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A-Level Mathematics 9709 Paper 1 (Pure 1, 10 Questions, 100 Marks, 60 Minutes)
A-Level Maths 9709 Paper 1
A-Level Mathematics 9709 Paper 1 Pure 1 practice paper (10 questions, 100 marks, 60 minutes): completing the square, perpendicular bisectors, sector and segment areas.
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Paper details Level: A-Level | Subject: Mathematics | Paper: 1 (Pure Mathematics 1) | Time: 60 minutes | Marks: 100 | Questions: 10
Section A Algebra, Coordinate Geometry and Trigonometry (46 marks)
Questions 1-2
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | 1. Let f(x) = 2x^2 – 8x + 5. (a) Express f(x) in the form a(x – b)^2 + c. (3 marks) (b) Hence state the coordinates of the vertex of y = f(x) and the range of f. (2 marks) (c) Solve f(x) = 0, giving your answers in simplest surd form. (3 marks) |
(a) f(x) = 2(x^2 – 4x) + 5 = 2[(x – 2)^2 – 4] + 5 = 2(x – 2)^2 – 3. (3 marks: factor 1, completing the square 1, constant 1) (b) Vertex (2, -3); range f(x) >= -3. (2 marks) (c) 2(x – 2)^2 = 3, so (x – 2)^2 = 3/2 and x = 2 +/- sqrt(6)/2, i.e. x = 3.2247 or 0.7753. (3 marks) Marking notes: vertex with no working scores at most 1; an un-simplified surd such as sqrt(1.5) loses 1; a missing +/- scores 0. |
| 2 | 2. The points A(1, 2) and B(7, 10) are given. (a) Find the length AB. (2 marks) (b) Find the coordinates of the midpoint M of AB. (1 mark) (c) Find the equation of the perpendicular bisector of AB, in the form ax + by = c. (4 marks) (d) Find where this perpendicular bisector meets the x-axis. (2 marks) |
(a) AB = sqrt((7 – 1)^2 + (10 – 2)^2) = sqrt(100) = 10. (2 marks) (b) M = (4, 6). (1 mark) (c) Gradient of AB = 8/6 = 4/3, so the perpendicular gradient is -3/4. Through (4, 6): y – 6 = -3/4 (x – 4) giving 3x + 4y = 36. (4 marks: gradient 2, point-slope 1, rearrangement 1) (d) Setting y = 0 gives x = 12, so (12, 0). (2 marks) Marking notes: using 4/3 directly loses (c) and (d); the condition m1*m2 = -1 must be stated. |
Questions 3-4
| # | Question | Answer and marking notes |
|---|---|---|
| 3 | 3. A sector OAB has radius 9 cm and angle AOB = 1.2 radians. (a) Find the length of the arc AB. (2 marks) (b) Find the perimeter of the sector. (2 marks) (c) Find the area of the sector. (2 marks) (d) Find the area of the segment bounded by the chord AB and the arc AB, correct to 3 decimal places. (4 marks) |
(a) Arc = r*theta = 9 * 1.2 = 10.8 cm. (2 marks) (b) Perimeter = 18 + 10.8 = 28.8 cm. (2 marks) (c) Area = 0.5 * 81 * 1.2 = 48.6 cm^2. (2 marks) (d) Triangle area = 0.5 * 81 * sin(1.2) = 37.7476 cm^2, so segment = 48.6 – 37.7476 = 10.852 cm^2. (4 marks: triangle formula 2, subtraction 1, units and rounding 1) Marking notes: the angle must be in radians; treating 1.2 as degrees earns formula marks only. |
| 4 | 4. (a) Expand (3 – 2x)^5 up to and including the term in x^3. (4 marks) (b) Hence find the coefficient of x^2 in (3 – 2x)^5(1 + x). (3 marks) (c) Using x = 0.1 in your expansion from (a), estimate (3 – 2x)^5 and compare with the exact value. (2 marks) |
(a) (3 – 2x)^5 = 243 – 810x + 1080x^2 – 720x^3 + … (4 marks: 1 per term) (b) Coefficient of x^2 = 1080 + (-810) = 270. (3 marks) (c) At x = 0.1: 243 – 81 + 10.8 – 0.72 = 172.08; exact value 2.8^5 = 172.10368, an error of about 0.0237 caused by discarding the x^4 and x^5 terms. (2 marks) Marking notes: binomial coefficients must come from nCr; dropping the minus sign or the power of 2 is the commonest error. |
Question 5
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | 5. (a) Show that 3sin^2(theta) – 2cos(theta) – 2 = 0 can be written as 3cos^2(theta) + 2cos(theta) – 1 = 0. (3 marks) (b) Solve 3sin^2(theta) – 2cos(theta) – 2 = 0 for 0 <= theta < 360 degrees. (5 marks) (c) Hence solve 3sin^2(2x) – 2cos(2x) – 2 = 0 for 0 <= x < 180 degrees. (2 marks) |
