The S6 Mathematics II 2024 National Examination, sat on 23 July 2024 under NESA, is one of the most calculation-intensive papers in the Advanced Level series, covering logarithms, trigonometry, complex numbers, vector spaces, sequences, integration, differential equations, matrices, conics, and probability distributions. This paper is set for the MCB, MCE, MEG, MPC, MPG, and PCM combinations. Below are fully worked solutions to every question, with each step of the working shown so you can follow the method, not just the final answer.
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Paper Overview: S6 Mathematics II 029
The paper is divided into two sections. Section A is compulsory and worth 55 marks across 15 questions covering algebra, trigonometry, functions, sequences, and calculus fundamentals. Section B requires candidates to attempt only 3 of 5 longer questions worth 45 marks, covering Maclaurin series, differential equations, matrices, conics, and probability distributions. Students are permitted geometrical instruments and silent, non-programmable calculators, with 3 hours to complete the paper.
Section A: Attempt All Questions (55 marks)
Question 1
a) A binary operator or a binary operation combines two elements to give a unique third element.
Answer: True. This is the defining property of a binary operation on a set — it takes two elements from the set and produces exactly one result, which may or may not belong to the same set depending on closure.
b) A compound proposition identically true for all possible truth values of its components is called a contradiction.
Answer: False. A proposition that is always true regardless of the truth values of its components is called a tautology, not a contradiction. A contradiction is always false under every possible truth assignment.
Question 2
Solve in the set of real numbers: \(2\log x = \log(x+12)\)
Working: Using the power rule of logarithms, \(2\log x = \log(x^2)\), so:
$$\log(x^2) = \log(x+12) \implies x^2 = x + 12$$ $$x^2 - x - 12 = 0 \implies (x-4)(x+3) = 0$$This gives \(x = 4\) or \(x = -3\). Since the logarithm of a negative number is undefined, \(x = -3\) is rejected.
Answer: \(x = 4\).
Question 3
Find the value of the real number \(n\) in \(P(x) = x^3 + nx^2 + 3x - 1\), if \(P(x)\) leaves a remainder of 1 when divided by \(x-2\).
Working: By the Remainder Theorem, \(P(2) = 1\):
$$2^3 + n(2)^2 + 3(2) - 1 = 1$$ $$8 + 4n + 6 - 1 = 1 \implies 4n + 13 = 1 \implies 4n = -12$$Answer: \(n = -3\).
Question 4
If \(\tan A = \dfrac{a}{a+1}\) and \(\tan B = \dfrac{1}{2a+1}\), prove that \(A+B = \dfrac{\pi}{4}\).
Working: Using the tangent addition formula:
$$\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$$Substituting:
$$\tan(A+B) = \dfrac{\dfrac{a}{a+1} + \dfrac{1}{2a+1}}{1 - \dfrac{a}{a+1}\cdot\dfrac{1}{2a+1}}$$Combining the numerator over a common denominator \((a+1)(2a+1)\) gives \(a(2a+1) + (a+1) = 2a^2 + 2a + 1\), and the denominator becomes \((a+1)(2a+1) - a = 2a^2+2a+1\). Since the numerator and denominator are identical:
$$\tan(A+B) = 1$$Since \(\tan\left(\dfrac{\pi}{4}\right) = 1\), and taking \(A+B\) in the principal range, this proves \(A+B = \dfrac{\pi}{4}\), as required.
Question 5
The graphs show two periodic functions \(f(x)\) and \(h(x)\) with wave-like curves oscillating between \(-1\) and \(1\).
a) Write the period of each function.
Answer: Reading from the graph, \(f(x)\) completes one full cycle over an interval of \(\pi\), so its period is \(\pi\). The function \(h(x)\) oscillates faster, completing a cycle over a shorter interval, consistent with a period of \(\dfrac{\pi}{2}\) based on the spacing shown.
b) State any difference referring to those functions.
Answer: The two functions differ in frequency — \(h(x)\) completes more oscillations over the same interval than \(f(x)\), meaning \(h(x)\) has a shorter period and higher frequency, even though both functions share the same amplitude of 1.
