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Cambridge AS & A Level Mathematics (9709), Paper 1 Pure Mathematics 1, May/June 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_s21_qp_119709_s21_ms_119709/11/M/J/21
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 equation of a curve is such that = 4 + 32x3 . It is given that the curve passes through the point 1 dx x 2 , 4 . Find the equation of the curve. [4]
Mark scheme
1 1 B1 B1 OE. Accept unsimplified. [ y =] − 3 + 8 x 4 [+ c] x 1 M1 1 4 = –8 + +c Substituting , 4 into an integrated expression 2 2 1 23 A1 OE. Accept −x−3 ; must be 8; y = must be seen in y=− 3 + 8x4 + x 2 working. 4
Question
The sum of the first 20 terms of an arithmetic progression is 405 and the sum of the first 40 terms is 1410. Find the 60th term of the progression. [5]
Mark scheme
2 10(2a + 19d) = 405 B1 20(2a + 39d) = 1410 B1 Solving simultaneously two equations obtained from using the correct sum M1 Reach a = or d = formulae [a = 6, d = 1.5] Using the correct formula for 60th term with their a and d M1 60th term = 94.5 A1 189 OE, e.g. 2 5
Question
(a) Find the first three terms in the expansion of 3 − 2x5 in ascending powers of x. [3] (b) Hence find the coefficient of x2 in the expansion of 4 + x2 3 − 2x5 . [3]
Mark scheme
3(a) 243 B1 −810x B1 +1080x2 B1 3 3(b) (4 + x)2 = 16 + 8x + x2 B1 Coefficient of x2 is 16 × 1080 + 8 × (−810) + 243 M1 Allow if at least 2 pairs used correctly 11043 A1 Allow 11043x2 3
Question
y 6 4 2 x − 14 π 0 1 4π 1 2π 3 4π π 5 4π 3 2π 7 4π −2 −4 −6 The diagram shows part of the graph of y = a tan x − b + c. Given that 0 < b < π, state the values of the constants a, b and c. [3]
Mark scheme
4 a=2 B1 π B1 2π b= or 4 8 c=1 B1 3
Question
The fifth, sixth and seventh terms of a geometric progression are 8k, −12 and 2k respectively. Given that k is negative, find the sum to infinity of the progression. [4]
Mark scheme
5 (−12)2 = 8k × 2k M1 Forming an equation in k k = −3 A1 Using correct formula for S∞ [r = 0.5, a = −384] M1 With −1 < r < 1 S∞ = −768 A1 Alternative method for Question 5 2k M1 r2 = 8k r = [ ± ] 0.5 A1 Using correct formula for S∞ [r = 0.5, a = −384] M1 −1 < r < 1 S∞ = −768 A1 4
Question
The equation of a curve is y = 2k − 3x2 − kx − k − 2, where k is a constant. The line y = 3x − 4 is a tangent to the curve. Find the value of k. [5] 1 − 2 sin2 1
Mark scheme
6 ( 2k − 3) x 2 − kx − ( k − 2 ) = 3x − 4 *M1 Equating curve and line ( 2k − 3) x 2 − ( k + 3) x − ( k − 6 ) [ = 0] DM1 Forming a 3-term quadratic ( k + 3)2 + 4 ( 2k − 3 )( k − 6 ) [ = 0] DM1 Use of discriminant (dependent on both previous M marks) 9k 2 − 54k + 81[ = 0] [leading to k 2 − 6k + 9 = 0 ] M1 Simplifying and solving their 3-term quadratic in k k=3 A1 Alternative method for Question 6 ( 2k − 3) x 2 − kx − ( k − 2 ) = 3x − 4 *M1 Equating curve and line k +3 3 + 6x DM1 Differentiating and solving for x or k 2 ( 2k − 3) x − k = 3 ⇒ x = or k = 4k − 6 4x − 1 k +3 2 k +3 k +3 DM1 Substituting their x into equation or their Either ( 2k − 3) −k − ( k − 2) = 3 −4 4 k − 6 4 k − 6 4k − 6 3x 2 + 3x − 6 3x + 6 k= 2 or k = into derivative 3x 2 + 3x − 6 3x 2 + 3x − 6 2x − x −1 2x + 1 Or 4 x 2 − 6 x − 2 =3 equation (dependent on both previous M marks) 2x − x −1 2x − x −1 9k 2 − 54k + 81[ = 0] [leading to k 2 − 6k + 9 = 0] M1 Simplifying and solving their 3-term quadratic in k (or solving for x) k=3 A1 SC If M0, B1 for differentiating, equating to 3 and solving for x or k 5
