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Mathematics 9709/13 May/June 2019: question paper and mark scheme

Cambridge AS & A Level Mathematics (9709), Paper 1, May/June 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.

9709_s19_qp_139709_s19_ms_139709/13/M/J/19

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Paper
Paper 1
Total marks
75
Time
1 hour 45 minutes
Code on the paper
9709/13/M/J/19
9709_s19_ms_13.pdf

9709/13 mark scheme, question by question

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 15 marks

Question

The function f is defined by f x = x2 − 4x + 8 for x ∈ >. (i) Express x2 − 4x + 8 in the form x − a2 + b. [2] (ii) Hence find the set of values of x for which f x < 9, giving your answer in exact form. [3] @ A 1 5

Mark scheme

1(i)  ( x − 2 ) 2  [ +4 ] B1 DB1 2nd B1 dependent on 2 inside bracket   2 1(ii) ( x − 2 ) < 5 → − 5 < x − 2 and/or x − 2 < 5 M1 Allow e.g. x − 2 < ± 5 , x ‒ 2 = ± 5 and decimal equivalents for √5 2 For M1, ft from their(i). Also allow √13 instead of √5 for clear slip 2− 5 < x<2+ 5 A1A1 A1 for each inequality – allow two separate statements but there must be 2 inequalities for x. Non-hence methods, if completely correct, score SC 1/3. Condone - [3]

Question 25 marks

Question

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

Mark scheme

2(i) −5 5 1 5 1 B1B1B1 B1 for each correct term + 3− 5 (or −5 x −1 + x −3 − x −5 ) +5 +1 x 8 x 32 x 8 32 SCB1 for both & x 32 x5 3 2(ii) 1 × 20 + 4 × their ( −5 ) = 0 M1A1 Must be from exactly 2 terms SCB1 for 20 + 20 = 40 2

Question 36 marks

Question

A 8 cm E 1 5 0 rad B D C The diagram shows triangle ABC which is right-angled at A. Angle ABC = 15 0 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] 48

Mark scheme

3(i) 3π B1 Angle EAD = Angle ACD = or 54° or 0.942 soi 10 π or Angle DAC = or 36° or 0.628 soi 5 3π π M1 Angles used must be correct AD = 8sin( ) or 8cos( ) 10 5 (AD =) 6.47 A1 Alternative method for question 3(i)  3π  B1 Angles used must be correct 8sin   8  10  or 11. 01 AB = or AB = ( ) π π  tan   sin   5 5 π M1 AD = 11.0 (1) sin oe 5 (AD =) 6.47 A1 3 3(ii) π π M1 19.7(4) ( theirAD ) × their  −  1 2 Area sector = 2 2 5 1 π 1  3π   3π  M1 Or e.g. ½ theirAD × 82 − theirAD 2 . Area ∆ADC = × 8 × theirAD × sin or × 8cos   × 8sin   2 5 2 10    10  15.2(2) (Shaded area =) 35.0 or 34.9 A1 3

Question 45 marks

Question

The function f is defined by f x = for 3 ≤ x ≤ 7. The function g is defined by g x = 2x − 4 for x−1 a ≤ x ≤ b, where a and b are constants. (i) Find the greatest value of a and the least value of b which will permit the formation of the composite function gf. [2] It is now given that the conditions for the formation of gf are satisfied. (ii) Find an expression for gf x. [1] (iii) Find an expression for gf−1 x. [2]

Mark scheme

4(i) Max(a) is 8 B1 Allow a = 8 or a - 8 Min(b) is 24 B1 Allow b = 24 or b . 24 2 SCB1 for 8 and 24 seen 4(ii) 96 100 − 4 x B1  48  gf(x) = − 4 or gf(x) = 2  − 4 is insufficient x −1 x −1  x −1  Apply ISW 1 4(iii) 96 96 96 M1 FT from their(ii) provided (ii) involves algebraic fraction. y= − 4→ y + 4 = → x −1 = Allow sign errors x −1 x −1 y+4 96 A1 100 + x ( gf )−1 ( x ) = +1 OR . Must be a function of x. Apply ISW x+4 x+4 2

Question 57 marks

Question

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. [2] 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. [4]

Mark scheme

5(i) x B1 Allow ‒ or + 0.02. Allow n used  2 + ( x − 1)( − / +0.02 )  or 1.01x − 0.01x 2 or 0.99 x + 0.01x 2 oe 2 1 5(ii) Equate to 13 then either simplify to a 3-term quadratic equation or M1 Expect n2 ‒ 101n + 1300 (=0) or 0.99 x + 0.01x 2 = 13 . Allow x used find at least 1 solution (need not be correct) to an unsimplified quadratic 5(iii) a (1 − r n ) M1 Use of with a = 1, r = 0.92, n = 20 soi 1− r (=) 10.1 A1 Use of ( S∞ = ) a with a = 1, r = 0.92 M1 (1) (1 − 0.92n) = 13 → 0.92n = −0.04 OR oe 1− r 1 − 0.92 S ∞ = 12.5 so never reaches target or < 13 A1 Conclusion required – 'Shown' is insufficient No solution so never reaches target or < 13 4

Question 67 marks

Question

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 . [4]

Mark scheme

6(i) MF = ‒4i + 2j + 7k B1 1 6(ii) FN = 2i ‒ j B1 1 6(iii) MN = ‒2i + j + 7k B1 FT on their (MF + FN) 1 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| = 4 2 + 2 2 + 7 2 × 22 + 12 + 7 2 *DM1 Product of modulus. At least one methodically correct + / −59 DM1 All linked correctly. Note 69 × 54 = 9 46 cos FMN = 69 × 54 FMN = 14.9° or 0.259 A1 Do not allow if exactly 1 vector is reversed – even if adjusted finally 4

