Mathematics 9709/32 February/March 2022: question paper and mark scheme

Mark scheme

1 State or imply non-modular inequality ( 2 x + 3 ) > 32 ( x + 2 ) , or 2 2 B1 corresponding quadratic equation, or pair of linear equations Make a reasonable attempt at solving a 3-term quadratic, or solve two linear M1 Quadratic formula or (5x + 9)(x +3) equations for x 9 A1 OE Obtain critical values x = – 3 and x = − 5 9 9 A1 [Do not condone ⩽ for < in the final answer.] State final answer −3 < x < − or x > −3 and x < − No ISW 5 5 Alternative method for question 1 Obtain critical value x = – 3 from a graphical method, or by solving a linear B1 2x + 3 = 3(x + 2) x = − 3 equation or linear inequality 9 B2 Obtain critical value x = − similarly 5 9 9 B1 [Do not condone ⩽ for < in the final answer.] State final answer −3 < x < − or x > −3 and x < − No ISW 5 5 4

Mark scheme

2 Show a circle with centre – 2 + 3i B1 Must see (− 2, 3) or appropriate marks on axes Show a circle of radius 2 and centre not at the origin. B1 Show correct half line from the origin B1 3π π or seen, or half line that approximately bisects 4 4 π angle . 2 Shade the correct region. B1 4 N.B. Maximum 3 out of 4 if any errors seen.

Mark scheme

3 State or imply n ln x + 2 ln y = ln C B1 1 M1 Using lnx and lny values Substitute values of ln y and ln x, or equate gradient of line to ± n , but not ± n, 2 and solve for n 4 A1 Obtain n = 0.8[0] or 0.8[00] or 5 Solve for C M1 Using lnx and lny values in equation of correct form, that is lnC not C. Allow C = e2.668. Obtain C = 14.41 A1 Must be 2 d.p. Alternative method for question 3 Obtain two correct equations in n and C by substituting x and y values in the B1 (2.886)n × (2.484)2 = C and (1.363)n × (3.353)2 = C given equation Solve for n M1 Using x and y values Obtain n = 0.8[0] or 0.8[00] or 4/5 A1 æ 2.886 ö÷n æ 2.484 ö÷2 çç ´ç = 1 leading to n = 0.7995 çè 1.363 ÷÷ø èçç 3.353 ø÷÷ Solve for C M1 Using x and y values Obtain C = 14.41 A1 Must be 2 d.p. 5

Mark scheme

4 dx dy 1 B1 State = sin θ or = − sin θ + sin 2θ dθ dθ 2 dy dy dx M1 Use = ÷ dx dθ dθ Obtain correct answer in any form A1 1 − sin θ + sin 2θ e.g. 2 sin θ dy M1 sin 2θ = 2sin θ cos θ Use double angle correctly to obtain in terms of θ dx 1 A1 AG. Obtain the given answer with no errors seen − 2sin2 θ Requires correct cancellation of ALL sin θ terms and 2 1 cos θ = 1 − 2sin2 θ seen 2 SC For incorrect signs, consistent throughout max. B0, M1, A0, M1, A1 5

Mark scheme

5 Use correct tan ( A + B ) formula and obtain an equation in tan α and tan β M1 tan + tan =2 1 tan tan Substitute throughout for tan α or for tan β M1 3tan + tan =2 1 3tan² Obtain 3tan 2 β + 2 tan β − 1 = 0 or tan 2α + 2 tan α − 3 = 0 A1 OE e.g. 6 tan 2 β + 4 tan β − 2 = 0 2 4 or tan 2α + tan α − 2 = 0 3 3 Solve a 3-term quadratic and find an angle M1 Obtain answer α = 45 ° , β = 18.4 ° A1 π or 0.785, 0.322 4 Obtain answer α =108.4 ° , β =135 ° A1 3π 1.89 , or 2.36. Answer in radians, max. A1A0 or 4 vice versa. Ignore answers outside [0°, 180°] 6 SC: If A0A0 allow SC B1 for both α’s or both β’s

Mark scheme

6 Substitute and obtain a correct equation in x and y B1 (x + iy)2 + 2i(x − iy) = 1 Use i 2 = −1 at least once and equate real and imaginary parts M1 Obtain two correct equations, e.g. x 2 − y 2 + 2 y = 1 and 2 xy + 2 x = 0 A1 Solve for x or for y M1 Using y = – 1, obtain answer w = – 2 – i only A1 A0 if w = 2 – i as well Using x = 0, obtain answer w = i A1 6

Mark scheme

7(a) Sketch a relevant graph, e.g. y = 4 − x 2 B1 Needs (0, 4) or marks on axis and (2, 0) or (π, 0) 1 B1 Needs (0, 1) or mark on axis and (π, 0) Sketch a second relevant graph, e.g. y = sec x , and justify the given statement Asymptote NOT required, but must NOT reach x = π. 2 é 3π ù Sec graph must exist over at least interval ê0, ú and êë 4 úû quadratic graph over [0, 2.5]. 2 7(b) Calculate the value of a relevant expression or values of a pair of relevant M1 Need all 4 values or the 2 values correct for M1. expressions at x = 1 and x = 2. Angles in degrees score M0. Complete the argument with correct calculated values A1 2 7(c) Use the iterative process correctly at least twice M1 Obtain final answer 1.60 A1 Must be 2 d.p. Show sufficient iterations to 4 d.p.to justify 1.60 to 2 d.p. or show there is a sign A1 change in the interval (1.595, 1.605) 3

