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• Question #1: If f(x) = (2x + 5)/(4x-7), what value does f(x) approach as x gets infinitely larger?

  (a) 1/2

  (b) 5/7

  (c) 2/7

  (d) 5/4

  (e) 1

  • Answer: As x gets infinitely larger, lim[(2x + 5)/(4x-7)] = lim[(2 + 5/x) / (4 - 7/x)] = (2 + 0) / (4 - 0) = 2/4 = 1/2

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• Question #2: In the figure below, what is the length of the arc AB, if O is the center of the circle and triangle OAB is equilateral? The radius of the circle is 9.

  (a) ¶

  (b) 2 · ¶

  (c) 3 · ¶

  (d) 4 · ¶

  (e) ¶/2

  • Answer: OAB equilateral means the AOB angle is 60^o The ratio between the AOB angle and 360^o is equal to the ratio between the length of the arc AB and the circumference of the circle 60^o/360^o = arc_AB / (2 ¶ · 9) arc_AB = 3 · ¶

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• Question #3: In the figure below, quadrilateral ABCD has AB parallel with CD. What is the area of triangle ABD? (a) 3 (b) 4 (c) 6 (d) 8 (e) 9

  • Answer: The area of triangle ABD is equal to (1/2)AB·DE, where DE is the altitude from D to AB. Area_ABD = (1/2)4·3 = 6.

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• Question #4: What is the distance in space between the points with coordinates (-1, 5, 3) and (2, 6, -1)?

  (a) √20

  (b) √24

  (c) 5

  (d) √26

  (e) 7

  • Answer: The coordinates of the 2 points are: x₁ = -1, y₁ = 5, z₁ = 3, x₂ = 2, y₂ = 6, z₂ = -1, The distance in space between the points with coordinates (-1, 5, 3) and (2, 6, -1) squared is: d² = [(x₁ - x₂)² + (y₁ - y₂)² + (z₁ - z₂)² ]= [ (-1 + (-2))² + (5 - 6)² + (3 - (-1)) ² ] = 9 + 1 + 16 = 26 d = √26

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• Question #5: In the standard (x, y) coordinate plane, the graph of (x – 2)² + (y + 2)² = 16 is a circle. What is the area enclosed by this circle, expressed in square coordinate units?

  (a) 4 · ¶

  (b) 8 · ¶

  (c) 16 · ¶

  (d) 18 · ¶

  (e) 20 · ¶

  • Answer: We compare the equation of the circle, (x – 2)² + (y + 2)² = 16 with the general equation of a circle: (x – x₁)² + (y - y₁)² = R², where R is the radius of the circle. We notice that the radius of our circle is √16 = 4 The area of the circle is ¶ · 4² = 16 · ¶

Question #6: |3 - 5| - |2 - 1| + |-5| =

• (a) 5

  (b) 6

  (c) 7

  (d) 9

  (d) 4

  • Answer: |3 - 5| = |-2| = 2 |2 - 1| = |1| = 1 |-5| = 5 |3 - 5| - |2 - 1| + |-5| = 2 - 1 + 5 = 6

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• Question #7: If a, b and c are the sides of any triangle, which of the following inequalities is not true?

  (a) a·b > 0

  (b) a + b > c

  (c) a + c/2 >b

  (d) b + c > a

  (e) (a + b)·(b + c) > a·c

  • Answer: The first answer is true, since the product of 2 positive reals will be positive. The second and the fourth answers will also be true, since the sum of 2 sides of a triangle is always higher than the third side. The fifth answers is also true because it is just a multiplication of the second and fourth inequalities of positive terms. Answer three should be the one that is not true, and we can verify this result with an example: an isosceles triangle with a = 3, c = 3, b= 10 will satisfy the inequality.

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• Question #8: For any x such that 0 < x < ¶/2, the expression (1 - sin²x)/cos(x) + (1 - cos²x)/sin(x) is equivalent to:

  (a) sin(x) (b) cos(x) (c) sin(x) - cos(x) (d) sin(x) + cos(x) (e) 2sin(x)

  • Answer: (1 - sin²x)/cos(x) + (1 - cos²x)/sin(x) = cos²x/cos(x) + sin²x/sin(x) = sin(x) + cos(x).

