转载请注明来自澳际留学

• Question 1: If the value of tan(-2a) is 2, what is the value of tan(2a)?

  (a) -2

  (b) 2

  (c) -4

  (d) 4

  (e) 3

  • We know that tan(θ) = -tan(-θ), any θ, so tan(2a) = - 2

---

• Question #2: What is sec(a)sin(a) + csc(a)cos(a) , if a = 45^o?

  (a) 2

  (b) 1

  (c) 1/2

  (d) -1/2

  (e) ¶/2

  • Answer: sec(a)sin(a) + csc(a)cos(a) = [1/cos(a)]sin(a) + [1/sin(a)]cos(a) = sin(a)/cos(a) + cos(a)/sin(a) = (√2/2)/(√2/2) + (√2/2)/(√2/2) = 1 + 1 = 2

---

• Question #3: What is sin²(a)·cos²(a)·[1 + tan²(a)]?

  (a) tan²(a)

  (b) cos²(a)

  (c) sin²(a)

  (d) sec²(a)

  (e) csc²(a)

  • Answer: sin²(a)·cos²(a)·[1 + tan²(a)] = sin²(a)·cos²(a)·[1 + sin²(a)/cos²(a)] = sin²(a)·cos²(a)·[(cos²(a) + sin²(a))/cos²(a)] = sin²(a)·cos²(a)·[1/cos²(a)] = sin²(a).

---

• Question #4: What are the polar coordinates of the point with (-5, 5) rectangular coordinates?

  (a) (5√2, 45^o)

  (b) (5, 45^o)

  (c) (5, 135^o)

  (d) (5√2, 135^o)

  (e) (-5, 135^o)

  • Answer: r² = 5² + 5², so r = 5√2. tan(θ) = -5/5 = -1 and θ = 135^o.

---

• Question #5: What is the volume of the geometric solid produced by the equilateral triangle in the figure below when it is rotated 360^o about the altitude m? (a) ¶m³/9

  (b) ¶m²/9

  (c) ¶m³/4

  (d) ¶m³

  (e) m³/9

  • Answer: The solid produced by the triangle rotation will be a cone with a radius equal to half the side of the triangle. We apply Pythagoras Theorem in one of the 2 right triangles created by the altitude m:

    r² + m² = (2r)² r = m/√3 The volume of the cone: V_cone = ¶m(m/√3)²/3 V_cone = ¶m³/9

---

• Question #6: Solve for x in the following equation: x² – 10x + 10 = 0.

  (a) -5 +/- sqrt(15)

  (b) 5 +/- sqrt(15)

  (c) 5

  (d) -5

  (d) 5 +/- sqrt(5)

  • Answer: x = (10 +/- sqrt(100 - 40))/2 x = 5 +/- sqrt(15)

• Question #7: If log(x) = 2 and log(y) = 5, what is log(x²y³)?

  (a) 11

  (b) 15

  (c) 17

  (d) 19

  (e) 29

  • Answer: log(x²y³) = log(x²) + log(y³) = 2log(x) + 3log(y) = 4 + 15 = 19

---

• Question #8: What is the greatest common factor of 17 and 81?

  (a) 1

  (b) 17

  (c) 3

  (d) 9

  (e) 27

  • Answer: 17 = 17·1 81 = 3·3·3·3·1 The correct answer is (a).

---

• Question #9: In the (x,y) plane, which of the following statements are true?

  I. Line y + x = 5 is perpendicular to line y - x = 5. II. Lines y + x = 5 and y - x = 5 intersect each other on the y axis. III. Lines y + x = 5 and y - x = 5 intersect each other on the x axis.