(a) Using sin^2 = 1 – cos^2: 3 – 3cos^2 – 2cos – 2 = 0, so -3cos^2 – 2cos + 1 = 0 and hence 3cos^2 + 2cos – 1 = 0. (3 marks) (b) (3cos – 1)(cos + 1) = 0 so cos(theta) = 1/3 or cos(theta) = -1, giving theta = 70.5, 289.5 or 180 degrees. (5 marks) (c) 2x = 70.5, 180 or 289.5, so x = 35.3, 90 or 144.7 degrees. (2 marks) Marking notes: only 180 degrees scores at most 2; the case cos(theta) = -1 must be handled explicitly. |
Section B Calculus, Series and Circles (54 marks)
Questions 6-7
| # | Question | Answer and marking notes |
|---|---|---|
| 6 | 6. The curve C has equation y = x^3 – 6x^2 + 9x + 2. (a) Find dy/dx. (2 marks) (b) Find the coordinates of the stationary points of C. (4 marks) (c) Determine the nature of each stationary point. (3 marks) (d) Find the equation of the tangent to C at x = 2. (2 marks) |
(a) dy/dx = 3x^2 – 12x + 9. (2 marks) (b) 3x^2 – 12x + 9 = 0 gives (x – 1)(x – 3) = 0, so x = 1 or x = 3, with y = 6 and y = 2. The stationary points are (1, 6) and (3, 2). (4 marks) (c) d2y/dx2 = 6x – 12, which is -6 at x = 1 (maximum) and +6 at x = 3 (minimum). (3 marks) (d) At x = 2, y = 4 and the gradient is -3, so y = -3x + 10. (2 marks) Marking notes: x-values without y-coordinates lose 2; the nature must be justified by the second derivative or a sign table. |
| 7 | 7. (a) Find the integral of 6x^2 – 4x + 3 with respect to x. (3 marks) (b) Find the area enclosed by y = 6x^2 – 4x + 3, the x-axis and the lines x = 0 and x = 2. (4 marks) (c) Find the area enclosed between the curve y = 6x^2 – 4x + 3 and the line y = 2x + 3 for 0 <= x <= 2. (3 marks) |
(a) Integral = 2x^3 – 2x^2 + 3x + C. (3 marks: 1 per term including +C) (b) Area = [2x^3 – 2x^2 + 3x] from 0 to 2 = 16 – 8 + 6 = 14 square units. (4 marks) (c) Curve minus line = 6x^2 – 6x, area = [2x^3 – 3x^2] from 0 to 2 = 16 – 12 = 4 square units. (3 marks) Marking notes: an area must be positive; a missing substitution of the limits loses 1. |
Questions 8-10
| # | Question | Answer and marking notes |
|---|---|---|
| 8 | 8. An arithmetic progression has first term 7 and common difference 3. (a) Find the 20th term. (2 marks) (b) Find the sum of the first 20 terms. (2 marks) A geometric progression has first term 8 and common ratio 3/4. (c) Find the sum to infinity. (3 marks) (d) Find the smallest n for which the sum of the first n terms exceeds 30. (4 marks) |
(a) u20 = 7 + 19*3 = 64. (2 marks) (b) S20 = 10*(7 + 64) = 710. (2 marks) (c) S = 8/(1 – 3/4) = 32. (3 marks) (d) 32[1 – (3/4)^n] > 30 gives (3/4)^n < 1/16, so n > 9.64 and n = 10. Check: S9 = 29.597 and S10 = 30.198. (4 marks) Marking notes: (d) needs the inequality and a check on both sides; a bare n = 10 scores at most 2. |
| 9 | 9. Let f(x) = x^2 – 4x + 7 for x >= 2. (a) Find f-1(x) and state its domain. (4 marks) (b) Solve |2x – 5| = x + 1. (4 marks) (c) Solve the inequality |2x – 5| < x + 1. (4 marks) |
(a) y = (x – 2)^2 + 3 gives x = 2 + sqrt(y – 3) (positive root since x >= 2), so f-1(x) = 2 + sqrt(x – 3) with domain x >= 3. (4 marks: completing the square 1, choice of root 1, formula 1, domain 1) (b) 2x – 5 = x + 1 gives x = 6; 2x – 5 = -(x + 1) gives x = 4/3. Both satisfy the original equation. (4 marks) (c) We need x + 1 > 0 and -(x + 1) < 2x – 5 < x + 1, giving x < 6 and x > 4/3, so 4/3 < x < 6. (4 marks) Marking notes: (b) missing the second case scores at most 2; (c) must handle the implicit condition x + 1 > 0. |
| 10 | 10. The circle C has equation x^2 + y^2 – 6x + 8y – 24 = 0. (a) Find the centre and radius of C. (3 marks) (b) Show that the line y = 2x – 1 meets C at two distinct points, stating the quadratic obtained. (3 marks) (c) The point P(3, 3) lies on C. Find the tangent to C at P. (4 marks) |
(a) (x – 3)^2 + (y + 4)^2 = 49, so centre (3, -4) and radius 7. (3 marks) (b) Substituting gives (x – 3)^2 + (2x + 3)^2 = 49, i.e. 5x^2 + 6x – 31 = 0. The discriminant is 6^2 + 620 = 656 > 0, so there are two distinct intersections. (3 marks) (c) The radius to P is vertical (x = 3), so the tangent is horizontal: y = 3. (4 marks) Marking notes: (a) requires completing the square; (b) needs the discriminant conclusion, the equation alone scores 0. |
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