Mathematics 9709/31 October/November 2023: question paper and mark scheme

Mark scheme

1 Use correct quotient or product rule *M1 (1 − 3 x)2 x − x 2 ( −3) 2 x − 3 x 2 Obtain correct derivative in any form A1 e.g. = or (1 − 3 x) 2 (1 − 3 x )2 3x2 (1 − 3x ) + 2x (1 − 3x ) . −2 −1 Equate derivative to 8 and solve for x DM1 75x2 − 50x + 8 = (15x − 4)(5x − 2) . 2 4 A1 Exact values required. Obtain answers x = and 5 15 4 16 A1 Allow A1 for one correct point. Obtain answers y = − and 5 45 5

Mark scheme

2 Show points representing 2i and –2 + i B1 Can be implied if the correct perpendicular is drawn. Show perpendicular bisector of their (2i and –2 + i) B1FT Show correct half–line of gradient 1 from point (–1, 0) B1 Should pass through (0, 1). Correct loci and shade correct region B1 4

Mark scheme

3 State or imply that ln y = ln a + x ln b B1 Carry out a completely correct method for finding ln a or ln b M1 3.7 = ln a + ln b and 6.46 = ln a + 2.2ln b leading to ln a = 1.4 , ln b = 2.3 . Obtain value a = 4.06 A1 Obtain value b = 9.97 A1 SC B1 for a = e1.4 and b = e2.3 . Alternative Method for Question 3 e3.7 = ab1 and e6.46 = ab2.2 B1 Divide to obtain e2.76 = b1.2 and state or imply 2.76 = 1.2lnb M1 Obtain value a = 4.06 A1 Obtain value b = 9.97 A1 4

Mark scheme

4(a) Multiply numerator and denominator by a + 5i M1 OE Use i2 = –1 M1 At least once. 3a − 10 2a + 15 A1 Obtain answer + i a 2 + 25 a 2 + 25 Alternative Method for Question 4(a) Multiply x + iy by a – 5i and use i2 = –1 M1 Compare real and imaginary parts M1 3 = ax + 5y, 2 = ay – 5x. 3a − 10 2a + 15 A1 Obtain answer + i a 2 + 25 a 2 + 25 3 4(b) State or imply Im(a) ÷ Re(a) = 1 M1 Or Im(a) = Re(a) or equivalent for their u. Obtain answer a = 25 A1 2

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5(a) Use correct trig formulae and obtain an equation in sin x and cos x *M1 Allow one sign error. Obtain a correct equation in any form A1 e.g 2cos x sin π6 = −2sin x sin π3 . Substitute exact trig ratios and obtain an expression for tan x DM1 Allow one sign error. 1 A1 Or exact equivalent. Obtain answer tan x = − 3 4 5(b) 5π B1 Obtain answer, e.g. x = 6 11π B1FT FT first answer + π (provided 0 ⩽ first answer ⩽ π). Obtain second answer, e.g. x = and no others in the interval 6 Or FT first answer – π (provided π ⩽ first answer ⩽ 2π). Ignore any answers outside interval. 2

Mark scheme

6(a) State correct derivative of x or y with respect to t B1 dx 1 − 2 1 dy 1 = t , = . dt 2 dt t dy dy dt M1 Use correct chain rule. Use = × dx dt dx dy 2 A1 −1 2 t Obtain answer = Or simplified equivalent e.g 2t 2 or . dx t t 3 6(b) dy 1 M1 State or imply their = dx 2 Obtain t =4 A1 Or equivalent. Obtain answer (7, ln 16) A1 Or exact equivalent. Can state the two components separately. 3 tanθ dθ = x + 3 dx .

