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Question: If 2^x < 100 and x is an integer, how many of the 2^x + 2 integers will be divisible by 3 and by 2?

(a) 1

(b) 2

(c) 3

(d) 4

(d) 5

• Answer: 2⁶ = 64 and 2⁷ = 128. If 2^x < 100, then the highest x is x = 6
  Possible values for x: 0 , 1 , 2 , 3 , 4, , 5 , 6. 2^x + 2 can take the values 3 , 4 , 6 , 10 , 18 , 34 , 66
  Out of these values, only 6 , 18 and 66 are divisible by 3 and by 2. The correct answer is (c).

Question: The inequality |2x – 1| > 5 must be true in which one of the following cases?
I. x < -5
II. x > 7
III. x > 0

(a) II only

(b) I, II and II

(c) I and II only

(d) I and III only

(e) I only

• Answer: |2x – 1| > 5,
  -5 > 2x – 1 or 2x – 1 > 5
  -4 > 2x or 2x > 6
  -4 > 2x results in x < -2
  2x > 6 results in x > 3
  I answer is true, II answer is also true, but III answer is false, so the correct answer is (c)

Question: What is the closest approximation of the solution of the equation 2^{x – 1} = 3^{x + 1}?

(a) -4.42

(b) -5.81

(c) -3.22

(d) 4.93

(e) 3.33

• log(2^{x – 1}) = 3log(^{x + 1})
  (x – 1)log2 = (x + 1)log3
  x(log2 – log3) = log3 + log2
  x = (log3 + log2)/(log2 – log3)
  x is aprox. = -4.42

Question: What is the range of (x – y) if 3 < x < 4 and -2 < y< -1?

(a) 4< x-y <5

(b) 1< x-y <3

(c) 1< x-y <5

(d) 4< x-y <6

(e) 3< x-y <6

• Answer: We can determine the range of -y:
  1<-y<2
  We determine the range of x-y by adding the ranges of x and -y:
  Therefore, 4< x-y <6

Question: For some positive real number ‘a’, the first 3 terms of a geometric progression are a – 1, a + 3 and 3a + 1. What is the numerical value of the fourth term?

(a) 25

(b) 36

(c) 32

(d) 100

(e) 9

• Answer:
  a + 3 = k(a – 1)
  3a + 1 = k(a + 3)

  (a + 3)(a + 3) = (3a + 1)(a – 1)
  a² + 6a + 9 = 3a² – 2a – 1
  2a² – 8a -10 = 0
  a² – 4a -5 = 0
  the solutions of this equation are 5 and -1. The only positive solution is 5, so the progression is 4, 8, 16. The fourth term will be 16·2 = 32

Question: 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: 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: 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 can’t take the values x = 2 and x = -2, so the domain of the function is (-∞ , -2) U ( -2 , 0]