Question 6
Solve the equation: \(3\cos x + 3 = 2\sin^2 x, \quad 0 \le x \le 2\pi\)
Working: Using the identity \(\sin^2 x = 1 - \cos^2 x\):
$$3\cos x + 3 = 2(1-\cos^2 x) \implies 3\cos x + 3 = 2 - 2\cos^2 x$$ $$2\cos^2 x + 3\cos x + 1 = 0$$Factoring this quadratic in \(\cos x\):
$$(2\cos x + 1)(\cos x + 1) = 0$$This gives \(\cos x = -\dfrac{1}{2}\) or \(\cos x = -1\).
For \(\cos x = -1\): \(x = \pi\).
For \(\cos x = -\dfrac{1}{2}\): \(x = \dfrac{2\pi}{3}\) or \(x = \dfrac{4\pi}{3}\).
Answer: \(x = \dfrac{2\pi}{3}, \pi, \dfrac{4\pi}{3}\).
Question 7
Determine the domain of definition for \(f(x) = \ln\left(\dfrac{e^x}{1-e^x}\right)\)
Working: For the logarithm to be defined, the argument must be strictly positive:
$$\frac{e^x}{1-e^x} > 0$$Since \(e^x > 0\) for all real \(x\), the sign of the fraction depends entirely on the denominator, so we need \(1 - e^x > 0\), which gives \(e^x < 1\), and therefore \(x < 0\).
Answer: Domain is \(x \in (-\infty, 0)\).
Question 8
Show that \(\left(i+\sqrt{3}\right)^{96} = 2^{96}\)
Working: Write \(z = \sqrt{3} + i\) in polar form. The modulus is:
$$|z| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = 2$$The argument is \(\arg(z) = \arctan\left(\dfrac{1}{\sqrt{3}}\right) = \dfrac{\pi}{3}\), so \(z = 2\left(\cos\dfrac{\pi}{3} + i\sin\dfrac{\pi}{3}\right)\). By De Moivre's Theorem:
$$z^{96} = 2^{96}\left(\cos\frac{96\pi}{3} + i\sin\frac{96\pi}{3}\right) = 2^{96}\left(\cos 32\pi + i\sin 32\pi\right)$$Since \(32\pi\) is a multiple of \(2\pi\), \(\cos 32\pi = 1\) and \(\sin 32\pi = 0\), giving:
$$z^{96} = 2^{96}(1 + 0i) = 2^{96}$$as required.
Question 9
Consider \(A = \{(-2x, 0, 5x), x \in \mathbb{R}\}\). Show that \((\mathbb{R}, A, +)\) is a sub-vector space of \(\mathbb{R}^3\).
Working: To prove \(A\) is a subspace of \(\mathbb{R}^3\), three conditions must hold:
1) Contains the zero vector: setting \(x=0\) gives \((0,0,0) \in A\). ✓
2) Closed under addition: let \(u=(-2x_1, 0, 5x_1)\) and \(v=(-2x_2, 0, 5x_2)\) be elements of \(A\). Then:
$$u+v = (-2x_1-2x_2,\ 0,\ 5x_1+5x_2) = (-2(x_1+x_2),\ 0,\ 5(x_1+x_2))$$which is of the same form with \(x = x_1+x_2 \in \mathbb{R}\), so \(u+v \in A\). ✓
3) Closed under scalar multiplication: for \(\lambda \in \mathbb{R}\), \(\lambda u = (-2\lambda x_1, 0, 5\lambda x_1)\), which is again of the same form with \(x=\lambda x_1 \in \mathbb{R}\), so \(\lambda u \in A\). ✓
Since all three conditions hold, \(A\) is a sub-vector space of \(\mathbb{R}^3\).
Question 10
In a geometric progression, the sixth term is 8 times the third term, and the sum of the seventh and eighth terms is 192.
a) Find the common ratio.
Working: Let the first term be \(a\) and common ratio \(r\). The condition \(T_6 = 8T_3\) gives:
$$ar^5 = 8ar^2 \implies r^3 = 8 \implies r = 2$$Answer: \(r = 2\).
b) Find the first term.