Mark scheme

7 Separate variables correctly B1 x 2 Condone missing integral signs or missing dx, dθ. Can be implied by later work. Obtain term –ln(cos θ) B1 Or equivalent e.g. ln(sec θ). Obtain term of the form a ln x2 + 3( ) M1 ( ) 1 A1 Obtain term ln x 2 + 3 2 Use x = 1, 𝜃 = 0 to evaluate a constant or as limits in a solution containing M1 If they have rearranged then the constant must be of the ( ) terms of the form a ln x2 + 3 and b ln(cosθ ) correct form. Obtain correct answer in any form A1 1 2 ln ( x + 3) = − ln cosθ + ln 2 . 2 4 A1 Or equivalent e.g. x 2 = 4sec2 x − 3 . Obtain final answer x 2 = −3 cos 2θ lns removed. 7

Mark scheme

8(d) Use the iterative process correctly at least once M1 Obtain final answer 1.43 A1 Show sufficient iterations to at least 4 d.p. to justify 1.43 to 2 d.p. or show A1 e.g. 1, 1.3864, 1.4297, 1.4341, … there is a sign change in the interval (1.425, 1.435) 1.5, 1.4210, 1.4332, 14.344, 1.4345, … Condone recovery and small differences in the final figure in the iteration 2, 1.4848, 1.4395, 1.4350, 1.4346, 1.4345, … 3

Mark scheme

9(a) Use the correct product rule *M1 Condone error in chain rule. Obtain correct derivative in any form A1 x2 x2 dy x2 − − e.g. =− e 4 +e 4 . dx 2 Equate derivative to zero and solve for x DM1 − 1 A1 Or exact equivalent. Obtain answer 2, 2e 2 Can state the components separately. 4 9(b) 1 − 1 B1 Or equivalent e.g. du = 2xdx . State or imply dx = u 2 du 1 2 Alternative substitution: u = − x 2 . 4 Substitute for x and dx M1 1 − u 1 A1 OE Obtain correct integral e 4 du 2 1 − u 1 − x2 M1 u = 9 and u = 0 or x = 3 and x = 0. Use correct limits in an integral of the form ae 4 or ae 4 − 9 A1 Or exact equivalent. Obtain answer 2 − 2e 4 Alternative Method for Question 9(b) 1 − x2 1 − x2 M1 Recognition used. xe 4 dx = ae 4 a negative A1 a=−2 A1 1 − x2 M1 x = 3 and x = 0. Use correct limits in an integral of the form ae 4 − 9 A1 Or exact equivalent. Obtain answer 2 − 2e 4 5

Mark scheme

10(a) A B C B1 State or imply the form + + 1 − 2 x 2 + x ( 2 + x )2 Use a correct method for finding a coefficient M1 A(2 + x)2 + B(1 – 2x)(2 + x) + C(1 – 2x) = 24x + 13. Obtain one of A = 4, B = 2 and C = –7 A1 If errors in equating still allow A marks for A and C. Obtain a second value A1 Obtain the third value A1 A Dx + E Mark the form + , where A = 4, D = 2 1 − 2x ( 2 + x )2 and E = –3, B1 M1 A1 A1 A1 as above. If there are extra term in partial fractions, that is 4 unknowns A, B, D and E then B0 unless recover at end, e.g. by setting B = 0. If B set to any value other than 0 and all coefficients correctly found to their new values then allow all A marks, but still B0 for partial fraction expression. Hence A1 for each coefficient, but nothing for coefficient set to specific value. Another case of extra term in partial fraction expression, namely + F, mark as above but need F = 0 to recover B1. 5 10(b) Use a correct method to find the first two terms of the expansion of M1 Symbolic coefficients are not sufficient for the M1. −1 −2 (1 − 2x )−1 , ( 2 + x )−1 , ( 2 + x )−2 , 1 + x x or 1 + 2 2 Obtain correct un-simplified expansions up to the term in x 2 of each partial A1 FT ( −1)( −2 ) A 1 + ( −1)( −2 x ) + ( −2 x )2 + .. A = 4. fraction 2 A1 FT B x ( −1)( −2 ) x 2 1 + ( −1) + + ... B = 2. 2 2 2 2 C x ( −2 )( −3) x A1 FT 2 1 + ( −2 ) + + ... C = –7 4 2 2 2 = 4(1 + 2x + 4x2) + 2/2(1 – x/2 + x2/4) – 7/4(1 – x + 3x2/4) = (4 + 1 – 7/4) + (8 – 1/2 + 7/4) x + (16 + 1/4 – 21/16) x2 The FT is on A, B, C. 13 37 239 2 A1 OE Obtain final answer + x+ x (Dx + E)/4 [1 + (–2)(x/2) +(–2)(–3)(x/2)2/2 …] 4 4 16 D = 2 E = –3 The FT is on A, D, E. Maclaurin’s Series f(0) = 13/4 B1 f’(0) = 37/4 B1 f”(0) = 239/8 B1. 13 37 239 2 + x+ x /2 or equivalent M1 A1. 4 4 8 37 239 2 If 1 + x+ x /2 then M0 A0 unless their f(0) 4 8 actually is 1. For the A, D, E form of fractions, give M1 A1FT A1FT for the expanded partial fractions, then, if D ≠ 0, M1 for multiplying out fully, and A1 for the final answer. If final answer has been multiplied throughout (e.g. by 16) then A0 at the end 5 10(c) 1 B1 OE x< 2 1

