with base area B and height h:
1. If 7x – 3x = 7x + 3x – 24, then what is the value of x ?
2. If w = 2, then what is the value of (w – 3)(w2 – 3)?
3. If a 0, then which of the following is equivalent to
4. The figure above is a circle whose center is O. The sum of the lengths of the two darkened arcs is what fraction of the circumference of the circle?
5. If 3a + 2b = 2a + 3b, then b =
6. In the figure above, if P and R lie on line l, what is the value of a ?
7. What is the slope of the line whose equation is 3x + 4y = 24?
8. If , then g(–0.2) =
9. If x2 – y2 = x – y and x < y, what is the average (arithmetic mean) of x and y ?
10. If f (x) = 4x4 – 4, for what value of x is f (x) = 4?
11. On January 1, 2000, John had 2,000 baseball cards and Bob had 1,000 baseball cards. If John adds 150 cards per year to his collection and Bob adds 220 cards per year to his collection, what is the earliest year in which Bob will have more cards than John?
12. The length of a rectangle is three times its width, and the perimeter of the rectangle is three times the perimeter of a square whose area is 100. What is the area of the rectangle?
13. Which of the following numbers is a counterexample to the statement, “All numbers that are divisible by 2 and 4 are divisible by 8”?
14. What is the distance between the points whose coordinates are (–2, 5) and (4, –3)?
15. On a map of the United States, 0.3 inches represents 75 miles. On the map, how many inches are there between two cities that are actually 420 miles from one another?
16. In right triangle ABC in the figure above, if sin A = 0.6, what is sin B ?
17. On the final exam in Mrs. Johnson’s math class, the average (arithmetic mean) grade of the 25 students was 90. If the average grade of the 15 girls in the class was 94, what was the average grade of the 10 boys?
Note: Figure not drawn to scale.
18. If in the figure above, AB = BC = CD = AC, what is the measure of
Note: Figure not drawn to scale.
19. In the figure above, what is the value of ?
20. If which of the following numbers is NOT in the domain of f ?
21. What is the area of a triangle whose sides are 5, 5, and 6?
22. A and B have coordinates (–3, 1) and (6, 13), respectively. How many points on line are twice as far from A as from B ?
(E) More than 3
23. In the figure above, the length of each edge of the cube is 2. If P, Q, R, and S are the centers of faces ABFE, BCGF, DCGH, and ADHE, respectively, what is the perimeter of quadrilateral PQRS ?
24. Five actors are auditioning for the 3 parts available in a play. Each part will be performed by one of these 5 actors, and no actor will perform more than one part. In how many ways can the part assignments be made?
25. If x and y are positive integers and 7x + 11y = z, which of the following could be a factor of z ?
(B) I only
(C) III only
(D) I and III only
(E) I, II, and III
26. In similar triangles ABC and DEF, and are corresponding sides and AB = 5 and DE = 7. If the area of triangle ABC is 10, what is the area of triangle DEF ?
27. If 310041005100 = 220a1550b, then a + b =
28. How many four-digit numbers are there in which the tens digit is 8 and none of the digits is 9?
29. In a trapezoid, which of the following could be true?
I. The diagonals are congruent
II. Two adjacent sides are congruent
III. Two adjacent angles are congruent
(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II, and III
30. Which of the following is an equation of a line that has the same x-intercept as the line whose equation is y = 3x – 6?
(A) y = 3x – 4
(B) y = 2x – 6
(C) y = 6x – 3
(D) y = x + 2
(E) y = 2x – 4
31. In the diagram above, square ABCD is inscribed in a circle and a circle is inscribed in square ABCD. What is the ratio of the area of the outer circle to the area of the inner circle?
32. Which of the following is the equation of the circle whose center is at (2, –2) and whose radius is 2?
(A) (x – 2)2 + (y + 2)2 = 2
(B) (x – 2)2 + (y + 2)2 = 4
(C) (x + 2)2 – (y – 2)2 = 2
(D) (x + 2)2 + (y – 2)2 = 4
(E) (x – 2)2 – (y + 2)2 = 4
33. If x > 0 and , then x =
34. If , then sin2 + 3cos2 tan2 + 4cos2 =
(D) 3tan2 + 5
(E) 5cos2 + sin2
35. What is the value of log816?
Note: Figure not drawn to scale.