Working: Using \(T_7 + T_8 = 192\):
$$ar^6 + ar^7 = 192 \implies ar^6(1+r) = 192$$ $$a(2)^6(1+2) = 192 \implies a(64)(3) = 192 \implies 192a = 192$$Answer: \(a = 1\).
c) Find the sum of the fifth to eleventh terms.
Working: This is the sum from \(T_5\) to \(T_{11}\), which is 7 terms starting at \(T_5 = ar^4\):
$$S = ar^4\cdot\frac{r^7-1}{r-1} = (1)(2)^4\cdot\frac{2^7-1}{2-1} = 16 \times 127 = 2032$$Answer: 2032.
Question 11
Each student in a class gives each other a handshake. If there are 66 handshakes in total, how many students are in the class?
Working: The number of handshakes among \(n\) students, where each pair shakes hands exactly once, is given by \(\binom{n}{2}\):
$$\frac{n(n-1)}{2} = 66 \implies n(n-1) = 132 \implies n^2 - n - 132 = 0$$Factoring: \((n-12)(n+11) = 0\), giving \(n=12\) (rejecting the negative root).
Answer: 12 students.
Question 12
Evaluate \(\displaystyle\int \sqrt{1-\sin x}\, dx\)
Working: Using the identity \(1-\sin x = \left(\sin\dfrac{x}{2} - \cos\dfrac{x}{2}\right)^2\), which follows from expanding \(\sin^2\frac{x}{2}+\cos^2\frac{x}{2}-2\sin\frac{x}{2}\cos\frac{x}{2} = 1-\sin x\), the square root becomes:
$$\sqrt{1-\sin x} = \left|\sin\frac{x}{2}-\cos\frac{x}{2}\right|$$On an interval where \(\sin\frac{x}{2} - \cos\frac{x}{2} \ge 0\), integrating directly gives:
$$\int \sqrt{1-\sin x}\, dx = -2\cos\frac{x}{2} - 2\sin\frac{x}{2} + C$$Question 13
Determine whether \(\displaystyle\int_{-\infty}^{0} xe^x\, dx\) is convergent or divergent.
Working: Using integration by parts with \(u=x, dv=e^x dx\), so \(du=dx, v=e^x\):
$$\int xe^x\, dx = xe^x - \int e^x\, dx = xe^x - e^x + C = e^x(x-1) + C$$Evaluating the limit as the lower bound approaches \(-\infty\):
$$\int_{-\infty}^{0} xe^x\, dx = \left[e^x(x-1)\right]_{-\infty}^{0} = (1)(0-1) - \lim_{t\to-\infty} e^t(t-1) = -1 - 0 = -1$$since \(e^t \to 0\) faster than \(t\) grows as \(t \to -\infty\).
Answer: The integral is convergent and equals \(-1\).
Question 14
Find the mean value with respect to \(x\) of the function \((5x^2-4x)\) for \(1 \le x \le 3\).
Working: The mean value of a function over \([a,b]\) is:
$$\bar{f} = \frac{1}{b-a}\int_a^b f(x)\,dx = \frac{1}{3-1}\int_1^3 (5x^2-4x)\,dx$$ $$= \frac{1}{2}\left[\frac{5x^3}{3} - 2x^2\right]_1^3 = \frac{1}{2}\left[(45-18)-\left(\frac{5}{3}-2\right)\right] = \frac{1}{2}\left[27+\frac{1}{3}\right]$$Answer: \(\bar{f} = \dfrac{41}{3}\).
Question 15
If the focus of a standard ellipse is at \((1,0)\) and the corresponding directrix has the equation \(x=4\), find its equation.
Working: For a standard ellipse centered at the origin, the focus is at \((ae, 0)\) and the corresponding directrix is \(x=\dfrac{a}{e}\). This gives two equations:
$$ae = 1 \qquad \text{and} \qquad \frac{a}{e} = 4$$Multiplying these two equations gives \(a^2 = 4\), so \(a=2\), and then \(e = \dfrac{1}{a} = \dfrac{1}{2}\). Using \(b^2 = a^2(1-e^2)\):
$$b^2 = 4\left(1-\frac{1}{4}\right) = 4\times\frac{3}{4} = 3$$Answer: \(\dfrac{x^2}{4} + \dfrac{y^2}{3} = 1\).
