Cambridge A Level Mathematics 9709 — 2021 May/June Paper 3 · Variant 3

2 Solve the equation 4x 3 Give your answer correct to 3 decimal places. [5] = + 4−x. ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 State or imply 2 3 1 0 − −= u u B1 Solve for u or 4x M1 Obtain root ( ) 1 3 13 2 + , or decimal in [3.30, 3.31] A1 Use correct method for finding x from a positive root M1 Obtain answer x = 0.862 and no other A1 5

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3 The parametric equations of a curve are x t ln t 2 , y t = + + = −1 e−2t, where t > −2. dy (a) Express in terms of t, simplifying your answer. [5] dx ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the exact y-coordinate of the stationary point of the curve. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(a) State d d x t = 1 + 1 2 + t B1 Use product rule M1 Obtain d d y t = ( ) 2 2 e 2 1 e − − − − t t t A1 OE Use d d y x = d d y t ÷ d d x t M1 Obtain correct answer in any simplified form, e.g. ( )( ) 3 2 2 3 − + + t t t 2 e−t A1 5 3(b) Equate derivative to zero and solve for t M1 Obtain t = 3 2 and obtain answer y = 3 1 e 2 −, or exact equivalent A1 2

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5 (a) By first expanding tan , show that the equation tan 1 tan may be expressed as 2 21 + 21 41 = 1 2 0. [4] tan41 + tan21 −7 = ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Hence solve the equation tan 1 tan for [3] 2 41 = 1, 0Å < 1 < 180Å. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) Use double angle formula to express tan 4θ in terms of tan 2θ M1 Use double angle formula to express result in terms of tan θ M1 Obtain a correct equation in tan θ in any form A1 Obtain the given answer A1 4 Question Answer Marks Guidance 5(b) Solve for tan θ and obtain a value of θ M1 Obtain answer, e.g. 53.5° A1 Obtain second answer, e.g. 126.5° and no other in the interval A1 Ignore answers outside the given interval. Treat answers in radians as a misread. 3

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6 (a) By sketching a suitable pair of graphs, show that the equation cot 1 1 has exactly one 2x = + e−x [2] root in the interval 0 x < ≤π. (b) Verify by calculation that this root lies between 1 and 1.5. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ @ A 1 (c) Use the iterative formula 2 to determine the root correct to 2 decimal tan−1 1 xn+1 = e−xn places. Give the result of each iteration to 4 +decimal places. 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Mark scheme: 6(a) Sketch a relevant graph, e.g. 1 cot 2 = y x B1 Sketch a second relevant graph, e.g. 1 e− = + x y , and justify the given statement B1 2 6(b) Calculate values of a relevant expression or pair of expressions at x = 1 and x = 1.5 M1 Complete the argument correctly with correct calculated values A1 2 6(c) Use the iterative formula correctly at least once M1 Obtain final answer 1.34 A1 Show sufficient iterations to 4 d.p. to justify 1.34 to 2 d.p. or show there is a sign change in the interval (1.335, 1.345) A1 3

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7 y P x O M N For the curve shown in the diagram, the normal to the curve at the point P with coordinates x, y meets the x-axis at N. The point M is the foot of the perpendicular from P to the x-axis. The curve is such that for all values of x in the interval 0 1 the area of triangle PMN is equal to tan x. ≤x < 2π, MN dy (a) (i) Show that [1] y dx. = ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ 1 dy (ii) Hence show that x and y satisfy the differential equation 2y2 tan x. [2] dx = ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ (b) Given that y 1 when x 0, solve this differential equation to find the equation of the curve, expressing y in= terms of x.= [6] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a)(i) Justify the given statement MN y = d d y x B1 1 7(a)(ii) Express the area of PMN in terms of y and d d y x and equate to tan x M1 Obtain the given equation correctly A1 2 7(b) Separate variables and integrate at least one side M1 Obtain term 3 1 6 y A1 Obtain term of the form lncos ± x M1 Evaluate a constant or use x = 0 and y = 1 in a solution containing terms 3 ay and lncos ± x , or equivalent M1 Obtain correct answer in any form, e.g. 3 1 1 lncos 6 6 = − + y x A1 Obtain final answer 3 (1 6ln cos ) = − y x A1 OE 6

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8 y M x O ln x The diagram shows the curve y and its maximum point M. = x4 (a) Find the exact coordinates of M. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ a ln x 1 (b) By using integration by parts, show that for all a 1, dx 9. [6] x4 < > Ô 1 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 8(a) Use quotient or product rule M1 Obtain correct derivative in any form A1 Equate derivative to zero and solve for x M1 Obtain x = 4 e and y = 1 4e , or exact equivalents A1 4 8(b) Commence integration and reach 3 3 ln . − − +  ax x b x 1 x dx *M1 Obtain 3 3 1 1 ln . 3 3 − − − +  x x x 1 x dx A1 OE Complete integration and obtain 3 3 1 1 ln 3 9 − − − − x x x A1 Substitute limits correctly, having integrated twice DM1 Obtain answer 3 3 1 1 1 ln 9 3 9 − − − − a a a A1 OE Justify the given statement A1 6

More questions on Integration

9 The quadrilateral ABCD is a trapezium in which AB and DC are parallel. With respect to the origin O, the position vectors of A, B and C are given by OA 2j 3k, OB i 3j k and −−¿ −−¿ = −i + + = + + OC 2i 2j −−¿ = + −3k. (a) Given that DC AB, find the position vector of D. [3] −−¿ 3−−¿ = ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) State a vector equation for the line through A and B. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Find the distance between the parallel sides and hence find the area of the trapezium. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 9(a) State or imply 2 2 = + − AB i j k B1 OE Carry out a correct method to find  OD M1 Obtain answer 4 3 − −+ i j k A1 OE 3 9(b) State ( ) 2 3 2 2 λ =−+ + + + − r i j k i j k B1FT OE. The FT is on  AB . 1 9(c) For a general point P on AB, state  CP or  DP in component form, e.g. ( ) 3 2 , , 6 2 λ λ λ = − − −+  CP *M1 Equate a relevant scalar product to zero or equate derivative of  CP to zero or use Pythagoras in a relevant triangle and solve for λ DM1 Obtain λ = 2 A1 Show the perpendicular is of length 3 A1 Carry out a correct method to find the area of ABCD and obtain the answer 18 A1 Alternative method for Question 9(c) Use a scalar product to find the projection CN (or DN) of BC (or AD) on CD *M1 Obtain CN = 3 (or DN = 3) A1 Use Pythagoras to obtain BN (or AN) DM1 Question Answer Marks Guidance 9(c) cont’d Obtain answer 3 A1 Carry out a correct method to find the area of ABCD and obtain the answer 18 A1 5

More questions on Vectors

10 (a) Verify that 2i is a root of the equation z4 3z2 2z 12 0. [3] −1 + + + + = ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the other roots of this equation. [7] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 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Mark scheme: 10(a) Substitute 1 2i −+ and attempt expansions of the 2z and 4z terms M1 Use 2i 1 =− at least once M1 Complete the verification correctly A1 3 10(b) State second root 1 2i −− A1 Carry out a method to find a quadratic factor with zeros 1 2i −± M1 Obtain 2 2 3 + + z z A1 Commence division and reach partial quotient 2 + z kz M1 Obtain second quadratic factor 2 2 4 − + z z A1 Solve a 3-term quadratic and use 2i 1 = − M1 Obtain roots 1 3i + and 1 3i − A1 7

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