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Cambridge AS & A Level Mathematics (9709), Paper 2 Pure Mathematics, February/March 2020. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_m20_qp_229709_m20_ms_229709/22/F/M/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 2 sin 1 + 30Å + 5 cos 1 = 2 sin 1 for 0Å < 1 < 90Å. [4]
Mark scheme
1 Express first term as 2sin θ cos30 + 2cosθ sin 30 B1 Divide by cosθ to produce linear equation in tan θ M1 Obtain tan θ = 2−6 3 or 22.39… A1 Obtain 87.4 A1 Or greater accuracy 87.44297… 4
Question
(a) Find the quotient when 4x3 + 17x2 + 9x is divided by x2 + 5x + 6, and show that the remainder is 18. [3] (b) Hence solve the equation 4x3 + 17x2 + 9x − 18 = 0. [3] 3a It is given that Ô dx = ln 72 . 2 2x − 5
Mark scheme
2(a) Carry out division as far as 4x + k M1 Obtain quotient 4 x − 3 A1 Confirm remainder is 18 A1 AG necessary detail needed 3 2(b) State or imply equation is (4 x − 3)( x 2 + 5 x + 6) = 0 B1FT Following their quotient from part (a) Attempt solution of cubic equation to find three real roots M1 Obtain −3, − 2, 34 A1 3
Question
a Find the value of the positive constant a. [6]
Mark scheme
3 Integrate to obtain k ln(2 x − 5) *M1 For non-zero constant k Apply limits to obtain ln(6a − 5) − ln(2a − 5) = ln 72 A1 Apply subtraction law for logarithms *M1 OE 6a − 5 7 A1 OE without logarithms Obtain equation = 2a − 5 2 Solve equation for a DM1 25 A1 Obtain a = 2 6
Question
A curve has equation 3x2 − y2 − 4 ln 2y + 3 = 26. Find the equation of the tangent to the curve at the point 3, −1. [6]
Mark scheme
4 dy B1 Differentiate − y 2 to obtain −2 y dx −8 dy B1 Differentiate −4ln(2 y + 3) to obtain 2 y + 3 dx Attempt differentiation of all terms M1 dy Dependent on appearance of at least one dx dy M1 Substitute x = 3, y = −1 to find numerical value of dx dy A1 Obtain =3 dx Obtain equation y = 3x − 10 A1 OE 6
Question
(a) Sketch, on the same diagram, the graphs of y = x + 2k and y = 2x − 3k , where k is a positive constant. Give, in terms of k, the coordinates of the points where each graph meets the axes. [3] (b) Find, in terms of k, the coordinates of each of the two points where the graphs intersect. [4] (c) Find, in terms of k, the largest value of t satisfying the inequality 2 + 2k ≥ 2t+1 − 3k . t [2]
Mark scheme
5(a) Draw two V-shaped graphs with one vertex on negative x-axis and M1 one vertex on positive x-axis Draw correct graphs related correctly to each other A1 State correct coordinates −2k , 2k , 32 k , 3k A1 Either given on axes or stated separately 3 5(b) State or imply non-modulus equation ( x + 2k ) 2 = (2 x − 3k )2 or pair B1 of linear equations Attempt solution of 3-term quadratic equation or pair of linear M1 equations Obtain x = 13 k , x = 5k A1 Obtain y = 73 k , y = 7k A1 If A0A0, award A1 for one pair of correct coordinates 4 5(c) Relate 2t to larger value of x from part (b) M1 ln(5k ) A1 log10 (5k ) Apply logarithms to obtain t = OE such as or log 2 (5k ) ln 2 log10 2 2
Question
A curve has equation y = x3 e0.2x where x ≥ 0. At the point P on the curve, the gradient of the curve is 15. _ 75e−0.2x (a) Show that the x-coordinate of P satisfies the equation x = 15 + x . [4] (b) Use the equation in part (a) to show by calculation that the x-coordinate of P lies between 1.7 and 1.8. [2] (c) Use an iterative formula, based on the equation in part (a), to find the x-coordinate of P correct to 4 significant figures. Give the result of each iteration to 6 significant figures. [3]
Mark scheme
6(a) Differentiate using the product rule *M1 Obtain 3 x 2 e 0.2 x + 0.2 x 3e 0.2 x A1 OE Equate first derivative to 15 and rearrange to x = ... DM1 75e −0.2 x A1 AG – necessary detail needed Confirm x = 15 + x 4 6(b) 75e −0.2 x M1 Consider sign of x − or equivalent for 1.7 and 1.8 15 + x Obtain −0.08... and 0.03... or equivalents and justify conclusion A1 2 6(c) Use iterative process correctly at least once M1 Answer required to exactly 4 sf Obtain final answer 1.771 A1 Show sufficient iterations to 6 sf to justify answer or show a sign A1 change in the interval [1.7705, 1.7715] 3
Question
y A x O The diagram shows part of the curve with equation y = 4 sin2 x + 8 sin x + 3, where x is measured in radians. The curve crosses the x-axis at the point A and the shaded region is bounded by the curve and the lines x = 0 and y = 0. (a) Find the exact x-coordinate of A. [2] (b) Find the exact gradient of the curve at A. [3] (c) Find the exact area of the shaded region. [5] 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) Solve equation y = 0 to find value of x M1 7 A1 Obtain π 6 2 7(b) Attempt first derivative using chain rule M1 OE dy A1 OE Obtain = 8sin x cos x + 8cos x dx Substitute value from part (a) to find gradient −2 3 A1 Or exact equivalent 3 7(c) Express integrand in the form k1 + k 2 cos 2 x + k3 sin x *M1 Obtain correct 5 − 2cos 2 x + 8sin x A1 OE. Allow unsimplified Integrate to obtain 5x − sin 2 x − 8cos x A1 Apply limits 0 and their value from part (a) correctly DM1 35 7 A1 Obtain π+ 3 + 8 or exact equivalent 6 2 5
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Mathematics 9709/22 for February/March 2020 is out of 50 marks, with 1 hour 15 minutes to answer it.
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