Cambridge A Level Mathematics 9709 — 2020 May/June Paper 3 · Variant 3

1 Solve the inequality 2x 3 x 2 . [4] −1 > + ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 Make reasonable attempt at solving a 3-term quadratic, or solve two linear equations for x M1 Obtain critical values x = –7 and x = –1 A1 State final answer –7 < x < –1 A1 Alternative method for question 1 Obtain critical value x = –1 from a graphical method, or by solving a linear equation or linear inequality B1 Obtain critical value x = –7 similarly B2 State final answer –7 < x < –1 [Do not condone ⩽ for < in the final answer.] B1 4

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6 (a) By sketching a suitable pair of graphs, show that the equation x5 2 x has exactly one real = + root. [2] ........................................................................................................................................................ ........................................................................................................................................................ (b) Show that if a sequence of values given by the iterative formula 4x5 2 n + xn+1 = 5x4 n −1 converges, then it converges to the root of the equation in part (a). 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(c) Use the iterative formula with initial value x1 1.5 to calculate the root correct to 3 decimal = places. Give the result of each iteration to 5 decimal places. 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Mark scheme: 6(a) Sketch a second relevant graph, e.g. y = x + 2 and justify the given statement B1 2 6(b) State a suitable equation, e.g. 5 4 4 2 5 1 x x x + = − B1 Rearrange this as x5 = 2 + x or commence working vice versa B1 2 6(c) Use the iterative formula correctly at least once M1 Obtain final answer 1.267 A1 Show sufficient iterations to 5 d.p. to justify 1.267 to 3 d.p., or show there is a sign change in the interval (1.2665, 1.2675) A1 3

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8 Relative to the origin O, the points A, B and D have position vectors given by −−¿OA i 2j k, −−¿OB 2i 5j 3k and −−¿OD 3i 2k. = + + = + + = + A fourth point C is such that ABCD is a parallelogram. (a) Find the position vector of C and verify that the parallelogram is not a rhombus. 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(b) Find angle BAD, giving your answer in degrees. 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(c) Find the area of the parallelogram correct to 3 significant figures. 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Mark scheme: 8(a) State or imply AB or AD in component form B1 Use a correct method for finding the position vector of C M1 Obtain answer 4i + 3j + 4k, or equivalent A1 Using the correct process for the moduli, compare lengths of a pair of adjacent sides, e.g. AB and AD M1 Show that ABCD has a pair of unequal adjacent sides A1 Alternative method for question 8(a) State or imply AB  or AD  in component form B1 Use a correct method for finding the position vector of C M1 Obtain answer 4i + 3j + 4k, or equivalent A1 Use the correct process to calculate the scalar product of AC  and BD  , or equivalent M1 Show that the diagonals of ABCD are not perpendicular A1 5 8(b) Use the correct process to calculate the scalar product of a pair of relevant vectors, e.g. AB  and AD  M1 Using the correct process for the moduli, divide the scalar product by the product of the moduli of the two vectors and evaluate the inverse cosine of the result M1 Obtain answer 100.3° A1 3 Question Answer Marks 8(c) Use a correct method to calculate the area, e.g. calculate AB.AC sin BAD M1 Obtain answer 11.0 (FT on angle BAD) A1 FT 2

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9 (a) The complex numbers u and w are such that u 2i and uw 6. −w = = Find u and w, giving your answers in the form x iy, where x and y are real and exact. [5] + ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) On a sketch of an Argand diagram, shade the region whose points represent complex numbers z satisfying the inequalities z 0 z and Re z [5] −2 −2i ≤2, ≤arg ≤14π ≤3.

Mark scheme: 9(a) Eliminate u or w and obtain an equation w or u M1 Obtain a quadratic in u or w, e.g. u2 – 2iu – 6 = 0 or w2 + 2iw – 6 = 0 A1 Solve a 3-term quadratic for u or for w M1 Obtain answer 5 i u = + , 5 i w = − A1 Obtain answer 5 i u = − + , 5 i w = − − A1 5 9(b) Show the point representing 2 + 2i B1 Show a circle with centre 2 + 2i and radius 2 (FT is on the position of 2 + 2i) B1 FT Show half-line from origin at 45° to the positive x-axis B1 Show line for Re z = 3 B1 Shade the correct region B1 5

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10 r h A A tank containing water is in the form of a hemisphere. The axis is vertical, the lowest point is A and the radius is r, as shown in the diagram. The depth of water at time t is h. At time t 0 the tank is = full and the depth of the water is r. At this instant a tap at A is opened and water begins to flow out at a rate proportional to h. The tank becomes empty at time t 14. = The volume of water in the tank is V when the depth is h. It is given that V 1 3rh2 . = 3π −h3 (a) Show that h and t satisfy a differential equation of the form dh B 1 3 , dt = − 2 2 2rh −h where B is a positive constant. 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(b) Solve the differential equation and obtain an expression for t in terms of h and r. 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Mark scheme: 10(a) State or imply d d V k h t = − B1 State or imply 2 d 2π π d V rh h h = − , or equivalent B1 Use d d d . d d d V V h t h t = M1 Obtain the given answer correctly A1 4 10(b) Separate variables and attempt integration of at least one side M1 Obtain terms 3 5 2 2 4 2 3 5 rh h − and –Bt A3, 2, 1, 0 Use t = 0, h = r to find a constant of integration c M1 Use t = 14, h = 0 to find B M1 Obtain correct c and B, e.g. 5 5 2 2 14 1 , 15 15 c r B r = = A1 Obtain final answer 3 5 2 2 14 20 6 h h t r r     = − +         , or equivalent A1 8

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