Mathematics 9709/32 May/June 2019: question paper and mark scheme

Mark scheme

1 State unsimplified term in x 2 , or its coefficient in the expansion of B1 Symbolic binomial coefficients are not sufficient for the B marks 1  1 × −2  (1 + 3x ) 3  3 3 ( 3 x )2   2    State unsimplified term in x 3 , or its coefficient in the expansion of B1 1  1 × −2 × −5  (1 + 3x ) 3 3 3 3 ( 3 x )3   6    Multiply by (3 – x) to give 2 terms in x 3 , or their coefficients M1 (3× 106 + 1) Ignore errors in terms other than x 3 3 × x 3coeff − x 2 coeff and no other term in x3 Obtain answer 6 A1 Not 6x 3 4

Mark scheme

2 State or imply u − u − 12 ( = 0 ) , or equivalent in 3x B1 ( ) 2 2 Need to be convinced they know 32 x = 3x Solve for u, or for 3x , and obtain root 4 B1 Use a correct method to solve an equation of the form 3x = a M1 Need to see evidence of method. Do not penalise an attempt where a >0 to use the negative root as well. e.g. x ln 3 = ln a, x = log 3 a If seen, accept solution of straight forward cases such as 3x = 3 , x = 1 without working Obtain final answer x = 1.26 only A1 The Q asks for 2 dp 4

Mark scheme

3 Use correct trig formulae to obtain an equation in tan θ or *M1 1 − tan 2 θ cot 2 θ − 1 2 equivalent (e.g all in sin θ or all in cos θ) = 2 tan θ . Allow = 2 tan θ 2 cot θ cot θ Obtain a correct simplified equation A1 1 5 5 tan 2 θ = 1 or sin 2 θ = or cos 2 θ = 6 6 Solve for θ DM1 Dependent on the first M1 Obtain answer 24.1° (or 155.9°) A1 One correct in range to at least 3 sf Obtain second answer A1 FT 180° − their 24.1° and no others in range. Correct to at least 3 sf. Accept 156° but not 156.0 Ignore values outside range If working in tan θ or cos θ need to be considering both square roots to score the second A1 Mark 0.421, 2.72 as a MR, so A0A1 5

Mark scheme

4 Use correct quotient rule ( ) M1 Allow use of correct product rule on x × 1 + ln x −1 Obtain correct derivative in any form A1 dy (1 + ln x ) − x × x  1 1  1 = =  −  dx (1 + ln x ) 2  1 + ln x (1 + ln x )2    1 M1 1 (1 + ln x ) 2 Equate derivative to and obtain a quadratic in ln x or (1+ ln x) Horizontal form. Accept ln x = 4 4 Reduce to (ln x) 2 − 2 ln x + 1 = 0 A1 or 3-term equivalent. Condone lnx 2 if later used correctly Solve a 3-term quadratic in ln x for x M1 Must see working if solving incorrect quadratic Obtain answer x = e A1 Accept e1 1 A1 Exact only with no decimals seen before the exact value. Obtain answer y = e 2 e1 e Accept but not 2 1 + ln e 7

