Mathematics 9709/31 May/June 2023: question paper and mark scheme

Mark scheme

2(a) y B1 Show a recognizable sketch graph of y = 2 x + 3 . (Ignore any attempt to sketch y = 3 x + 8 ). Straight lines. Vertex in approximately correct position 3 on x axis. Symmetry. x -3 2 1 2(b) Find x-coordinate of intersection with y = 3x + 8 M1 Obtain x = − 11 A1 5 State final answer x > − 11 only A1 ( x > −2.2 ) Do not condone ⩾ for >. 5 Alternative Method 1 Solve the linear inequality 3x + 8 > − ( 2 x + 3) , or corresponding linear M1 equation Obtain critical value x = − 11 A1 5 State final answer x > − 11 only A1 ( x > −2.2 ) Do not condone ⩾ for >. 5 Alternative Method 2 Solve the quadratic inequality ( 3x + 8) > ( 2 x + 3) , or corresponding (M1) 2 2 5 x 2 + 36 x + 55 . quadratic equation Obtain critical value x = − 11 (A1) Ignore -5 if seen. 5 State final answer x > − 11 only (A1) ( x > −2.2 ) Do not condone ⩾ for >. 5 3

Mark scheme

3 State unsimplified term in x3 , or its coefficient, in the expansion of (1 + 4x ) 2 1 B1 1 × −1 × −3 2 2 2 ( 4 x )3 ( = 4 ) Must expand binomial 6 coefficient. State unsimplified term in x 2 , or its coefficient, in the expansion of (1 + 4x ) 2 1 B1 1 × −1 2 2 ( 4 x )2 ( = −2 ) Must expand binomial coefficient. 2 Multiply by ( 3 + x ) and combine terms in x3 , or their coefficients M1 ( 3 × 4 − 1× 2) Allow if they expanded with x rather than 4x. Obtain answer 10 A1 Accept 10x3 4

Mark scheme

4(a) Use correct double angle formulae M1 e.g. 2sin θ cos θ + cos 2 θ − sin 2 θ = 2sin 2 θ Obtain cos 2 θ + 2sin θ cos θ − 3sin 2 θ = 0 from full and correct working A1 AG Check conclusion is complete and matches the working. 2 4(b) Factorise to obtain ( cosθ − sin θ )( cos θ + 3sin θ ) = 0 B1 OE Solve a quadratic in sin θ and cos θ to obtain a value for θ . M1 tan θ = 1 or tan θ = − 13 . Obtain one correct value e.g. 45° A1 Obtain a second correct value e.g. 161.6° and no others in the interval A1 Mark answers in radians (0.785 and 2.82) as a misread. Accept awrt 161.6 . Alternative Method 1 Obtain 3 tan 2 θ − 2 tan θ − 1 = 0 B1 Solve a 3 term quadratic in tan θ to obtain a value for θ . M1 tan θ = 1 or tan θ = − 13 . Obtain one correct value e.g. 45° A1 Obtain a second correct value e.g. 161.6° and no others in the interval A1 Mark answers in radians (0.785 and 2.82) as a misread. Alternative Method 2 Obtain ( cosθ + sin θ ) = ( 2sin θ ) B1 2 2 Solve to obtain a value for θ . M1 tan θ = 1 or tan θ = − 13 . Obtain one correct value e.g. 45° A1 Obtain a second correct value e.g. 161.6° and no others in the interval A1 Mark answers in radians (0.785 and 2.82) as a misread. 4

Mark scheme

5(a) dy B1 State or imply 2 xy + x 2 as derivative of x 2 y Accept partial: → 2xy . dx x dy B1 State or imply 2ay as derivative of ay 2 Accept partial: → x 2 − 2ay . dx y dy M1 Equate attempted derivative to zero and solve for dx dy 2 xy A1 AG Obtain answer = from correct working dx 2ay − x 2 4 5(b) State or imply 2ay − x 2 = 0 *M1 Substitute into equation of curve to obtain equation in x and a or in y and a DM1 x4 x4 e.g. 2ay 2 − ay 2 = 4a 3 or − = 4a 3 . 2a 4a Obtain one correct point A1 e.g. ( 2a, 2a ) . Obtain second correct point and no others A1 e.g. ( −2a, 2a ) . 4 SC: Allow A1 A0 for x = ±2 a or for y = 2a.

Mark scheme

6(a) Obtain a vector for one side of the parallelogram B1 2 −1 ρ ρ e.g. AB = 2 or BC = −5 . −1 −6 ρ ρ ρ ρ Correct method to obtain ± OD M1 e.g. OD = OA + BC . ρ ρ ρ ρ MO if use AB = CD or BC = DA . ρ Obtain OD = i − 4 j − 3k A1 Any equivalent form. Accept coordinates. 3 ρ ρ 6(b) Using the correct process, evaluate the scalar product BA.BC M1 ( 2 + 10 − 6 ) Scalar product of two relevant vectors. OE Using the correct process for the moduli, divide the scalar product by the M1 2 + 10 − 6 product of the moduli. . 9 × 62 Obtain answer 2 A1 62 . 62 ISW Or simplified equivalent i.e. 31 3 6(c) B1 FT Follow their cos θ . State or imply sin θ = 58 62 Use correct method to find the area of ABCD M1 e.g. 2 × 1 BA × BC sin θ . Condone decimals. 2 Correct unsimplified expression for the area A1 FT e.g. 2 × 1 × 3 × 62 × sin θ . Condone decimals. 2 Follow their sides and angle. Obtain answer 3 58 A1 Correct only. 4