36. In the circle above, the ratio of the measure of arc to the measure of arc to the measure of arc is 3 to 4 to 5. What is the measure of angle P ?
37. If f (x) = 3x + 2 and g(x) = 2x + 3, then what is the y-intercept of the line whose equation is y = g (f (x))?
38. If b2 – 4ac = 13, how many times does the graph of y = ax 2 + bx + c intersect the x-axis?
(E) More than 3
39. For what value of k will the graphs of 3x + 4y + 5 = 0 and kx + 6y + 7 = 0 NOT intersect?
40. For which of the following equations is the sum of the roots equal to twice the product of the roots?
(A) x 2 – 5x + 10 = 0
(B) x 2 + 5x – 10 = 0
(C) x 2 + 10x + 5 = 0
(D) x 2 – 10x – 5 = 0
(E) x 2 – 10x + 5 = 0
41. If the length of a rectangle is three times its width, what is the sine of the angle that the diagonal makes with the longer side?
42. The figure above shows the graphs of y = g(x) and y = h(x). What is h(g(4)) – g(h(4))?
43. If in the figure above, O is the center of the circle, what is the area of sector OPR?
44. If logb2 = x and logb3 = y, what is logb72?
(A) 2x + 3y
(B) 3x + 2y
(E) x3y 2
45. A jar contains only red and blue marbles, 40 of which are red and 75 of which are blue. How many red marbles must be added to the jar so that when one marble is drawn from the jar, the probability of drawing a red one will be ?
46. Spheres 1 and 2 have radii of 3 and 9, respectively. If the surface area of sphere 3 is the average (arithmetic mean) of the surface areas of spheres 1 and 2, what is the radius of sphere 3?
47. While speaking to Jane, Adam made the following true statement, “If I win the lottery, I will buy you a car.” Which of the following statements could be true?
I. Adam won the lottery and did not buy Jane a car.
II. Adam did not win the lottery and bought Jane a car.
III. Adam won the lottery or Adam bought Jane a car.
(B) III only
(C) I and III only
(D) II and III only
(E) I, II, and III
48. If b and c are real numbers and 1 + i is a root of the equation x 2 + bx + c = 0, what is the value of b + c ?
49. If the volume of a sphere is equal to the volume of a cube, what is the ratio of the edge of the cube to the radius of the sphere?
50. The graph of y = g (x) shown above is the result of shifting the graph of y = f (x) 3 units to the right and 2 units up. If the equation of g (x) is g (x) = x 2 – 2x + 2, what is the equation of f (x)?
(A) f (x) = x 2 + 4x + 3
(B) f (x) = x 2 + 4x + 15
(C) f (x) = x 2 + 6x + 19
(D) f (x) = x 2 – 8x + 15
(E) f (x) = x 2 – 8x + 19
MODEL TEST 2
For many of the questions in this model test, an alternative solution, indicated by two asterisks (**), follows the first solution. Almost always the first solution is the direct mathematical one and the other is based on one of the tactics discussed in the “Tactics” chapter.
See Important Tactics for an explanation of the symbol ⇒, which is used in several answer explanations.
1. (E) Combine like terms: 4x = 10x – 24
Subtract 4x from and add 24 to each side: 24 = 6x
Divide both sides by 6: x = 4
**You could use TACTIC 2 and backsolve. However, on a problem this easy, you should not.
2. (C) Do not FOIL. Just replace each w by 2:
(w – 3)(w2 – 3) = (2 – 3)(22 – 3) = (–1)(1) = –1.
3. (A) To divide by a fraction, multiply by its reciprocal:
**Use TACTIC 3: Let a = 2. Then:
Only Choice A is equal to 12 when a = 2.
4. (C) Since mAOB + mAOC = 180°, the measures of central angles AOB and COD are each 40°, as are the measures of their intercepted arcs. Therefore, arcs AB and CD are each of the circle. The sum of their lengths is of the circle.
**If you get stuck on a question like this, use TACTIC 5: since the diagram is drawn to scale, you can trust it. Clearly, and are too small, and is too big. Eliminate choices A, B, and E. Since and are very close ( = 0.22 and = 0.25), you would have to guess between C and D.
5. (D) Subtract 2a and 2b from each side of the given equation:
**Since this question has two variables but only one equation, there is one extra variable. By TACTIC 3, you can plug in any number for either variable. Say, a = 2. Then:
3(2) + 2b = 2(2) + 3b 6 + 2b = 4 + 3b b = 2. So b = a.