Question
(a) Prove the identity 1 − tan2 1. [2] 1 − sin2 1 1 − 2 sin2 1 (b) Hence solve the equation = 2 tan4 1 for 0Å ≤ 1 ≤ 180Å. [3] 1 − sin2 1
Mark scheme
7(a) cos 2θ − sin 2θ 1 − sin 2θ sin 2θ sin 2θ + cos 2θ M1 May start with 1 − tan 2θ Reach 2 or 2 − 2 or 2 − 2tan 2θ cos θ 1 − sin θ cos θ cos θ 2 2sin θ cos2θ or sec2 θ − 2 or 2 − sec2 θ or cos θ cos 2θ 1 − tan 2θ A1 AG, must show sufficient stages 2 7(b) 1 − tan 2θ = 2tan 4θ ⇒ 2tan 4θ + tan 2θ −1 [= 0] M1 Forming a 3-term quadratic in tan 2θ or e.g. u tan 2θ = 0.5 or −1 leading to tanθ = [ ± ] 0.5 M1 θ = 35.3° and 144.7° (AWRT) A1 Both correct. Radians 0.615, 2.53 scores A0. 3
Question
P Q C S R The diagram shows a symmetrical metal plate. The plate is made by removing two identical pieces from a circular disc with centre C. The boundary of the plate consists of two arcs PS and QR of the original circle and two semicircles with PQ and RS as diameters. The radius of the circle with centre C is 4 cm, and PQ = RS = 4 cm also. (a) Show that angle PCS = 23 π radians. [2] (b) Find the exact perimeter of the plate. [3] (c) Show that the area of the plate is 20 3 π + 8 3 cm2. [5]
Mark scheme
8(a) 2 π M1 Either Let midpoint of PQ be H: sin HCP = ⇒ Angle HCP = 4 6 4 π Or sin PSQ = ⇒ Angle PSQ = 8 6 π Or using cosine rule: angle PCQ = 3 π Or by inspection: triangle PCQ or PCT is equilateral so angle PCQ = 3 π π 2 A1 AG Angle PCS = π − − = π 6 6 3 2 8(b) 2π 8π M1 Length of two arcs PS and QR Perimeter = 2 × 4 × or 8π − 3 3 +2π × 2 M1 Adding circumference of two semicircles 28π A1 Must be a single term 3 3 8(c) 1 2 π 8π M1 Uses correct formula for sector Area sector CPQ = ×4 × = 2 3 3 Area of segment of large circle beyond CPQ M1 Attempts to find area of segment 8π 1 2 π 8π = − × 4 × sin = − 4 3 3 2 3 3 Area of small semicircle = π × 2 or area of small circle = π × 22 M1 Area of plate = Large circle – [2 ×] small semicircle – [2 ×] segment area M1 8π 20π A1 AG π × 42 − π × 22 − 2 × − 4 3 = +8 3 3 3 Alternative method for Question 8(c) 1 2π 16π M1 Uses correct formula for sector Area of sector PCS = × 42 × = 2 3 3 1 2 π M1 Uses correct formula for triangle Area of triangle PCQ = × 4 × sin = 4 3 2 3 Area of small semicircle = π × 2 or area of circle = π × 22 M1 Area of plate = [2 ×] large sector + [2 ×] triangle – [2 ×] small semicircle M1 16π 20π A1 AG 2 3 ( ) 2 + 2 4 3 − π×2 = 3 +8 3 5
Question
Functions f and g are defined as follows: f x = x − 22 − 4 for x ≥ 2, g x = ax + 2 for x ∈ >, where a is a constant. (a) State the range of f. [1] (b) Find f −1 x. [2] (c) Given that a = − 53 , solve the equation f x = g x. [3] (d) Given instead that ggf −1 12 = 62, find the possible values of a. [5]
Mark scheme
9(a) Range of f is f(x) ⩾ −4 B1 Allow y, f or ‘range’ or [ −4, ∞ ) 1 9(b) y = (x – 2)2 – 4 ⇒ (x – 2)2 = y + 4 ⇒ x – 2 = + ( y + 4 ) or ± ( y + 4 ) M1 May swap variables here f −1 ( x ) = ( x + 4) + 2 A1 2 9(c) 5 5 7 M1 Equating and simplifying to a 3-term quadratic (x – 2)2 – 4 = – x + 2 ⇒ x2 – 4x + 4 – 4 = – x + 2 [⇒ x 2 − x − 2 = 0] 3 3 3 7 ± 7 2 − 4 ( 3)( −6 ) M1 Solving quadratic ( 3x + 2 )( x − 3) [ = 0] or OE 6 x = 3 only A1 3 9(d) f-1 (12) = 6 M1 Substitute 12 into their f–1 (x) and evaluate g(f–1 (12)) = 6a + 2 M1 Substitute their ‘6’ into g(x) g(g(f–1 (12))) = a(6a + 2) + 2 = 62 M1 Substitute the result into g(x) and = 62 6a2 + 2a – 60 [= 0] M1 Forming and solving a 3-term quadratic 10 A1 a= − or 3 3 Alternative method for Question 9(d) g(f–1 (x)) = a ( x + 4 + 2 ) + 2 or gg(x) = a ( ax + 2 ) + 2 M1 Substitute their f-1 (x) or g(x) into g(x) g(g(f–1 (x))) = a a ( ( x + 4 + 2) + 2) + 2 M1 Substitute the result into g(x) g(g(f–1 (12))) = a(6a + 2) + 2 = 62 M1 Substitute 12 and = 62 6a2 + 2a – 60 [= 0] M1 Forming and solving a 3-term quadratic 10 A1 a= − or 3 3 5
