Cambridge A Level Mathematics 9709 — 2019 Oct/Nov Paper 3 · Variant 1

1 Given that ln 1 e2y x, express y in terms of x. [3] + = ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 State 2 1 e e y x + = B1 Make y the subject M1 Rearrange to 2 e ... y = and use logs Obtain answer ( ) 1 ln e 1 2 x y = − A1 OE 3

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

Mark scheme: 2 State or imply non-modular inequality ( ) ( ) 2 2 2 2 3 4 1 x x − > + , or corresponding quadratic equation, or pair of linear equations ( ) ( ) 2 3 4 1 x x − =± + B1 2 12 44 7 0 x x + + < Make reasonable attempt at solving a 3-term quadratic, or solve two linear equations for x M1 Correct method seen, or implied by correct answers Obtain critical values x = 7 2 − and x = 1 6 − A1 State final answer 7 1 2 6 x − < < − A1 Alternative method for question 2 Obtain critical value x = 7 2 − from a graphical method, or by inspection, or by solving a linear equation or an inequality B1 Obtain critical value 1 6 x = − similarly B2 State final answer 7 1 2 6 x − < < − B1 4

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4 The number of insects in a population t weeks after the start of observations is denoted by N. The population is decreasing at a rate proportional to Ne−0.02t. The variables N and t are treated as dN continuous, and it is given that when t 0, N 1000 and = = dt = −10. (i) Show that N and t satisfy the differential equation dN [1] dt = −0.01e−0.02tN. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Solve the differential equation and find the value of t when N 800. [6] = ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) State what happens to the value of N as t becomes large. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(i) State d d N t = 0.02 e t k N − and show k = – 0.01 B1 OE ( ) 10 1 1000 k − = × × 1 4(ii) Separate variables correctly and integrate at least one side B1 0.02 1 d 0.01e d t N t N − = − ∫ ∫ Obtain term ln N B1 OE Obtain term 0.02 0.5e t − B1 OE Use N = 1000, t = 0 to evaluate a constant, or as limits, in a solution with terms ln a N and 0.02 e t b − , where ab ≠ 0 M1 Obtain correct solution in any form e.g. ( ) 0.02 ln ln1000 0.5 e 1 t N − − = − A1 1 ln1000 6.41 2 − = Substitute N = 800 and obtain t = 29.6 A1 6 4(iii) State that N approaches 1000 e B1 Accept 606 or 607 or 606.5 1

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7 Two lines l and m have equations r ai 2j 3k i 3k and r 2i j 2k 2i k = + + + , −2j + = + + + - −j + respectively, where a is a constant. It is given that the lines intersect. (i) Find the value of a. 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(ii) When a has this value, find the equation of the plane containing l and m. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(i) Express general point of l or m in component form e.g. (a + λ, 2 – 2λ, 3 + 3λ) or (2 + 2µ, 1 – µ, 2 + µ) B1 Equate at least two pairs of corresponding components and solve for λ or for µ M1 Obtain either λ = – 2 or µ = – 5 or 1 3 a λ = or 2 1 3 a µ = − or ( ) 1 4 5 a λ = − or ( ) 1 3 7 5 a µ = − A1 Obtain a = – 6 A1 4 7(ii) Use scalar product to obtain a relevant equation in a, b and c, e.g. a – 2b + 3c = 0 B1 Obtain a second equation, e.g. 2a – b + c = 0 and solve for one ratio M1 Obtain a : b : c = 1 : 5 : 3 A1 OE Substitute a relevant point and values of a, b, c in general equation and find d M1 Obtain correct answer x + 5y + 3z = 13 A1FT OE. The FT is on a from part (i), if used Alternative method for question 7(ii) Attempt to calculate vector product of relevant vectors, M1 e.g. ( ) ( ) 2 3 . 2 − + −+ i j k i j k Obtain two correct components A1 Obtain correct answer, e.g. i + 5j + 3k A1 Substitute a relevant point and find d M1 Obtain correct answer x + 5y + 3z = 13 A1FT OE. The FT is on a from part (i), if used

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10 (a) The complex number u is given by u i. Showing all necessary working and without = −3 − 2ï10 using a calculator, find the square roots of u. Give your answers in the form a ib, where the + numbers a and b 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 arg z and Im z where Im z denotes the −3 −i ≤3, ≥140 ≥2, imaginary part of the complex number z. [5]

Mark scheme: 10(a) Square a + ib and equate real and imaginary parts to – 3 and 2 10 − respectively *M1 Obtain 2 2 3 a b − = − and 2 2 10 ab = − A1 Eliminate one unknown and find an equation in the other DM1 Obtain 4 2 3 10 0 a a + − = , or 4 2 3 10 0 b b − − = , or horizontal 3-term equivalent A1 Obtain answers ( ) 2 5i ± − , or exact equivalent A1 5 10(b) Show point representing 3 + i in relatively correct position B1 Show a circle with radius 3 and centre not at the origin B1 Show correct half line from the origin at 1 4π to the real axis B1 Show horizontal line y = 2 B1 Shade the correct region B1 5 i 3 Re(z) Im(z) shaded π 4 Im(z) = 2

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