6. (C) Since QRS is an exterior angle of PQR, its measure is the sum of the two opposite interior angles (see KEY FACT H2). So 150 = a + 2a = 3a a = 50.
**In the diagram above, 150 + b = 180 b = 30. Then
30 + a + 2a = 180 3a + 30 = 180 3a = 150 a = 50.
7. (B) Rewrite the given equation in y = mx + b form.
So the slope, m, is – (see KEY FACT L8).
**Find two points on the given line and use the slope formula. For example, when x = 0, y = 6, and when y = 0, x = 8. Therefore, (0, 6) and (8, 0) are points on the line, and the slope of the line is .
8. (D) Just replace x by –0.2:
Then enter this on your calculator, being sure to put parentheses around the entire numerator:
(100 (–0.2) + 12) ÷ (–0.2)3 = 1,000.
9. (B) Since x 2 – y 2 = (x + y)(x – y), if x 2 – y 2 = x – y, then x + y = 1. So the average of x and y is . Note that it is irrelevant that x < y.
**x = 0, y = 1 is a fairly obvious solution, in which case . This is not the only solution. In fact, there are infinitely many solutions. For example, if x = –9 and y = 10, x 2 – y 2 = 81 – 100 = –19 and x – y = –9 – 10 = –19. The average of x and y, however, is always the same.
10. (C) f (x) = 4x4 – 4 = 4 4x4 = 8 x4 = 2 x = 1.189.
**Use TACTIC 2: backsolve. Try choice C. f (1.19) = 4(1.19)4 = 4.02. That is so close to 4 that it must be the answer. The small difference is due to rounding in the answer choices. The real value of x is closer to 1.1892, and 4(1.1892)4 – 4 = 3.9998.
11. (C) In x years, John will have 2,000 + 150x cards and Bob will have 1,000 + 220x cards. Solve the inequality:
1,000 + 220x > 2,000 + 150x 70x > 1,000 x > 14.28.
So the smallest integer value for x is 15. Note that at the end of 2014, after 14 years, John will have 2,000 + 14(150) = 4,100 cards and Bob will have 1,000 + 14(220) = 4,080 cards. John still has more. In the following year, 2015, Bob will finally have more cards than John.
12. (E) Draw and label a diagram.
Since the area of the square is 100, each side is 10 and the perimeter is 40. Therefore, the perimeter of the rectangle, which equals 8w, is 3 × 40 = 120. So 8w = 120 w = 15 and 3w = 45. Finally, the area of the rectangle is 15 × 45 = 675.
13. (D) Only 12, choice D, is divisible by both 2 and 4 but not by 8. Choices A and C are not divisible by 4, and choices B and E are divisible by 8.
14. (E) Use the distance formula (KEY FACT L3):
**Draw a diagram, and use the Pythagorean theorem.
d 2 = 62 + 82 = 36 + 64 = 100.
So d = 10.
15. (A) All scale drawings are direct proportions. Set up the ratio and cross multiply:
16. (E) Since sin and since sin let
BC = 3 and AB = 5. Then AC = 4 (ABC is a 3-4-5 right triangle), and sin B = = 0.80.
**By the Pythagorean identity (KEY FACT M2)
sin2A + cos2A = 1 (0.6)2 + cos2A = 1 0.36 + cos2A = 1
cos2A = 0.64 cos A = 0.8
Since cos A = sin B, sin B = 0.8.
17. (A) The 25 students earned a total of 25 × 90 = 2,250 points. Since the 15 girls earned a total of 15 × 94 = 1,410 points, the 10 boys earned 2,250 – 1,410 = 840 points. So the average grade of the 10 boys was = 84.
**Set this up as a weighted average (see KEY FACT O2). Let x be the average grade of the boys. Then:
18. (D) Since AB = BC = AC, ACB is equilateral, and so the measure of each angle is 60°. Since mBCA + mACD = 180°, mACD = 120°. Since AC = CD, ACD is isosceles. Therefore, the congruent base angles each measure 30°. Finally, mADC + mCDE = 180°, and so mCDE = 150°.
19. (B) In ABC, = tan 28° = 0.53.
** sin 28° = b =11 sin 28° = 5.164.
cos 28° = a =11cos 28° = 9.712.
