Cambridge A Level Mathematics 9709 — 2019 May/June Paper 1 · Variant 3

2 (i) In the binomial expansion of 2x −1 5 , the first three terms are 32x5 −40x3 + 20x. Find the 2x remaining three terms of the expansion. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ @ A5 (ii) Hence find the coefficient of x in the expansion of 1 + 4x2 2x −1 . [2] 2x ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) 3 5 5 5 1 8 32 x x x − (or 1 3 5 5 1 5 8 32 x x x − − − − + − ) B1B1B1 B1 for each correct term SCB1 for both 5 5 1 & 32 x x + + 3 2(ii) ( ) 1 20 4 5 their × + × − = 0 M1A1 Must be from exactly 2 terms SCB1 for 20 + 20 = 40 2

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3 A 8 cm E 150 rad B D C The diagram shows triangle ABC which is right-angled at A. Angle ABC = 150 radians and AC = 8 cm. The points D and E lie on BC and BA respectively. The sector ADE is part of a circle with centre A and is such that BDC is the tangent to the arc DE at D. (i) Find the length of AD. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the area of the shaded region. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(i) Angle EAD = Angle ACD = 3π 10 or 54° or 0.942 soi or Angle DAC = π 5 or 36° or 0.628 soi AD = 8sin( 3π 10 ) or 8cos( π 5 ) M1 Angles used must be correct (AD =) 6.47 A1 Alternative method for question 3(i) ( ) 3 8sin 8 10 or or 1 1. 01 π tan sin 5 5 AB AB π π       = =             B1 Angles used must be correct ( ) 11.0 1 sin 5 AD π = oe M1 (AD =) 6.47 A1 3 3(ii) Area sector = ( ) 2 1 2 2 5 theirAD their π π   × −     M1 19.7(4) Area 1 8 sin 2 5 ADC theirAD π ∆ = × × × or 1 3 3 8cos 8sin 2 10 10 π π     × ×         M1 Or e.g. ½ 2 2 8 theirAD theirAD × − . 15.2(2) (Shaded area =) 35.0 or 34.9 A1 3

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5 Two heavyweight boxers decide that they would be more successful if they competed in a lower weight class. For each boxer this would require a total weight loss of 13 kg. At the end of week 1 they have each recorded a weight loss of 1 kg and they both find that in each of the following weeks their weight loss is slightly less than the week before. Boxer A’s weight loss in week 2 is 0.98 kg. It is given that his weekly weight loss follows an arithmetic progression. (i) Write down an expression for his total weight loss after x weeks. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) He reaches his 13 kg target during week n. Use your answer to part (i) to find the value of n. 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Boxer B’s weight loss in week 2 is 0.92 kg and it is given that his weekly weight loss follows a geometric progression. (iii) Calculate his total weight loss after 20 weeks and show that he can never reach his target. 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Mark scheme: 5(i) ( )( ) 2 1 / 0.02 2 x x + − −+    or 2 2 1.01 0.01 or 0.99 0.01 x x x x − + oe 1 Question Answer Marks Guidance 5(ii) Equate to 13 then either simplify to a 3-term quadratic equation or find at least 1 solution (need not be correct) to an unsimplified quadratic M1 Expect n2 ‒ 101n + 1300 (=0) or 2 0.99 0.01 13 x x + = . Allow x used 16 A1 Ignore 85.8 or 86 2 5(iii) Use of ) (1 1 n a r r − − with a = 1, r = 0.92, n = 20 soi M1 (=) 10.1 A1 Use of ( ) 1 a S r ∞= − with a = 1, r = 0.92 M1 OR ( ) ) 1 (1 0.92 13 0.92 0.04 1 0.92 n n − = → = − − oe S∞= 12.5 so never reaches target or < 13 A1 Conclusion required – 'Shown' is insufficient No solution so never reaches target or < 13 4

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6 F N 2 E D 4 7 C 4 k j M A i 8 B The diagram shows a solid figure ABCDEF in which the horizontal base ABC is a triangle right-angled at A. The lengths of AB and AC are 8 units and 4 units respectively and M is the mid-point of AB. The point D is 7 units vertically above A. Triangle DEF lies in a horizontal plane with DE, DF and FE parallel to AB, AC and CB respectively and N is the mid-point of FE. The lengths of DE and DF −−→ −−→ −−→ are 4 units and 2 units respectively. Unit vectors i, j and k are parallel to AB, AC and AD respectively. −−−→ (i) Find MF in terms of i, j and k. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ −−→ (ii) Find FN in terms of i and j. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ −−−→ (iii) Find MN in terms of i, j and k. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iv) Use a scalar product to find angle FMN. 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Mark scheme: 6(i) B1 1 6(ii) FN = 2i ‒ j B1 1 6(iii) MN = ‒2i + j + 7k B1 FT on their (MF + FN) 1 Question Answer Marks Guidance 6(iv) MF.MN = 8 + 2 + 49 = 59 *M1 MF.MN or FM.NM but allow if one is reversed (implied by ‒59) |MF| × |MN| = 2 2 2 2 2 2 4 2 7 2 1 7 + + × + + *DM1 Product of modulus. At least one methodically correct / 59 cos 69 54 FMN + − = × DM1 All linked correctly. Note 69 54 × = 9 46 FMN = 14.9° or 0.259 A1 Do not allow if exactly 1 vector is reversed – even if adjusted finally 4

