Mathematics 9709/32 May/June 2023: question paper and mark scheme

Mark scheme

1 State or imply non-modular inequality (5 x − 3) 2 < 22 (3x − 7) 2 , or B1 11x 2 − 138 x + 187 > 0 . corresponding quadratic equation, or pair of linear equations (5x – 3) = ± 2(3x – 7) Solve a 3-term quadratic, or solve two linear equations for x M1 If no working is shown, the M1 is implied by the correct roots for an incorrect quadratic. 17 A1 Accept 1.55 or better. Obtain critical values x = and x = 11 11 17 A1 Strict inequality required. State final answer x < , x > 11 In set notation, allow notation for open sets but not for 11 ( ) closed sets e.g. accept −∞, 17 (11, ∞ ) or 11 ( −∞, 17 [ ]11, ∞ ) but not ( −∞, 17 ] [11, ∞ ) . 11 11 Allow ‘or’ but not ‘and’. Accept . Final A0 for 17 > x > 11 . 11 Exact values expected but ISW if exact inequalities seen followed by decimal approx. Alternative Method for Question 1 Obtain critical value x = 11 from a graphical method, or by inspection, or B1 by solving a linear equation or an inequality 17 B2 Accept decimal value. Obtain critical value x = similarly 11 17 B1 Strict inequality required. State final answer x < , x > 11 See notes above. 11 4

Mark scheme

2 Use law of the logarithm of a power, quotient or product M1 Must be used correctly on a correct term. e.g. M1 for 2 ln x = ln x2 but M0 for 2 ln x – ln 2 = 2 ln x . 2 2 2 M0 for ln(2x – 3) = ln 2x – ln 3 = ln 2 + 2 ln x – ln 3. Remove logarithms and obtain a correct equation in x A1 x2 e.g. 2 x 2 − 3 = . 2 Obtain final answer x = 2 only A1 If x = − 2 is mentioned, it must be rejected. 3 Im(z)

Mark scheme

3(a) B1 Show a circle with centre –3 + 2i. Allow for a curved figure with ‘centre’ in roughly the correct position. 2i Accept marks or numbers on axes, coordinates of centre shown. bold distance is the length required for part (b) B0B1 available for axes the wrong way round (and M1 A1 in part (b)). Re(z) -3 Show a circle with radius 2 B1 FT FT centre not at the origin. Allow ‘near miss’ on x axis. Different scales on axes require an ellipse for B1 B1. Scales on the axes and any label of the radius must be consistent for B1 B1. Correct circle shaded scores B1 B0. 2 3(b) Carry out a correct method for finding the least value of |z| M1 e.g. distance of centre from origin – radius or find point of intersection of circle and 3y = -2x and use Pythagoras. If they subtract the wrong way round M0. If their diagram is a reflection or a rotation of the correct diagram, M1 A1 is available (requires equivalent work). Any other circle M0. A1 Obtain answer 13 − 2 or 17 − 4 13 26 36 Or exact equivalent e.g. 17 − . 3 13 Correct solution only. Allow A1 if exact answer seen and then decimal given. 2

Mark scheme

4 x *M1 x x Use correct double angle formula to obtain an equation in cos only e.g. 2 2cos 2 − 1 − cos = 1 . 2 2 2 x A1 x x Obtain a 3 term quadratic in cos , e.g. 4cos 2 − cos − 3 = 0 . 2 2 2 Allow 4cos u − cos u − 3 = 0 . Condone 2x = x . 2 x 3 x A1 Allow answer in u e.g. ( 4 cos u + 3)( cos u − 1) and condone Obtain cos = − and cos = 1 2 4 2 x =x. 2 Solve for the original x DM1 Must see evidence of doubling, not halving. Obtain x = 0 and 4.84 and no others in the interval A1 Ignore any answers outside interval. Accept AWRT 4.84. Accept 1.54π. Must be in radians. 277.2 indicates M1 but is A0. Alternative Method for Question 4 Use correct double angle formula to obtain an equation in cos x only *M1 cos x + 1 e.g. 2cos x − 1 = . 2 Obtain a 3 term quadratic in cos x , A1 e.g. 8cos 2 x − 9 cos x + 1 = 0 . 1 A1 Obtain cos x = and cos x = 1 8 Solve for x DM1 Obtain answers x = 0 and 4.84 and no others in the interval A1 Ignore any answers outside interval. Accept AWRT 4.84. Must be in radians. 277.2 is A0. 5