20. (E) The domain of a function that is an algebraic fraction consists of all real numbers except those for which the denominator equals 0.
3 – 2x = 0 3 = 2x x = 1.5
**Use TACTIC 2: backsolve. Use your calculator to evaluate each choice until you get an error message. Be sure to enter the numbers correctly, putting the entire numerator and the entire denominator in parentheses.
When x = 1.5, (1.52 – 1) ÷ (3 – 2(1.5)) gives you an error message; none of the other choices do.
**On a graphing calculator, enter . Then go to TABLE and test the five choices.
When 1.5 is in the x-column, the y-column has ERR.
21. (B) In isosceles triangle ABC, height bisects base , so AD = 3. You should immediately see that ABD is a 3-4-5 right triangle, so BD = 4. Of course, if you do not notice that, you can use the Pythagorean theorem to find BD. So the area of ABC = (6)(4) = 12.
**If you happen to know Heron’s formula for the area of a triangle, ,where a, b, and c are the lengths of the three sides and s is one-half the perimeter, you can use it here. Since
22. (C) The coordinates are actually irrelevant. If A and B are any two points on a line, there are exactly 2 points P, Q that are twice as far from A as from B.
Assume AB = d. If P is d from B on the opposite side of B as A, then BP = d and AP = 2d. So AP = 2(BP). If Q is between A and B and if AQ = d and QB = d, then AQ = 2(BQ). In the actual example, P is (15, 25) and Q is (3, 9).
23. (B) If you draw segments and perpendicular to the top edges of the cube, then it should be clear that quadrilaterals PQRS and WXYZ are congruent.
XCY is an isosceles right triangle whose legs are 1 and whose hypotenuse, , is . Similarly, each side of quadrilateral XYZW is , and so the perimeter is 4 .
24. (D) The easiest way to answer this question is to use the counting principle (see KEY FACT O2). The director can choose any of the 5 actors to play the first role, any of the 4 remaining actors for the second role, and any of the 3 remaining actors for the third role. Therefore, there are 5 × 4 × 3 = 60 different ways to make the assignment.
**You can always systematically list the possibilities until you see a pattern. Assume the actors are A, B, C, D, and E, and list the possibilities in alphabetical order.
ABC ACB ADB AEB
ABD ACD ADC AEC
ABE ACE ADE AED
These are the 12 possible ways to assign the actors if the first role is performed by actor A. Similarly, there are 12 ways to assign the actors if the first role is played by B, and so on. Each of the five actors could be assigned the first role, and so there are 5 × 12 = 60 possible assignments.
25. (E) Any positive integer, n, could be a factor of z. Simply let x = y = n. Then z = 7n + 11n = 18n, which is clearly divisible by n. Statements I, II, and III are all true.
**By trial and error or systematic reasoning, you can find values of x and y for each choice. For example,
26. (E) By KEY FACT H15, since the ratio of similitude is , the ratio of the areas of the two triangles is . So if x represents the area of triangle
**Sketch and label the two triangles.
Since the area of ABC is 10, we have 10 = (5)h h = 4. Since the heights of similar triangles are in the same ratio as their sides, the height of DEF is = 5.6. So the area of DEF is (7)(5.6) = 19.6.
27. (B) By the laws of exponents (KEY FACT A11), 310041005100 = (3 × 4 × 5)100 = 60100 = 4100 × 15100 = (22)10015100 = 2200 × 15100.
So 20a = 200 a = 10 and 50b = 100 b = 2. Therefore, a + b = 12.
28. (D) Use the counting principle (see KEY FACT O2). There are 8 choices for the thousands digit (any of the ten digits except 0 and 9). There are 9 choices each for the hundreds digit and for the ones digit (any digit except 9). Of course, there is only 1 choice for the tens digit (it must be 8). Therefore, there are 8 × 9 × 1 × 9 = 648 numbers that satisfy the given condition.
Number of choices
29. (E) Read carefully. None of these statements must be true, but the question asks which statement could be true. In any isosceles trapezoid, like ABCD, the diagonals are congruent (statement I is true).
In trapezoid RQST, and (statements II and III are true). So statements I, II, and III are true.
30. (E) The x-intercepts of any graph are the points on the graph whose y-coordinates are 0 (see KEY FACT L1). To find where the line y = 3x – 6 crosses the x-axis, let y = 0:
0 = 3x – 6 3x = 6 x = 2.