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• Question #9: A line has parametric equations x = t + 3 and y = t + 9. The slope of the line is:

  (a) 3

  (b) 9

  (c) 1

  (d) 4

  (e) 5

  • Answer: From the first equation, x = t + 3, t = x - 3. We substitute t in the second equation: y = t + 9 = x - 3 + 9 = x + 6 We compare y = x + 6 with the equation of any line: y = mx + n, where m is the slope and n the offset and the slope of y = x + 6 is 1.

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• Question #10:

  x	-5	0	3	7
  y	2.01	5	8.64	17.92

  Which of the following equations best models the data in the table above?

  (a) y = 5· 1.2^x - 2

  (b) y = 4· 1.2^x

  (c) y = 5· 1.3^x

  (d) y = 5· 1.2^x - 1

  (e) y = 5· 1.3^x 1

  • Answer: The easiest way to solve this problem is to check the results of the 5 answers for x = 0: (a) y = 5· 1.2⁰ - 2 = 3

    (b) y = 4· 1.2⁰ = 4

    (c) y = 5· 1.3⁰ = 5

    (d) y = 5· 1.2⁰ - 1 = 4

    (e) y = 5· 1.3⁰ + 1 = 6 and (c) must be the correct answer. We can verify at this point the results of y = 5· 1.3^x for x = -5, x = 3 and x = 7

SAT2数学Level 2试题2 含答案SAT2数学Level 2试题2 含答案

转载请注明来自澳际留学

• Question #1: If f(x) = (2x + 5)/(4x-7), what value does f(x) approach as x gets infinitely larger?

  (a) 1/2

  (b) 5/7

  (c) 2/7

  (d) 5/4

  (e) 1

  • Answer: As x gets infinitely larger, lim[(2x + 5)/(4x-7)] = lim[(2 + 5/x) / (4 - 7/x)] = (2 + 0) / (4 - 0) = 2/4 = 1/2

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• Question #2: In the figure below, what is the length of the arc AB, if O is the center of the circle and triangle OAB is equilateral? The radius of the circle is 9.

  (a) ¶

  (b) 2 · ¶

  (c) 3 · ¶

  (d) 4 · ¶

  (e) ¶/2

  • Answer: OAB equilateral means the AOB angle is 60^o The ratio between the AOB angle and 360^o is equal to the ratio between the length of the arc AB and the circumference of the circle 60^o/360^o = arc_AB / (2 ¶ · 9) arc_AB = 3 · ¶

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• Question #3: In the figure below, quadrilateral ABCD has AB parallel with CD. What is the area of triangle ABD? (a) 3 (b) 4 (c) 6 (d) 8 (e) 9

  • Answer: The area of triangle ABD is equal to (1/2)AB·DE, where DE is the altitude from D to AB. Area_ABD = (1/2)4·3 = 6.

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• Question #4: What is the distance in space between the points with coordinates (-1, 5, 3) and (2, 6, -1)?

  (a) √20

  (b) √24

  (c) 5

  (d) √26

  (e) 7

  • Answer: The coordinates of the 2 points are: x₁ = -1, y₁ = 5, z₁ = 3, x₂ = 2, y₂ = 6, z₂ = -1, The distance in space between the points with coordinates (-1, 5, 3) and (2, 6, -1) squared is: d² = [(x₁ - x₂)² + (y₁ - y₂)² + (z₁ - z₂)² ]= [ (-1 + (-2))² + (5 - 6)² + (3 - (-1)) ² ] = 9 + 1 + 16 = 26 d = √26

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• Question #5: In the standard (x, y) coordinate plane, the graph of (x – 2)² + (y + 2)² = 16 is a circle. What is the area enclosed by this circle, expressed in square coordinate units?

  (a) 4 · ¶

  (b) 8 · ¶

  (c) 16 · ¶

  (d) 18 · ¶

  (e) 20 · ¶

  • Answer: We compare the equation of the circle, (x – 2)² + (y + 2)² = 16 with the general equation of a circle: (x – x₁)² + (y - y₁)² = R², where R is the radius of the circle. We notice that the radius of our circle is √16 = 4 The area of the circle is ¶ · 4² = 16 · ¶

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