Mathematics 9709/31 October/November 2022: question paper and mark scheme

Mark scheme

1(a) Show a recognisable sketch graph of y = 2 x + 1 B1 y 1 1 x - 2 Ignore y = 3x + 5 if also drawn on the sketch. 1 1(b) Find x-coordinate of intersection with y = 3x + 5 M1 6 A1 Obtain x = − 5 6 A1 Do not condone ≤ for < in the final answer. State final answer x < − only 5 Alternative method 1 for question 1(b) Solve the linear inequality 3 x + 5 < − ( 2 x + 1) , or corresponding equation M1 Must solve the relevant equation. 6 A1 Ignore – 4 if seen. Obtain critical value x = − 5 6 A1 State final answer x < − only 5 Alternative method 2 for question 1(b) Solve the quadratic inequality ( 3x + 5) < ( 2 x + 1) , or corresponding M1 2 2 5 x 2 + 26 x + 24 < 0 equation 6 A1 Ignore – 4 if seen. Obtain critical value x = − 5 6 A1 State final answer x < − only 5 3

Mark scheme

2 Show a circle with radius 3 and centre the origin B1 Im(z) Show the line x = – 2 B1 π B1 Show the correct half line for 4 Shade the correct region B1 -3 -2 0 Re(z) For the vertical line and the circle, allow the B1 marks if all you see is the relevant part. 4

Mark scheme

3 Use law of logarithm of a product or power M1 One correct application of a log law. Obtain a correct linear equation in any form, e.g. ( 3 x − 1) ln 2 = ln 5 − x ln 3 A1 Solve for x M1 As far as x = … with only minor slips in processing. ln10 A1 Obtain answer x = ln 24 Alternative method for question 3 Use laws of indices to split at least one exponential term M1 23 x e.g. or an arrangement with 8 x 2 Obtain 24 = 10 x A1 OE Solve for x M1 ln10 A1 Obtain answer x = ln 24 4

Mark scheme

4 Use correct tan ( A + B ) formula and obtain an equation in tan x or an M1 tan x + tan 45° 2 e.g. = Allow if 2 in denominator equation in cos x and sin x 1 − tan x tan 45° tan x sin x cos 45° + cos x sin 45° 2cos x or = . cos x cos 45° − sin x sin 45° sin x Obtain correct 3 term equation tan 2 x + 3tan x − 2 = 0 , or equivalent A1 or 3sin x cos x = 2cos 2 x − sin 2 x Solve a 3-term quadratic in tan x and obtain a value for x M1 Obtain answer, e.g. 29.3 ° A1 29.316… Obtain second answer, e.g. 105.7 ° and no other A1 105.583…. Ignore answers outside the given interval. Treat answers in radians as a misread. 5

Mark scheme

5(a) 1 πi B1 State or imply u 2 = 4e 2 1 4 6 πi B1 + B1 For the modulus and the argument. Obtain answer v = e 3 3 5(b) State n = 6 B1 1

Mark scheme

6(a) Express cos4θ in terms of cos2θ and/or sin 2θ B1 Express cos2θ in terms of cosθ and/or sinθ B1 Anywhere Expand to obtain a correct expression in terms of cosθ B1 ( ) ( ) 2 e.g. 2 2cos θ − 1 − 1 + 4 2cos θ − 1 + 3 2 2 Reduce correctly to cos 4θ + 4cos 2θ + 3 ≡ 8cos 4 θ B1 AG 4 6(b) Use the identity and carry out method to calculate a root M1 8cos 4 θ − 3 = 4 Obtain answer, e.g. 14.7 ° A1 Obtain second answer, e.g.165.3 ° , and no other in the given interval A1 FT Ignore answers outside the given interval. Treat answers in radians as a misread. 3

Mark scheme

7(a) Use correct product or quotient rule M1 Obtain correct derivative in any form A1 dy cos 2 x + 2 x sin x cos x e.g. = or dx cos 4 x dy = sec x + 2 x sec x tan x 2 2 dx cos a + 2a sin a A1 AG Equate derivative at x = a to 12 and obtain a = cos −1 3 12 3 7(b) Evaluate a relevant expression or pair of expressions at a = 0.9 and a = 1 M1 Must be calculated in radians. Complete the argument correctly with correct calculated values A1 cos 0.9 = 0.622 > 0.553 0.9 < 0.985 0.0846 > 0 e.g. or or cos1 = 0.540 < 0.570 1 > 0.964 −0.0358 < 0 or could be looking at values of the gradient 8.46 & 14.1 2 7(c) cosan + 2an sinan M1 Must be working in radians. Use the process an +1 = cos −1 3 correctly at least once 12 Obtain final answer 0.97 A1 Show sufficient iterations to 4 d.p. to justify 0.97 to 2 d.p., or show there A1 e.g. 0.95, 0.9743, 0.9694, 0.9704 is a sign change in the interval (0.965, 0.975) 3 dx = ke

