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Cambridge AS & A Level Mathematics (9709), Paper 3, May/June 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_s19_qp_339709_s19_ms_339709/33/M/J/19
Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.
Question
Use logarithms to solve the equation 53−2x = 4 7x , giving your answer correct to 3 decimal places. [4] 10
Mark scheme
1 Use law of the logarithm of a product or quotient M1 Use law of the logarithm of power twice M1 Obtain a correct linear equation in x, e.g. A1 ( 3 − 2x) ln5 = ln4 + xln7 Obtain answer x = 0.666 A1 4
Question
Show that Ó 4 1 02 − 8. x2 cos 2x dx = 32 [5] 0 1 − cos 21 + sin 21
Mark scheme
2 Commence integration and reach M1* 2 a x sin 2 x + b ∫ x sin 2 x d x 1 2 A1 Obtain x sin 2 x − ∫ x sin 2 x dx , or equivalent 2 Complete the integration and obtain A1 1 2 1 1 x sin 2 x + x cos 2 x − sin 2 x , or equivalent 2 2 4 Use limits correctly, having integrated twice DM1 Obtain given answer correctly A1 5
Question
Let f 1 = . 1 + cos 21 + sin 21 (i) Show that f 1 = tan 1. [3] 10 (ii) Hence show that Ó f 1 d1 = 12 ln 23 . 4 [4] 10 6 1 + e−x
Mark scheme
3(i) Use double angle formulae and express entire fraction in terms of sinθ and cos θ M1 Obtain a correct expression A1 Obtain the given answer A1 3 3(ii) State integral of the form ± lncosθ M1* Use correct limits correctly and insert exact values for the trig ratios DM1 1 3 A1 Obtain a correct expression, e.g. − ln + ln 2 2 Obtain the given answer following full and exact working A1 4
Question
The equation of a curve is y = , for x > 0. 1 − e−x dy (i) Show that is always negative. [3] dx (ii) The gradient of the curve is equal to −1 when x = a. Show that a satisfies the equation e2a − 4ea + 1 = 0. Hence find the exact value of a. [4]
Mark scheme
4(i) Use the quotient or product rule M1 Obtain correct derivative in any form A1 2e − x A1 Reduce to – 2 , or equivalent, and explain why this is always negative (1 − e ) −x 3 4(ii) Equate derivative to – 1 and obtain the given equation B1 State or imply u 2 − 4u + 1 = 0 , or equivalent in e a B1 Solve for a M1 ( ) Obtain answer a = ln 2 + 3 and no other A1 4
Question
The variables x and y satisfy the differential equation dy x + 1y = y2 + 5. dx It is given that y = 2 when x = 0. Solve the differential equation obtaining an expression for y2 in terms of x. [7]
Mark scheme
5 Separate variables correctly and integrate at least one side B1 Obtain term ln ( x +1) B1 ( Obtain term of the form a ln y 2 + 5 ) M1 1 A1 Obtain term ln( y 2 + 5) 2 Use y = 2, x = 0 to determine a constant, or as limits, in a solution containing terms M1 aln(y +5) and b ln ( x +1) , where ab ≠ 0 2 Obtain correct solution in any form A1 Obtain final answer y = 9 ( x + 1) − 5 A1 2 2 7
Question
y a x O b The diagram shows the curve y = x4 − 2x3 − 7x − 6. The curve intersects the x-axis at the points a, 0 and b, 0, where a < b. It is given that b is an integer. (i) Find the value of b. [1] (ii) Hence show that a satisfies the equation a = − 13 2 + a2 + a3 . [4] (iii) Use an iterative formula based on the equation in part (ii) to determine a correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3]
Mark scheme
6(i) State b = 3 B1 1 6(ii) Commence division by x – b and reach partial quotient x 3 + kx 2 M1 Obtain quotient x 3 + x 2 + 3 x + 2 A1 There being no remainder Equate quotient to zero and rearrange to make the subject a M1 Obtain the given equation A1 4 6(iii) M1 Use the iterative formula an +1 = − 1 3 ( ) 2 + an 2 + an3 correctly at least once Obtain final answer –0.715 A1 Show sufficient iterations to 5 d.p. to justify –0.715 to 3 d.p., or show there is a sign A1 change in the interval (–0.7145, –0.7155) 3
Question
The curve y = sin x + 13 0 cos x has two stationary points in the interval 0 ≤ x ≤ 0. dy (i) Find . [2] dx the formula for cos A + B, show that, at the stationary points on the curve, (ii) By considering cos 2x + 13 0 = 0. [2] (iii) Hence find the exact x-coordinates of the stationary points. [3]
Mark scheme
