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Cambridge AS & A Level Mathematics (9709), Paper 1 Pure Mathematics 1, May/June 2020. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_s20_qp_119709_s20_ms_119709/11/M/J/20
Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.
Question
The sum of the first nine terms of an arithmetic progression is 117. The sum of the next four terms is 91. Find the first term and the common difference of the progression. [4] @ A @ A 1 1 5 2 8
Mark scheme
1 9 B1 117 = ( 2 a + 8d ) 2 Either 91 = S4 with ‘a’ as a + 4d or 117 + 91 = S13 M1M1 (M1 for overall approach. M1 for Sn) Simultaneous Equations → a = 7, d = 1.5 A1 4
Question
The coefficient of in the expansion of kx + + 1− is 74. x x x Find the value of the positive constant k. [5]
Mark scheme
2 5 1 2 8 B1B1 kx + + 1 − x x 5 1 Coefficient in kx + = 10 × k² x (B1 for 10. B1 for k²) 2 8 B2,1,0 Coefficient in 1 − = 8 × −2 x 10k² − 16 = 74 → k = 3 B1 5
Question
Each year the selling price of a diamond necklace increases by 5% of the price the year before. The selling price of the necklace in the year 2000 was $36 000. (a) Write down an expression for the selling price of the necklace n years later and hence find the selling price in 2008. [3] (b) The company that makes the necklace only sells one each year. Find the total amount of money obtained in the ten-year period starting in the year 2000. [2]
Mark scheme
3(a) $36 000 × (1.05)n B1M1 (B1 for r = 1.05. M1 method for rth term) $53 200 after 8 years. A1 3 3(b) S10 = 36000 (1.05 − 1) 10 M1 (1.05 − 1) $453 000 A1 2
Question
y y = f x x O π The diagram shows the graph of y = f x, where f x = 32 cos 2x + 12 for 0 ≤ x ≤ π. (a) State the range of f. [2] A function g is such that g x = f x + k, where k is a positive constant. The x-axis is a tangent to the curve y = g x. (b) State the value of k and hence describe fully the transformation that maps the curve y = f x on to y = g x. [2] (c) State the equation of the curve which is the reflection of y = f x in the x-axis. Give your answer in the form y = a cos 2x + b, where a and b are constants. [1]
Mark scheme
4(a) −1 ⩽ f(x) ⩽ 2 B1 B1 2 4(b) k=1 B1 Translation by 1 unit upwards parallel to the y-axis B1 2 4(c) 3 1 B1 y=− cos2 x − 2 2 1
Question
The equation of a line is y = mx + c, where m and c are constants, and the equation of a curve is xy = 16. (a) Given that the line is a tangent to the curve, express m in terms of c. [3] (b) Given instead that m = −4, find the set of values of c for which the line intersects the curve at two distinct points. [3]
Mark scheme
5(a) x ( mx + c ) = 16 → mx 2 + cx − 16 = 0 B1 Use of b² − 4ac = c² + 64m M1 −c ² A1 Sets to 0 → m = 64 3 5(b) x ( −4 x + c ) = 16 M1 Use of b² − 4ac → c² − 256 c > 16 and c < −16 A1 A1 3
Question
Functions f and g are defined for x ∈ > by f : x → 12 x − a, g : x → 3x + b, where a and b are constants. (a) Given that gg 2 = 10 and f −1 2 = 14, find the values of a and b. [4] (b) Using these values of a and b, find an expression for gf x in the form cx + d , where c and d are constants. [2] 1 + sin 1 cos 1 2
Mark scheme
6(a) 3( 3x + b ) + b = 9 x + 4b → 10 = 18 + 4b M1 b = −2 A1 Either f(14) = 2 or f−1(x) = 2(x + a) etc. M1 a=5 A1 4 6(b) 1 M1 gf(x) = 3 x − 5 − 2 2 3 A1 gf(x) = x − 17 2 2
Question
