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

1 (a) Express x2 −8x + 11 in the form x + p 2 + q where p and q are constants. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Hence find the exact solutions of the equation x2 −8x + 11 = 1. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(a) B1 If p and q-values given after their completed square expression, mark the expression and ISW. … 5  or q = − 5 B1 2 1(b) (x – 4)2 − 5 = 1 so (x – 4)2 = 6 so  4 6  x M1 Using their p and q values or by quadratic formula 4 6   x or 8 24 2  A1 Or exact equivalent. No FT; must have  for this mark. ISW decimals 1.55, 6.45 if exact answers seen. If M0, SC B1 possible for correct answers. 2

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2 The thirteenth term of an arithmetic progression is 12 and the sum of the first 30 terms is −15. Find the sum of the first 50 terms of the progression. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 12 12 a d     30 2 30 1 15 2    a d B1 For correct equation in a and d. If using   2  n a l , must replace l with an expression involving a and d. a = 72, d = 5  B1 Both values correct SOI.       50 50 S 2 49 2   theira theird M1 Using sum formula with their a and d values obtained via a valid method. S50 = 2525  A1 5

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k2 5 3 The coefficient of x4 in the expansion of 2x2 + is a. The coefficient of x2 in the expansion of x 2kx −1 4 is b. (a) Find a and b in terms of the constant k. 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(b) Given that a + b = 216, find the possible values of k. 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Mark scheme: 3(a) 4 x term is    2 2 3 2 10 2        k x x M1 For selecting the term in 4 x . 4 4 4 80 80 k x a k   A1 For correct value of a. Allow 4 4 8 0 k x . [ 2 x term is [6 ](2kx)2  1 = 24k2x2 ] 2 2 4  b k B1 For correct value of b. Allow 2 2 2 4 k x . 3 3(b)   4 2 4 2 80 24 216 0 10 3 27 0            k k k k M1 Forming a 3-term equation in k (all terms on one side) with their a and b and no x’s.     2 2 2 3 5 9 0    k k [⇒ 2 3 9 or ] 2 5   k M1 Attempt to solve 3-term quartic (or quadratic in another variable) by factorisation, formula or completing the square – see guidance.  3 2  k A1 OE e.g. 6 2  , 1 .5  , AWRT 1 .22  Omission of  A0. Additional answers A0. If M1 M0, SC B1 can be awarded for correct final answer, max 2/3. 3

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5 C D A 1 r B The diagram shows a sector ABC of a circle with centre A and radius r. The line BD is perpendicular to AC. Angle CAB is 1 radians. (a) Given that 1 = 16π, find the exact area of BCD in terms of r. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 3 (b) Given instead that the length of BD is r, find the exact perimeter of BCD in terms of r. 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Mark scheme: 5(a) Sector area = 2 2 1 π π 2 6 12           r r 2 1 2  r with θ in radians SOI. B0 if using a value for r. BD = π 1 sin 6 2        r r and AD = π 3 cos r 6 2        r so triangle area = 1 π π 1 1 3 sin cos r 2 6 6 2 2 2                 r r r or 1 π π 1 3 1 cos sin 2 6 6 2 2 2                 r r r r B1 SOI Finding triangle area. Decimals B0 unless exact values seen in working. Area of BCD = 2 2 1 3 π 12 8  r r B1 OE e.g. 2 π 3 4 3 2          r with π cos 6 and π sin 6 evaluated. Must be exact, in terms of 2 r . ISW 3 Question Answer Marks Guidance 5(b) Angle BAC = 1 3 π 2 sin 3                  r r B1 SOI by length of AD, CD or arc, or by perimeter. Length AD = π 1 cos 3 2        r r [so length CD = 1 2r] M1 SOI Finding length by Pythagoras, or by trigonometry with their angle BAC, provided π 6  BAC . Length of arc BC = π 3  r M1 SOI Using  r with in radians. Condone π 6  . Perimeter of BCD = 3 1 π 2 2 3   r r r A1 OE e.g. 3 1 π 2 3          r with e.g. π cos 3 evaluated. Must be exact, in terms of r. ISW 4

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7 y y = 12x + 1 B 1 y = 3x −2 2 A x O 1 The diagram shows the curve with equation y = 3x −2 2 and the line y = 12x + 1. The curve and the line intersect at points A and B. (a) Find the coordinates of A and B. 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(b) Hence find the area of the region enclosed between the curve and the line. 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Mark scheme: 7(a)   2 1 2 2 1 1 1 3 2 1 3 2 1 1 2 2 4                x x x x x x M1 Equating curve and line, attempt to square; 2 1 1 4  x M0       2 2 1 2 3 0 8 12 0 6 2 0 4                 x x x x x x M1 Forming and solving a 3TQ by factorisation, formula or completing the square – see guidance. (2, 2) and (6, 4) A1 A1 A1 for each point, or A1 A0 for two correct x-values. If M0 for solving, SC B2 possible: B1 for each point or B1 B0 for two correct x-values. 4 Question Answer Marks Guidance 7(b) Area =     6 1 2 2 1 3 2 1 [d ] 2                  x x x *M1 For intention to integrate and subtract (M0 if squared).   6 3 2 2 2 2 1 3 2 9 4 x x x                 B1 B1 B1 for each bracket integrated correctly (in any form).    3 3 2 2 2 1 2 1 16 36 6 4 4 2 9 4 9 4                                       DM1 ( F(their 6) – F(their 2)) with their integral. Allow 1 sign error. 4 9 A1 AWRT 0.444. SC1 B1 for 4 9 if *M1 B1 B1 DM0. SC2 B1 for 4 9 if *M1 B0 B0 DM0, provided limits stated. Alternative method for question 7(b) Area =     6 1 2 2 3 2 [ d ]     x x area of trapezium (or triangle + rectangle) *M1 For intention to integrate and subtract (M0 if squared).   6 3 2 2 2 2 4 3 2 4 9 2 x                 or     6 3 2 2 2 2 4 3 2 2 4 9 2                   x B1 B1 FT B1 for bracket integrated correctly (in any form). B1 FT for using correct formula with their values.    3 3 2 2 2 2 16 4 12 9 9                DM1 (F(their 6) – F(their 2)) using their integral. Allow 1 sign error. Question Answer Marks Guidance 7(b) 4 9 A1 AWRT 0.444. SC1 B1 for 4 9 if *M1 B1 B1 DM0. SC2 B1 for 4 9 if *M1 B0 B0 DM0, provided limits stated. 5

