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Cambridge AS & A Level Mathematics (9709), Paper 2 Pure Mathematics 2, May/June 2020. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_s20_qp_219709_s20_ms_219709/21/M/J/20
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
Solve the equation ln x + 1 − ln x = 2 ln 2. [3]
Mark scheme
1 Use correct logarithm property to produce one term on LHS M1 Use correct process to obtain equation without logarithms M1 x +1 1 A1 Obtain = 4 or equivalent and hence x = x 3 3
Question
The polynomial p x is defined by p x = 6x3 + ax2 + 9x + b, where a and b are constants. It is given that x − 2 and 2x + 1 are factors of p x. Find the values of a and b. [5]
Mark scheme
2 Substitute x = 2 and equate to zero M1 1 M1 Substitute x = − and equate to zero 2 1 21 A1 Obtain 4a + b + 66 = 0 and a + b − = 0 or equivalents 4 4 Solve a relevant pair of linear simultaneous equations DM1 (Dependent on at least one M mark) Obtain a = −19, b = 10 A1 5
Question
A curve has parametric equations x = et − 2e−t , y = 3e2t + 1. Find the equation of the tangent to the curve at the point for which t = 0. [5]
Mark scheme
3 dx dy B1 State = et + 2e −t , = 6e2t dt dt dy dy dx *M1 Use = / either in terms of t or after substitution of t = 0 dx dt dt Obtain gradient of tangent is 2 A1 Attempt equation of tangent with numerical gradient and coordinates DM1 Obtain y = 2 x + 6 or equivalent A1 5
Question
(a) Sketch, on the same diagram, the graphs of y = 3x + 2a and y = 3x − 4a , where a is a positive constant. Give the coordinates of the points where each graph meets the axes. [3] (b) Find the coordinates of the point of intersection of the two graphs. [3] (c) Deduce the solution of the inequality 3x + 2a < 3x − 4a . [1]
Mark scheme
4(a) Draw two V-shaped graphs with one vertex on negative x-axis and one vertex on positive x-axis M1 Draw correct graphs related correctly to each other A1 2 4 A1 State correct coordinates − a, 2a, a, 4a 3 3 3 4(b) Solve linear equation with signs of 3x different or solve non-modulus equation (3x + 2a ) 2 = (3x − 4a) 2 M1 Obtain x = 13 a A1 Obtain y = 3a A1 3 4(c) 1 B1FT State x < a (FT from part (b)) 3 1
Question
y M x O The diagram shows part of the curve with equation y = x3 cos 2x. The curve has a maximum at the point M . 3 (a) Show that the x-coordinate of M satisfies the equation x = 1.5x2 cot 2x. [3] (b) Use the equation in part (a) to show by calculation that the x-coordinate of M lies between 0.59 and 0.60. [2] (c) Use an iterative formula, based on the equation in part (a), to find the x-coordinate of M correct to 3 significant figures. Give the result of each iteration to 5 significant figures. [3]
Mark scheme
5(a) Differentiate using the product rule to obtain ax 2 cos 2 x − bx 3 sin 2 x M1 Obtain 3 x 2 cos 2 x − 2 x 3 sin 2 x A1 Equate first derivative to zero and confirm x = 3 1.5 x 2 cot 2 x AG A1 3 5(b) Consider sign of x − 3 1.5 x 2 cot 2 x or equivalent for 0.59 and 0.60 M1 Obtain −0.009... and 0.005... or equivalents and justify conclusion A1 2 5(c) Use iteration correctly at least once M1 Obtain final answer 0.596 A1 Show sufficient iterations to 5 sf to justify answer or show sign change in interval [0.5955, 0.5965] A1 3
Question
(a) Prove that sin 21 cosec 1 − sec 1 8 cos 1 + 14 π . [5] (b) Solve the equation sin 21 cosec 1 − sec 1 = 1 for 0 < 1 < 12 π. Give the answer correct to 3 significant figures. [2] (c) Find Ó sin x cosec 21 x − sec 21 x dx. [3]
Mark scheme
6(a) Express left-hand side in terms of sin θ and cos θ M1 Obtain 2cos θ − 2sin θ A1 Attempt to express a cos θ + b sin θ in R cos(θ + β ) form M1 Confirm R = 8 AG A1 1 A1 Carry out necessary trigonometry and confirm π AG 4 5 6(b) 1 1 M1 Carry out correct process to find θ from cos θ + π = 4 8 Obtain 0.424 A1 2 6(c) 1 1 1 1 B1 Express integrand as 8 cos x + π or as 2cos x − 2sin x 2 4 2 2 1 1 1 1 M1 Integrate to obtain k sin x + π or k1 sin x + k2 cos x 2 4 2 2 1 1 1 1 A1 Obtain correct 2 8 sin x + π or 4sin x + 4cos x 2 4 2 2 3
Question
(a) Find the quotient when 9x3 − 6x2 − 20x + 1 is divided by 3x + 2, and show that the remainder is 9. [3] 9x3 − 6x2 − 20x + 1 6 (b) Hence find Ô dx, giving the answer in the form a + ln b where a and b are 3x + 2 1 integers. [5] (c) Find the exact root of the equation 9e9y − 6e6y − 20e3y − 8 = 0. [4] 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
7(a) Carry out division at least as far as 3 x 2 + kx M1 Obtain quotient 3 x 2 − 4 x − 4 A1 Confirm remainder is 9 AG A1 3 7(b) Integrate to obtain at least k1 x 3 and k2 ln(3 x + 2) terms *M1 Obtain x3 − 2 x 2 − 4 x + 3ln(3x + 2) A1FT (FT from quotient in part (a)) Apply limits correctly DM1 Apply appropriate logarithm properties correctly M1 Obtain 125 + ln 64 A1 5 7(c) State or imply 9 x3 − 6 x 2 − 20 x − 8 = (3 x + 2)(3x 2 − 4 x − 4) B1FT (FT from quotient in part (a)) Attempt to solve cubic eqn to find positive value of x (or of e3 y ) M1 Use logarithms to solve equation of form e3 y = k where k > 0 M1 1 A1 Obtain ln 2 or exact equivalent 3 4
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On this page, free: the mark scheme for AS & A Level Mathematics 9709/21, May/June 2020 (9709_s20_ms_21), and the question paper (9709_s20_qp_21), to read or download, with each question's marking points written out underneath.
Mathematics 9709/21 for May/June 2020 is out of 50 marks, with 1 hour 15 minutes to answer it.
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