Of the five choices, only in E, y = 2x – 4, is x = 2 when y = 0.
**Graph y = 3x – 6 and each of the five choices on your calculator. Only y = 2x – 4 crosses the x-axis at the same point that y = 3x – 6 does.
31. (B) Pick a value for the side of the square, say 2. Then , the diagonal of the square, is 2 . However, is also a diameter of the larger circle. So the radius of that circle is and the area is . Diameter of the smaller circle is the same length as side of the square. So the smaller circle has a diameter of 2 and, therefore, a radius of 1. The area of the small circle is Finally, the ratio of the areas is .
**If you get stuck on a question like this, do not leave it out. Trust the diagram, and make your best estimate. At the very least, eliminate absurd choices. For example, the ratio of the area of the larger circle to the area of the smaller circle must be a fraction greater than 1. However, choices D and E are less than 1.
32. (B) By KEY FACT L10, the equation of a circle whose center is at (h, k) and whose radius is r is (x – h)2 + (y – k)2 = r 2. So the equation is (x – 2)2 + (y – (–2))2 = 22 or (x – 2)2 + (y + 2)2 = 4.
**If you forgot the formula, you could have made a sketch and tested one or two points. Since (2, 0) is a point on the circle, x = 2, y = 0 must satisfy the equation. Only choice B is true when x = 2 and y = 0.
33. (C) Rewrite 4 as 22 and 8 as 23, and then use the laws of exponents (KEY FACT A11).
So, x 2 + 2x + 3 = 3x + 9 x 2 – x – 6 = 0 (x – 3)(x + 2) = 0
Therefore, x = 3 or x = –2. Since it is given that x > 0, x = 3.
**Use TACTIC 2: backsolve starting with choice C. Replace x by 3 and use your calculator to verify that (28)(45) = 86.
34. (C) Simplifying the given expression requires the use of the only two identities you need to know (KEY FACT M1 and KEY FACT M2):
So sin2 + 3cos2 tan2 + 4cos2 = sin2 + 3sin2 + 4cos2 = 4sin2 + 4cos2.
Now factor out a 4: 4(sin2 + cos2) = 4(1) = 4.
**Plug in any number for and carefully enter the entire expression in your calculator, making sure to use parentheses properly. For example, if = 30:
(sin(30))2 + 3(cos(30))2(tan(30))2 + 4(cos(30))2 = 4.
35. (C) Remember that a logarithm is an exponent (see KEY FACT A15). log816 is the exponent, x, to which base 8 is raised to equal 16:
log816 = x 8x = 16
Since 8 = 23 and 16 = 24, we have 24 = 8x= (23)x = 23x. So 4 = 3x x = .
**If in the above solution you get that 8x= 16 but cannot solve the equation, you can estimate. Since 81 = 8 and 82 = 64, if 8x= 16, x must be between 1 and 2. Only choice C is between 1 and 2.
**If you know the change of base formula for logarithms (KEY FACT A16), you can evaluate this on your calculator:
You can even evaluate this without a calculator:
36. (C) Let the degree of arcs and be 3x, 4x, and 5x, respectively. Since the sum of the measures of the three arcs is 360°:
3x + 4x + 5x = 360° 12x = 360° x = 30°.
So the measure of arc is 4 × 30 = 120°. By KEY FACT J10, the measure of angle P is one-half the measure of arc
37. (D) Remember that g(x) = 2x + 3 means that g (anything) = 2(that thing) + 3. Then y = g (f (x)) = 2(f (x)) + 3 = 2(3x + 2) + 3 = (6x + 4) + 3 = 6x + 7.
y = 6x + 7 is the equation of a line whose slope is 6 and whose y-intercept is 7.
**Let x = 0, then f (0) = 3(0) + 2 = 2, and g (2) = 2(2) + 3 = 7. So (0, 7) is a point on the line whose equation is y = g (f (x)). In fact, it is the y-intercept.
38. (C) By KEY FACT E2, if the discriminant b2 – 4ac is positive, the quadratic equation ax 2 + bx + c = 0 has two different roots, and so the graph of y = ax2 + bx + c crosses the x-axis two times.