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8 (a) The curve y = sin x is transformed to the curve y = 4 sin 12x −30Å . Describe fully a sequence of transformations that have been combined, making clear the order in which the transformations are applied. 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(b) Find the exact solutions of the equation 4 sin 12x −30Å = 2 2 for 0Å ≤x ≤360Å. 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Mark scheme: 8(a)  30 0        60 0       B2,1,0 B2 for fully correct, B1 with two elements correct. { } indicates different elements. Accept angle in radians. (3){Stretch} {factor 2} {in x-direction} B2,1,0 B2 for fully correct, B1 with two elements correct. { } indicates different elements. (4) Stretch factor 4 in y-direction and correct order B1 Stretch, y-direction and factor and correct order. Correct order is either (1) then (3) or (3) then (2). (4) can be anywhere in the sequence. 5 8(b)   1 1 2 4sin 30 2 2 sin 45 2 2                   x M1 SOI     1 30 45 or 1 35 2 45 30 or 2 135 30 2       x x x M1 SOI. The M marks are independent. x = 150°, x = 330° A1 Both exact values, condone 5π 11π , 6 6 . A0 if extra solutions in the interval. Ignore other solutions outside   0, 360  . 3

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9 The equation of a circle is x2 + y2 + 6x −2y −26 = 0. (a) Find the coordinates of the centre of the circle and the radius. Hence find the coordinates of the lowest point on the circle. 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(b) Find the set of values of the constant k for which the line with equation y = kx −5 intersects the circle at two distinct points. 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Mark scheme: 9(a) 2 2 3 1 26 9 1 36      x y 2 2 2 2 0     x y gx fy c , centre   ,  g f and radius 2 2   g f c . SOI by correct answer. Centre (−3, 1) B1 Radius 6 B1 So lowest point is (−3, −5) A1 FT FT on their centre and their radius. 4 9(b) Intersects when     2 2 5 6 2 5 26 0 x kx x kx        or     2 2 3 5 1 36 x kx      *M1 Substituting 5 y kx   into their circle equation or rearranging and equating y. 2 2 2 10 25 6 2 10 26 0 x k x kx x kx         or 2 2 2 6 9 12 36 36 x x k x kx       leading to   2 2 2 6 12 9 0 k x x x kx      or       2 2 1 6 12 9 0 k x k x      DM1 A1 Rearranging to 3-term quadratic (terms grouped, all on one side). Allow 1 error. Correct quadratic (need to see 9 as constant term).     2 2 6 12 4 1 9 [ 0]      k k 2 2 leading to 144 144 36 36 36 0 k k k          DM1 Using discriminant   2 4 0 b ac   with their values. Allow if in square root. [108k2 – 144k = 0 leading to] k = 0 or k = 4 3 A1 Need not see method for solving. 4 0, 3 k k   A1 Do not accept 4 0 3   k . 6

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d2y 910 The equation of a curve is such that = 6x2 −4 . The curve has a stationary point at −1, . dx2 x3 2 (a) Determine the nature of the stationary point at −1, 9 . [1] 2 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the equation of the curve. 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(c) Show that the curve has no other stationary points. 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(d) A point A is moving along the curve and the y-coordinate of A is increasing at a rate of 5 units per second. Find the rate of increase of the x-coordinate of A at the point where x = 1. 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Mark scheme: 10(a)     2 2 2 2 3 2 d 4 d 6 1 0 minimum 10 minimum d d 1        y y x x or B1 Sub 1  x into 2 2 d d y x , correct conclusion. WWW 1 10(b)   3 2 d 2 2 d    y x c x x *M1 Integrating 2 2 d d y x (at least one term correct). 0 = −2 + 2 + c leading to c = [0] DM1 Substituting d 1, 0 d   y x x (need to see) to evaluate c. DM0 if simply state 0  c or omit c.   4 1 2 2     y x their c x k x A1 FT Integrated. FT their non-zero value of c if DM1 awarded. 9 1 2 2 2   k leading to k = [2] DM1 Substituting x = –1, y = 9 2 to evaluate k (dep on *M1). 4 1 2 2 2    y x x A1 OE e.g. 1 2  x or 4 2 . A0 (wrong process) if c not evaluated but correct answer obtained. 5 10(c) 3 2 d 2 2 0 d    y x x x M1 Their d 0 d  y x . Leading to 5 1 x  M1 Reaching equation of the form 5 x a  . So only stationary point is when x = −1 A1 1  x and stating e.g. ‘only’ or ‘no other solutions. 3 Question Answer Marks Guidance 10(d) At x = 1,  d 4 d  y x *M1 Substituting 1  x into their d d y x. d d d 1 5 d d d 4     x x y t y t DM1 OE Using chain rule correctly SOI. 5 4 A1 OE e.g. 1.25. 3

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