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Cambridge AS & A Level Mathematics (9709), Paper 3 Pure Mathematics 3, May/June 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_s21_qp_339709_s21_ms_339709/33/M/J/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
coefficients. [4]
Mark scheme
1 State correct first two terms 1 + 2x B1 State a correct unsimplified version of the x 2 or x3 term M1 Symbolic binomial coefficients are not sufficient for the M mark. Obtain the next term −x 2 A1 4 3 A1 Obtain the final term x 3 4
Question
Solve the equation 4x = 3 + 4−x . Give your answer correct to 3 decimal places. [5]
Mark scheme
2 State or imply u 2 − 3u − 1 = 0 , or equivalent in 4 x B1 Solve for u or 4 x M1 1 A1 Obtain root 2 ( ) 3 + 13 , or decimal in [3.30, 3.31] Use correct method for finding x from a positive root M1 Obtain answer x = 0.862 and no other A1 5
Question
The parametric equations of a curve are x = t + ln t + 2, y = t − 1e−2t , where t > −2. dy (a) Express in terms of t, simplifying your answer. [5] dx (b) Find the exact y-coordinate of the stationary point of the curve. [2] 15 − 6x Let f x = 1 + 2x 4 − x
Mark scheme
3(a) dx 1 B1 State =1+ dt t+2 Use product rule M1 dy A1 OE Obtain = e −2t − 2 ( t − 1) e −2t dt dy dy dx M1 Use = ÷ dx dt dt Obtain correct answer in any simplified form, A1 ( 3 − 2t )( t + 2 ) −2t e.g. e t +3 5 3(b) Equate derivative to zero and solve for t M1 3 1 A1 Obtain t = and obtain answer y = e −3 , or exact equivalent 2 2 2
Question
. (a) Express f x in partial fractions. [3] 0a1 (b) Hence find Ó f x dx, giving your answer in the form ln 2 , where a and b are integers. [4] 1 b
Mark scheme
4(a) A B M1 State or imply the form + and use a correct method to 1 + 2x 4− x find a constant Obtain one of A = 4 and B = – 1 A1 Obtain the second value A1 3 4(b) Integrate and obtain terms 2ln (1 + 2 x ) + ln ( 4 − x ) B1FT The FT is on A and B. +B1FT Substitute limits correctly in an integral of the form M1 a ln (1 + 2 x ) + b ln ( 4 − x ) , where ab ≠ 0 50 A1 Obtain final answer ln 27 4
Question
(a) By first expanding tan 21 + 21, show that the equation tan 41 = 12 tan 1 may be expressed as tan4 1 + 2 tan2 1 − 7 = 0. [4] (b) Hence solve the equation tan 41 = 12 tan 1, for 0Å < 1 < 180Å. [3]
Mark scheme
5(a) Use double angle formula to express tan 4θ in terms of tan 2θ M1 Use double angle formula to express result in terms of tan θ M1 Obtain a correct equation in tan θ in any form A1 Obtain the given answer A1 4 5(b) Solve for tan θ and obtain a value of θ M1 Obtain answer, e.g. 53.5° A1 Obtain second answer, e.g. 126.5° and no other in the interval A1 Ignore answers outside the given interval. Treat answers in radians as a misread. 3
Question
(a) By sketching a suitable pair of graphs, show that the equation cot 21 x = 1 + e−x has exactly one root in the interval 0 < x ≤ π. [2] (b) Verify by calculation that this root lies between 1 and 1.5. [2] A @ (c) Use the iterative formula xn+1 = 2 tan−1 1 1 + e−xn to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]
Mark scheme
6(a) 1 B1 Sketch a relevant graph, e.g. y = cot x 2 Sketch a second relevant graph, e.g. y = 1 + e− x , and justify the B1 given statement 2 6(b) Calculate values of a relevant expression or pair of expressions at M1 x = 1 and x = 1.5 Complete the argument correctly with correct calculated values A1 2 6(c) Use the iterative formula correctly at least once M1 Obtain final answer 1.34 A1 Show sufficient iterations to 4 d.p. to justify 1.34 to 2 d.p. or show A1 there is a sign change in the interval (1.335, 1.345) 3
Question
