Cambridge A Level Mathematics 9709 — 2022 May/June Paper 1 · Variant 3

1 4 1 The coefficient of x3 in the expansion of p is 144. + px Find the possible values of the constant p. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 3 3 1 4 1  C p x p B1 OE soi Can be seen in an expansion. 2 4 144  p B1 OE Correct with correct power of p and only one p term. 1 6  p B1 B1 OE ± 2 12 etc. Allow ±0.167 for B1 B1. SC B1 for 1 36  B1 only, 4

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2 y 1 0 1 0 π 2π 3π 4π −1 −2 −3 = sin q1 + r y p −4 −5 The diagram shows part of the curve with equation y p sin r, where p, q and r are constants. = q1 + (a) State the value of p. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) State the value of q. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) State the value of r. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(a) [p =] 3 B1 1 2(b) [q =] 1 2 B1 1 2(c) [r =] ‒2 B1 1

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3 An arithmetic progression has first term 4 and common difference d. The sum of the first n terms of the progression is 5863. 11726 (a) Show that n d [1] −1 = n −8. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Given that the nth term is 139, find the values of n and d, giving the value of d as a fraction. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(a)     8 1 5863 leading to 8 1 11726 2 n n d n n d             11726 leading to 1 8 n d n    B1 Must show a useful intermediate step. WWW AG. 1 3(b)   11726 4 1 139 leading to 8 135 n d n      *M1 OE Use of correct un formula with expression from (a) or Sn formula to eliminate d. n = 11726 143 = 82 A1 11726 81 8 82 d   DM1 Substitute their n into a correct un or Sn formula 5 3 d  A1 Accept 138 81 OE fraction only If M0 DM0 scored them SC B1 B1 for correct n and d values only. 4

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@ A 4 (a) The curve with equation y x2 2x is translated by −1 . = + −5 3 Find the equation of the translated curve, giving your answer in the form y ax2 bx c. [3] = + + ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) The curve with equation y x2 2x is transformed to a curve with equation y 4x2 4x = + −5 = + −5. Describe fully the single transformation that has been applied. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a)     2 1 2 1 5 3      x x , or     2 1 1 6 3   x M1 for dealing with 1 0        and M1 for dealing with 0 3    .   2 4 1    y x x A1 Answer only given full marks. 3 Question Answer Marks Guidance 4(b) {Stretch}{x direction or horizontally or y-axis invariant}{ factor ½} B2, 1, 0 Additional transformation B0. 2

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2 5 (a) Solve the equation 6 y 0. [4] + y −7 = ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 2 (b) Hence solve the equation 6 tan x 0 for [3] + tan x −7 = 0Å ≤x ≤360Å. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) 1/2 6 2 7   y y [= 0] 1 1 2 2 2 1 3 2               y y [= 0] or e.g.    2 1 3 2   u u [= 0] DM1 Or use of formula or completing the square. 1/2 1 2 [ ] , 2 3 y  A1 Answers only SC B1 if DM1 not scored.  1 4 , 4 9 y  A1 Answers only SC B1 if DM1 not scored. 4 5(b) Use of tan x = their y values M1 Must have at least 2 values of y from part (a).  x 14[.0], 24[.0],  x 194[.0], 204[.0] A1 A1 FT FT for 180 + angle (twice). AWRT 3

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7 y B 0, 2 P O x C x −2 2 + y + 4 2 = 20 The diagram shows the circle with equation x 2 y 4 2 20 and with centre C. The point B −2 + + = has coordinates 0, 2 and the line segment BC intersects the circle at P. (a) Find the equation of BC. 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(b) Hence find the coordinates of P, giving your answer in exact form. 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Mark scheme: 7(a) 2 3   y x B2, 1, 0 OE forms 4 3 2 or 2 3 0       y x y x . 2 7(b)     2 2 2 2 3 4 20      x x *M1 OE Sub line equation into equation of circle to eliminate y. 10(x – 2)2 = 20 or [10](x2 – 4x + 2)[= 0] A1 OE Accept (10x2 – 40x + 20).   4 16 8 2 2 or 2      x x DM1 Correctly solving their quadratic. 2 2   x A1 OE only solution. Answer only SC B1 If DM1 not scored. 3 2 4   y A1 OE only solution. Answer only SC B1 If DM1 not scored. 5