Section B: Attempt Only Three Questions (45 marks)
Question 16: Maclaurin Series of ln(1+x)
Consider \(f(x) = \ln(1+x)\).
a) Determine the domain and range of \(f\).
Answer: Domain: \(x > -1\), i.e. \(x \in (-1, \infty)\). Range: all real numbers, \(y \in \mathbb{R}\), since \(\ln(1+x)\) takes every real value as \(x\) ranges over its domain.
b) Write down the first four non-zero terms of the Maclaurin expansion of \(\ln(1+x)\).
$$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots$$c) Without using L'Hôpital's rule, evaluate \(\displaystyle\lim_{x\to 0}\frac{\ln(1+x)}{x}\)
Working: Dividing the Maclaurin series by \(x\):
$$\frac{\ln(1+x)}{x} = 1 - \frac{x}{2} + \frac{x^2}{3} - \cdots$$As \(x \to 0\), every term after the first vanishes, giving the limit equal to 1.
d) From the results in b) above, find an approximation to \(\displaystyle\int_0^1 \frac{\ln(1+x)}{x}\,dx\), correct to 4 decimal places.
Working: Using the series from part (c), \(\dfrac{\ln(1+x)}{x} = 1 - \dfrac{x}{2} + \dfrac{x^2}{3} - \dfrac{x^3}{4} + \cdots\), and integrating term by term:
$$\int_0^1 \left(1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}\right)dx = \left[x - \frac{x^2}{4} + \frac{x^3}{9} - \frac{x^4}{16}\right]_0^1 = 1 - \frac{1}{4} + \frac{1}{9} - \frac{1}{16}$$Answer: \(\approx 0.7986\).
e) Write down the power series for \(\ln(1+x)\).
$$\ln(1+x) = \sum_{n=1}^{\infty} (-1)^{n+1}\frac{x^n}{n}$$f) Determine the radius and interval of convergence of the series \(\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^n}{n}\)
Working: Applying the ratio test to the general term \(a_n = \dfrac{x^n}{n}\):
$$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty}\left|\frac{x^{n+1}/(n+1)}{x^n/n}\right| = |x|$$The series converges when \(|x| < 1\), giving a radius of convergence \(R=1\). Checking the endpoints: at \(x=1\), the series becomes the alternating harmonic series, which converges; at \(x=-1\), it becomes the negative harmonic series, which diverges.
Answer: Interval of convergence: \(-1 < x \le 1\).
Question 17: Differential Equation
Find the general solution of \(\dfrac{d^2y}{dx^2} - 2\dfrac{dy}{dx} + y = \dfrac{e^x}{x^2+1}\)
Working — Homogeneous solution: The auxiliary equation is:
$$r^2 - 2r + 1 = 0 \implies (r-1)^2 = 0 \implies r = 1 \text{ (repeated root)}$$giving \(y_h = (C_1 + C_2 x)e^x\).
Particular solution via variation of parameters: With \(y_1 = e^x\) and \(y_2 = xe^x\), the Wronskian is \(W = y_1y_2' - y_2y_1' = e^{2x}\). Then:
$$u_1' = \frac{-y_2 g(x)}{W} = \frac{-x}{x^2+1}, \qquad u_2' = \frac{y_1 g(x)}{W} = \frac{1}{x^2+1}$$where \(g(x) = \dfrac{e^x}{x^2+1}\). Integrating:
$$u_1 = -\frac{1}{2}\ln(x^2+1), \qquad u_2 = \arctan x$$The particular solution is \(y_p = u_1y_1 + u_2y_2\):
$$y_p = e^x\left(x\arctan x - \frac{1}{2}\ln(x^2+1)\right)$$Answer:
$$y = (C_1 + C_2x)e^x + e^x\left(x\arctan x - \frac{1}{2}\ln(x^2+1)\right)$$Question 18: System of Equations via Matrix Method
Solve using the matrix method:
$$\begin{cases} 2x+y+z=5 \\ x+y+z=4 \\ x-y+2z=1 \end{cases}$$Working: Writing in matrix form \(AX=B\) and solving (via Cramer's rule or Gaussian elimination), subtracting the second equation from the first gives \(x=1\). Substituting into the second and third equations:
$$1+y+z=4 \implies y+z=3, \qquad 1-y+2z=1 \implies -y+2z=0 \implies y=2z$$Substituting \(y=2z\) into \(y+z=3\): \(3z=3 \implies z=1\), so \(y=2\).
