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HKDSE Mathematics (Compulsory Part) Practice Paper: Quadratic Equations, Functions and Graphs (12 questions, 100 marks, 75 minutes)
HKDSE Mathematics: Quadratic Equations and Functions
Twelve questions on quadratic equations, discriminants, vertex form, roots and inequalities, with full worked answers and marking points.
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Paper details Level: S6 | Subject: Mathematics | Time: 75 minutes | Marks: 100 | Questions: 12
Section A (32 marks)
Solving equations, discriminants, completing the square and the sum/product of roots: show every step and simplify.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | Solve the following quadratic equations. (a) x^2 – 5x + 6 = 0 (b) 2x^2 + 3x – 2 = 0 (c) x^2 + 4x + 1 = 0 (give the answers in surd form) |
(a) x^2 – 5x + 6 = (x – 2)(x – 3) = 0, so x = 2 or x = 3. (b) Discriminant = 3^2 – 4(2)(-2) = 9 + 16 = 25; x = (-3 ± 5)/4, so x = 1/2 or x = -2. (c) Discriminant = 4^2 – 4(1)(1) = 12; x = (-4 ± √12)/2 = -2 ± √3. |
| 2 | Let k be a constant. (a) If x^2 + kx + 9 = 0 has two equal real roots, find the two possible values of k. (b) Determine the nature of the roots of 2x^2 – 4x + 5 = 0 and explain your answer. |
(a) Equal real roots require the discriminant to be zero: k^2 – 4(1)(9) = 0, so k^2 = 36 and k = 6 or k = -6. (b) Discriminant = (-4)^2 – 4(2)(5) = 16 – 40 = -24, which is less than 0, so the equation has no real roots; the two roots form a pair of conjugate complex numbers. |
| 3 | It is given that y = x^2 – 6x + 5. (a) By completing the square, express y in the form (x – h)^2 + k. (b) Write down the vertex and the axis of symmetry of the graph. (c) Find the minimum value of y and the corresponding value of x. |
(a) y = x^2 – 6x + 5 = (x^2 – 6x + 9) – 9 + 5 = (x – 3)^2 – 4. (b) The vertex is (3, -4) and the axis of symmetry is the line x = 3. (c) Since (x – 3)^2 is at least 0, y is at least -4; the minimum value is -4, attained at x = 3. |
| 4 | Let α and β be the two roots of 3x^2 – 5x – 2 = 0. (a) Find α + β and αβ. (b) Find the value of α^2 + β^2. (c) Find the quadratic equation whose roots are α + 1 and β + 1, with integral coefficients. |
(a) α + β = 5/3 and αβ = -2/3. (b) α^2 + β^2 = (α + β)^2 – 2αβ = 25/9 + 4/3 = 37/9. (c) Sum of new roots = 5/3 + 2 = 11/3; product of new roots = αβ + (α + β) + 1 = -2/3 + 5/3 + 1 = 2; the equation is x^2 – (11/3)x + 2 = 0, that is 3x^2 – 11x + 6 = 0. |
Section B (36 marks)
Function graphs, quadratic inequalities, graph features and word problems: explain your reasoning clearly.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | It is given that f(x) = 2x^2 – 8x + 3. (a) Express f(x) in the form a(x – h)^2 + k. (b) Describe how the graph of y = f(x) is obtained from the graph of y = 2x^2 by translation. (c) Solve f(x) = 3. |
(a) f(x) = 2(x^2 – 4x) + 3 = 2[(x – 2)^2 – 4] + 3 = 2(x – 2)^2 – 5. (b) Translate the graph of y = 2x^2 two units to the right and five units downwards; the vertex moves from (0, 0) to (2, -5). (c) 2(x – 2)^2 – 5 = 3 gives (x – 2)^2 = 4, so x – 2 = ±2 and x = 4 or x = 0. |
| 6 | Solve the following quadratic inequalities. (a) x^2 – x – 6 <= 0 (b) 2x^2 + 5x - 3 > 0 |
