Mathematics 9709/33 May/June 2022: question paper and mark scheme

Mark scheme

1 State or imply non-modular inequality 22 ( 3x + a ) < ( 2 x + 3a ) , or B1 e.g. ( 6 x + 2a )2 = ( 2 x + 3a )2 or 32x2 + 12xa − 5a2 = 0 2 2 corresponding quadratic equation, or pair of linear equations 2(3x + a) = (2x + 3a) and − 2(3x + a) = (2x + 3a) Solve 3-term quadratic, or solve two linear equations for x M1 Apply general rules for solving quadratic equation by formula or by factors. Instead of x = {formula}, have {formula} = 0 and try to solve for a then M0 1 5 A1 Obtain critical values x = a and x = − a 4 8 5 1 A1 Do not condone ⩽ for < in the final answer. State final answer − a < x < a or − 0.625a < x < 0.25a Do not ISW. 8 4 5 1 5 1 SC Set a to value, (say a = 1), after initial B1 gained, then or x > − a and x < a or x > − a x < a 5 1 8 4 8 4 − <x< B1 maximum 2 out of 4. 8 4 Alternative method for question 1 1 B1 Obtain critical value x = a from a graphical method, or by solving a 4 linear equation or linear inequality 5 B2 Obtain critical value x = − a similarly 8 5 1 B1 Do not condone ⩽ for < in the final answer. State final answer − a < x < a or − 0.625a < x < 0.25a Do not ISW. 8 4 5 1 5 1 SC Set a to value, (say a = 1), after initial B1 gained, then or x > − a and x < a or x > − a x < a 5 1 8 4 8 4 − <x< B1 maximum 2 out of 4. 8 4 4

Mark scheme

2 Use correct cos ( A − B ) formula to obtain an equation in cos θ and sin θ B1 cos θ cos 60 + sin θ sin 60 = 3 sin θ Use trigonometric formula and substitute values for cos 60 and sin 60 to M1 1 3 obtain an equation in tan θ (or cosθ or sin θ ) Allow and interchanged. 2 2 1 3 + tan θ = 3 tan θ 2 2 1 3 3 cos2 θ = 3 − (1 − cos θ) 2 3− 4 2 2 1 3 3 2 (1 − sin2 θ) = 3 − 3− sin θ 4 2 2 1 6+ 3 A1 OE Obtain tan θ = or tan θ = or 0.2343, 6− 3 33 3 1 3 cosθ = 2 2 or 0.9736 or sin θ = or 0.2281 10 − 3 3 10 −3 3 Obtain answer, e.g. θ = 13.2° A1 May be more accurate, allow value rounding to 13. 2°. θ = 13.1867° . Obtain second answer, e.g. θ = 193.2° and no others in the given interval A1 FT May be more accurate. Allow value rounding to 193.2°. FT is on previous value of θ, must have scored M1. Note if θ is negative (e.g. − 13.2 ): −13.2 + 180 = 166.8 A0 but −13.2 + 360 = 346.8 A1 FT. Ignore answers outside the given interval. Treat answers in radians as a misread. 0.23015, 3.3717. 2 Alternative method for question 2 – using Rcos(θ ± α ) or equivalent Use correct cos(A – B) formula to obtain an equation in cos θ and sin θ B1 cos θ cos 60 + sin θ sin 60 = 3 sin θ Correct method for finding tan α from pcosθ + qsinθ = 0 M1 q tan α = ± p Correct value of α A1 76.8° or 1.34 radians (may be more accurate). Obtain answer, e.g. θ = 13.2° A1 May be more accurate, allow value rounding to 13. 2°. θ = 13.1867° . Obtain second answer, e.g. θ = 193.2° and no others in the given interval A1 FT May be more accurate. Allow value rounding to 193.2°. FT is on previous value of θ, must have scored M1. Note if θ is negative (e.g. − 13.2 ): −13.2 + 180 = 166.8 A0 but −13.2 + 360 = 346.8 A1 FT. Ignore answers outside the given interval. Treat answers in radians as a misread. 0.23015, 3.3717. 5

Mark scheme

3(a) Use law of logarithm of a power M1 log3(2x + 1) = 1 + log3(x − 1)2 Use log 3 3 = 1 B1 log3(2x + 1) = log 3 3 + 2log3(x − 1) 2x + 1 2x + 1 log3 = log 3 or = 3 ( x − 1)2 3 ( x − 1)2 SC For candidates scoring M0 B0 due to combining logs before dealing with coefficient 2, and confusing coefficients, allow log3 (…) = c leading to (...) = 3c B1. Obtain 3 x 2 − 8 x + 2 = 0 or 1.5 x 2 − 4 x + 1 = 0 A1 OE 3 terms only and = 0 required. 3 3(b) Solve 3-term quadratic equation from part 3(a) or restart to find y M1 4 ± 10 y= or y = 1.1937… or y = 0.1396… 6 (x = 2.3874 or x = 0.2792) x May solve for x but must find y = to gain M1. 2 Obtain answer 1.19 A1 CAO. 2 dp required. 2

