General Instructions
- The question paper has 19 questions divided into four sections, carrying a total of 40 marks.
- Section A: Questions 1–8, one mark each (8 marks).
- Section B: Questions 9–12, two marks each (8 marks).
- Section C: Questions 13–16, three marks each (12 marks).
- Section D: Questions 17–19, four marks each (12 marks).
- All questions are compulsory. Write the full working wherever marks are awarded for method, not just the final answer.
- Use of a calculator is not permitted; questions that ask for roots to a given accuracy expect working by hand.
Section A – Multiple Choice Questions
8 questions × 1 mark = 8 marks
Each question below has four options (A, B, C, D). Choose the one correct answer.
If 3 is a root of the quadratic equation \(x^2 - px + 3 = 0\), the value of \(p\) is:
The roots of the equation \(px^2 - qx + r = 0\) are real and equal when:
The mean proportional between 4 and 9 is:
The remainder when \(x^3 - 2x^2 + x + 5\) is divided by \(x - 1\) is:
Which term of the AP \(15, 30, 45, \ldots\) is 300?
The general term of the GP \(15, 30, 60, 120, \ldots\) is:
If \(A = \begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}\) and \(I\) is the \(2 \times 2\) identity matrix, then \(2A - I\) equals:
The fourth proportional to 5, 8 and 15 is:
Section B – Short Answer Questions
4 questions × 2 marks = 8 marks
If \(x - 2\) is a factor of \(x^3 - kx - 12\), find the value of \(k\).
Solve the inequation \(2x + 4 \le 14\) where \(x \in \mathbb{W}\) (whole numbers). Write down the solution set.
Solve the quadratic equation \(x^2 - 8x + 15 = 0\) by factorisation.
Solve \(-3 < 2x + 1 \le 9\) where \(x \in \mathbb{R}\). Write the solution set using inequality notation.
Section C – Application Questions
4 questions × 3 marks = 12 marks
Solve the quadratic equation \(x^2 + 4x - 8 = 0\), giving the roots correct to one decimal place.
The 5th and 9th terms of an AP are 4 and \(-12\). Find (a) the first term, (b) the common difference, (c) the sum of the first 16 terms.
If \(A = \begin{bmatrix}2 & 3 \\ 4 & 5\end{bmatrix}\), find \(2A - 3I\), where \(I\) is the \(2 \times 2\) identity matrix.
If 2, \(x\) and 18 are in continued proportion, find the value of \(x\).
Section D – Structured / Long Answer Questions
3 questions × 4 marks = 12 marks
Given \(A = \begin{bmatrix}1 & 3 \\ 2 & 4\end{bmatrix}\), \(B = \begin{bmatrix}1 & 2 \\ 2 & 4\end{bmatrix}\), \(C = \begin{bmatrix}4 & 1 \\ 1 & 5\end{bmatrix}\) and \(I = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\), find the matrix \(A(B + C) - 14I\).
If \(\dfrac{(a + b)^3}{(a - b)^3} = \dfrac{64}{27}\), find the ratio \(a : b\).
Find the value of \(k\) for which \(3x^3 + 8x^2 - 15x + k\) has a factor \((x - 1)\). Hence factorise the polynomial completely.
Check your work: Section A (8) + Section B (8) + Section C (12) + Section D (12) = 40 marks. Attempt the whole paper before revealing the answers below.
Answer Key
Section A – Multiple Choice Questions
| Question | Answer |
|---|---|
| 1 | B (4) |
| 2 | B |
| 3 | A (6) |
| 4 | A (5) |
| 5 | A (20th) |
| 6 | B |
| 7 | B |
| 8 | A (24) |
Sections B to D – Final Answers
| Question | Answer |
|---|---|
| 9 | \(k = -2\) |
| 10 | \(\{0, 1, 2, 3, 4, 5\}\) |
| 11 | \(x = 3\) or \(x = 5\) |
| 12 | \(-2 < x \le 4\) |
| 13 | \(x = 1.5\) or \(x = -5.5\) |
| 14 | \(a = 20\); \(d = -4\); \(S_{16} = -160\) |
| 15 | \(\begin{bmatrix}1 & 6 \\ 8 & 7\end{bmatrix}\) |
| 16 | \(x = 6\) |
| 17 | \(\begin{bmatrix}0 & 30 \\ 22 & 28\end{bmatrix}\) |
| 18 | \(a : b = 7 : 1\) |
| 19 | \(k = 4\); \((x - 1)(3x - 1)(x + 4)\) |
Detailed Solutions
Section A – Multiple Choice Questions
Since 3 is a root, \(3^2 - 3p + 3 = 0\), so \(9 - 3p + 3 = 0\) and \(p = 4\).