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7 The coordinates of two points A and B are 1, 3 and 9, −1 respectively and D is the mid-point of AB. A point C has coordinates x, y , where x and y are variables. (i) State the coordinates of D. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) It is given that CD2 = 20. Write down an equation relating x and y. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) It is given that AC and BC are equal in length. Find an equation relating x and y and show that it can be simplified to y = 2x −9. 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(iv) Using the results from parts (ii) and (iii), and showing all necessary working, find the possible coordinates of C. 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Mark scheme: 7(i) B1 1 7(ii) ( ) ( ) 2 2 5 1 20 x y − + − = oe B1 FT on their D. Apply ISW, oe but not to contain square roots 1 Question Answer Marks Guidance 7(iii) ( ) ( ) ( ) ( ) 2 2 2 2 1 3 9 1 x y x y − + − = − + + soi M1 Allow 1 sign slip For M1 allow with √ signs round both sides but sides must be equated 2 2 2 2 2 1 6 9 18 81 2 1 x x y y x x y y − + + − + = − + + + + A1 2 9 y x = − www AG A1 Alternative method for question 7(iii) grad. of AB = ‒½ → grad of perp bisector = 1 ½ − − M1 Equation of perp. bisector is ( ) 1 2 5 y x −= − A1 2 9 y x = − www AG A1 3 7(iv) Eliminate y (or x) using equations in (ii) and (iii) *M1 To give an (unsimplified) quadratic equation 5x2 ‒50x + 105 (= 0) or 5(x‒5)2 = 20 or 5y2‒10y‒75 (= 0) or 5(y‒1)2 = 80 DM1 Simplify to one of the forms shown on the right (allow arithmetic slips) x = 3 and 7, or y = ‒3 and 5 A1 (3, ‒3), (7, 5) A1 Both pairs of x & y correct implies A1A1. SC B2 for no working 4

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10 y A 1 2 y = 3x + 4 x O 4 1 The diagram shows part of the curve with equation y = 3x + 4 2 and the tangent to the curve at the point A. The x-coordinate of A is 4. (i) Find the equation of the tangent to the curve at A. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find, showing all necessary working, the area of the shaded region. 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[Question 10 (iii) is printed on the next page.] (iii) A point is moving along the curve. At the point P the y-coordinate is increasing at half the rate at which the x-coordinate is increasing. Find the x-coordinate of P. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 10(i) ( ) 1 2 1 3 4 2 x −   +     ( ) 1 2 d 1 3 4 3 d 2 y x x −   = + ×     B1 Must have ‘ 3 × ’ At x = 4, d 3 d 8 y x = soi B1 Line through (4, their4) with gradient their 3 8 M1 If y ≠ 4 is used then clear evidence of substitution of x = 4 is needed Equation of tangent is ( ) 3 4 4 8 y x − = − or 3 5 8 2 y x = + A1 oe 5 Question Answer Marks Guidance 10(ii) Area under line 1 5 4 4 13 2 2   = + × =     B1 OR [ ] 4 2 0 3 5 3 5 3 10 13 8 2 16 2 x x x   + = + = + =     ∫ Area under curve: ( ) ( ) [ ] 3/2 1 2 3 4 3 4 3 3 / 2 x x   + ∫ + =  ÷     B1B1 Allow if seen as part of the difference of 2 integrals First B1 for integral without [ ] 3 ÷ Second B1 must have [ ] 3 ÷ 128 16 112 4 12 9 9 9 9 − = = M1 Apply limits 0 → 4 to an integrated expression Area = 13 ‒ 4 12 9 = 5 9 (or 0.556) A1 Alternative method for question 10(ii) Area for line = 1/2 × 4 × 3/2 = 3 B1 OR ( ) [ ] 4 2 5/2 1 1 1 8 20 4 20 16 25 3 3 3 3 y y   − = − = − + =   ∫ Area for curve = 3 2 4 ( 4) 9 3 y y y     ∫ − = −         ⅓ B1B1 64 16 8 8 32 9 3 9 3 9     − − − =         M1 Apply limits 2 → 4 to an integrated expression for curve Area = 32 3 9 − = 5 9 (or 0.556) A1 5 Question Answer Marks Guidance 10(iii) d 1 d 2 y x = B1 ( ) 1 2 3 3 4 2 x − + = 1 2 M1 Allow M1 for ( ) 1 2 3 3 4 2 x − + = 2. ( ) 1 2 3 4 3 x + = →3 4 9 x x + = → 5 = 3 oe A1 3

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