Mark scheme

5(a) Substitute 2 + yi in a3 – a2 – 2a and attempt expansions of a2 and a3 M1 a 2 = 4 + 4 yi − y 2 a3 = 8 + 12 yi − 6 y 2 − y 3i . ( ) If using a a 2 − a − 2 must then expand fully. Must see working. Use i2 = –1 M1 Seen at least once (e.g. in squaring). Obtain final answer –5y2 + (6y – y3)i A1 Or simplified equivalent e.g. 6 yi − 5 y 2 − y 3i . Do not ISW. 3 No evidence of working for the square or the cube can score SC B1 for the correct answer. 5(b) Equate their –5y2 to –20 and solve for y M1 Need to obtain a value for y. Available even if their y is not real. Obtain y = –2 A1 From correct work. Allow after incorrect f(a) if the real part was correct. Condone ±2 with positive not rejected. π A1 Correct only (must have rejected y positive). Obtain final answer arg a = − 4 OE e.g. − π4 ± 2nπ . Accept –0.785, 5.50 . Allow after incorrect f(a) if the real part was correct. Accept degrees. Do not ISW. 3

Mark scheme

6(a) Calculate the values of a relevant expression or pair of expressions at x = M1 Need to evaluate at both points, but M1 still available if one 0.5 and x = 1 value incorrect. Use of degrees is M0. Correct use of a smaller interval is M1. If using g(x) – f(x), there needs to be a clear indication of the comparison being made e.g. by listing values in a table. Embedded values 0.5 and 1 are not sufficient. 3.92 and 1.83 alone are not sufficient. Complete the argument correctly with conclusion about change of sign or A1 e.g. 3.92 > 1.5, 1.83 < 3 change of inequalities and with correct calculated values. or 2.42 > 0, –1.17 < 0. Can all be in symbols – an explanation in words is not required. 2 6(b) 1 1 M1 x 1 State x = x + 4 tan −1 Or rearrange cot = 3x as far as 2 x = 4 tan −1 3 3x 2 3x x A1 1 1 Rearrange to the given equation cot = 3x Or continue rearrangement to x = x + 4 tan −1 and 2 3 3x 1 1 state iterative formula of xn +1 = xn + 4 tan −1 3 3xn x 1 x Need intermediate step between = tan −1 and cot = 3x AG 2 3x 2 2 6(c) Use the iterative process correctly at least once M1 Obtain one value and substitute that back in to obtain a second value. Working in degrees is M0. Obtain final answer 0.79 A1 Must be to 2 d.p. Show sufficient iterations to at least 4 d.p. to justify 0.79 to 2 d.p. or show A1 e.g. 1, 0.7623, 0.8037, 0.7921, 0.7951, 0.7943, 0.7945 there is a sign change in the interval (0.785, 0.795) or 0.5, 0.9506, 0.7665, 0.8024, 0.7924, 0.7950, 0.7944, 0.7945 or 0.75, 0.8076, 0.7911, 0.7954, 0.7943, 0.7946, 0.7945 . Condone truncation. Allow recovery. Condone minor differences in the final d.p. 3 If they do the iteration in (b) but restate the conclusion here, no marks in (b) but could score 3/3 for (c).

Mark scheme

7(a) dy B1 dy f State or imply 6 y as the derivative of 3y2 Allow y′ for throughout. Accept = 6x + 4 y . dx dx x dy B1 f State or imply 4 x + 4 y as the derivative of 4xy Accept = 4x + 6 y . dx y dy M1 dy Equate derivative of LHS to zero and solve for Allow an extra in front of their differentiated equation. dx dx Allow if ‘= 0’ is implied but not seen. f dy Allow = − x dx f y dy 3x + 2 y A1 AG – must come from correct working. Obtain =− The position of the negative must be clear. dx 2x + 3y 4 7(b) dy *M1 Must be using the given derivative. Equate to –2 and solve for x in terms of y or for y in terms of x dx x A1 Seen or implied by correct later work. Obtain x = −4 y or y = − 4 x DM1 Allow unsimplified. Substitute their x = −4 y or their y = − in curve equation 4 1 4 A1 Or exact equivalent. Obtain y = ± or x = ± 7 7 4 1 Or x = and y = − or exact equivalent. 7 7 Obtain both pairs of values A1 4 1 Or x = − and y = or exact equivalent. 7 7 A1 A0 for incorrect final pairing. 5