Mark scheme

7 Correct separation of variables B1 2 sin 3 y dy = 4sec 2 x tan 2 x dx or equivalent. Condone missing integral signs or dx and dy. Integrate to obtain k sec 2 x M1 Obtain 2sec 2x A1 Use double angle formula and integrate to obtain py + q sin 6 y M1 Or two cycles of integration by parts. 1 1 A1 Obtain y − sin 6 y 2 12 π M1 Use y = 0 , x = in a solution containing terms λ sec 2x and μ sin 6 y to find 6 the constant of integration 1 1 A1 Or equivalent seen or implied by Obtain y − sin 6 y = 2sec 2 x − 4 2 12 π 1 − sin π = 2sec 2 x − 4 . 2 12 Obtain x = 0.541 A1 From correct working (not by using the calculator to integrate). 8

Mark scheme

8(a) A B C B1 A Dx + E State or imply the form + + Accept + . 2 x + 1 x + 2 ( x + 2 )2 2 x + 1 ( x + 2 )2 Use a correct method for finding a constant M1 Obtain one of A = 1, B = −2, C = 3 A1 For alternative form: A = 1, D = −2, E = −1 . Obtain a second value A1 Obtain the third value A1 5 8(b) −3 B1 FT The follow through is on their A, B, C. Integrate and obtain one of 12 ln ( 2 x + 1) , − 2ln ( x + 2 ) , x+2 Obtain a second term B1 FT If the alternative form is used, then either need to use integration by parts or split the fraction further. Obtain the third term B1 FT Substitute limits correctly in an integral with at least two terms of the form M1 The terms used need to have been obtained correctly. −3 Must be exact values, not decimals. 1 2 ln ( 2 x + 1) , − 2ln ( x + 2 ) and and subtract in correct order x+2 Obtain 1 – ln 3 A1 5

Mark scheme

9(a) *M1 OE Commence integration and reach pxe−2 x + q e−2 x dx A1 OE Obtain − 12 xe −2 x + 12 e −2 x dx Complete integration and obtain − 1 xe −2 x − 1 e −2 x A1 2 4 Use limits correctly and equate to 1 , having integrated twice DM1 − 1 a e −2 a − 1 e −2 a + 1 = 1 . 8 2 4 4 8 Obtain a = 1 ln ( 4a + 2 ) correctly A1 AG 2 5 9(b) Calculate the values of a relevant expression or pair of expressions at a = 0.5 M1 and a = 1 Justify the given statement with correct calculated values A1 e.g. 0.5 < 0.69…, 1 > 0.89… 0.193 > 0, –1.105 < 0 0.066 < 0.125, 0.148 > 0.125 if put limits in the integral. Condone if they use calculator for the definite integral. 2 9(c) Use the iterative process an +1 = 12 ln ( 4an + 2 ) correctly at least once. M1 Obtain final answer 0.84 A1 Show sufficient iterations to at least 4 d.p. to justify 0.84 to 2 d.p. or show A1 e.g. 0.75, 0.8047, 0.8261, 0.8343, 0.8373, 0.8385 that there is a sign change in (0.835, 0.845) 1, 0.8959, 0.8599, 0.8469, 0.8420, 0.8402 . 3

Mark scheme

10(a) Substitute x = −3 to obtain value of p ( −3) M1 Obtain p ( −3) = 0 and hence given result A1 Alternative method for Question 10(a) Divide p ( x ) by ( x + 3) to obtain quotient x 2 ± 2 x + ... M1 Obtain quotient x2 + 2x + 25, with zero remainder and hence given result A1 2 10(b) Substitute z = −1 + 2 6i and attempt expansions of z 2 and z 3 M1 z 2 = −23 − 4 6i, z 3 = −1 + 6 6i + 72 − 48 6i . Use i 2 = −1 M1 Seen at least once. Obtain p ( z ) = 0 and hence given result A1 SC B1 if there is no evidence of working for the square or the cube. Total 1/3. Alternative Method 1 Use roots z = −1 + 2 6i to form quadratic factor M1 z2 + 2z + 25. Divide p ( z ) by their quadratic factor M1 Obtain zero remainder and hence given result. A1 Alternative Method 2 Set their quadratic factor from (a) equal to zero M1 Solve for z M1 Need to see method here as answer is given. Obtain z = −1 + 2 6i (and z = −1 − 2 6i ) A1 Alternative Method 3 Substitute z = −1 + 2 6i into their quadratic factor and attempt expansion of z2 M1 Use i 2 = −1 M1 Obtain 0 and hence given result A1 3 10(c) State z1 = 3i and z2 = − 3i B1 Expand ( x + iy ) = −1 + 2 6i and compare real and imaginary parts M1 Allow for use of z 2 = −1 − 2 6i . 2 Obtain x 2 − y 2 = −1 and xy = 6 A1 Solve to obtain x and y M1 Obtain z3 = 2 + 3i and z4 = − 2 − 3i A1 Use z 2 = −1 − 2 6i to obtain z5 and z6 M1 Allow for use of z 2 = −1 + 2 6i . Obtain z5 = 2 − 3i and z6 = − 2 + 3i A1 7