Answer: \(x=1,\ y=2,\ z=1\).
Question 19: Conic in Polar Form
A conic is given by \(r = \dfrac{10}{3-2\cos\theta}\)
Working: Dividing numerator and denominator by 3 to match the standard form \(r = \dfrac{l}{1-e\cos\theta}\):
$$r = \frac{10/3}{1-(2/3)\cos\theta}$$a) Eccentricity: \(e = \dfrac{2}{3}\).
b) Identify the conic: Since \(0 < e < 1\), the conic is an ellipse.
c) Locate its directrix: The semi-latus rectum is \(l = \dfrac{10}{3}\), and since \(l = ed\), the directrix lies at a distance \(d = \dfrac{l}{e} = \dfrac{10/3}{2/3} = 5\) from the pole (focus), on the side corresponding to the negative cosine term.
d) Sketch the conic: The curve is an ellipse with its focus at the pole, elongated along the polar axis, with the nearest vertex to the pole occurring at \(\theta=0\) (where \(r\) is smallest) and the farthest vertex at \(\theta=\pi\) (where \(r\) is largest).
Question 20: Probability Density Function
A random variable \(x\) has probability density function \(F(x) = Ax(6-x)^2\) for \(0 \le x \le 6\), and 0 elsewhere.
a) Find the value of the constant \(A\).
Working: For a valid probability density function, the total integral must equal 1:
$$\int_0^6 Ax(6-x)^2\,dx = 1$$Expanding \(x(6-x)^2 = x(36-12x+x^2) = 36x-12x^2+x^3\) and integrating:
$$A\left[18x^2 - 4x^3 + \frac{x^4}{4}\right]_0^6 = A(648-864+324) = 108A = 1$$Answer: \(A = \dfrac{1}{108}\).
b) i) Find the mean.
$$E(x) = \int_0^6 x \cdot \frac{1}{108}x(6-x)^2\,dx = \frac{1}{108}\int_0^6 x^2(6-x)^2\,dx = \frac{12}{5} = 2.4$$ii) Find the variance.
Working: First find \(E(x^2)\):
$$E(x^2) = \frac{1}{108}\int_0^6 x^3(6-x)^2\,dx = \frac{36}{5}$$Then apply \(\text{Var}(x) = E(x^2) - [E(x)]^2\):
$$\text{Var}(x) = \frac{36}{5} - \left(\frac{12}{5}\right)^2 = \frac{36}{5} - \frac{144}{25} = \frac{36}{25} = 1.44$$iii) Find the standard deviation.
$$\sigma = \sqrt{1.44} = 1.2$$Study Tips for S6 Mathematics II
This paper illustrates a pattern worth internalizing: nearly every Section A question rewards recognizing the correct identity or formula immediately, rather than working through a problem from first principles under time pressure. Trigonometric identities, the remainder theorem, and standard series expansions like \(\ln(1+x)\) reappear across multiple exam series, so building genuine fluency with these — rather than memorizing individual past answers — pays off far beyond this single paper. Our A-Level subject combinations guide is useful if you're still deciding on your MCB, MCE, MEG, MPC, MPG, or PCM pathway, and our guide to scoring above 80% in S6 Mathematics covers independent revision strategies that pair well with past-paper practice like this.
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⚠️ Important Disclaimer: The solutions on this page are prepared by the Mathrone Academy team for revision and learning purposes only. This is not an official NESA marking scheme, REB-approved answer guide, or official Cambridge/Pearson mark scheme. While every effort has been made to ensure accuracy, answers and explanations may differ from the official examiners' marking guide. Always refer to your school teacher or the official examination board publications for authoritative marking guidance. Mathrone Academy accepts no responsibility for any discrepancies between these solutions and official results.