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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_129709_s20_ms_129709/12/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
(a) Find the coefficient of x in the expansion of x − . [2] x @ A 2 6 (b) Find the coefficient of x2 in the expansion of 2 + 3x2 x − . [3] x
Mark scheme
1(a) B1 ( 2 + 3x ) ( x − 2x ) 6 2 −2 Term in x² in ( ) 6 x − 2x = 15 x 4 × x Coefficient = 60 B1 2 1(b) −2 3 B2, 1 Constant term in ( x − 2x ) = 20 x 3 × (−160) 6 x B1FT Coefficient of x² in ( 2 + 3x ) ( x − 2x ) = 120 – 480 = −360 6 3
Question
(a) Express the equation 3 cos 1 = 8 tan 1 as a quadratic equation in sin 1. [3] (b) Hence find the acute angle, in degrees, for which 3 cos 1 = 8 tan 1. [2]
Mark scheme
2(a) 8sin θ M1 3cos θ = 8 tan θ → 3cosθ = cos θ 3(1 − sin 2 θ ) = 8 sin θ M1 3 sin² θ + 8 sin θ – 3 = 0 A1 3 2(b) (3 sin θ – 1)( sin θ + 3) = 0 → sin θ = ⅓ M1 θ = 19.5° A1 2
Question
A weather balloon in the shape of a sphere is being inflated by a pump. The volume of the balloon is increasing at a constant rate of 600 cm3 per second. The balloon was empty at the start of pumping. (a) Find the radius of the balloon after 30 seconds. [2] (b) Find the rate of increase of the radius after 30 seconds. [3]
Mark scheme
3(a) 4 M1 Volume after 30 s = 18000 πr ³ = 18000 3 r = 16.3 cm A1 2 3(b) dV B1 = 4πr ² dr dr dr dV 600 M1 = × = dt dV dt 4πr ² dr A1 = 0.181 cm per second dt 3
Question
The nth term of an arithmetic progression is 12 3n − 15. Find the value of n for which the sum of the first n terms is 84. [5]
Mark scheme
4 1st term is −6, 2nd term is −4.5 M1 (M1 for using kth terms to find both a and d) → a = −6, d = 1.5 A1 A1 Sn = 84 → 3n² − 27n – 336 = 0 M1 Solution n = 16 A1 5
Question
The function f is defined for x ∈ > by f : x → a − 2x, where a is a constant. (a) Express ff x and f −1 x in terms of a and x. [4] (b) Given that ff x = f −1 x, find x in terms of a. [2]
Mark scheme
5(a) ff(x) = a − 2 ( a − 2 x ) M1 ff(x) = 4 x − a A1 a−x M1 A1 f−1(x) = 2 4 5(b) a−x M1 4x − a = → 9 x = 3a 2 a A1 x= 3 2
Question
The equation of a curve is y = 2x2 + kx + k − 1, where k is a constant. (a) Given that the line y = 2x + 3 is a tangent to the curve, find the value of k. [3] It is now given that k = 2. (b) Express the equation of the curve in the form y = 2 x + a2 + b, where a and b are constants, and hence state the coordinates of the vertex of the curve. [3]
Mark scheme
6(a) 2 x 2 + kx + k − 1 = 2 x + 3 → 2 x 2 + ( k − 2 ) x + k − 4 = 0 M1 Use of b² − 4ac = 0 → (k – 2)² = 8(k – 4) M1 k=6 A1 3 6(b) 1 1 2x² + 2x + 1 = 2 x + ² + 1 − 2 2 1 1 B1 B1 a= ,b= 2 2 1 1 B1FT vertex − , 2 2 (FT on a and b values) 3
Question
A r C r 1 6 π rad O B 2r In the diagram, OAB is a sector of a circle with centre O and radius 2r, and angle AOB = 16 π radians. The point C is the midpoint of OA. (a) Show that the exact length of BC is r 5 − 2 3. [2] (b) Find the exact perimeter of the shaded region. [2] (c) Find the exact area of the shaded region. [3]
Mark scheme
7(a) π M1 BC² = r² + 4r² − 2r.2r × cos = 5r² − 2r²√3 6 BC = r (5 − 2 3 ) A1 2 7(b) 2π r M1 A1 Perimeter = 6 +r + r (5 − 2 3 ) 2 7(c) Area = sector – triangle 1 π M1 Sector area = 4r ² 2 6 π M1 Triangle area = ½ r. 2r sin 6 π 1 A1 Shaded area = r ² − 3 2 3