Mark scheme

5(i) State answer −1 − 3i B1 1 If − given as well at this point, still just B1 2 1 5(ii) Substitute x = −1 + 3i in the equation and attempt expansions of M1 Need to see sufficient working to be convinced that a calculator has not been used. x 2 and x3 Use i 2 = −1 correctly at least once M1 Allow for relevant use at any point in the solution Obtain k = 2 A1 Carry out a complete method for finding a quadratic factor with M1 Could use factor theorem from this point. Need to see zeros −1 + 3i and −1 − 3i working. M1 for correct testing of correct root or allow M1 for three unsuccessful valid attempts. Obtain x 2 + 2 x + 4 A1  1 Using factor theorem, obtain f  −  = 0  2 1 A1 Final answer Obtain root x = − , or equivalent, via division or 2 inspection 5(ii) Alternative method 1 Carry out a complete method for finding a quadratic factor with M1 Need to see sufficient working to be convinced that a zeros −1 + 3i and −1 − 3i (multiplying two linear factors or calculator has not been used. using sum and product of roots) Use i 2 = −1 correctly at least once M1 Allow for relevant use at any point in the solution Obtain x 2 + 2 x + 4 A1 Allow M1A0 for x 2 + 2 x + 3 Obtain linear factor kx + 1 and compare coefficients of x or x2 and M1 Can find the factor by inspection or by long division solve for k Must get to zero remainder Obtain k = 2 A1 1 A1 Final answer Obtain root x = − 2 1 Note: Verification that x = − is a root is worth no marks 2 without a clear demonstration of how the root was obtained 5(ii) Alternative method 2 Use equation for sum of roots of cubic and use equation for product M1 of roots of cubic Use i 2 = −1 correctly at least once M1 Allow for relevant use at any point in the solution 5 4 A1 Obtain − = −2 + γ , − = 4γ k k Solve simultaneous equations for k and γ M1 Obtain k = 2 A1 1 A1 Final answer Obtain root γ = − 2 6

Mark scheme

6(i) Correct use of trigonometry to obtain AB = 2r cos x B1 AG 1 6(ii) Use correct method for finding the area of the sector and the M1 1 1 2 1 × π r = ( 2r cos x ) 2 x 2 semicircle and form an equation in x 2 2 2 π A1 π Obtain x = cos −1 correctly AG Via correct simplification e.g. from cos 2 x = 16 x 16 x 2 6(iii) Calculate values of a relevant expression or pair of expressions at M1 x =1 1 → 1.11 f (1) = 1.11 x = 1 and x = 1.5 e.g. Accept x = 1.5 1.5 → 1.20 f (1.5 ) = 1.20 Must be working in radians π f ( x ) = x − cos −1 : f (1) = −0.111., f (1.5) = 0.3.. 16 x π f ( x) = cos x − : f (1) = 0.097., f (1.5) = −0.291. 16 x For 16 x cos 2 x − π f (1) = 1.529.., f (1.5) = −3.02.. Must find values. M1 if at least one value correct Correct values and complete the argument correctly A1 2 6(iv)  π  M1 1,1.11173, 1.13707, 1.14225, 1.14329, 1.14349, −1 Use xn +1 = cos   correctly at least twice 1.14354,1.14354  16 xn  1.25,1.16328,1.14742,1.14432,1.14370 Must be working in radians 1.5,1.20060,1.15447,1.14570,1.14397,1.14363 Obtain final answer 1.144 A1 Show sufficient iterations to at least 5 d.p. to justify 1.144 to 3 d.p. A1 or show there is a sign change in the interval (1.1435, 1.1445) 3

Mark scheme

7(i) Separate variables correctly and attempt integration of at least one B1 −y ∫ e dy = ∫ xe dx x side Obtain term −e − y B1 B0B1 is possible Commence integration by parts and reach xe ± ∫ e dx x x M1 B0B0M1A1 is possible Obtain xe x − e x A1 or equivalent B1B1M1A1 is available if there is no constant of integration Use x = 0, y = 0 to evaluate a constant, or as limits in a definite M1 Must see this step integral, in a solution with terms ae − y , bxe x and ce x , where abc ≠ 0 Obtain correct solution in any form A1 e.g. e − y = e x − xe x Rearrange as y = − ln (1 − x ) − x A1 or equivalent e.g. y = ln 1 e x (1 − x ) ISW 7 7(ii) Justify the given statement B1 e.g. require 1 − x > 0 for the ln term to exist, hence x < 1 Must be considering the range of values of x, and must be relevant to their y involving ln (1− x ) 1

