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Grade 12 Mathematics entrance exam, 2008 E.C. (2016)

Real questions from the Grade 12 university entrance exam in Mathematics, as students sat it in 2008 E.C. Every question comes with the correct answer and a worked explanation.

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  • 2008 E.C. (2016)

    Exam year

  • 57 questions with answers

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Answer each question yourself before opening the answer. The explanations point to the exact place in the textbook, so you know which page to reread when you miss one.

Question 1

Which one of the following is a one to one correspondence function from A=[0, 1] to B=[1, 2]?

  1. A.f(x)=xf(x) = x
  2. B.f(x)=1/3x3+1f(x) = 1/3x^{3} + 1
  3. C.f(x)=2x+1f(x) = 2x + 1
  4. D.f(x)=x2+1f(x) = x^{2} + 1
Show the answer and explanation

Answer: D f(x)=x2+1f(x) = x^{2} + 1

Why

The function f(x) = x^2 + 1 is continuous and strictly increasing on [0, 1], with f(0) = 1 and f(1) = 2, making it a bijection (one-to-one correspondence) from [0, 1] onto [1, 2].

Question 2

What is the solution set of 1−1x1−1x2=3x2−x1+1x\dfrac{1-\frac{1}{x}}{1-\frac{1}{x^{2}}} = 3x^{2} - \dfrac{x}{1+\frac{1}{x}} ? ?

  1. A.{−1, 1/3}\{-1,\ 1/3\}
  2. B.{1/3}\{1/3\}
  3. C.{3, −1/3}\{3,\ -1/3\}
  4. D.{−1/3}\{-1/3\}
Show the answer and explanation

Answer: B {1/3}\{1/3\}

Why

Simplifying the left-hand side gives x/(x+1) and the right-hand side gives 3x^2 - x^2/(x+1) = (3x^3 + 2x^2)/(x+1). Equating and solving yields 3x^2 - x = 0, so x = 0 or x = 1/3; however, x = 0 is not in the domain of 1/x, leaving {1/3} as the only valid solution.

Question 5

The planet Mercury's orbit around the sun is an ellipse with eccentricity 0.206, length of the major axis 1.16×1081.16 \times 10^{8} km and the sun at one focus. What is the max distance from Mercury to the sun?

  1. A.6.99×1086.99 \times 10^{8} km
  2. B.6.99×1076.99 \times 10^{7} km
  3. C.9.66×1079.66 \times 10^{7} km
  4. D.9.66×1089.66 \times 10^{8} km
Show the answer and explanation

Answer: B 6.99×1076.99 \times 10^{7} km

Why

The semi-major axis is a = (1.16 × 10^8)/2 = 5.8 × 10^7 km. The maximum distance (aphelion) is r_max = a(1 + e) = 5.8 × 10^7 × (1 + 0.206) ≈ 6.99 × 10^7 km.

Question 6

The graph of a hyperbola and the lines of its asymptotes are given as shown in the following figure. Which one of the following is an equation of the hyperbola? [Figure: A scanned hand-drawn Cartesian plane. The horizontal axis is drawn as a solid arrow to the right and labelled x; the vertical axis is a solid arrow upward and labelled y. Two solid curves form a hyperbola that opens vertically (up and down): the upper branch has its vertex on the positive y-axis roughly two units above the origin and sweeps up to the left and to the right; the lower branch has its vertex on the y-axis just below the origin and sweeps down to the left and to the right. Two dashed straight lines are drawn as the asymptotes; they cross each other on the y-axis at a point about one unit above the origin (the centre of the hyperbola), and each branch approaches them. The dashed line of positive slope, running from lower left to upper right, carries the printed label y = x + 1 near its upper-right end. The dashed line of negative slope, running from upper left to lower right, carries the printed label y = -x + 1 near its lower-right end. No numeric tick marks are shown on either axis.]

  1. A.y2−2y−x2=0y^{2} - 2y - x^{2} = 0
  2. B.y2−3y−x2=0y^{2} - 3y - x^{2} = 0
  3. C.x2−(y−1)2=1x^{2} - (y-1)^{2} = 1
  4. D.(x−1)2−y2=1(x-1)^{2} - y^{2} = 1
Show the answer and explanation

Answer: A y2−2y−x2=0y^{2} - 2y - x^{2} = 0

Why

The asymptotes intersect at the center (0, 1) and have slopes ±1. Since the hyperbola opens vertically, its equation is (y - 1)^2 - x^2 = 1, which expands to y^2 - 2y - x^2 = 0.

Question 7

If the truth value of a proposition p is False, then which one of the following compound proposition has a truth value True?

  1. A.¬p∧p\neg p \wedge p
  2. B.¬p⇒p\neg p \Rightarrow p
  3. C.¬(¬p∨p)\neg(\neg p \vee p)
  4. D.p⇒¬pp \Rightarrow \neg p
Show the answer and explanation

Answer: D p⇒¬pp \Rightarrow \neg p

Why

When p is False, ¬p is True. The conditional statement p ⇒ ¬p has a False hypothesis, which makes the implication vacuously True.

Question 9

Which one of the following compound propositions is a tautology?

  1. A.(q∨¬q)⇒p(q \vee \neg q) \Rightarrow p
  2. B.p⇒(p∨¬q)p \Rightarrow (p \vee \neg q)
  3. C.p∨(q∧¬q)p \vee (q \wedge \neg q)
  4. D.p⇒(q∧¬q)p \Rightarrow (q \wedge \neg q)
Show the answer and explanation

Answer: B p⇒(p∨¬q)p \Rightarrow (p \vee \neg q)

Why

By the rule of addition, p implies p ∨ ¬q regardless of the truth values of p and q, making p ⇒ (p ∨ ¬q) a tautology.

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Mathematics papers from other years

2018 E.C. (2026)2017 E.C. (2025)2016 E.C. (2024)2015 E.C. (2023)2014 E.C. (2022)2013 E.C. (2021)2011 E.C. (2019)2009 E.C. (2017)2007 E.C. (2015)2006 E.C. (2014)2005 E.C. (2013)

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Are these the real 2008 E.C. Mathematics entrance exam questions?

Yes. They come from the national university entrance exam in Mathematics that students sat in 2008 E.C. (2016). Our set holds 57 of its questions, each with its answer and an explanation that points back to the textbook.

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