(a) (x – 3)(x + 2) <= 0; the critical values are -2 and 3, so the solution is -2 <= x <= 3. (b) 2x^2 + 5x - 3 = (2x - 1)(x + 3) > 0; the critical values are -3 and 1/2, so the solution is x less than -3 or x greater than 1/2. |
| 7 | It is given that the curve is y = x^2 – 4x + 3. (a) Find the intercepts of the graph with the x-axis and the y-axis. (b) Find the vertex of the graph. (c) State the direction in which the graph opens and write down the minimum value of y. |
(a) Putting y = 0 gives (x – 1)(x – 3) = 0, so the x-intercepts are (1, 0) and (3, 0); putting x = 0 gives the y-intercept (0, 3). (b) y = (x – 2)^2 – 1, so the vertex is (2, -1). (c) The coefficient of x^2 is positive, so the graph opens upwards; the minimum value of y is -1, attained at x = 2. |
| 8 | A rectangular garden has a perimeter of 40 m and an area of 96 m^2. (a) If the width of the garden is x m, express its length in terms of x. (b) Set up a quadratic equation in x. (c) Find the length and the width of the garden. |
(a) Length + width = 40 ÷ 2 = 20, so the length is (20 – x) m. (b) Area = x(20 – x) = 96, that is x^2 – 20x + 96 = 0. (c) (x – 8)(x – 12) = 0, so x = 8 or x = 12; the garden measures 12 m by 8 m (check: perimeter 40 m, area 96 m^2). |
Section C (32 marks)
Simultaneous equations, equal-root conditions, maximum values and finding a parameter from the roots: show the full working.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | Solve the simultaneous equations y = x + 1 and y = x^2 – 1. | From the two equations, x^2 – 1 = x + 1, that is x^2 – x – 2 = 0, so (x – 2)(x + 1) = 0 and x = 2 or x = -1. When x = 2, y = 3; when x = -1, y = 0. The solutions are (2, 3) and (-1, 0). |
| 10 | The equation x^2 + (k – 2)x + 9 = 0 has two equal real roots, where k is a constant. (a) Set up an equation in k. (b) Find all possible values of k. (c) For each value of k, write down the corresponding equation and its repeated root. |
(a) Discriminant = (k – 2)^2 – 4(1)(9) = 0, so (k – 2)^2 = 36. (b) k – 2 = ±6, so k = 8 or k = -4. (c) When k = 8 the equation is x^2 + 6x + 9 = 0, that is (x + 3)^2 = 0, with repeated root x = -3; when k = -4 the equation is x^2 – 6x + 9 = 0, that is (x – 3)^2 = 0, with repeated root x = 3. |
| 11 | A shop sells a product: when the price is $p, (120 – 2p) items are sold each month. (a) Express the monthly revenue R in terms of p. (b) By completing the square, find the maximum value of R. (c) Find the price that maximises the revenue and the number of items sold at that price. |
(a) R = p(120 – 2p) = -2p^2 + 120p. (b) R = -2(p^2 – 60p) = -2[(p – 30)^2 – 900] = -2(p – 30)^2 + 1800; since (p – 30)^2 is at least 0, R is at most 1800, so the maximum revenue is $1,800. (c) The maximum is attained at p = 30, so the price is $30; then 120 – 60 = 60 items are sold and the revenue is 30 x 60 = $1,800. |
| 12 | Let α and β be the two roots of x^2 – 7x + k = 0, where k is a constant. It is given that α^2 + β^2 = 25. (a) Express α^2 + β^2 in terms of k. (b) Find k. (c) Hence find α and β. |
(a) α + β = 7 and αβ = k, so α^2 + β^2 = (α + β)^2 – 2αβ = 49 – 2k. (b) 49 – 2k = 25, so 2k = 24 and k = 12. (c) The equation is x^2 – 7x + 12 = 0, that is (x – 3)(x – 4) = 0, so α and β are 3 and 4 (check: 3 + 4 = 7 and 3^2 + 4^2 = 25). |
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