Mark scheme

8(i) A B C B1 State or imply the form + + 2 x + 1 2 x + 3 ( 2 x + 3)2 Use a correct method to find a constant M1 Obtain the values A = 1, B = – 1, C = 3 A1 A1 A1 A Dx + E Full marks for the three correct constants – do not actually [Mark the form + , where A = 1, D = – 2 and need to see the partial fractions 2 x + 1 ( 2 x + 3)2 E = 0, B1M1A1A1A1 as above.] 5 8(ii) Integrate and obtain terms B1 B1 B1 FT on A, B and C. 1 1 3 ln ( 2 x + 1) − ln ( 2 x + 3) − 2 2 2 ( 2 x + 3) [Correct integration of the A, D, E form of fractions gives 1 ln ( 2 x + 1) + x − 1 ln ( 2 x + 3) if integration by parts is used 2 2x + 3 2 for the second partial fraction.] Substitute limits correctly in an integral with terms a ln ( 2 x + 1) , M1 value for upper limit – value for lower limit 1 slip in substituting can still score M1 b ln ( 2 x + 3) and c / ( 2 x + 3) , where abc ≠ 0 Condone omission of ln (1) If using alternative form: cx / ( 2 x + 3 ) Obtain the given answer following full and correct working A1 Need to see at least one interim step of valid log work. AG 5

Mark scheme

9(i) Carry out correct method for finding a vector equation for AB M1 Obtain ( r = ) i + 2 j − k + λ ( 2i − j + 2k ) , or equivalent A1 Equate two pairs of components of general points on their AB and l M1  1 + 2λ   2 + µ  and solve for λ or forµ      2 − λ  =  1+ µ   −1 + 2λ  1 + 2µ      Obtain correct answer for λ or µ, e.g. λ = 0, µ = – 1 A1 Verify that all three equations are not satisfied and the lines fail to A1 Alternatives intersect ( ≠ is sufficient justification e.g. 2 ≠ 0 ) A λ µ B λ µ Conclusion needs to follow correct values ij 2 1 1 5 −1 1 1 5 3 3 ≠ 3 3 ≠ 3 3 3 3 ik 0 –1 2≠0 –1 –1 2≠0 jk 1 0 3≠ 2 0 0 3≠ 2 5 9(ii) 3 B1 State or imply midpoint has position vector 2i + j 2 Substitute in 2x – y + 2z = d and find d M1 Correct use of their direction for AB and their midpoint Obtain plane equation 4x – 2y + 4z = 5 A1 2   5 or equivalent e.g. r.  −1 = 2 2   Substitute components of l in plane equation and solve for µ M1 Correct use of their plane equation. 1 3 1 A1 Final answer Obtain µ = − and position vector i + j for the point P Accept coordinates in place of position vector 2 2 2 5

Mark scheme

10(i) State correct expansion of sin ( 3x + x ) or sin ( 3x − x ) B1 B0 If their formula retains ± in the middle 1 M1 Substitute expansions in (sin 4 x + sin 2 x) 2 1 A1 Must see the sin 4x and sin 2x or reference to LHS and Obtain sin 3 x cos x = ( sin 4 x + sin 2 x ) correctly RHS for A1 2 AG 3 10(ii) 1 1 B1 B1 Integrate and obtain − cos 4 x − cos 2 x 8 4 1 M1 In their expression Substitute limits x = 0 and x = π correctly 3 9 A1 From correct working seen. Obtain answer 16 4 10(iii) State correct derivative 2cos 4 x + cos 2 x B1 Using correct double angle formula, express derivative in terms of M1 cos 2x and equate the result to zero Obtain 4cos 2 2 x + cos 2 x − 2 = 0 A1 Solve for x or 2x (could be labelled x) M1 Must see working if solving an incorrect quadratic  −1 ± 33  The roots of the correct quadratic are -0.843 and 0.593  cos 2 x =  Need to get as far as x = ...  8  The wrong value of x is 0.468 and can imply M1 if correct quadratic seen Could be working from a quartic in cos x : 16 cos 4 x − 14 cos 2 x + 1 = 0 Obtain answer x = 1.29 only A1 5