  (a) I is the only true statement

  (b) II is the only true statement

  (c) I and II are both true

  (d) I and III are both true

  (e) II and III are both true

  • Answer: y + x = 5 can be written as y = -x + 5. The slope of this equation is m₁ = -1. y - x = 5 can be written as y = x + 5. The slope of this equation is m₂ = 1. m₂ = -1/m₁ so the 2 lines are perpendicular. We also need to find where the 2 lines intersect. If we add the 2 equations, 2·y = 10, y = 5 From the first equation, x = 5 - y = 5 - 5 = 0. In conclusion the lines intersect at (0, 5) and this point is on the y axis. In conclusion I and II statements are correct.

---

• Question #10: Find the domain of the function f(x) = √( -x) / [(x - 2)(x + 2)]:

  (a) (-∞ , -2) U ( -2 , 0)

  (b) (-∞ , -2) U ( -2 , 0]

  (c) (-∞ , 2) U ( 2 , 0]

  (d) (-∞ , 2) U ( 2 , 0)

  (e) (-∞ , -2) U ( -2 , 2)

  • Answer: From the numerator of the fraction, -x should be positive or equal to zero, so x<=0. x cannot take the values x = 2 and x = -2, so the domain of the function is (-∞ , -2) U ( -2 , 0]

SAT2数学Level 2试题3 含答案SAT2数学Level 2试题3 含答案

转载请注明来自澳际留学

• Question 1: If the value of tan(-2a) is 2, what is the value of tan(2a)?

  (a) -2

  (b) 2

  (c) -4

  (d) 4

  (e) 3

  • We know that tan(θ) = -tan(-θ), any θ, so tan(2a) = - 2

---

• Question #2: What is sec(a)sin(a) + csc(a)cos(a) , if a = 45^o?

  (a) 2

  (b) 1

  (c) 1/2

  (d) -1/2

  (e) ¶/2

  • Answer: sec(a)sin(a) + csc(a)cos(a) = [1/cos(a)]sin(a) + [1/sin(a)]cos(a) = sin(a)/cos(a) + cos(a)/sin(a) = (√2/2)/(√2/2) + (√2/2)/(√2/2) = 1 + 1 = 2

---

• Question #3: What is sin²(a)·cos²(a)·[1 + tan²(a)]?

  (a) tan²(a)

  (b) cos²(a)

  (c) sin²(a)

  (d) sec²(a)

  (e) csc²(a)

  • Answer: sin²(a)·cos²(a)·[1 + tan²(a)] = sin²(a)·cos²(a)·[1 + sin²(a)/cos²(a)] = sin²(a)·cos²(a)·[(cos²(a) + sin²(a))/cos²(a)] = sin²(a)·cos²(a)·[1/cos²(a)] = sin²(a).

---

• Question #4: What are the polar coordinates of the point with (-5, 5) rectangular coordinates?

  (a) (5√2, 45^o)

  (b) (5, 45^o)

  (c) (5, 135^o)

  (d) (5√2, 135^o)

  (e) (-5, 135^o)

  • Answer: r² = 5² + 5², so r = 5√2. tan(θ) = -5/5 = -1 and θ = 135^o.

---

• Question #5: What is the volume of the geometric solid produced by the equilateral triangle in the figure below when it is rotated 360^o about the altitude m? (a) ¶m³/9

  (b) ¶m²/9

  (c) ¶m³/4

  (d) ¶m³

  (e) m³/9

  • Answer: The solid produced by the triangle rotation will be a cone with a radius equal to half the side of the triangle. We apply Pythagoras Theorem in one of the 2 right triangles created by the altitude m:

    r² + m² = (2r)² r = m/√3 The volume of the cone: V_cone = ¶m(m/√3)²/3 V_cone = ¶m³/9

---

• Question #6: Solve for x in the following equation: x² – 10x + 10 = 0.

  (a) -5 +/- sqrt(15)

  (b) 5 +/- sqrt(15)

  (c) 5

  (d) -5

  (d) 5 +/- sqrt(5)

  • Answer: x = (10 +/- sqrt(100 - 40))/2 x = 5 +/- sqrt(15)

上12下

共2页

阅读全文

更多出国留学最新动态，敬请关注澳际教育手机端网站，并可拨打咨询热线：400-601-0022