Question
y 1, 6 3, 2 1, 2 6 y= x x O 6 The diagram shows part of the curve y = . The points 1, 6 and 3, 2 lie on the curve. The shaded x region is bounded by the curve and the lines y = 2 and x = 1. (a) Find the volume generated when the shaded region is rotated through 360Å about the y-axis. [5] (b) The tangent to the curve at a point X is parallel to the line y + 2x = 0. Show that X lies on the line y = 2x. [3]
Mark scheme
8(a) 36 *M1 Volume = π ∫ x²dy = π dy y² −36 A1 = π y Uses limits 2 to 6 correctly → (12π) DM1 Vol of cylinder = π.1².4 or 12.dy = [y] from 2 to 6 M1 Vol = 12π − 4π = 8π A1 5 8(b) dy −6 B1 = dx x² −6 M1 = −2 → x = 3 x² 6 A1 y= = 2 3 Lies on y = 2x 3 3
Question
Functions f and g are such that f x = 2 − 3 sin 2x for 0 ≤ x ≤ π, g x = −2f x for 0 ≤ x ≤ π. (a) State the ranges of f and g. [3] The diagram below shows the graph of y = f x. y y = f x x O π (b) Sketch, on this diagram, the graph of y = g x. [2] The function h is such that h x = g x + π for − π ≤ x ≤ 0. (c) Describe fully a sequence of transformations that maps the curve y = f x on to y = h x. [3]
Mark scheme
9(a) f(x) from −1 to 5 B1B1 g(x) from −10 to 2 B1FT (FT from part (a)) 3 9(b) B2, 1 2 9(c) Reflect in x-axis B1 Stretch by factor 2 in the y direction B1 0 B1 Translation by –π in the x direction OR translation by . −π 3
Question
The equation of a curve is y = 54x − 2x − 73 . dy d2 y (a) Find 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 of the stationary points. [2]
Mark scheme
10(a) dy B2,1 = 54 – 6(2x – 7)² dx d² y B2,1 FT = −24(2x – 7) dx ² (FT only for omission of ‘ ×2 ’ from the bracket) 4 10(b) dy M1 = 0 → ( 2x − 7) = 9 2 dx x = 5, y = 243 or x = 2, y = 135 A1 A1 3 10(c) d² y B1FT x=5 = −72 → Maximum dx ² (FT only for omission of ‘ ×2 ’ from the bracket) d² y B1FT x =2 = 72 → Minimum dx ² (FT only for omission of ‘ ×2 ’ from the bracket) 2
Question
The equation of a circle with centre C is x2 + y2 − 8x + 4y − 5 = 0. (a) Find the radius of the circle and the coordinates of C. [3] The point P 1, 2 lies on the circle. (b) Show that the equation of the tangent to the circle at P is 4y = 3x + 5. [3] The point Q also lies on the circle and PQ is parallel to the x-axis. (c) Write down the coordinates of Q. [2] The tangents to the circle at P and Q meet at T . (d) Find the coordinates of T . [3] 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) Express as ( x − 4 ) ² + ( y + 2 ) = 16 + 4 + 5 2 M1 Centre C(4, −2) A1 Radius = 25 = 5 A1 3 11(b) 4 B1FT P(1,2) to C(4, − 2) has gradient − 3 (FT on coordinates of C) 3 M1 Tangent at P has gradient = 4 3 A1 Equation is y − 2 = ( x − 1) or 4y = 3x + 5 4 3 11(c) Q has the same coordinate as P y = 2 B1 Q is as far to the right of C as P x = 3 + 3 + 1 = 7 Q (7, 2) B1 2 11(d) 3 B1FT Gradient of tangent at Q = − by symmetry 4 (FT from part (b)) 3 M1 Eqn of tangent at Q is y − 2 = − ( x − 7 ) or 4y + 3x = 29 4 17 A1 T (4, ) 4 3
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Mathematics 9709/12 for May/June 2020 is out of 75 marks, with 1 hour 50 minutes to answer it.
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