Mark scheme

11(a) Obtain 3i + 2j + k B1 Accept coordinates in place of position vector. 1 11(b) AM or AP correct soi B1 AM = 2j + k, or AP = –2i + j + 2k. Carry out correct process for evaluating the scalar product of AM and AP M1 or MA and PA : 0 + 2 + 2 . Using the correct process for the moduli, divide the scalar product by the M1 4 product of the moduli and obtain the inverse cosine of the result For their vectors. θ = cos −1 . 3 5 Obtain answer 53.4° or 0.932c A1 4 11(c) Find PQ (or QP ) for a general point Q on the line passing through O and M, B1 FT e.g. PQ = - (i + j + 2k) + μ(3i + 2j + k). Follow their M. Calculate the scalar product of PQ and a direction vector for the line passing *M1 through O and M and equate to zero 1 A1 Solve and obtain correct solution e.g. μ = − 2 Carry out method to calculate PQ DM1 .52 + 0 + 1.52 . 10 A1 Or exact equivalent. Obtain answer 2 Alternative Method 1 for Question 11(c) Find PQ (or QP ) for a general point Q on the line passing through O and M, B1 FT e.g. PQ = - (i + j + 2k) + μ(3i + 2j + k). Follow their M. Use a correct method to express PQ2 (or PQ) in terms of µ *M1 Obtain a correct equation in any form A1 e.g. PQ2 = (1 + 3μ)2 + (1 + 2μ)2 + (2 + μ)2 11(c) Carry out a complete method for finding its minimum DM1 1 e.g. 6(1 + 3μ ) + 4(1 + 2μ ) + 2(2 + μ ) = 0, μ = − . 2 10 A1 Or exact equivalent. Obtain answer 2 Alternative Method 2 for Question 11(c) Calling (0, 0, 0) A, state PA (or AP ) in component form, e.g. i + j + 2k B1 Use a scalar product to find the projection of PA (or AP ) on the line passing M1 through O and M 7 A1 OE Obtain correct answer 14 Use Pythagoras to find the perpendicular M1 2 7 d = AP − AQ = 1 + 1 + 2 − 2 2 2 . 14 10 A1 Or exact equivalent. Obtain answer 2 Alternative Method 3 for Question 11(c) Calling (0, 0, 0) A, state PA (or AP ) in component form, e.g. i + j + 2k B1 Calculate the vector product of PA and a direction vector for the line passing M1 through O and M Obtain correct answer, e.g. 3i – 5j + k A1 11(c) Divide modulus of the product by that of the direction vector M1 32 + 52 + 12 e.g. . 32 + 22 + 12 10 A1 Or exact equivalent. Obtain answer 2 5