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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_239709_s20_ms_239709/23/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
Given that 2y = 93x , use logarithms to show that y = kx and find the value of k correct to 3 significant figures. [3] Find the exact coordinates of the stationary point on the curve with equation y = 5xe 2 . 1x
Mark scheme
1 Apply logarithms to both sides and apply power law at least once M1 3ln 9 A1 Rearrange to the form y = x OE ln 2 Obtain k = 9.51 A1 3
Question
[5]
Mark scheme
2 1x 1x Differentiate using product rule to obtain ae 2 + bx e 2 *M1 1x 5 12 x A1 Obtain correct 5e 2 + x e OE 2 Equate first derivative to zero and solve for x DM1 Obtain x-coordinate –2 A1 Obtain y-coordinate −10e −1 A1 5
Question
The equation of a curve is cos 3x + 5 sin y = 3. Find the gradient of the curve at the point 19 π, 61 π . [5]
Mark scheme
3 Differentiate cos3 x to obtain −3sin 3 x B1 dy B1 Differentiate 5sin y to obtain 5cos y dx dy B1 Obtain −3sin 3x + 5cos y = 0 OE dx Substitute x and y values to find value of first derivative M1 3 A1 Obtain 5 5
Question
ln y −0.68, 3.02 ln x O 1.07, −1.53 The variables x and y satisfy the equation y = Ax−2p , where A and p are constants. The graph of ln y against ln x is a straight line passing through the points −0.68, 3.02 and 1.07, −1.53, as shown in the diagram. Find the values of A and p. [5]
Mark scheme
4 State or imply equation is ln y = ln A − 2 p ln x B1 Equate gradient of line to −2 p M1 Obtain −2 p = −2.6 and hence p = 1.3 A1 Substitute appropriate values to find ln A M1 Obtain ln A = 1.252 and hence A = 3.5 A1 Alternative method for question 4 State or imply equation is ln y = ln A − 2 p ln x B1 Substitute given coordinates to obtain 2 simultaneous equations and solve to obtain 3.5 p M1 Obtain 3.5 p = 4.55 and hence p = 1.3 A1 Substitute appropriate values to find ln A M1 Obtain ln A = 1.252 and hence A = 3.5 A1 5
Question
(a) Sketch, on the same diagram, the graphs of y = 2x − 3 and y = 3x + 5. [2] (b) Solve the inequality 3x + 5 < 2x − 3. [3]
Mark scheme
5(a) Draw V-shaped graph with vertex on positive x-axis B1 Draw (more or less) correct graph of y = 3x + 5 B1 2 5(b) State equation 3x + 5 = −(2 x − 3) or corresponding inequality B1 Attempt solution of linear equation / inequality where signs of 3 x and 2 x are different M1 2 A1 State answer x < − 5 Alternative method for question 5(b) Square both sides of equation / inequality and attempt solution of 3-term quadratic equation / inequality M1 2 A1 Obtain (eventually) only − 5 2 A1 State answer x < − 5 3
Question
The polynomial p x is defined by p x = 6x3 + ax2 − 4x − 3, where a is a constant. It is given that x + 3 is a factor of p x. (a) Find the value of a. [2] (b) Using this value of a, factorise p x completely. [3] (c) Hence solve the equation p cosec 1 = 0 for 0Å < 1 < 360Å. [3] a@ A It is given that Ô + 8x dx = 10, where a is a positive constant. 4 2x + 1
Mark scheme
6(a) Substitute x = −3 , equate to zero and attempt solution for a M1 Obtain a = 17 A1 2 6(b) Divide by x + 3 at least as far as the x term M1 Obtain 6 x 2 − x − 1 A1 Conclude ( x + 3)(3 x + 1)(2 x − 1) A1 3 6(c) Attempt solution of sin θ = k where –1 ⩽ k ⩽ 1 M1 Obtain 199.5 A1 Obtain 340.5 A1 3
Question
0 (a) Show that a = 2.5 − 0.5 ln 2a + 1. [4] (b) Using the equation in part (a), show by calculation that 1 < a < 2. [2] (c) Use an iterative formula, based on the equation in part (a), to find the value of a correct to 4 significant figures. Give the result of each iteration to 6 significant figures. [3]
Mark scheme
7(a) Integrate to obtain the form k1 ln(2 x + 1) + k2 x2 *M1 Obtain correct 2ln(2 x + 1) + 4 x2 A1 Use limits correctly and attempt rearrangement DM1 Confirm a = 2.5 − 0.5ln(2a + 1) AG A1 4 7(b) Consider sign of a − 2.5 − 0.5ln(2a + 1) or equivalent for 1 and 2 M1 Obtain −0.3... and 0.6... or equivalents and justify conclusion A1 2 7(c) Use iteration process correctly at least once M1 Obtain final answer 1.358 A1 Show sufficient iterations to 6 sf to justify answer or show a sign change in the interval [1.3575, 1.3585] A1 3
Question
(a) Show that 3 sin 21 cot 1 6 cos2 1. [2] (b) Solve the equation 3 sin 21 cot 1 = 5 for 0 < 1 < π. [3] 1π (c) Find the exact value of Ó 2 3 sin x cot 12 x dx. [5] 1π 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
8(a) cosθ B1 Use at least one of sin 2θ = 2sin θ cos θ and cot θ = sin θ Use both and conclude 6cos 2 θ AG B1 2 8(b) 5 M1 Attempt solution of cos 2 θ = to find at least one value 6 Obtain 0.421 A1 Obtain 2.72 A1 3 8(c) Express integrand in form a + b cos x M1 Obtain correct integrand 3 + 3cos x A1 Integrate to obtain px + q sin x *M1 Apply limits correctly DM1 3 3 A1 Obtain π+3− or exact equivalent 4 2 5
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Mathematics 9709/23 for May/June 2020 is out of 50 marks, with 1 hour 15 minutes to answer it.
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