Mark scheme

8(a) Separate variables correctly B1 1 dy = e −( 2 x +1) dx . 4 + 9 y2 Condone missing integral signs or dx and dy missing. 1 B1 1 −2 x Obtain term − e−2 x −1 OE e.g. − e . 2 2e 3y M1 Obtain term of the form a tan −1 2 1 −1 3 y A1 1 3 −1 3 y Obtain term tan OE e.g. × tan . 6 2 9 2 2 Use x = 1, y = 0 to evaluate a constant or as limits in a solution containing M1 If they rearrange before evaluating the constant, the ± ( 2 x +1) constant must be of the correct form. or derived from terms of the form atan −1 ( by ) and ce Obtain correct answer in any form A1 1 −1 3 y 1 −3 1 −( 2 x +1) e.g. tan = e − e . 6 2 2 2 A1 OE Allow with 3e −3 = 0.149.... Obtain final answer y = 2 3 ( tan 3e−3 − 3e −2 x −1 ) 7 8(b) B1 FT Or exact equivalent. State that y approaches 2 3 ( tan 3e−3 ) The FT is on correct work on a solution containing e–2x–1. Condone y = … Accept correct answer stated with minimal wording. 0.10032… is not exact so B0. 1

Mark scheme

9(a) A B C B1 A Dx + E State or imply the form + + Alternative form: + 1 + 2 x 2 − x ( 2 − x )2 1 + 2x ( 2 − x )2 Use a correct method for finding a coefficient M1 e.g. A(2 – x)2 + B(1 + 2x)(2 – x) + C(1 + 2x) = 2x2 + 17x – 17 and compare coefficients or substitute for x. A(2 – x)3 + B(1 + 2x)(2 – x) 2 + C(1 + 2x)(2 - x) = 2x2 + 17x – 17 scores M0. Obtain one of A = –4, B = –3 and C = 5 A1 Obtain a second value A1 Obtain the third value A1 Extra term in partial fractions, then B0 unless recover at end. Allow the marks for any constants found correctly. Missing terms in partial fractions, B0 but M1A1 is available for a correct method that obtains at least one correct constant (e.g. cover-up rule) Max 2/5. Ignore any substitution back into their original expression. If alternative form used: A = –4, D = 3 and E = –1. 5 9(b) 5 B1FT OE Integrate and obtain terms −2ln (1 + 2 x ) + 3ln ( 2 − x ) + The FT is on correct use of their A, B and C; or on A, D and 2− x B1FT E. If using the A, D, E form then B1 for the A term, but no B1FT further marks until partial fractions are used to split the second term or they use integration by parts to obtain Dx + E D − dx for the 2nd B1 and 3rd B1 for correct 2− x 2− x completion. B0FT, B0FT, B0FT if they place their A, B, C with incorrect denominators. Substitute limits correctly in an integral with two terms (obtained correctly) M1 Condone minor slips in substitution. c Exact substitution required. of the form a ln (1 + 2 x ) + b ln ( 2 − x ) + , where abc ≠ 0 2− x 5 A1 AG – evidence of some correct work to combine or Obtain answer − ln 72 after full and correct working 2 simplify logs is required e.g. allow from − ln 9 + ln 1 or 8 − ln 23 − ln 32 . 5