y P x O M N For the curve shown in the diagram, the normal to the curve at the point P with coordinates x, y meets the x-axis at N . The point M is the foot of the perpendicular from P to the x-axis. The curve is such that for all values of x in the interval 0 ≤ x < 12 π, the area of triangle PMN is equal to tan x. = MN dy (a) (i) Show that . [1] y dx = tan x. dy (ii) Hence show that x and y satisfy the differential equation 12 y2 [2] dx (b) Given that y = 1 when x = 0, solve this differential equation to find the equation of the curve, expressing y in terms of x. [6]
Mark scheme
7(a)(i) MN dy B1 Justify the given statement = y dx 1 7(a)(ii) dy M1 Express the area of PMN in terms of y and and equate to tan x dx Obtain the given equation correctly A1 2 7(b) Separate variables and integrate at least one side M1 1 3 A1 Obtain term y 6 Obtain term of the form ± ln cos x M1 Evaluate a constant or use x = 0 and y = 1 in a solution containing M1 terms ay 3 and ± ln cos x , or equivalent 1 3 1 A1 Obtain correct answer in any form, e.g. y = − ln cos x + 6 6 Obtain final answer y = 3 (1 − 6ln cosx) A1 OE 6
Question
y M x O The diagram shows the curve y = ln x and its maximum point M . x4 (a) Find the exact coordinates of M . [4] a (b) By using integration by parts, show that for all a > 1, Ô dx < 19 . ln x 4 [6] 1 x
Mark scheme
8(a) Use quotient or product rule M1 Obtain correct derivative in any form A1 Equate derivative to zero and solve for x M1 1 A1 Obtain x = 4 e and y = , or exact equivalents 4e 4 8(b) 1 *M1 Commence integration and reach ax −3 ln x + b x −3 . dx x 1 1 1 A1 OE Obtain − x −3 ln x + x −3 . dx 3 3 x 1 1 A1 Complete integration and obtain − x −3 ln x − x −3 3 9 Substitute limits correctly, having integrated twice DM1 1 1 −3 1 A1 OE Obtain answer − a ln a − a −3 9 3 9 Justify the given statement A1 6 ρ
Question
The quadrilateral ABCD is a trapezium in which AB and DC are parallel. With respect to the origin O, the position vectors of A, B and C are given by OA = −i + 2j + 3k, OB = i + 3j + k and −−¿ OC = 2i + 2j − 3k. (a) Given that DC = 3AB, find the position vector of D. [3] (b) State a vector equation for the line through A and B. [1] (c) Find the distance between the parallel sides and hence find the area of the trapezium. [5] (a) Verify that −1 + 2i is a root of the equation z4 + 3z2 + 2z + 12 = 0.
Mark scheme
9(a) State or imply AB = 2i + j − 2k B1 OE ρ Carry out a correct method to find OD M1 Obtain answer −4i − j + 3k A1 OE 3 ρ 9(b) State r = −i + 2 j + 3k + λ ( 2i + j − 2k ) B1FT OE. The FT is on AB . 1 ρ ρ 9(c) For a general point P on AB, state CP or DP in component form, *M1 ρ e.g. CP = ( 3 − 2λ , −λ , −6 + 2λ ) Equate a relevant scalar product to zero or equate derivative of DM1 ρ CP to zero or use Pythagoras in a relevant triangle and solve for λ Obtain λ = 2 A1 Show the perpendicular is of length 3 A1 Carry out a correct method to find the area of ABCD and obtain the A1 answer 18 Alternative method for Question 9(c) Use a scalar product to find the projection CN (or DN) of BC *M1 (or AD) on CD Obtain CN = 3 (or DN = 3) A1 Use Pythagoras to obtain BN (or AN) DM1 9(c) cont’d Obtain answer 3 A1 Carry out a correct method to find the area of ABCD and obtain the A1 answer 18 5
Question
[3] (b) Find the other roots of this equation. [7] 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(a) Substitute −1 + 2i and attempt expansions of the z 2 and z 4 terms M1 Use i 2 = −1 at least once M1 Complete the verification correctly A1 3 10(b) State second root −1 − 2i A1 Carry out a method to find a quadratic factor with zeros −1 ± 2i M1 Obtain z 2 + 2 z + 3 A1 Commence division and reach partial quotient z 2 + kz M1 Obtain second quadratic factor z 2 − 2 z + 4 A1 Solve a 3-term quadratic and use i 2 = −1 M1 Obtain roots 1 + 3i and 1 − 3i A1 7
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Mathematics 9709/33 for May/June 2021 is out of 75 marks, with 1 hour 50 minutes to answer it.
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