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Grade 12 Mathematics entrance exam, 2016 E.C. (2024)

Real questions from the Grade 12 university entrance exam in Mathematics, as students sat it in 2016 E.C. Every question comes with the correct answer and a worked explanation.

  • Natural science stream

    Who sits it

  • 2016 E.C. (2024)

    Exam year

  • 52 questions with answers

    In this set

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Question 1

Let pp and qq be propositions. Which one of the following propositions is equivalent to the proposition ¬p∨(p∧q)\neg p \lor (p \land q)?

  1. A.¬p∨¬q\neg p \lor \neg q
  2. B.p⇒qp \Rightarrow q
  3. C.¬p∧q\neg p \land q
  4. D.¬p⇒q\neg p \Rightarrow q
Show the answer and explanation

Answer: B p⇒qp \Rightarrow q

Why

Using the distributive law, we have ¬p∨(p∧q)≡(¬p∨p)∧(¬p∨q)≡True∧(¬p∨q)≡¬p∨q\neg p \lor (p \land q) \equiv (\neg p \lor p) \land (\neg p \lor q) \equiv \text{True} \land (\neg p \lor q) \equiv \neg p \lor q, which is equivalent to the conditional statement p⇒qp \Rightarrow q.

Question 2

Which one of the following is NOT true about the function f(x)=tan⁡−1xf(x) = \tan^{-1} x?

  1. A.The line x=−π2x = -\frac{\pi}{2} is a vertical asymptote to the graph of ff.
  2. B.The graph of ff is the image of the graph of y=tan⁡xy = \tan x after it is reflected with respect to the line y=xy = x.
  3. C.The graph of ff intersects the xx-axis at (0,0)(0, 0).
  4. D.The line y=π2y = \frac{\pi}{2} is a horizontal asymptote to the graph of ff.
Show the answer and explanation

Answer: A The line x=−π2x = -\frac{\pi}{2} is a vertical asymptote to the graph of ff.

Why

The domain of f(x)=tan⁡−1xf(x) = \tan^{-1} x is the entire set of real numbers R\mathbb{R}, and it has horizontal asymptotes at y=±π2y = \pm \frac{\pi}{2} but no vertical asymptotes. The line x=−π2x = -\frac{\pi}{2} is actually a vertical asymptote for y=tan⁡xy = \tan x, not for its inverse.

Question 3

Which one of the following statements is true?

  1. A.If xx and yy are integers and xx divides yy, then yy divides xx.
  2. B.If xx is an even integer and yy is an odd integer, then x2+y2x^2 + y^2 is odd.
  3. C.If xx and yy are even integers, then x+yx + y is an odd integer.
  4. D.If xx and yy are positive integers, then xy\frac{x}{y} is a positive integer.
Show the answer and explanation

Answer: B If xx is an even integer and yy is an odd integer, then x2+y2x^2 + y^2 is odd.

Why

If xx is even, then x2x^2 is even; if yy is odd, then y2y^2 is odd. The sum of an even number and an odd number, x2+y2x^2 + y^2, is always odd.

Question 4

Let pp, qq, and rr be propositions. If pp is True , qq is False , and rr is False , which one of the following propositions has truth value True ?

  1. A.(p∧¬q)⇔(p∨r)(p \land \neg q) \Leftrightarrow (p \lor r)
  2. B.(q∧r)⇔(p∧¬r)(q \land r) \Leftrightarrow (p \land \neg r)
  3. C.p⇒(q∨r)p \Rightarrow (q \lor r)
  4. D.(p∨q)⇒r(p \lor q) \Rightarrow r
Show the answer and explanation

Answer: A (p∧¬q)⇔(p∨r)(p \land \neg q) \Leftrightarrow (p \lor r)

Why

Substituting the truth values, we have p∧¬q≡True∧True≡Truep \land \neg q \equiv \text{True} \land \text{True} \equiv \text{True}, and p∨r≡True∨False≡Truep \lor r \equiv \text{True} \lor \text{False} \equiv \text{True}. Thus, (p∧¬q)⇔(p∨r)≡True⇔True(p \land \neg q) \Leftrightarrow (p \lor r) \equiv \text{True} \Leftrightarrow \text{True}, which is True.

Question 5

Which one of the following is the equation of the tangent line to the graph of f(x)=2x3−4x2+5x+3f(x) = 2x^3 - 4x^2 + 5x + 3 at the point (0,3)(0, 3)?

  1. A.3x+5y−3=03x + 5y - 3 = 0
  2. B.3x−5y−3=03x - 5y - 3 = 0
  3. C.5x−y+3=05x - y + 3 = 0
  4. D.5x+y+3=05x + y + 3 = 0
Show the answer and explanation

Answer: C 5x−y+3=05x - y + 3 = 0

Why

The derivative of f(x)f(x) is f′(x)=6x2−8x+5f'(x) = 6x^2 - 8x + 5, which gives a tangent slope of f′(0)=5f'(0) = 5 at x=0x = 0. The equation of the tangent line is y−3=5(x−0)y - 3 = 5(x - 0), which simplifies to 5x−y+3=05x - y + 3 = 0.

Question 6

Let v=AB⃗v = \vec{AB} and w=CD⃗w = \vec{CD} be vectors in the plane, where A(1,−1)A(1, -1), B(2,−2)B(2, -2), C(−1,3)C(-1, 3), and D(4,−3)D(4, -3). Which one of the following is equal to v−wv - w?

  1. A.5i−6j5i - 6j
  2. B.i−15ji - 15j
  3. C.−4i+5j-4i + 5j
  4. D.4i−5j4i - 5j
Show the answer and explanation

Answer: C −4i+5j-4i + 5j

Why

The vector v=B−A=(1,−1)=i−jv = B - A = (1, -1) = i - j and w=D−C=(5,−6)=5i−6jw = D - C = (5, -6) = 5i - 6j. Subtracting these gives v−w=(i−j)−(5i−6j)=−4i+5jv - w = (i - j) - (5i - 6j) = -4i + 5j.

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Are these the real 2016 E.C. Mathematics entrance exam questions?

Yes. They come from the national university entrance exam in Mathematics that students sat in 2016 E.C. (2024). Our set holds 52 of its questions, each with its answer and an explanation that points back to the textbook.

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