Real and equal roots require the discriminant to be zero. Here \(b = -q\), so \(b^2 - 4ac = q^2 - 4pr = 0\), i.e. \(q^2 = 4pr\).
Mean proportional \(m\) satisfies \(4 : m = m : 9\), so \(m^2 = 36\) and \(m = 6\).
By the remainder theorem, remainder \(= P(1) = 1 - 2 + 1 + 5 = 5\).
\(a = 15\), \(d = 15\), so \(T_n = 15n\). Putting \(15n = 300\) gives \(n = 20\).
\(a = 15\), \(r = 2\), so \(T_n = 15 \times 2^{n-1}\).
\(2A = \begin{bmatrix}4 & 4 \\ 6 & 8\end{bmatrix}\), \(I = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\). \(2A - I = \begin{bmatrix}1 & 4 \\ 6 & 7\end{bmatrix}\).
\(5 : 8 = 15 : x\) gives \(5x = 8 \times 15 = 120\), so \(x = 24\).
Section B – Short Answer Questions
\(P(2) = 2^3 - 2k - 12 = 8 - 2k - 12 = -4 - 2k = 0\). Hence \(k = -2\).
\(2x \le 10\), so \(x \le 5\). With \(x \in \mathbb{W}\), the solution set is \(\{0, 1, 2, 3, 4, 5\}\).
\(x^2 - 8x + 15 = (x - 3)(x - 5) = 0\).
Split into \(-3 < 2x + 1\) and \(2x + 1 \le 9\). Subtract 1: \(-4 < 2x \le 8\). Divide by 2: \(-2 < x \le 4\).
Section C – Application Questions
Use the quadratic formula, keeping the surd form until the last step.
\(a = 1, b = 4, c = -8\). \(D = 16 + 32 = 48\).
\(x = \dfrac{-4 \pm \sqrt{48}}{2} = -2 \pm 2\sqrt{3} \approx 1.464 \text{ or } -5.464\).
Write two equations in \(a\) and \(d\), solve, then use the sum formula.
\(a + 4d = 4\) and \(a + 8d = -12\). Subtracting: \(4d = -16\), so \(d = -4\) and \(a = 20\).
\(S_{16} = \dfrac{16}{2}\big[2(20) + 15(-4)\big] = 8(40 - 60) = -160\).
\(2A = \begin{bmatrix}4 & 6 \\ 8 & 10\end{bmatrix}\), \(3I = \begin{bmatrix}3 & 0 \\ 0 & 3\end{bmatrix}\).
\(2A - 3I = \begin{bmatrix}4 - 3 & 6 - 0 \\ 8 - 0 & 10 - 3\end{bmatrix}\).
Continued proportion: \(2 : x = x : 18\), so \(x^2 = 36\). Since \(x\) is positive, \(x = 6\).
Section D – Structured / Long Answer Questions
Add \(B + C\), multiply by \(A\) row × column, then subtract \(14I\).
\(B + C = \begin{bmatrix}5 & 3 \\ 3 & 9\end{bmatrix}\).
\(A(B + C) = \begin{bmatrix}1 & 3 \\ 2 & 4\end{bmatrix}\begin{bmatrix}5 & 3 \\ 3 & 9\end{bmatrix} = \begin{bmatrix}14 & 30 \\ 22 & 42\end{bmatrix}\).
\(A(B + C) - 14I = \begin{bmatrix}14 & 30 \\ 22 & 42\end{bmatrix} - \begin{bmatrix}14 & 0 \\ 0 & 14\end{bmatrix}\).
Take cube roots, then cross-multiply and simplify to a ratio.
\(\dfrac{a + b}{a - b} = \dfrac{\sqrt[3]{64}}{\sqrt[3]{27}} = \dfrac{4}{3}\).
\(3(a + b) = 4(a - b)\) gives \(3a + 3b = 4a - 4b\), so \(7b = a\).
Use the factor theorem to find \(k\), then divide and factorise fully.
\(P(1) = 3 + 8 - 15 + k = -4 + k = 0\), so \(k = 4\).
Polynomial: \(3x^3 + 8x^2 - 15x + 4 = (x - 1)(3x^2 + 11x - 4)\).
\(3x^2 + 11x - 4 = (3x - 1)(x + 4)\).