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Cambridge AS & A Level Mathematics (9709), Paper 2 Pure Mathematics 2, October/November 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_w21_qp_219709_w21_ms_219709/21/O/N/21
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
Find the exact value of Ó 4e2x − 2e−x dx. [4] −1
Mark scheme
1 Integrate to obtain 2e 2 x B1 Integrate to obtain 2e− x B1 Apply limits correctly to integral of the form k1e 2 x + k2 e− x M1 k1 ≠ 4 . Condone one error. Obtain 2e 4 − 2e A1 or exact equivalent. 4
Question
(a) Sketch, on the same diagram, the graphs of y = 3x and y = x − 3. [2] (b) Find the coordinates of the point where the two graphs intersect. [3] (c) Deduce the solution of the inequality 3x < x − 3. [1]
Mark scheme
2(a) Draw V-shaped graph with vertex on positive x-axis B1 Must be straight lines. Draw straight line through origin with positive gradient greater than gradient B1 Must have the first B1. of first graph, together with a V shaped graph 2 2(b) Solve linear equation with signs of 3x and x different or solve non-modulus M1 equation (3 x) 2 = ( x − 3) 2 Obtain x = 34 A1 Obtain y = 94 A1 And no other point. 3 2(c) State x < 34 B1 FT Following their (single) x-coordinate from part (b). 1
Question
y 2.58, 9.00 1.03, 6.36 ln x O The variables x and y satisfy the equation ay = kx, where a and k are constants. The graph of y against ln x is a straight line passing through the points 1.03, 6.36 and 2.58, 9.00, as shown in the diagram. Find the values of a and k, giving each value correct to 2 significant figures. [5]
Mark scheme
3 State or imply equation is y ln a = ln k + ln x B1 1 M1 Or eliminate lnk from simultaneous equations. Equate gradient of line to ln a 1 2.64 A1 AWRT Obtain = or equivalent and hence a = 1.8 ln a 1.55 Substitute appropriate values to find lnk M1 Obtain ln k = 2.7... and hence k = 15 A1 AWRT Alternative method for question 3 a 6.36 = ke1.03 and a 9 = ke 2.58 B1 For both. ( Elimination of k to obtain an equation in a only a 2.64 = e1.55 ) M1 Must have previous B1. Use of a correct method to obtain a M1 Allow for a = e0.59 . a = 1.8 A1 k = 15 A1 5
Question
The curve with equation y = xe2x + 5e−x has a minimum point M . (a) Show that the x-coordinate of M satisfies the equation x = 13 ln 5 − 13 ln 1 + 2x. [5] (b) Use an iterative formula, based on the equation in part (a), to find the x-coordinate of M correct to 3 significant figures. Use an initial value of 0.35 and give the result of each iteration to 5 significant figures. [3]
Mark scheme
4(a) Attempt use of product rule to differentiate xe 2 x M1 Obtain e 2 x + 2 xe 2 x − 5e − x A1 Equate first derivative to zero and multiply by e x to obtain an equation M1 involving e3 x Obtain e3 x (1 + 2 x) = 5 or equivalent A1 Confirm given result x = 13 ln 5 − 13 ln(1 + 2 x) with sufficient detail A1 AG 5 4(b) Use iteration process correctly at least once M1 Need 0.35 and 2 correct values. Obtain final answer 0.357 A1 Answer required to exactly 3sf. Allow recovery. Show sufficient iterations to 5sf to justify answer or show sign change in A1 interval [0.3565, 0.3575] 3
Question
y x O B A The diagram shows the curve with parametric equations 2t − 3 x = ln 2t + 3, y= . 2t + 3 The curve crosses the y-axis at the point A and the x-axis at the point B. dy 6 (a) Show that = . [4] dx 2t + 3 (b) Find the gradient of the curve at A. [2] (c) Find the gradient of the curve at B. [2]
Mark scheme
5(a) dx 2 B1 Obtain = dt 2t + 3 dy M1 Use quotient rule, or equivalent, to find dt dy 2(2t + 3) − 2(2t − 3) A1 OE Obtain = dt (2t + 3) 2 dy 6 A1 AG Divide to confirm = dx 2t + 3 4 5(b) Attempt to find value of t corresponding to x = 0 M1 Obtain t = −1 and hence gradient is 6 A1 2 5(c) Attempt to find value of t corresponding to y = 0 M1 Obtain t = 32 and hence gradient is 1 A1 2
Question
The polynomials f x and g x are defined by f x = 4x3 + ax2 + 8x + 15 and g x = x2 + bx + 18, where a and b are constants. (a) Given that x + 3 is a factor of f x, find the value of a. [2] (b) Given that the remainder is 40 when g x is divided by x − 2, find the value of b. [2] (c) When a and b have these values, factorise f x − g x completely. [3] (d) Hence solve the equation f cosec 1 − g cosec 1 = 0 for 0 < 1 < 2π. [3]
Mark scheme
6(a) Substitute x = −3 , equate to zero and attempt solution for a M1 Allow attempt at synthetic division, must be a complete method. Allow one sign error carried through. Allow attempt at algebraic long division, must be complete with the remainder equated to zero. Obtain a = 13 A1 2 6(b) Substitute x = 2 , equate to 40 and attempt solution for b M1 Obtain b = 9 A1 2 6(c) Identify x + 3 as factor of f ( x) − g( x) B1 May be implied by synthetic division. If working backwards from solutions from a calculator then B0 M0. Attempt, by division or equivalent, to find quadratic factor M1 k ( x + 3)( 2 x + 1)( 2 x − 1) where k ≠ 1 gets B1 M1. Obtain ( x + 3)(2 x − 1)(2 x + 1) A1 3 6(d) Attempt correct process to find at least 1 value from cosecθ = k where M1 Allow for 199.5o or −19.5o . k < −1 Obtain 3.48 or 5.94 A1 Obtain a second correct solution A1 And no others within the range. 3
Question
(a) By first expanding cos 21 + 1, show that cos 31 4 cos3 1 − 3 cos 1. [3] 5 3 5 (b) Find the exact value of 2 cos3 18 π − 2 cos 18 π . [2] (c) Find Ó 12 cos3 x − 4 cos3 3x dx. [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) State cos2θ cosθ − sin 2θ sin θ B1 Attempt correct relevant identities to express in terms of cosθ only M1 M0 if moving terms from side to side. Confirm 4cos3 θ − 3cosθ with sufficient detail A1 AG 3 7(b) Use identity with θ = 185 π M1 Obtain 12 cos 56 π and hence − 14 3 A1 2 7(c) Express integrand in form k1 (cos3 x + 3cos x) + k2 (cos9 x + 3cos3 x) M1 Obtain correct integrand 9cos x − cos9x A1 OE (allow unsimplified). Integrate to obtain form k3 sin x + k4 sin 9 x M1 Obtain correct 9sin x − 19 sin 9 x A1 Now simplified; condone missing ... + c . 4
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Mathematics 9709/21 for October/November 2021 is out of 50 marks, with 1 hour 15 minutes to answer it.
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