Cambridge A Level Mathematics 9709 — 2023 May/June Paper 3 · Variant 1

1 Solve the equation 3e2x 5. −4e−2x = Give the answer correct to 3 decimal places. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 2 2 2 3 e 5 e 4 0 x x    2 2 5 73 1 5 73 e , ln 6 2 6 x x             M1 Use correct method to solve for x. x = 0.407 A1 Only 3

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2 (a) Sketch the graph of y 2x 3 . [1] = + (b) Solve the inequality 3x 8 2x 3 . [3] + > + ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

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

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4 (a) Show that the equation sin cos 2 can be expressed in the form 21 + 21 = sin21 2 sin cos 0. [2] cos21 + 1 1 −3 sin21 = ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Hence solve the equation sin cos 2 for [4] 21 + 21 = sin21 0Å < 1 < 180Å. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) Use correct double angle formulae M1 e.g. 2 2 2 2sin cos cos sin 2sin         Obtain 2 2 cos 2sin cos 3sin 0        from full and correct working A1 AG Check conclusion is complete and matches the working. 2 Question Answer Marks Guidance 4(b) Factorise to obtain    cos sin cos 3sin 0        B1 OE Solve a quadratic in sinand costo obtain a value for .  M1 1 3 tan 1 tan or    . Obtain one correct value e.g. 45 A1 Obtain a second correct value e.g. 161.6 and no others in the interval A1 Mark answers in radians (0.785 and 2.82) as a misread. Accept awrt 161.6 . Alternative Method 1 Obtain 2 3tan 2tan 1 0     B1 Solve a 3 term quadratic in tan to obtain a value for .  M1 1 3 tan 1 tan or    . Obtain one correct value e.g. 45 A1 Obtain a second correct value e.g. 161.6 and no others in the interval A1 Mark answers in radians (0.785 and 2.82) as a misread. Alternative Method 2 Obtain     2 2 cos sin 2sin      B1 Solve to obtain a value for .  M1 1 3 tan 1 tan or    . Obtain one correct value e.g. 45 A1 Obtain a second correct value e.g. 161.6 and no others in the interval A1 Mark answers in radians (0.785 and 2.82) as a misread. 4

More questions on Trigonometry

5 The equation of a curve is x2y 4a3, where a is a non-zero constant. −ay2 = dy 2xy (a) Show that . 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(b) Hence find the coordinates of the points where the tangent to the curve is parallel to the y-axis. 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Mark scheme: 5(a) State or imply 2 d 2 d y xy x x  as derivative of 2 x y 2xy x  . State or imply d 2 d y ay x as derivative of 2 ay B1 Accept partial: 2 2 x ay y    . Equate attempted derivative to zero and solve for d d y x M1 Obtain answer 2 d 2 d 2 y xy x ay x   from correct working A1 AG 4 5(b) State or imply 2 2 0 ay x   *M1 Substitute into equation of curve to obtain equation in x and a or in y and a DM1 e.g. 2 2 3 2 4 ay ay a   or 4 4 3 4 2 4 x x a a a   . Obtain one correct point A1 e.g.   2 , 2 a a . Obtain second correct point and no others A1 e.g.   2 , 2 a a  . 4 SC: Allow A1 A0 for 2 x a  or for y = 2a.

More questions on Differentiation

6 Relative to the origin O, the points A, B and C have position vectors given by ` a ` a ` a −−¿OA 21 , −−¿OB 43 and −−¿OC 3 . = = = −2 3 2 −4 The quadrilateral ABCD is a parallelogram. (a) Find the position vector of D. 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(b) The angle between BA and BC is 1. Find the exact value of cos [3] 1. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Hence find the area of ABCD, giving your answer in the form p q, where p and q are integers. 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Mark scheme: 6(a) Obtain a vector for one side of the parallelogram B1 e.g. 2 2 1 AB             or 1 5 6 BC               . Correct method to obtain OD   M1 e.g. OD OA BC      . MO if use AB CD    or BC DA    . Obtain 4 3 OD   i j k  A1 Any equivalent form. Accept coordinates. 3 6(b) Using the correct process, evaluate the scalar product . BA BC  M1   2 10 6   Scalar product of two relevant vectors. OE Using the correct process for the moduli, divide the scalar product by the product of the moduli. M1 2 10 6 9 62    . Obtain answer 2 62 A1 ISW Or simplified equivalent i.e. 62 31 . 3 Question Answer Marks Guidance 6(c) State or imply 58 sin 62  B1 FT Follow their cos. Use correct method to find the area of ABCD M1 e.g. 1 2 sin 2 BA BC    . Condone decimals. Correct unsimplified expression for the area A1 FT e.g. 1 2 3 62 sin 2     . Condone decimals. Follow their sides and angle. Obtain answer 3 58 A1 Correct only. 4

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7 The variables x and y satisfy the differential equation dy 4 tan 2x cos 2x , dx = sin2 3y where 0 1 It is given that y 0 when x 1 ≤x < 4π. = = 6π. Solve the differential equation to obtain the value of x when y 1 Give your answer correct to = 6π. 3 decimal places. 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Mark scheme: 7 Correct separation of variables B1 2 sin 3 d 4sec2 tan 2 d y y x x x    or equivalent. Condone missing integral signs or dx and dy. Integrate to obtain sec2 k x M1 Obtain 2sec2x A1 Use double angle formula and integrate to obtain sin6 py q y  M1 Or two cycles of integration by parts. Obtain 1 1 sin6 2 12 y y  A1 Use 0 y  , π 6 x  in a solution containing terms sec2x  and sin6y  to find the constant of integration M1 Obtain 1 1 sin6 2sec2 4 2 12 y y x    A1 Or equivalent seen or implied by π 1 sin π 2sec2 4 2 12 x          . Obtain 0.541 x  A1 From correct working (not by using the calculator to integrate). 8

More questions on Integration

a 19 The constant a is such that xe−2x dx 8. Ó 0 = (a) Show that a 1 ln 4a 2 . [5] = 2 + ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Verify by calculation that a lies between 0.5 and 1. 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(c) Use an iterative formula based on the equation in (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: 9(a) Commence integration and reach 2 2 e e d x x px q x    Obtain 2 2 1 1 2 2 e e d x x x x     A1 OE Complete integration and obtain 2 2 1 1 e e 2 4 x x x     A1 Use limits correctly and equate to 1 , 8 having integrated twice DM1 2 2 1 1 1 1 e e 2 4 4 8 a a a       . Obtain   1 ln 4 2 2 a a   correctly A1 AG 5 9(b) Calculate the values of a relevant expression or pair of expressions at a = 0.5 and a = 1 M1 Justify the given statement with correct calculated values A1 e.g. 0.5 < 0.69…, 1 > 0.89… 0.193 > 0, –1.105 < 0 0.066 < 0.125, 0.148 > 0.125 if put limits in the integral. Condone if they use calculator for the definite integral. 2 9(c) Use the iterative process   1 1 2 ln 4 2 n n a a   correctly at least once. M1 Obtain final answer 0.84 A1 Show sufficient iterations to at least 4 d.p. to justify 0.84 to 2 d.p. or show that there is a sign change in (0.835, 0.845) A1 e.g. 0.75, 0.8047, 0.8261, 0.8343, 0.8373, 0.8385 1, 0.8959, 0.8599, 0.8469, 0.8420, 0.8402 . 3

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