Cambridge A Level Mathematics 9709 — 2022 May/June Paper 3 · Variant 3

1 Find, in terms of a, the set of values of x satisfying the inequality 2 3x a 2x 3a , + < + where a is a positive constant. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 State or imply non-modular inequality 2 2 2 2 3 2 3    x a x a , or corresponding quadratic equation, or pair of linear equations 2 2 6 2 2 3    x a x a or 32x2 + 12xa  5a2 = 0 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 Obtain critical values x = 1 4 a and x = 5 8  a A1 State final answer 5 1 8 4    a x a or  0.625a < x < 0.25a or 5 8  x a and 1 4  x a or 5 8  x a  1 4  x a A1 Do not condone ⩽ for < in the final answer. Do not ISW. SC Set a to value, (say a = 1), after initial B1 gained, then 5 1 8 4    x B1 maximum 2 out of 4. Alternative method for question 1 Obtain critical value x = 1 4 a from a graphical method, or by solving a linear equation or linear inequality B1 Obtain critical value x = 5 8  a similarly B2 State final answer 5 1 8 4    a x a or  0.625a < x < 0.25a or 5 8  x a and 1 4  x a or 5 8  x a  1 4  x a B1 Do not condone ⩽ for < in the final answer. Do not ISW. SC Set a to value, (say a = 1), after initial B1 gained, then 5 1 8 4    x B1 maximum 2 out of 4. 4

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14 The curve y tan x has two stationary points in the interval 0 = e−4x ≤x < 2π. dy (a) Obtain an expression for and show it can be written in the form sec2x a b sin 2x e−4x, where + dx a and b are constants. 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(b) Hence find the exact x-coordinates of the two stationary points. 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Mark scheme: 4(a) Use correct product rule or quotient rule, and attempt at chain rule M1 ke4x tan x + e4x sec2 x or 4 2 4 4 2 e sec tan ( e ) (e )  x x x x x k Need to see d(tan x)/dx = sec2 x (formula sheet) and attempt at ke4x, where k ≠ 1. Obtain correct derivative in any form A1  4e4x tan x + e4x sec2 x or 4 2 4 4 2 e sec tan (4e ) (e )  x x x x x Use trigonometric formulae to express derivative in the form ke4x sin x cos x sec2 x + ae4x sec2 x or ke4x sin cos cos cos x x x x + ae4x sec2 x or sec2 x(ke4x sin x cos x + ae4x) Allow 2 1 cos x instead of sec2 x M1 Need to use 2 tan sin cos sec  x x x x or sin cos tan . cos cos  x x x x x OE. M1 is independent of previous M1, but expression must be of appropriate form. 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 Question Answer Marks Guidance 4(b) Equate derivative to zero and use correct method to solve for x M1 sin 2x = 1 2 , hence x = 1 2 sin–1 1 2 or x = tan–1(2 ± 3 ) Allow M1 for correct method for non-exact value. Obtain answer, e.g. x = 1 π 12 A1 [0.262 M1 A0] Obtain second answer, e.g. 5 π 12 and no other in the given interval A1 FT FT – 2 2  their x if exact values; x must be 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

More questions on Differentiation

5 The complex number 3 is denoted by u. −i (a) Show, on an Argand diagram with origin O, the points A, B and C representing the complex numbers u, u* and u* respectively. −u State the type of quadrilateral formed by the points O, A, B and C. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ u* (b) Express in the form x iy, where x and y are real. [3] u + ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ u* 3 1(c) By considering the argument of , or otherwise, prove that tan−1 2 tan−1 . [2] 3 u 4 = ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

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

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1 16 The parametric equations of a curve are x y ln tan t, where 0 t = cos t, = < < 2π. dy cos t (a) Show that [5] dx = sin2t. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the equation of the tangent to the curve at the point where y 0. 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Mark scheme: 6(a) Use chain rule at least once M1 Needs d 1 d (tan ) d tan d  y t t t t or d d x t = (  1) 2 d (cos ) (cos ) d t t t  . BOD if + and ( 1) (  1) not seen. d d x t = sec t tan t (from List of Formulae MF19) M1 A1. If d d x t =  sec t tan t M1 A0. Obtain d d x t = sec tan t t A1 OE e.g. sin t (cos t)–2 . If e.g. d d x t = sec tan x x or sec tan   or sec tan t x , condone recovery on next line. Obtain d d y t = 2 sec tan t t A1 OE e.g. 1 sin cos t t . If e.g. d d y t = 2 sec tan x x or 2 sec tan  , condone recovery on next line. Only penalise notation errors once in d d x t and d d y t if no recovery. Use d d y x = d d y t ÷ d d x t M1 Allow even if previous M0 scored, but must be using derivatives. Obtain given answer 2 cos sin t t A1 AG After d d y x = d d y t ÷ d d x t used, any notation error A0. Must cancel cos t correctly. 5 Question Answer Marks Guidance 6(b) State or imply t = 1 π 4 when y = 0 B1 Form the equation of the tangent at y = 0 or find c M1 x = 2 , d d y x = 2 and y = 0, their coordinates and gradient used in y = mx + c. Obtain answer 2 2   y x A1 OE e.g. 2 ( 2   y x ) ISW. Allow y = 1.41x  2[.00] or 1.41(x  1.41). 3