Question
The equation of a circle is x2 + y2 − 4x + 6y − 77 = 0. (a) Find the x-coordinates of the points A and B where the circle intersects the x-axis. [2] (b) Find the point of intersection of the tangents to the circle at A and B. [6]
Mark scheme
10(a) When y = 0 x2 − 4 x − 77 = 0 [⇒ ( x + 7 )( x − 11) = 0 or ( x − 2 ) = 81 ] 2 M1 Substituting y = 0 So x-coordinates are −7 and 11 A1 2 10(b) Centre of circle C is (2, −3) B1 1 1 M1 For either gradient (M1 sign error, Gradient of AC is − or Gradient of BC is M0 if x-coordinate(s) in numerator) 3 3 Gradient of tangent at A is 3 or Gradient of tangent at B is −3 M1 For either perpendicular gradient Equations of tangents are y = 3x + 21, y = −3x + 33 A1 For either equation Meet when 3x + 21 = −3x + 33 M1 OR: (centre of circle has x coordinate 2) so x coordinate of point of intersection is 2 Coordinates of point of intersection (2, 27) A1 Alternative method for Question 10(b) dy B1 Implicit differentiation: 2 y seen dx dy dy M1 Fully differentiated = 0 with at least one term 2x − 4 + 2 y +6 =0 dx dx involving y differentiated correctly Gradient of tangent at A is 3 or Gradient of tangent at B is −3 M1 For either gradient Equations of tangents are y = 3x + 21, y = −3x + 33 A1 For either equation Meet when 3x + 21 = −3x + 33 M1 OR: (centre of circle has x coordinate 2) so x coordinate of point of intersection is 2 Coordinates of point of intersection (2, 27) A1 6
Question
The equation of a curve is y = 2 3x + 4 − x. (a) Find the equation of the normal to the curve at the point 4, 4, giving your answer in the form y = mx + c. [5] (b) Find the coordinates of the stationary point. [3] (c) Determine the nature of the stationary point. [2] (d) Find the exact area of the region bounded by the curve, the x-axis and the lines x = 0 and x = 4. [4] 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) dy −0.5 B1 B1 B1 All correct with 1 error, B2 if all correct = 3 ( 3x + 4 ) − 1 dx 1 *M1 Substituting x = 4 into a differentiated expression and Gradient of tangent = − and Gradient of normal = 4 4 using m1 m2 =−1 Equation of line is (y – 4) = 4(x – 4) or evaluate c DM1 With (4, 4) and their gradient of normal So y = 4x – 12 A1 5 11(b) 3 (3x + 4 ) −0.5 −1 = 0 M1 dy Setting their =0 dx Solving as far as x = M1 dy −0.5 Where contains a ( bx + c ) a, b, c any values dx 5 5 0.5 5 13 A1 x= , y = 2 3× + 4 − = 3 3 3 3 3 11(c) d2 y 9 −1.5 M1 dy dy = − ( 3x + 4 ) Differentiating their OR checking to find +ve dx 2 2 dx dx 5 and -ve either side of their x = 3 5 d2 y A1 At x = is negative so the point is a maximum 3 dx 2 2 11(d) 4 1 B1 B1 B1 for each correct term (unsimplified) Area = 2 ( 3 x + 4 ) − x dx = ( 3 x + 4 ) − x 2 0.5 1.5 9 2 4 1 2 4 1.5 256 32 M1 Substituting limits 0 and 4 into an expression obtained (16 ) − ( 4 ) − ( 4 ) = 1.5 −8− by integrating y 9 2 9 9 9 8 A1 152 16 Or 9 9 4
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Mathematics 9709/11 for May/June 2021 is out of 75 marks, with 1 hour 50 minutes to answer it.
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