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8 y 1 2 2 + 4x−1 y = x A 1, 5 B 16, 5 x O 1 5 intersects the curve at the The diagram shows the curve with equation y x 2 2. The line y = + 4x−1 = points A 1, 5 and B 16, 5 . (a) Find the equation of the tangent to the curve at the point A. 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(b) Calculate the area of the shaded region. 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Mark scheme: 8(a) 1/2 3/2 d ½ 2 d           y x x x At x = 1, d 1 3 2 d 2 2    y x M1 Substitute x = 1 into a differentiated y. Equation of tangent is   3 5 1 2    y x A1 WWW Or 3 13 2 2   y x . 4 Question Answer Marks Guidance 8(b) 3/2 1/2 8 3 / 2  x x B1 OE Integrate to find area under curve, allow unsimplified versions. 128 2 32 8 3 3                      M1 Apply limits 1 → 16 to an integrated expression. Area under line = 15  5 = 75 B1 Or by 16 1 5d  x . Required area = 75 ‒ 66 = 9 A1 4

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9 D B 1.8 rad 6 cm 6 cm C A The diagram shows triangle ABC with AB BC 6cm and angle ABC 1.8 radians. The arc CD is = = = part of a circle with centre A and ABD is a straight line. (a) Find the perimeter of the shaded region. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the area of the shaded region. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 9(a) 6sin0.9 2 AC or 2 2 2 6 6 2 6 6cos1.8     AC M1 OE Correct working in degrees is acceptable throughout. AC = 9.40 A1 SOI Accept 9.39 – 9.41, may be used but not seen for A1. Angle CAB = ½(π ‒ 1.8) M1 SOI Expect 0.6708 (or 0.671). Arc CD = their 9.40  their 0.6708 M1 Expect 6.306 (or 6.31), do not accept 6 for their AC or 1.8 for CAB. [Perimeter = 6 + 3.40 + 6.306 =] 15.7 A1 Accept 15.69 – 15.72. 5 Question Answer Marks Guidance 9(b) Sector ADC ‒ ABC = ½  their 9.402  their 0.6708 – ½  62  sin 1.8 M1 M1 Accept correct use of their answers from part (a). [29.64 ‒ 17.53 =] 12.1 A1 AWRT 3

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11 The point P lies on the line with equation y mx c, where m and c are positive constants. A curve = + has equation y . There is a single point P on the curve such that the straight line is a tangent to = −mx the curve at P. (a) Find the coordinates of P, giving the y-coordinate in terms of m. [6] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The normal to the curve at P intersects the curve again at the point Q. (b) Find the coordinates of Q in terms of m. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 11(a) 2 0 m mx c mx cx m x       M1 All x terms in the numerator. OE e.g.  mx cx m . 2 2 2 4 0 4 0 b ac c m      M1 OE b2 – 4ac = 0 is implied by c2 – 4m2 = 0.  2  c m A1 SOI. Allow  at this stage. 2 mx [  2 ] 2 0 2 1 0 mx m x x       M1 Sub c = +2m Ignore substitution of -2m.   2 1 0 1 x x     only A1  y m only or (‒1, m) only A1 Alternative method to question 11(a) 2 d d  y m x x M1 As this is a method mark a sign error is allowed. 2  m m x 2 1 x   M1 A1 Equating their d d y x and m and attempt to solve. x = ±1 or 1  x A1 If 1 x  and y m  are the only answers offered here award the final M1 A1. Selecting x = –1 as the only answer and attempt to find y M1  y m or (‒1, m) A1 6 Question Answer Marks Guidance 11(b) Equation of normal is   1 1     y m x m *M1 Through their P with gradient 1  m , OE e.g. 2 1 1    m y x m m . Allow use of the gradient of the curve as   2 1 their m x        with their P. Coordinates of P must be in terms of m only.     2 2 2 1 1 0 x m m x x m m m m x           DM1 OE Equating their normal equation to the equation of the curve and removing x from the denominator.     2 2 1 0 x x m x m      A1 or   2 2 2 2 4 2 2 1 1 1 1 2 4 2 2          m m m m m m x m 2 1     m y m m A1 or 2 1 ,        m m , ignore the coordinates of P. 4

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