Mark scheme

4(a) Use correct product rule or quotient rule, and attempt at chain rule M1 e 4 x sec 2 x − tanx(ke 4 x ) ke−4x tan x + e−4x sec2 x or (e 4 x ) 2 2 Need to see d(tan x)/dx = sec x (formula sheet) and attempt at ke−4x, where k ≠ 1. Obtain correct derivative in any form A1 e 4 x sec 2 x − tanx (4e 4 x ) − 4e−4x tan x + e−4x sec2 x or (e 4 x ) 2 Use trigonometric formulae to express derivative in the form M1 tan x sin x cos x Need to use = sin x cos x or tan x = . OE. sin x cos x 2 sec x cos x cos x ke−4x sin x cos x sec2 x + ae−4x sec2 x or ke−4x + ae−4x sec2 x or cos x cos x M1 is independent of previous M1, but expression must be of sec2 x(ke−4x sin x cos x + ae−4x) appropriate form. 1 Allow instead of sec2 x cos 2 x Obtain correct answer with a = 1 and b = – 2 A1 At least one line of trigonometric working is required from −4e−4x tan x + e−4x sec2 x to given answer sec2 x(1 – 2 sin 2x) e−4x with elements in any order. If only error: 4 sin x cos x = 4 sin 2x M1 A1 M1 A0. 4 4(b) Equate derivative to zero and use correct method to solve for x M1 1 1 1 sin 2x = , hence x = sin–1 or x = tan–1(2 ± 3 ) 2 2 2 Allow M1 for correct method for non-exact value. 1 A1 [0.262 M1 A0] Obtain answer, e.g. x = π 12 5 A1 FT π – their 2 x π Obtain second answer, e.g. π and no other in the given interval FT if exact values; x must be < . 12 2 2 Ignore answers outside the given interval. Treat answers in degrees as a misread. 15°, 75°. SC No values found for a and b in 4(a) but chooses values in 4(b): max M1 for x . 3

Mark scheme

5(a) Show u and u* in relatively correct positions. Must have sense of scale on B1 u = 3 − i, u* = 3 + i Ignore labels. axes Show u* – u in a relatively correct position. Must have sense of scale on B1 2i. Scale only on Imaginary axis is sufficient for this mark. axes State that OABC is a parallelogram [independent of previous marks] B1 Ignore ‘quadrilateral’. Allow ‘trapezium’ from correct work. 3 5(b) Multiply their numerator and the given denominator by 3 + i and attempt M1 Can have missing term and arithmetic errors but need to evaluate either i2 = −1 once, seen or implied. Obtain numerator 8 + 6i or denominator 10 A1 4 3 8 6 A1 Correct answer with no working scores 0/3. State final answer + i or + i or 0.8 + 0.6i 5 5 10 10 Alternative method for question 5(b) Obtain two equations in x and y, and attempt to solve for x or for y M1 3 = 3x + y and 1 = − x + 3y 4 8 3 6 A1 Obtain x = or or 0.8 y = or or 0.6 5 10 5 10 4 3 8 6 A1 Correct answer with no working scores 0/3. State final answer + i or + i or 0.8 + 0.6i 5 5 10 10 3 5(c) u* M1 State or imply arg = arg u* − arg u or 2arg u* u Justify the given statement correctly A1 AG u* 3 1 arg = tan–1 , arg u* = tan−1 and u 4 3 1 1 arg u = tan–1 − (or arg u = − tan–1 ), needed if use first 3 3 expression in M1; u* 3 1 or arg = tan–1 and arg u* = tan−1 , needed if use u 4 3 second expression in M1. Alternative method for question 5(c) 1 M1 2 tan A 1 3 Use tan 2A formula with tan A = tan 2 A = 2 , tan A = , hence tan 2A = . 3 1 − tan A 3 4 Justify the given statement correctly A1 AG 3 u* 1 So 2A = tan−1 = arg and A = tan−1 = arg u* 4 u 3 u* hence arg = 2 arg u*. u 2