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8 At time t days after the start of observations, the number of insects in a population is N. The variation dN 3 in the number of insects is modelled by a differential equation of the form kN 2 cos 0.02t, where dt = k is a constant and N is a continuous variable. It is given that when t 0, N 100. = = (a) Solve the differential equation, obtaining a relation between N, k and t. 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(b) Given also that N 625 when t 50, find the value of k. 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(c) Obtain an expression for N in terms of t, and find the greatest value of N predicted by this model. 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Mark scheme: 8(a) Separate variables correctly B1 3 2 dN N = (k cos 0.02t) dt Allow without integral signs. Obtain term – 2 N B1 OE Ignore position of k . Obtain term 50sin0.02t B1 OE Ignore position of k . Use t = 0, N = 100 to evaluate a constant, or as limits, in a solution containing terms a N and sin0.02 b t , where 0  ab M1 0.2 e.g. 0.2 or          c c k Obtain correct solution in any form, e.g. – 2 N = 50 sin0.02 0.2  k t A1 OE ISW e.g.   2 1 25 sin0.02 0.1   N k t 1 2 1 2 sin0.02 0.02 5     k N t 2 1 50 sin0.02 5   k t N   1 1 1 50sin0.02 2 10   k t N 1 2 20 50sin 50 100        N t kN k 5 8(b) Use the substitution N = 625 and t = 50 in expression of appropriate form to evaluate k M1 Expression must contain a + bksin 0.02t , ( N )±n, where n =  1, 1, 3 or 5 and a and b are constants 0  ab or (a + bksin 0.02t) ±2 and (N) ±n. Allow with k replaced by 1 k , 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 Question Answer Marks Guidance 8(c) Rearrange and obtain N =   2 4(0.2 0.142 607 sin 0.02 )  t Substitution for k required M1 Anything of the form N = c(d  ek sin 0.02t)2, where c, d and e are constants 0  cde 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 2 10 1 0.7125sin0.02 1 0.0713sin0.02 0.1            N N t t 2 100 0.6 sin0.02 1 sin1               N t 2 1 3 1 sin0.02 50sin1 10           N t 2 0.06 sin0.02 0.1 sin1          N t 2 800 80 57 0.02        N sin t 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

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9 With respect to the origin O, the point A has position vector given by −−¿OA i 5j 6k. The line l has = + + vector equation r 4i k 2j 3k . = + + , −i + + (a) Find in degrees the acute angle between the directions of OA and l. 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(b) Find the position vector of the foot of the perpendicular from A to l. 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(c) Hence find the position vector of the reflection of A in l. 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Mark scheme: 9(a) Using the correct process find the scalar product of direction vectors of l and OA Using the correct process for the moduli, divide the scalar product by the product of the moduli and find the inverse cosine of the result M1 Their scalar product ÷ [(12 + 52 + 62)((1)2 + 22 + 32)]. Angle = 1 27 cos 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, e.g. (3 – λ, – 5 + 2λ, – 5 + 3λ) B1 Note: (4, 1, 0) or (4, 1, 1), for 4i + k is not MR, but M1 possible. Either equate scalar product of AP and direction vector of l to zero and solve for λ or use Pythagoras in a relevant triangle and solve for λ M1 (3 – λ, – 5 + 2λ, – 5 + 3λ).( 1, 2, 3) = 0 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 Question Answer Marks Guidance 9(c) Set up a correct method for finding the position vector of the reflection of A in l M1 For all methods, allow a sign error in one component only: OA′ = OP*  (OP*  OA)    2, 4, 7 2, 4, 7 1, 5, 6   their their or OA′ = OP*  (OA  OP*)    2, 4, 7 1, 5, 6 2, 4, 7   their their or OA′ = OA + 2(OP*  OA)       1 2 2 1 5 2 4 5 6 2 7 6                their their their or midpoint OP* = (OA + OA′)/2 with their λ value substituted. 1 2 2   x their 5 4 2   y their 6 7 2   z their Obtain answer 3i + 3j + 8k or 8 3 3        i j A1 OE Condone coordinates 3, 3, 8    x y z A1. No method shown and correct answer 2/2. 2

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a 10 The constant a is such that x2 ln x dx 4. Ó 1 = @ A1 35 3 (a) Show that a . [5] = 3 ln a −1 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Verify by calculation that a lies between 2.4 and 2.8. 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(c) Use an iterative formula based on the equation in part (a) to determine a correct to 2 decimal places. Give the result of each iteration to 4 decimal places. 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Mark scheme: 10(a) Commence integration and reach 3 3 ln   ax x b x . 1 x dx Obtain 3 3 1 1 ln . 3 3   x x x 1 x dx A1 OE Allow omission of dx. Complete integration and obtain 3 3 1 1 ln 3 9  x x x A1 Allow  1 3 3 1 3       x . Use limits correctly and equate to 4, having integrated twice DM1 3 3 1 1 ln 3 9  a a a  (0  1 9 ) = 4 allow one sign error OR one numerical error, but 0 may be absent or expressed as 3 ln1 3 a . Allow  1 3 3 1 3       ax and  1 3 1 3       . Obtain given result correctly A1 1 3 35 3ln 1        a a AG 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 a = 2.4 and a = 2.8 All values must be correct for M1 (numerical question) M1 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 Question Answer Marks Guidance 10(c) Use the iterative process 1  n a 1 3 35 3ln 1        n a correctly at least twice M1 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 a sign change in (2.635, 2.645) A1 2.635 (35/(3lna  1))1/3 a = 0.0029(4) > 0 2.645 (35/(3lna  1))1/3 a =  0.012 < 0 3

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