(a) Prove the identity + . [3] cos 1 1 + sin 1 cos 1 1 + sin 1 cos 1 3 (b) Hence solve the equation + = , for 0 ≤ 1 ≤ 2π. [3] cos 1 1 + sin 1 sin 1
Mark scheme
7(a) (1 + sin θ )2 + cos 2 θ M1 cos θ (1 + sin θ ) 2 + 2sin θ 2 M1A1 Use of sin 2 θ + cos 2 θ = 1 → → . cos θ (1 + sin θ ) cosθ 3 7(b) 2 3 M1 = → tanθ = 1.5 cosθ sinθ θ = 0.983 or 4.12 A1 (FT on second value for 1st value + π) A1FT 3
Question
15 cm B A C 6 cm O X In the diagram, ABC is a semicircle with diameter AC, centre O and radius 6 cm. The length of the arc AB is 15 cm. The point X lies on AC and BX is perpendicular to AX . Find the perimeter of the shaded region BXC. [6]
Mark scheme
8 Angle AOB = 15 ÷ 6 = 2.5 radians B1 Angle BOC = π – 2.5 (FT on angle AOB) B1FT BC = 6(π – 2.5) (BC = 3.850) M1 sin(π – 2.5) = BX ÷6 (BX = 3.59) M1 Either OX = 6cos(π – 2.5) or Pythagoras (OX = 4.807) M1 XC = 6 – OX (XC = 1.193) → P = 8.63 A1 6
Question
The equation of a curve is y = 3 − 2x3 + 24x. dy d2 y (a) Find expressions for and 2 . [4] dx dx (b) Find the coordinates of each of the stationary points on the curve. [3] (c) Determine the nature of each stationary point. [2]
Mark scheme
9(a) dy B1B1 = 3(3−2x)² × −2 + 24 = −6 ( 3 − 2 x ) + 24 2 dx (B1 without ×−2. B1 for ×−2) d² y B1FT = −12 ( 3 − 2 x ) ×−2 = 24(3 – 2x) B1 dx² (B1FT from without – 2) 4 9(b) dy M1 = 0 when 6 ( 3 − 2 x ) = 24 → 3 − 2 x = ±2 2 dx x = ½, y = 20 or x = 2½, y = 52 A1A1 (A1 for both x values or a correct pair) 3 9(c) d² y B1FT If x = ½, = 48 Minimum dx ² d² y B1FT If x = 2½, = −48 Maximum dx² 2
Question
The coordinates of the points A and B are −1, −2 and 7, 4 respectively. (a) Find the equation of the circle, C, for which AB is a diameter. [4] (b) Find the equation of the tangent, T , to circle C at the point B. [4] (c) Find the equation of the circle which is the reflection of circle C in the line T . [3]
Mark scheme
10(a) Centre is (3, 1) B1 Radius = 5 (Pythagoras) B1 Equation of C is ( x − 3) + ( y − 1) = 25 2 2 M1 A1FT (FT on their centre) 4 10(b) Gradient from (3, 1) to (7, 4) = ¾ (this is the normal) B1 4 M1 Gradient of tangent = − 3 4 M1A1 Equation is y − 4 = − ( x − 7 ) or 3 y + 4 x = 40 3 4 10(c) B is centre of line joining centres → (11, 7) B1 Radius = 5 M1 New equation is ( x − 11) + ( y − 7 ) = 25 2 2 A1FT (FT on coordinates of B) 3
Question
y A 2y + x = 8 C B 8 y= x+2 x O 8 The diagram shows part of the curve y = and the line 2y + x = 8, intersecting at points A and B. x+2 The point C lies on the curve and the tangent to the curve at C is parallel to AB. (a) Find, by calculation, the coordinates of A, B and C. [6] (b) Find the volume generated when the shaded region, bounded by the curve and the line, is rotated through 360Å about the x-axis. [6] Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
Mark scheme
11(a) 8 M1 Simultaneous equations = 4 − ½x x+2 x = 0 or x = 6 → A (0, 4) and B (6, 1) B1A1 −8 1 B1 At C = − → C (2, 2) ( x + 2) ² 2 M1A1 (B1 for the differentiation. M1 for equating and solving) 6 11(b) x³ M1 Volume under line = π ( − 12 x + 4 ) ²dx = π − 2 x² + 16 x = (42π) A2,1 12 (M1 for volume formula. A2,1 for integration) 8 2 −64 A1 Volume under curve = π dx = π x+2 = (24π) x + 2 Subtracts and uses 0 to 6 → 18π M1A1 6
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Mathematics 9709/11 for May/June 2020 is out of 75 marks, with 1 hour 50 minutes to answer it.
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