Mark scheme

8(a) Commence division and reach quotient of the form 2x ± 1 M1 Or by inspection 8x3 + 4x2 + 2x + 7 = (4x2 + 1)(2x ± 1) +r Obtain (quotient) 2x + 1 A1 Obtain (remainder) 6 A1 3 8(b) Obtain terms x 2 + x B1 OE Obtain term of the form a tan −1 2 x M1 Obtain term 3 tan −1 2 x A1 OE 1 M1 æ 1 ö÷2 1 Use x = 0 and x = as limits in a solution containing a term of the form çç ÷ + + a π , need π seen or implied 2 çè 2 ÷ø 2 4 4 a tan −1 2 x 3 A1 ISW, Answers in degrees score A0. Obtain final answer (1 + π ), or exact equivalent 4 5

Mark scheme

9 Correctly separate variables and integrate at least one side M1 To obtain aln y or bln (x + 1) + cln(3x + 1) Obtain term ln y from integral of 1/y B1 A B M1 State or imply the form + and use a correct method to find a x + 1 3x + 1 constant 1 3 A1 Obtain A = − and B = 2 2 1 1 1 1 A1 FT The FT is on the values of A and B. Obtain terms − ln ( x + 1) + ln ( 3 x + 1) or − ln ( 2 x + 2 ) + ln ( 6 x + 2 ) or + A1 2 2 2 2 combination of these terms FT Use x = 1 and y = 1 to evaluate a constant, or expression for a constant, (decimal *M1 1 1 1 equivalent of ln terms allowed) or as limits, in a solution containing terms e.g. ln y = − ln ( x + 1) + ln ( 3 x + 1) − ln 2 2 2 2 a ln y , b ln ( x + 1) and c ln ( 3x + 1) , where abc ≠ 0 Obtain an expression for y or y2 and substitute x = 3 DM1 Do not accept decimal equivalent of ln terms 1 5 10 A1 ISW. Must be simplified and exact, do not allow 1.118 Obtain answer y = 5 or or 1 ln 5 2 4 8 or e 2 4. 9

Mark scheme

10(a) Obtain direction vector −i − 3 j + k B1 OE Use a correct method to form a vector equation M1 Obtain answer r = 2i + j + k + λ ( −i − 3 j + k ) A1 Need r or r on LHS or r = i 2 j + 2k + λ ( −i − 3 j + k ) 3 10(b) Carry out the correct process for evaluating the scalar product of the direction M1 (−1, − 3, 1) . (1, −3, −2) = − 1 + 9 − 2 vectors. Using the correct process for the moduli, divide the scalar product by the product M1 1+ 9 2 cos −1 ( 1 + 9 + 1)(1 + 9 + 4 )) of the moduli and find the inverse cosine of the result for any 2 vectors Obtain answer 61.1 ° A1 61.086° 3 10(c) Express general point of AB or l in component form, e.g. (2 – λ, 1 – 3λ, 1 + λ) or B1 (1 + µ, 2 – 3µ, – 3 – 2µ) Equate at least two pairs of components and solve for λ or for µ M1 1 14 2 11 A1 Obtain a correct answer for λ or µ, e.g. λ = 6, , or − ; µ = – 5, or − 3 9 3 9 Verify that all three equations are not satisfied, and the lines do not intersect A1 Express general point of AB or l in component form, 4 e.g. (1 – λ*, − 2 – 3λ*, 2 + λ*) or (1 + µ*, 2 – 3µ*, – 3 – 2µ*)

Mark scheme

11(a) Use correct product rule or chain rule M1 Obtain correct derivative in any form A1 cos x.cos 2x − sin x.2sin 2x Equate derivative to zero and use a correct double angle formula *M1 If chain rule used then derivative set to 0 gains M1 since correct double angle formula has already been used. Obtain an equation in one trigonometric variable DM1 Allow following from coefficient errors in differentiation only Obtain 6sin 2 x = 1 , 6cos 2 x = 5 or 5tan 2 x = 1 A1 One of these 3 expressions Obtain final answer x = 0.421 A1 Must be 3s.f. 6 11(b) State or imply du = − sin x dx B1 Using double angle formula, express integral in terms of u and du M1 Use cos2x = 2cos2x − 1 2 3 A1 Integrate and obtain ± u − u 3 1 M1 Require both limits substituted twice in au + bu 3 for Use limits u = 1, u = in an integral of the form au + bu 3 , where ab ≠ 0 2 M1. Do not condone decimals. A1 ISW Obtain 1 3 ( 2 − 1) or 13 2 13 or 23 12 13 or simplified equivalent 5