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Mathematics entrance exam, 2005 E.C. (2013)

Real questions from the Grade 12 university entrance exam in Mathematics, as students sat it in 2005 E.C. Every question comes with the correct answer and a worked explanation.

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Both streams
Who sits it
2005 E.C. (2013)
Exam year
60 questions with answers
In this set

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

Question 1

If \(\{a_n\}\) is a sequence such that \(a_1 = 2\), and \(a_{n+1} = a_n + 4\) for all \(n \geq 1\), then \(\sum_{n=1}^{35} a_n\) is equal to :

A.2460
B.2458
C.2450
D.2442

Show the answer and explanation

Answer: C 2450

Why

The rule \(a_{n+1} = a_n + 4\) means every term is 4 more than the one before it, so this is an arithmetic sequence with \(a_1 = 2\) and \(d = 4\). From the 1st term to the 35th there are 34 steps, so \(a_{35} = 2 + 34(4) = 138\). The sum of an arithmetic sequence is \(S_n = \frac{n}{2}(a_1 + a_n)\), so \(S_{35} = \frac{35}{2}(2 + 138) = 35 \times 70 = 2450\). The common slip is to count 35 steps and write \(a_{35} = 142\), which gives a total of 2520. A total that is not on the list is your signal to recount the steps.

Question 2

If \(f(x) = x^2 + 2\ln x\), then \(\lim_{h \to 0} \dfrac{f(2+h)-f(2)}{h}\) ?

Show the answer and explanation

Answer: A 5

Why

That limit is the definition of the derivative of \(f\) at \(x = 2\), so find \(f'\) and put 2 into it. For \(f(x) = x^2 + 2\ln x\), the derivative is \(f'(x) = 2x + \frac{2}{x}\). At \(x = 2\) this gives \(4 + 1 = 5\). Option B, 4, is what you get if you differentiate only the \(x^2\) part and forget that \(2\ln x\) contributes \(\frac{2}{x}\).

Question 3

If \(f(x) = e^{2x} + x - 3\cos x\), then what is \(f''(x)\)?

A.\(e^{2x} + 1 - 3\sin(x)\)
B.\(e^{2x} + 1 + 3\sin(x)\)
C.\(4e^{2x} - 3\cos(x)\)
D.\(4e^{2x} + 3\cos(x)\)

Show the answer and explanation

Answer: D \(4e^{2x} + 3\cos(x)\)

Why

Differentiate twice. The first derivative is \(f'(x) = 2e^{2x} + 1 + 3\sin x\): the chain rule pulls a factor of 2 out of \(e^{2x}\), and the derivative of \(-3\cos x\) is \(+3\sin x\). Differentiating again, the constant 1 disappears and \(3\sin x\) becomes \(3\cos x\), so \(f''(x) = 4e^{2x} + 3\cos x\). Option C keeps a minus sign in front of the cosine. That minus sign was already used up at the first step, when \(-3\cos x\) turned into \(+3\sin x\).

Question 4

Which one of the following intervals does \(f(x) = x^4 + 4x\) increase?

A.\((-\infty, -1]\)
B.\((-\infty, 0]\)
C.\([-1, \infty)\)
D.\((-\infty, \infty)\)

Show the answer and explanation

Answer: C \([-1, \infty)\)

Why

A function increases on the interval where its derivative is not negative. Here \(f'(x) = 4x^3 + 4 = 4(x^3 + 1)\), and \(x^3 + 1 \geq 0\) exactly when \(x^3 \geq -1\), which means \(x \geq -1\). So \(f\) increases on \([-1, \infty)\). Option A is the mirror image of the answer: on \((-\infty, -1]\) the derivative is negative, so the function is falling there.

Question 5

Which one of the following is the simplest form of \(\left|3+4i\right| - \dfrac{25i}{3+4i}\)

A.5 ? 5i
B.5 + 5i
C.1 + 3i
D.1 � 3i

Show the answer and explanation

Answer: D 1 � 3i

Why

Handle the two pieces separately. The modulus is \(|3+4i| = \sqrt{9+16} = 5\). For the fraction, multiply top and bottom by the conjugate \(3-4i\): \(\frac{25i(3-4i)}{(3+4i)(3-4i)} = \frac{75i - 100i^2}{25} = \frac{100 + 75i}{25} = 4 + 3i\). So the expression is \(5 - (4+3i) = 1 - 3i\). Option C, \(1 + 3i\), comes from subtracting the real part but leaving the imaginary part with the sign it had before the subtraction.

Question 6

If \(Z = \cos(\pi/10) + i\sin(\pi/10)\), then what is the value of \(Z^5\)?

A.\(\pi/2 + \pi/2\ i\)
B.\(1/2 + 1/2\ i\)
C.\(i\)
D.\(1 + i\)

Show the answer and explanation

Answer: C \(i\)

Why

The number is already in the form \(\cos\theta + i\sin\theta\) with \(\theta = \frac{\pi}{10}\), so use de Moivre's rule: raising to the power 5 multiplies the angle by 5. That gives \(Z^5 = \cos\frac{5\pi}{10} + i\sin\frac{5\pi}{10} = \cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\). Since \(\cos\frac{\pi}{2} = 0\) and \(\sin\frac{\pi}{2} = 1\), the value is \(i\). Option B comes from raising the two numbers \(\cos\frac{\pi}{10}\) and \(\sin\frac{\pi}{10}\) to the power 5 one by one, instead of multiplying the angle by 5.

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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)

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Are these the real 2005 E.C. Mathematics entrance exam questions?

Yes. They come from the national university entrance exam in Mathematics that students sat in 2005 E.C. (2013). Our set holds 60 of its questions, each with its answer and an explanation that points back to the textbook.

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