Mark scheme

10(a) 1 *M1 Allow with incorrect chain rule. Use the product rule correctly to obtain p( x + 5)(3 − 2 x) n + q(3 − 2 x) 2 BOD over sign errors unless an incorrect rule is quoted. Obtain correct derivative in any form A1 − 1 1 e.g. −( x + 5)(3 − 2 x) 2 + (3 − 2 x) 2 . Equate derivative to zero and obtain a linear equation DM1 Allow with surd factor e.g. ( 3 − 2 x )− ( − ( x + 5 ) + ( 3 − 2 x ) ) = 0 . 1 2 Obtain a correct linear equation. A1 e.g. –(x + 5) + 3 – 2x = 0. 2 13 39 A1 2 13 13 2 2197 Obtain answer − , . Or exact equivalent e.g. − , or − , . 3 9 3 3 3 3 27 Accept with x, y stated separately. ISW Alternative Method for Question 10(a) Obtain y2 and differentiate *M1 Ignore their left hand side i.e. their d y 2 . dx Obtain correct derivative in any form A1 e.g. −6 x 2 − 34 x − 20 . Equate derivative to zero and solve for x DM1 2 A1 Ignore –5 if seen. Obtain − 3 2 13 39 A1 2 13 13 2 2197 Obtain answer − , only Or exact equivalent e.g. − , or − , . 3 9 3 3 3 3 27 ISW 5 10(b) 13 u 1 *M1 OE Need to see -2 or -½ used. Condone if du missing or Use the given substitution and reach a − u 2 du the integral sign is missing. 2 2 Allow M1A0 for complete substitution into x 3 − 2 x dx to obtain first term of the line below. 1 A1 1 3−u 1 13 u Obtain correct integral − − u 2 du OE e.g. − u du + 5 u du . 2 2 2 2 2 Ignore limits at this stage. Condone if du missing. 3 B1 SOI e.g. by u = 13 and 0. In any order and at any stage. x = –5 and 2 Use correct limits the right way round in an integral of the form DM1 26 3 2 5 a u2 − u2 3 5 169 169 A1 or exact equivalents. Obtain answer 13 or a = 15 15 5

Mark scheme

11(a) Carry out correct method for finding a vector equation for AB M1 Obtain [r =] i + 2j – 2k + λ (i – 3j + 3k) A1 OE e.g. r = 2i − j + k + λ ( −i + 3j − 3k ) . Equate two pairs of components of general points on their AB and l M1 1 + λ 1 + 2μ and evaluate λ or μ 2 − 3λ = −1 − 3μ . −2 + 3λ 3 + 4 μ Obtain correct answer for λ or μ, e.g. λ = –1, μ = –2 A1 Correct value from two correct component equations. Verify that all three equations are not satisfied and the lines fail to intersect A1 Conclusion needs to follow correct values. (≠ is sufficient justification e.g. 0 ≠ –3). Hybrid versions are possible e.g. using j and k to get one parameter and then i to obtain the other. or e.g. solving two pairs of simultaneous equations and showing that the results are not the same. Alternatives: A λ μ B λ μ ij 2 1 4≠7 ij 1 1 4≠7 ik 5 5/2 –13≠- ik 4 5/2 –13≠- 17/2 17/2 jk –1 –2 0 ≠ –3 jk –2 –2 0 ≠ –3 5 ρ ρ 11(b) Find AP for a general point P on l, e.g. –3j + 5k + μ(2i – 3j + 4k) B1 Or equivalent e.g. PA = −2 μ i + ( 3μ + 3) j − ( 4 μ + 5 ) k . ρ Calculate scalar product of their AP and a direction vector for l and equate M1 e.g. 4μ + ( 9 + 9 μ ) + ( 20 + 16 μ ) = 0 . the result to zero ρ M0 if using OP . M0 if using parallel line through A. Obtain μ= –1 A1 Obtain answer –i + 2j – k A1 Accept coordinates in place of position vector. Alternative Method for Question 11(b) ρ ρ Find AP for a general point P on l, e.g. –3j + 5k + μ(2i – 3j + 4k) B1 Or equivalent e.g. PA = −2 μ i + ( 3μ + 3) j − ( 4 μ + 5 ) k . Use Pythagoras and differentiate with respect to μ to obtain value of μ corresponding to minimum distance. (No need to prove it is a minimum) M1 d dμ ( ) 4 μ 2 + 9 ( μ + 1) + ( 4 μ + 5 ) = 0 . 2 2 Obtain μ= –1 A1 Obtain answer –i + 2j – k A1 Accept coordinates in place of position vector. 4