Mark scheme

6(a) Use chain rule at least once M1 dy 1 d Needs = (tan t ) dt tan t dt dx d or = ( − 1) (cos −2 t ) (cos t ) . dt dt BOD if + and (− 1) ( − 1) not seen. dx = sec t tan t (from List of Formulae MF19) M1 A1. dt dx If = − sec t tan t M1 A0. dt dx A1 OE e.g. sin t (cos t)–2 . Obtain = sec t tan t dt dx If e.g. = sec x tan x or secθ tan θ or sec t tan x , condone dt recovery on next line. dy sec 2t A1 1 Obtain = OE e.g. . dt tan t sint cos t dy sec 2 x sec 2θ If e.g. = or , condone recovery on next dt tan x tan θ line. dx dy Only penalise notation errors once in and if no dt dt recovery. dy dy dx M1 Allow even if previous M0 scored, but must be using Use = ÷ derivatives. dx dt dt cos t A1 dy dy dx Obtain given answer AG After = ÷ used, any notation error A0. sin 2 t dx dt dt Must cancel cos t correctly. 5 6(b) 1 B1 State or imply t = π when y = 0 4 Form the equation of the tangent at y = 0 or find c M1 dy x= 2 , = 2 and y = 0, their coordinates and gradient dx used in y = mx + c. Obtain answer y = 2 x − 2 A1 OE e.g. y = 2 ( x − 2 ) ISW. Allow y = 1.41x − 2[.00] or 1.41(x − 1.41). 3

Mark scheme

7(a) A Bx + C B1 A Bx + C A C State or imply the form + If 1 − + or + B0 x−2 2x2 + 3 x−2 2 2x + 3 x−2 2 x2 + 3 then M1 A1 (for A = 3) still possible. Use a correct method for finding a constant M1 Obtain one of A = 3, B = – 1 and C = 6 A1 Allow all A marks obtained even if method would give errors if equations solved in a different order. Obtain a second value A1 Obtain the third value A1 5 7(b) Use correct method to find the first two terms of the expansion of M1 Symbolic binomial coefficients not sufficient for the M1. −1 −1 ( x − 2 )−1 , 1 − 1 x , 2 x 2 + 3 or 1 + 2 x 2 ( ) −1 2 3 Obtain correct unsimplified expansions, up to the term in x2, of each A1 FT The FT is on A, B and C. partial fraction A1 FT A x (−1)(−2) x 2 − 1 − − + − + ... 2 2 2 2 Bx + C 2 x 2 1 − + ... 3 3 ( ) with expansion to M1 −1 Extract the coefficient 3 correctly from 2 x 2 + 3 C Bx C2 1 2 + ± x 2 or C + Bx ± C x 2 3 3 3 3 3 3 2 2 1± x then multiply by Bx + C up to the terms in x2, where BC ≠ 0 Allow a slip in multiplication for M1. 3 Allow miscopies in B and C from 7(a). 1 13 41 2 A1 Do not ISW. Obtain final answer − x− x 2 12 24 5

Mark scheme

8(a) Separate variables correctly B1 dN 3 = (k cos 0.02t) dt Allow without integral signs. N2 2 B1 OE Ignore position of k . Obtain term – N Obtain term 50sin 0.02t B1 OE Ignore position of k . Use t = 0, N = 100 to evaluate a constant, or as limits, in a solution M1 −0.2 a e.g. c = −0.2 or c = k containing terms and b sin 0.02t , where ab ≠ 0 N 2 A1 OE ISW Obtain correct solution in any form, e.g. – = 50k sin 0.02t − 0.2 1 N 1 − k 1 e.g. N = −2 N 2 = sin0.02t − ( 25ksin0.02t − 0.1) 2 0.02 5 2 1 1 1 1 50ksin0.02t = − + = − k ( 50sin0.02t ) + N 5 N 2 10 1 2 N 20 50sin t = − + 50 kN 100 k 5 8(b) Use the substitution N = 625 and t = 50 in expression of appropriate form M1 Expression must contain a + bksin 0.02t , ( N )±n, where to evaluate k n = − 1, 1, 3 or 5 and a and b are constants ab ≠ 0 or (a + bksin 0.02t) ±2 and (N) ±n. Allow with k replaced by 1k , error due to k(N −3/2) when separating variables in 8(a). If invert term by term when 3 terms shown then M0. Obtain k = 0.00285[2148] A1 Must evaluate sin1. Degrees k = 0.138 M1 A0. 2 8(c) Rearrange and obtain N = 4(0.2 − 0.142 ( 607 ) sin 0.02t ) −2 M1 Anything of the form N = c(d − ek sin 0.02t)−2, where c, d and Substitution for k required e are constants cde ≠ 0 and value of k substituted. Allow with k replaced by 1/k, error due to k(N −3/2) when separating variables in 8(a). OE ISW e.g. 2 10 1 N = − N= 0.7125sin0.02t − 1 ( −0.0713sin0.02t + 0.1)2 100 1 N= 2 N= 2 0.6 3 1 sin1 sin0.02t − 1 × sin0.02t + −50sin1 10 −2 2 0.06 800 N = − sin0.02t + 0.1 N = sin1 80 − 57 sin0.02t Do not need to substitute for sin(0.02t) = 1, but must substitute for k. Accept answers between 1209 and 1215 A1 ISW Substitute sin 0.02t = 1 or t = 50 sin−1 1 or 78.5 or 25π. Answer with no working (rubric) 0/2. SC N = … not seen but correct numerical answer B1 1/2. 2

