中文（繁體）
中文（简体）
English
DSE Mathematics (Compulsory Part) Practice Paper: Quadratic Equations and Functions (12 Questions, 50 Marks, 75 Minutes)
DSE Mathematics: Quadratic Equations and Functions
DSE Mathematics (Compulsory Part): 12 original questions, 50 marks, 75 minutes on quadratics, discriminants, completing the square and inequalities.
Download Question paper PDF | Answer key PDF
Paper details Level: S6 | Subject: Mathematics | Time: 75 minutes | Marks: 50 | Questions: 12
Section A: Short Questions (4 marks each)
Answer all questions. Show the working and the steps of calculation.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | Solve x² − 5x + 6 = 0, and find the sum and the product of its two roots. (4 marks) | Factorising gives (x − 2)(x − 3) = 0, so x = 2 or x = 3. Sum of roots = 2 + 3 = 5; product of roots = 2 × 3 = 6. Marking: 1 mark for factorisation, 1 mark for each root, 1 mark for the sum and product. |
| 2 | Consider the equation 2x² + 4x + k = 0. (a) Find its discriminant in terms of k. (b) If the equation has a repeated root, find k and that root. (4 marks) | (a) Discriminant Δ = 4² − 4 × 2 × k = 16 − 8k. (b) A repeated root means Δ = 0, so 16 − 8k = 0 and k = 2. The equation becomes 2x² + 4x + 2 = 2(x + 1)² = 0, so the repeated root is x = −1. |
| 3 | Express x² + 8x + 11 in the form a(x − h)² + k, and state the vertex and the minimum value of the graph. (4 marks) | x² + 8x + 11 = (x² + 8x + 16) − 16 + 11 = (x + 4)² − 5. The vertex is (−4, −5) and the minimum value is −5. Marking: 2 marks for completing the square, 1 for the vertex, 1 for the minimum value. |
| 4 | Given y = x² − 6x + 5. (a) Find the x-intercepts of the graph. (b) Find the axis of symmetry and the vertex. (4 marks) | (a) Put y = 0: x² − 6x + 5 = (x − 1)(x − 5) = 0, so the graph cuts the x-axis at (1, 0) and (5, 0). (b) The axis of symmetry is x = 6 ÷ 2 = 3; then y = 9 − 18 + 5 = −4, so the vertex is (3, −4). |
| 5 | Let α and β be the roots of 3x² − 7x + 2 = 0. (a) Find α + β and αβ. (b) Find the quadratic equation with integral coefficients whose roots are 1/α and 1/β. (4 marks) | (a) α + β = 7/3 and αβ = 2/3. (b) 1/α + 1/β = (α + β) ÷ αβ = (7/3) ÷ (2/3) = 7/2, and 1/α × 1/β = 1 ÷ (2/3) = 3/2. The required equation is x² − (7/2)x + 3/2 = 0, that is 2x² − 7x + 3 = 0 after multiplying by 2. |
| 6 | A ball is thrown vertically upwards from the ground. Its height above the ground after t seconds is h = −5t² + 20t metres, where 0 ≤ t ≤ 4. (a) Find the greatest height reached by the ball. (b) Find the times when the ball is 15 m above the ground. (4 marks) | (a) h = −5(t² − 4t) = −5[(t − 2)² − 4] = −5(t − 2)² + 20, so the height is greatest at t = 2, namely 20 m. (b) Put −5t² + 20t = 15, giving t² − 4t + 3 = 0, that is (t − 1)(t − 3) = 0, so t = 1 s or t = 3 s. |
Section B: Application Questions (4 marks each)
For equations and inequalities, show the factorisation or formula used and state the answers clearly.
| # | Question | Answer and marking notes |
|---|---|---|
| 7 | Solve the inequality x² − 3x − 10 ≤ 0. (4 marks) | x² − 3x − 10 = (x − 5)(x + 2) ≤ 0. The critical values are −2 and 5. The parabola opens upwards, so the expression is negative between the roots and the solution is −2 ≤ x ≤ 5. Marking: 2 marks for factorisation, 2 marks for the range. |
| 8 | Determine the number of real roots of 4x² − 12x + 9 = 0, and solve the equation. (4 marks) | Discriminant Δ = (−12)² − 4 × 4 × 9 = 144 − 144 = 0, so the equation has one repeated real root. Since 4x² − 12x + 9 = (2x − 3)² = 0, we get x = 3/2. |
| 9 | A rectangle has perimeter 28 cm and area 45 cm². Find its length and its width. (4 marks) | Let the width be w cm; then the length is (14 − w) cm. From the area, w(14 − w) = 45, that is w² − 14w + 45 = 0, so (w − 5)(w − 9) = 0 and w = 5 or w = 9. Hence the length is 9 cm and the width is 5 cm. |
Section C: Long Questions (14 marks in total)
Show full working in the long questions, and give units in applied answers.
| # | Question | Answer and marking notes |
|---|---|---|
| 10 | The line y = 2x + 3 meets the parabola y = x² − 2x + 1 at two points P and Q. (a) Show that the two graphs have two points of intersection. (b) Find the x-coordinates of P and Q, correct to 3 significant figures. (4 marks) | (a) Put x² − 2x + 1 = 2x + 3, giving x² − 4x − 2 = 0. Its discriminant is Δ = 16 + 8 = 24 ＞ 0, so there are two distinct real roots and hence two points of intersection. (b) By the formula, x = (4 ± √24) ÷ 2 = 2 ± √6, so x ≈ 4.45 or x ≈ −0.449 (3 s.f.). |
| 11 | Given f(x) = 2x² − 12x + 13. (a) Express f(x) in the form a(x − h)² + k. (b) Find the minimum value of f(x) and the corresponding value of x. (4 marks) | (a) f(x) = 2(x² − 6x) + 13 = 2[(x − 3)² − 9] + 13 = 2(x − 3)² − 5. (b) Since (x − 3)² ≥ 0, f(x) ≥ −5; the minimum value is −5, attained at x = 3. Marking: 2 marks for completing the square, 1 for the minimum, 1 for the value of x. |
| 12 | A shop sells notebooks. When the price is x dollars per notebook, the monthly profit P in dollars is P = −2x² + 80x − 600, where 10 ≤ x ≤ 30. (a) Find the price that gives the greatest monthly profit, and find that profit. (b) Find the range of prices for which the monthly profit is positive. (6 marks) | (a) P = −2(x² − 40x) − 600 = −2[(x − 20)² − 400] − 600 = −2(x − 20)² + 200. Since −2(x − 20)² ≤ 0, the profit is greatest when x = 20, and the greatest profit is $200. (b) P ＞ 0 means −2x² + 80x − 600 ＞ 0; dividing by −2 reverses the inequality, giving x² − 40x + 300 ＜ 0, that is (x − 10)(x − 30) ＜ 0, so 10 ＜ x ＜ 30. Marking: 2 marks for completing the square, 2 for the maximum profit, 2 for the range. |
Find a tutor — freeTutor Registration →
A TutorZone practice paper; original questions and passages, not publisher material.