7(i) Use product rule M1 Obtain correct derivative in any form A1 2 7(ii) Equate derivative to zero and use correct cos(A + B) formula M1 Obtain the given equation A1 2 7(iii) Use correct method to solve for x M1 1 A1 Obtain answer, e.g. x = π 12 7 A1 Obtain second answer, e.g. π , and no other 12 3
Question
Throughout this question the use of a calculator is not permitted. The complex number u is defined by 4i u= . 1 − ï3i (i) Express u in the form x + iy, where x and y are real and exact. [3] (ii) Find the exact modulus and argument of u. [2] (iii) On a sketch of an Argand diagram, shade the region whose points represent complex numbers z satisfying the inequalities z < 2 and z − u < z . [4] 2x 5 − x
Mark scheme
8(i) Multiply numerator and denominator by 1 + 3i , or equivalent M1 4i − 4 3 and 3 + 1 A1 Obtain final answer − 3 + i A1 3 8(ii) State that the modulus of u is 2 B1 5 B1 State that the argument of u is π (or 150°) 6 2 8(iii) Show a circle with centre the origin and radius 2 B1 Show u in a relatively correct position B1 FT Show the perpendicular bisector of the line joining u and the origin B1 FT Shade the correct region B1 4
Question
Let f x = . 3 + x 1 − x 2 (i) Express f x in partial fractions. [5] (ii) Hence obtain the expansion of f x in ascending powers of x up to and including the term in x3 . [5]
Mark scheme
9(i) A B C B1 State or imply the form + + 3+ x 1− x (1− x )2 Use a correct method for finding a constant M1 Obtain one of A = –3, B = –1, C = 2 A1 Obtain a second value A1 Obtain the third value A1 A Dx + E Mark the form + , where A = –3, 3+ x (1 − x)2 D = 1 and E = 1, B1M1A1A1A1 as above. 5 9(ii) Use a correct method to find the first two terms of the expansion of ( 3 + x ) , M1 −1 1 (1 + x)−1 , (1− x) or (1− x) −1 −2 3 Obtain correct unsimplified expansions up to the term in x 3 of each partial fraction A1 FT on A A1 FT on B A1 FT on C 10 44 190 3 A1 For the A, D, E form of fractions give Obtain final answer x + x2 + x M1A1ftA1ft for the expanded partial fractions, 3 9 27 then, if D ≠ 0, M1 for multiplying out fully, and A1 for the final answer. 5 JJJG Find PQ for a general point Q on l, e.g. −3i + 6k + µ ( 2i − j − 2k)
Question
The line l has equation r = i + 2j + 3k + - 2i − j − 2k. (i) The point P has position vector 4i + 2j − 3k. Find the length of the perpendicular from P to l. [5] (ii) It is given that l lies in the plane with equation ax + by + 2z = 13, where a and b are constants. Find the values of a and b. [6] Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
Mark scheme
10(i) B1 JJJG M1 Calculate scalar product of PQ and a direction vector for l and equate the result to zero Solve for µ and obtain µ = 2 A1 JJJG M1 Carry out a complete method for finding the length of PQ Obtain answer 3 A1 Alternative method for question 10(i) JJJG JJJG Calling the point (1, 2, 3) A, state A P (or PA ) in component form, e.g. 3i – 6k B1 JJJG JJJG Use a scalar product with a direction vector for l to find the projection of A P (or PA ) M1 on l 18 A1 Obtain correct answer in any form, e.g. 9 Use Pythagoras to find the perpendicular M1 Obtain answer 3 A1 10(i) Alternative method for question 10(i) JJJG JJJG B1 State A P (or PA) in component form Calculate a vector product with a direction vector for l M1 Obtain correct answer, e.g. 6i – 6j – 3k A1 Divide modulus of the product by that of the direction vector M1 Obtain answer 3 A1 5 10(ii) Substitute coordinates of a general point of l in the plane equation and equate constant M1 terms Obtain a correct equation, e.g. a + 2b + 6 = 13 A1 Equate the coefficient of µ to zero M1 Obtain a correct equation, e.g. 2a – b – 4 = 0 A1 Substitute (1, 2, 3) in the plane equation M1 Obtain a correct equation, e.g. a + 2b + 6 = 13 A1 Alternative method for question 10(ii) Find a second point on l and obtain an equation in a and/or b M1 Obtain a correct equation, e.g. 5a – 2 = 13 A1 Equate scalar product of a direction vector for l and a vector normal for the plane to M1 zero Obtain a correct equation, e.g. 2a – b – 4 = 0 A1 Solve for a or for b M1 Obtain a = 3 and b = 2 A1 6
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Mathematics 9709/33 for May/June 2019 is out of 75 marks, with 1 hour 45 minutes to answer it.
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