39. (C) The two given equations are each the equation of a line. If two lines do not intersect, they are parallel. So by KEY FACT L7, they have equal slopes. By solving for y to put 3x + 4y + 5 = 0 into slope-intercept form, we get . So the slope is – . Similarly, by rewriting kx + 6y + 7 = 0 as , we see that its slope is – . So
40. (E) By KEY FACT E3, if ax2 + bx + c = 0, the sum of the roots is and the product of the roots is . In choice E, the product of the roots is and the sum of the roots is In none of the other choices is the sum of the roots twice the product of the roots.
41. (A) Of course, you should draw a diagram and label it. You could label the sides w and 3w, but it is even easier to label them 1 and 3. Now you have two choices:
(1) Since tan = , = tan–1 = 18.430. Then sin = sin(18.430) = 0.316.
(2) Use the Pythagorean theorem to find AC.
42. (A) g(4) is the y-component of the point on the graph of y = g(x) where x = 4. Since the point (4, 3) is on the graph of y = g(x), g (4) = 3. So h (g (4)) = h(3) = –2. Similarly, since (4, –3) is a point on the graph of y = h(x), h(4) = –3. Then g (h (4)) = g (–3) = 1. Finally, h (g (4)) – g (h (4)) = –2 – 1 = –3.
43. (D) The area of circle O is . Since sector OPR is of the circle, its area is
44. (B) By using the laws of logarithms (KEY FACT A18), we get logb72 = logb(8)(9) = logb8 + logb9 = logb23 + logb32 = 3logb2 + 2logb3 = 3x + 2y.
**Let b = 10 and use your calculator. log 2 = 0.301, log 3 = 0.477, and log 72 = 1.857. Which of the answer choices is 1.857 when x = 0.301 and y = 0.477? Only choice B.
45. (C) If after adding the red marbles, the probability of drawing a red one is , the probability of drawing a blue one will be . This means that of the marbles are blue. Since there are 75 blue marbles in the jar, the jar must contain 4 × 75 = 300 marbles. So 300 – 75 = 225 of the marbles are red. Since there were originally 40 red marbles, 225 – 40 = 185 red marbles had to be added to the jar.
46. (C) By KEY FACT K8, the formula for the surface area of a sphere is . (Remember that this is one of the five formulas given to you on the test.)
S1 = 4 (3)2 = 4 (9) = 36 .
S2 = 4 (9)2 = 4 (81) = 324 .
So the average of the surface areas of spheres 1 and 2 is . Then the surface area of sphere 3 is 180 , and so Finally,
47. (D) Statement I contradicts the given true statement that if Adam wins the lottery, he will buy Jane a car. So I is false. Neither statement II nor statement III must be true, but each of them could be true. The original statement did not say what would happen if Adam did not win the lottery, only what would happen if he did win it. Even without winning the lottery, Adam might choose to buy Jane a car. Only statements II and III are true.
48. (C) If 1 + i is a root of a quadratic equation with real coefficients, then so is its conjugate, 1 – i. So the sum of the roots is (1 + i ) + (1 – i ) = 2 and the product of the roots is (1 + i )(1 – i ) = 1 – i 2 = 1 –(–1) = 2. By KEY FACT E3, the sum of the roots of ax 2 + bx + c = 0 is and the product of the roots is . Since a = 1, we have
**As above, if 1 + i is a root of x 2 + bx + c = 0, then so is 1 – i. Therefore, (x – (1 + i )) and (x – (1 – i )) are the factors of x 2 + bx + c.
(x – (1 + i )) (x – (1 – i )) = ((x – 1) – i ) ((x – 1) + i ) =
(x – 1)2 – i 2 = (x 2 – 2x + 1) – (–1) = x 2 – 2x + 2
So b = –2 and c = 2, and therefore b + c = 0.
49. (D) The formula for the volume of a sphere is (KEY FACT K8).
(Remember that this formula is given to you on the first page of the test.) The formula for the volume of a cube is V = e 3 (KEY FACT K1). So:
50. (A) If the graph of y = g(x) is the result of shifting the graph of y = f (x) 3 units right and 2 units up, shifting the graph of y = g (x) 2 units down and 3 units left will bring us back to the graph of y = f (x). So by KEY FACT N4:
f (x) = g(x + 3) – 2 =
((x + 3)2 – 2(x + 3) + 2) – 2 = ((x 2 + 6x + 9) – 2x – 6 + 2 – 2) = x 2 + 4x + 3.