Mark scheme

9(a) Using the correct process find the scalar product of direction vectors of l M1 (1, 5, 6).(− 1, 2, 3) = −1.1 + 5.2 + 6.3 = – 1 + 10 + 18 and OA Using the correct process for the moduli, divide the scalar product by the M1 Their scalar product ÷ [√(12 + 52 + 62)√((−1)2 + 22 + 32)]. product of the moduli and find the inverse cosine of the result 27 Angle = cos −1 . 62 14 Obtain answer 23.6°. A1 AWRT 23.6°. 23.5889°. Radians 0.412 scores A0 (0.4117…). 3 9(b) Taking a general point P on l, state AP (or PA) in component form, B1 Note: (4, 1, 0) or (4, 1, 1), for 4i + k is not MR, but M1 e.g. (3 – λ, – 5 + 2λ, – 5 + 3λ) possible. Either equate scalar product of AP and direction vector of l to zero and M1 (3 – λ, – 5 + 2λ, – 5 + 3λ).( −1, 2, 3) = 0 solve for λ or use Pythagoras in a relevant triangle and solve for λ −3 −10 −15 + λ + 4 λ + 9 λ = 0 or let OQ = (4, 0, 1) so AQ = (3, − 5, − 5), QP = (– λ, 2λ, 3λ), AP = (3 – λ, –5 + 2λ, –5 + 3λ) hence 32 + (−5)2 + (−5)2 = (3 – λ)2 + (–5 + 2λ)2 + (–5 + 3λ)2 + (–λ)2 + (2λ)2 + (3λ)2 Other alternative approaches are possible, e.g. minimise AP or AP2, either by completing the square or by differentiating. Obtain λ = 2 A1 λ = 2 State that the position vector OP* of the foot is 2i + 4j + 7k A1 OE Condone coordinates. 4 9(c) Set up a correct method for finding the position vector of the reflection of M1 For all methods, allow a sign error in one component only: A in l OA′ = OP* + (OP* − OA) their ( 2, 4, 7 ) + ( their 2, 4, 7 − 1, 5, 6) or OA′ = OP* − (OA − OP*) their ( 2, 4, 7 ) − (1, 5, 6 − their 2, 4, 7 ) or OA′ = OA + 2(OP* − OA) 1 + 2 ( their 2 − 1) 5 + 2 ( their 4 − 5 ) 6 + 2 ( their 7 − 6 ) or midpoint OP* = (OA + OA′)/2 with their λ value substituted. 1+ x 5+ y 6+ z = their 2 = their 4 = their 7 2 2 2 8 A1 OE Obtain answer 3i + 3j + 8k or 3 i + j + Condone coordinates 3 x = 3, y = 3, z = 8 A1. No method shown and correct answer 2/2. 2

Mark scheme

10(a) 1 *M1 OE Allow omission of dx. Commence integration and reach ax3 ln x + b x3 . dx x 1 3 1 1 A1 OE Allow omission of dx. Obtain x ln x − x3 . dx 3 3 x 1 3 1 A1 1 1 3 Complete integration and obtain x ln x − x3 Allow − x . 3 9 3 3 Use limits correctly and equate to 4, having integrated twice DM1 1 3 1 1 a ln a − a 3 − (0 − ) = 4 allow one sign error OR one 3 9 9 a3 numerical error, but 0 may be absent or expressed as ln1 . 3 1 1 1 1 Allow − ax3 and − . 3 3 3 3 Obtain given result correctly A1 1 35 3 a= AG 3lna − 1 After substitution, any errors even if corrected A0. Need to see at least one line of working between substitution and the given answer. 5 10(b) Calculate the values of a relevant expression or pair of expressions at M1 a = 2.4 and a = 2.8 All values must be correct for M1 (numerical question) Justify the given statement with correct calculated values A1 2.4 < 2.7(8) and 2.8 > 2.5(6) sign change here insufficient OR –0.3(8) and 0.2(4) < 0, > 0 or change of sign. 2 10(c) 1 M1 35 3 Use the iterative process an +1 = correctly at least twice 3ln an − 1 Obtain final answer a = 2.64 A1 Must be 2 dp. Show sufficient iterations to 4 dp to justify 2.64 to 2 dp, or show there is A1 2.635 (35/(3lna − 1))1/3− a = 0.0029(4) > 0 a sign change in (2.635, 2.645) 2.645 (35/(3lna − 1))1/3− a = − 0.012 <0 3