Question 79 marks

Question

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. [3] (iv) Using the results from parts (ii) and (iii), and showing all necessary working, find the possible coordinates of C. [4] dy

Mark scheme

7(i) D = (5, 1) B1 1 7(ii) ( x − 5) + ( y − 1) = 20 oe B1 FT on their D. 2 2 Apply ISW, oe but not to contain square roots 1 7(iii) ( x − 1) + ( y − 3) = ( 9 − x ) + ( y + 1) soi M1 Allow 1 sign slip 2 2 2 2 For M1 allow with √ signs round both sides but sides must be equated x 2 − 2 x + 1 + y 2 − 6 y + 9 = x 2 − 18 x + 81 + y 2 + 2 y + 1 A1 y = 2 x − 9 www AG A1 Alternative method for question 7(iii) −1 M1 grad. of AB = ‒½ → grad of perp bisector = −½ Equation of perp. bisector is y − 1 = 2 ( x − 5 ) A1 y = 2 x − 9 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 DM1 Simplify to one of the forms shown on the right (allow arithmetic 5(y‒1)2 = 80 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

Question 88 marks

Question

A curve is such that = 3x2 + ax + b. The curve has stationary points at −1, 2 and 3, k. Find dx the values of the constants a, b and k. [8]

Mark scheme

8 f ′ ( −1) = 0 → 3 − a + b = 0 f ′ ( 3) = 0 → 27 + 3a + b = 0 M1 Stationary points at x = ‒1 & x = 3 gives sim. equations in a & b a = ‒6 A1 Solve simultaneous equation b = ‒9 A1 Hence f ′ ( x ) = 3 x 2 − 6 x − 9 → f ( x ) = x 3 − 3 x 2 − 9 x ( + c ) B1 FT correct integration for their a,b (numerical a, b) 2 = −1 − 3 + 9 + c M1 Sub x = ‒1, y = 2 into their integrated f(x). c must be present c = ‒3 A1 FT from their f(x) f ( 3) = k → k = 27 − 27 − 27 − 3 M1 Sub x = 3, y = k into their integrated f(x) (Allow c omitted) k = −30 A1 8

Question 910 marks

Question

y y = f x x O 1 0 20 The function f : x → p sin2 2x + q is defined for 0 ≤ x ≤ 0, where p and q are positive constants. The diagram shows the graph of y = f x. (i) In terms of p and q, state the range of f. [2] (ii) State the number of solutions of the following equations. (a) f x = p + q [1] (b) f x = q [1] (c) f x = 12 p + q [1] (iii) For the case where p = 3 and q = 2, solve the equation f x = 4, showing all necessary working. [5]

Mark scheme

9(i) q - f ( x) - p + q B1B1 B1 each inequality – allow two separate statements Accept < , (q, p + q), [q, p + q] Condone y or x or f in place of f(x) 2 9(ii) (a) 2 B1 π 3π Allow , 4 4 (b) 3 B1 π Allow 0, ,π 2 (c) 4 B1 π 3π 5π 7π Allow , , , 8 8 8 8 3 9(iii) 2 M1 3sin 2 2 x + 2 = 4 → sin 2 2 x = soi 3 2 2 A1 OR Implied by at least one correct value for x. Allow sin–1 form Sin2x = (±)0.816(5). Allow sin 2 x = (± ) or 2 x = sin −1 (±) 3 3 (2x =) at least two of 0.955(3), 2.18(6), 4.09(7) , 5.32(8) A1 Can be implied by corresponding values of x below Allow for at least two of 0.304π, 0.696π, 1.30(4)π, 1.69(6)π OR at least two of 54.7(4)°, 125.2(6)°, 234.7(4)°, 305.2(6)° (x =) 0.478, 1.09, 2.05, 2.66. A1A1 Allow 0.152π, 0.348π, 0.652π, 0.848π SC A1 for 2 or 3 correct. SC A1 for all of 27.4º, 62.6º, 117.4º, 152.6º 2 Sin2x = ± → x = 0.365,1.21,1.94,2.78 scores SC M1A0A0A1 3 5 9 9 9 9 4 5 A1 Area = 13 ‒ 12 = (or 0.556) 9 9 Alternative method for question 10(ii) Area for line = 1/2 × 4 × 3/2 = 3 B1 4 1 1 1 ∫ 3 (8 y − 20 ) = 3 4 y − 20 = 3 [ −16 + 25] = 3 2 OR 5/2  y3   4 y  B1B1 Area for curve = ∫ ⅓( y 2 − 4) =   −   9  3   64 16   8 8  32 M1 Apply limits 2 → 4 to an integrated expression for curve  − − −  =  9 3   9 3 9 32 5 A1 Area = − 3 = (or 0.556) 9 9 5

Question 1013 marks

Question

y A 1 y = 3x + 4 2 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. [5] [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] 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

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

Question 16

Question

In the question paper above.

Mark scheme

16 A1 Ignore 85.8 or 86 2

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On this page, free: the mark scheme for AS & A Level Mathematics 9709/13, May/June 2019 (9709_s19_ms_13), and the question paper (9709_s19_qp_13), to read or download, with each question's marking points written out underneath.

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Mathematics 9709/13 for May/June 2019 is out of 75 marks, with 1 hour 45 minutes to answer it.

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