Mark scheme

8(a) Separate variables correctly B1 1 −0.1t dt x Obtain term ln x B1 xe −0.1t −0.1t Obtain term −10ke B1 Not from dt Use x = 20, t = 0 to evaluate a constant or as limits in a solution containing M1 terms alnx, be−0.1t where ab ≠ 0 ( ) Obtain ln x = 10k 1 − e−0.1t + ln 20 A1 or equivalent ISW 5 8(b) Use x = 40, t = 10 to find k or 10k M1 Available for their function of the correct structure even if they found no constant in (a). Obtain 10k = 1.09654 A1 ln 2 or equivalent e.g. 10k = 1 − e−1 State that x tends to 59.9 A1 Need a number, not an expression for that value 3 sf or better 59.87595….. 3

Mark scheme

9(a) Use correct product or quotient rule *M1 e.g. dx = −e 3 − 13 ( 3 − x ) e 3 Obtain correct derivative in any form A1 dy −x −x Equate their derivative to zero and solve for x DM1 Obtain x = 6 A1 Obtain y = −3e −2 A1 Or exact equivalent. 5 9(b) 1 − x 1 − x *M1 Commence integration and reach a ( 3 − x ) e 3 + b e 3 dx, where ab ≠ 0 1 − x 1 − x A1 Obtain −3( 3 − x ) e 3 − 3 e 3 dx, or equivalent −3e 3 ( 3 − x ) + 9e 3 1 − x A1 −x −x Complete integration and obtain 3xe 3 , or equivalent Substitute limits x = 0 and x = 3, having integrated twice DM1 9 A1 Obtain answer , or exact equivalent e 5

Mark scheme

10(a) A B C B1 State or imply the form + + 1 + x 2 + x ( 2 + x )2 Use a correct method to find a constant M1 Obtain one of A= 3, B = – 1 and C = – 2 A1 SR after B0 can score M1A1 for one correct value Obtain a second value A1 Obtain the third value A1 A Dx + E + , where A = 3, D = – 1 and 1+ x ( 2 + x )2 E = – 4, is awarded B1 M1 A1 A1 A1 as above. 5 10(b) Use a correct method to find the first two terms of the expansion of M1 For the A, D, E form of fractions, award M1 A1FT A1FT −1 −2 for the expanded partial fractions, then if D ≠ 0, M1 for (1 + x )−1 , ( 2 + x )−1 , 1 + 1 x , ( 2 + x )−2 or 1 + 1 x multiplying out fully, and A1 for the final answer. 2 2 Obtain correct unsimplified expansions up to the term in x 2 of each partial fraction A3 FT ( 3 1 − x + x 2 ..... ) ( − 12 1 − 2x + x4 ..... 2 ) ( − 24 1 − x + 43 x 2 ) 9 5 A1 Obtain final answer 2 − x + x 2 4 2 5

Mark scheme

11(a) State OM = 2i + 2 j or equivalent B1 Can be implied by MB = −2i + 2 j or MA = 2i − 2 j . Obtain MD = – 2i – 2j + 3k B1 Use a correct method to find ON M1 e.g. OC + 2 CB 3 Obtain answer 3j + k A1 4 11(b) Use the correct process for evaluating the scalar product of MD and ON M1 Using the correct process for the moduli, divide the scalar product by the M1 −6 + 3 product of the moduli and reach the inverse cosine of the result cos −1 10 17 Obtain final answer 103.3° A1 3 11(c) Taking a general point P of ON to have position vector λ ( 3 j + k ) , form M1 −2 0 an equation in λ by either equating the scalar product of ON and MP to e.g. −2 + 3λ . 3 = 0 zero, or applying Pythagoras to triangle OMP, or equating the derivative λ 1 of MP to zero 3 A1 Solve and obtain λ = 5 Substitute for λ and calculate MP M1 MP = −2i − 15 j + 53 k 22 A1 AG Obtain 5 Alternative method for question 11(c) Use a scalar product to find the projection OQ of OM on OM M1 6 A1 Obtain OQ = 10 Use Pythagoras in triangle OMQ to find MQ M1 22 A1 AG Obtain 5 4