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Grade 12 Mathematics entrance exam, 2015 E.C. (2023)

Real questions from the Grade 12 university entrance exam in Mathematics, as students sat it in 2015 E.C. Every question comes with the correct answer and a worked explanation.

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  • 2015 E.C. (2023)

    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

If f(x)=x+2x+2f(x) = \frac{\sqrt{x + 2}}{x + 2} and g(x)=1x−2g(x) = \frac{1}{x} - 2 then what is the value of f(g(x))f(g(x)) for all x>0x > 0 ?

  1. A.√x - 2
  2. B.√x + 2
  3. C.√x
  4. D.1/√x
Show the answer and explanation

Answer: C √x

Why

Substituting g(x)=1x−2g(x) = \frac{1}{x} - 2 into f(x)f(x) gives f(g(x))=(1/x−2)+2(1/x−2)+2=1/x1/x=xx=xf(g(x)) = \frac{\sqrt{(1/x - 2) + 2}}{(1/x - 2) + 2} = \frac{\sqrt{1/x}}{1/x} = \frac{x}{\sqrt{x}} = \sqrt{x} for all x>0x > 0.

Question 2

Which of the following is Not the property of f(x)=x2/5f(x) = x^{2/5}?

  1. A.f(xy) = f(x)f(y)
  2. B.decreasing on (-∞,0]
  3. C.the range is real number
  4. D.the graph is symmetric with respect to the y-axis
Show the answer and explanation

Answer: C the range is real number

Why

Since f(x)=x2/5=(x1/5)2f(x) = x^{2/5} = (x^{1/5})^2 is always non-negative for all real numbers xx, its range is [0,∞)[0, \infty), not the set of all real numbers.

Question 4

Which one of the following function is one to one correspondence function from A = [0,1] to B = [2,3]?

  1. A.f(x)=x+2
  2. B.f(x)=2x+2
  3. C.f(x)=x+3
  4. D.f(x)=2x+1
Show the answer and explanation

Answer: A f(x)=x+2

Why

For a function to be a bijection from [0,1][0, 1] to [2,3][2, 3], it must map the endpoints to the endpoints. For f(x)=x+2f(x) = x + 2, we have f(0)=2f(0) = 2 and f(1)=3f(1) = 3, and since it is strictly increasing and linear, it is a one-to-one correspondence.

Question 6

Suppose f(x)=Q(x)x(x2−1)f(x) = \frac{Q(x)}{x(x^2 - 1)} where Q(x)Q(x) is a quadratic function. Which one of the following is necessarily true about the graph ff ?

  1. A.x=0, x=1 and x=-1 are vertical asymptote of the graph f
  2. B.The graph of f does not intersect with its horizontal asymptote.
  3. C.The vertical asymptote of the graph of f is only x=-1, if Q(x)=x^2-x
  4. D.The vertical asymptote of the graph of f is only x=1, if Q(x)=2x^2
Show the answer and explanation

Answer: C The vertical asymptote of the graph of f is only x=-1, if Q(x)=x^2-x

Why

If Q(x)=x2−x=x(x−1)Q(x) = x^2 - x = x(x-1), the common factors xx and x−1x-1 in the numerator and denominator cancel out, leaving f(x)=1x+1f(x) = \frac{1}{x+1} (with holes at x=0x=0 and x=1x=1). Thus, the only vertical asymptote is x=−1x = -1.

Question 7

What is the solution set of 2x=x−2x2−2x=1−2x−23x−2\frac{2}{x} = \frac{x - 2}{x^2 - 2x} = 1 - \frac{2x - 2}{3x - 2} ? (Note: The equation has two equals signs, likely a typo; probably it is 2x=x−2x2−2x\frac{2}{x} = \frac{x - 2}{x^2 - 2x} and also equals the other expression. We solve as two equations.)

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

Answer: D {1}

Why

By correcting the typo where the first minus sign was written as an equals sign, the equation 2x−x−2x2−2x=1−2x−23x−2\frac{2}{x} - \frac{x-2}{x^2-2x} = 1 - \frac{2x-2}{3x-2} simplifies to 1x=x3x−2\frac{1}{x} = \frac{x}{3x-2}, which yields x=1x = 1 and x=2x = 2. Since x=2x = 2 is excluded from the domain of the original expression, the only valid solution is x=1x = 1.

Question 8

Given line 4y=3x+244y = 3x + 24 and another line ℓ\ell parallel to this line at a distance of 4 units from it. A possible equation for ℓ\ell is

  1. A.y = (3/4)x
  2. B.y = (3/4)x + 1
  3. C.y = (3/4)x + 6
  4. D.y = (3/4)x - 6
Show the answer and explanation

Answer: B y = (3/4)x + 1

Why

The distance between parallel lines 3x−4y+24=03x - 4y + 24 = 0 and 3x−4y+C=03x - 4y + C = 0 is given by ∣C−24∣32+(−4)2=4\frac{|C - 24|}{\sqrt{3^2 + (-4)^2}} = 4, which simplifies to ∣C−24∣=20|C - 24| = 20. This gives C=4C = 4, resulting in the line 3x−4y+4=03x - 4y + 4 = 0, or y=34x+1y = \frac{3}{4}x + 1.

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Mathematics papers from other years

2018 E.C. (2026)2017 E.C. (2025)2016 E.C. (2024)2014 E.C. (2022)2013 E.C. (2021)2011 E.C. (2019)2009 E.C. (2017)2008 E.C. (2016)2007 E.C. (2015)2006 E.C. (2014)2005 E.C. (2013)

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Are these the real 2015 E.C. Mathematics entrance exam questions?

Yes. They come from the national university entrance exam in Mathematics that students sat in 2015 E.C. (2023). Our set holds 57 of its questions, each with its answer and an explanation that points back to the textbook.

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