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Cambridge AS & A Level Mathematics (9709), Paper 3 Pure Mathematics 3, 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_319709_w21_ms_319709/31/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
(a) Express 5 sin x − 3 cos x in the form R sin x − !, where R > 0 and 0 < ! < 12 π. Give the exact value of R and give ! correct to 2 decimal places. [3] (b) Hence state the greatest and least possible values of 5 sin x − 3 cos x2 . [2]
Mark scheme
2(a) State R = 34 B1 Use trig formulae to find α M1 3 3 5 tan α = or sin α = or cos α = . 5 34 34 Obtain α = 0.54 A1 30.96° scores M1A0. 3 2(b) State greatest value 34 B1 FT Their R 2 . State least value 0 B1 2
Question
The curve with equation y = x e1−2x has one stationary point. (a) Find the coordinates of this point. [4] (b) Determine whether the stationary point is a maximum or a minimum. [2] Using the substitution u = x, find the exact value of
Mark scheme
3(a) Use correct product rule M1 Obtain correct derivative in any form A1 dy 1− 2 x =e − 2 xe1− 2 x dx Equate derivative to zero and solve for x M1 1 1 A1 Obtain x = and y = 2 2 4 3(b) Use a correct method for determining the nature of a stationary point M1 d2 y e.g. 2 = −2e1− 2 x − 2 (1 − 2 x ) e1− 2 x dx Show that it is a maximum point A1 2
Question
∞ Ô 1 x + 1 x dx. [6] 3
Mark scheme
4 du 1 1 B1 State that = or du = dx dx 2 x 2 x Substitute throughout for x and dx M1 2 A1 Obtain a correct integral with integrand 2 u +1 Integrate and obtain term of the form k tan −1 u M1 ( 2 tan u ) −1 Use limits 3 and ∞ for u or equivalent and evaluate trig. A1 π π e.g. 2 − Must be working in radians. 2 3 1 A1 Or equivalent single term. Obtain answer π 3 6
Question
(a) Show that the equation cot 21 + cot 1 = 2 can be expressed as a quadratic equation in tan 1. [3] (b) Hence solve the equation cot 21 + cot 1 = 2, for 0 < 1 < π, giving your answers correct to 3 decimal places. [3] When a + bx 1 + 4x, where a and b are constants, is expanded in ascending powers of x, the
Mark scheme
5(a) Use correct trig formulae and express equation in terms of tan θ M1 Obtain a correct equation in tan θ in any form A1 1 − tan 2 θ 1 e.g. + =2 2 tan θ tan θ Reduce to tan 2θ + 4 tan θ − 3 = 0 , or 3-term equivalent A1 3 5(b) Solve a 3-term quadratic for tan θ and calculate θ M1 ( tan θ = −2 ± 7 ) Obtain answer, e.g. 0.573 A1 Must be 3 d.p. Obtain second answer, e.g. 1.783 and no other A1 Ignore answers outside the given interval. Treat answers in degrees as a misread. ( 32.9°,102.1° ) 3
Question
coefficients of x and x2 are 3 and −6 respectively. Find the values of a and b. [6]
Mark scheme
6 State or imply 1 + 2x as first terms of the expansion of 1 + 4 x B1 Allow for correct unsimplified expression. State or imply −2 x2 as third term of the expansion of 1 + 4 x B1 Allow for correct unsimplified expression. Form an expression for the coefficient of x or coefficient of x 2 in the M1 All relevant terms considered. expansion of ( a + bx ) 1 + 4 x and equate to given coefficient Obtain 2a + b = 3, or equivalent A1 One correct equation. Obtain– 2a + 2b = – 6 or equivalent A1 Second correct equation. Obtain answer a = 2 and b = – 1 A1 6
Question
(a) Given that y = ln ln x, show that = dy 1 . [1] dx x ln x The variables x and t satisfy the differential equation x ln x + t = 0. dx dt It is given that x = e when t = 2. (b) Solve the differential equation obtaining an expression for x in terms of t, simplifying your answer. [7] (c) Hence state what happens to the value of x as t tends to infinity. [1] a The constant a is such that Ô dx = 6. ln x
Mark scheme
7(a) Show sufficient working to justify the given answer B1 1 7(b) Correct separation of variables B1 e.g. − 1dt = t x ln1 x dx Obtain term ln ( ln x ) B1 Obtain term − ln t B1 Evaluate a constant or use x = e and t = 2 as limits in an expression involving M1 ln ( ln x ) Obtain correct solution in any form, e.g. ln(ln x) = − ln t + ln 2 A1 Use log laws to enable removal of logarithms M1 2 A1 Obtain answer x = e t , or simplified equivalent 7 7(c) k B1 State that x tends to 1 coming from x = e t 1
Question
x 1 A @ (a) Show that a = exp + 2 . 1 [5] a [exp x is an alternative notation for ex .] (b) Verify by calculation that a lies between 9 and 11. [2] (c) Use an iterative formula based on the equation in part (a) to determine a correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]
Mark scheme
8(a) 1 *M1 Commence integration and reach a x ln x + b x . dx, or equivalent x 1 A1 Obtain 2 x ln x − 2 x . dx, or equivalent x Obtain integral 2 x ln x − 4 x , or equivalent A1 Substitute limits and equate result to 6 DM1 1 A1 Obtain given answer from full and correct working. Rearrange and obtain a = exp + 2 a 5 8(b) Calculate the values of a relevant expression or pair of expressions at M1 9 < 10.31 a = 9 and a = 11 e.g. or 1.31 > 0, − 1.01 < 0 11 > 9.99 Complete the argument correctly with correct values A1 2 8(c) 1 M1 Use the iterative process an +1 = exp( + 2) correctly at least once an Obtain answer 10.12 A1 Show sufficient iterations to 4dp to justify 10.12 to 2dp, or show there is a A1 e.g. 10, 10.1374, 10.1156, 10.1190, .... sign change in the interval (10.115, 10.125) 9, 10.3123, 10.0886, 10.1233, 10.1178, ... 11, 9.9893, 10.1391, 10.1153, 10.1191, .... 3
Question
Two lines l and m have equations r = 3i + 2j + 5k + s 4i − j + 3k and r = i − j − 2k + t −i + 2j + 2k respectively. (a) Show that l and m are perpendicular. [2] (b) Show that l and m intersect and state the position vector of the point of intersection. [5] (c) Show that the length of the perpendicular from the origin to the line m is 31 5. [4]
Mark scheme
9(a) Use correct method to evaluate the scalar product of relevant vectors M1 ( −4 − 2 + 6 ) Obtain answer zero and deduce the given statement A1 Need a conclusion or a statement in advance that the scalar product will be zero. 2 9(b) Express general point of l or m in component form, B1 e.g. (3 + 4s, 2 – s, 5 + 3s) or (1 – t, – 1 +2t, – 2 + 2t) Equate at least two pairs of components and solve for s or for t M1 Obtain correct answer s = – 1 and t = 2 A1 Verify that all three equations are satisfied A1 State position vector of the intersection – i + 3j +2k, or equivalent A1 Can come from 1 correct value and no contradictory statement. 5 9(c) Taking a general point P on m, form an equation in t by either equating a *M1 1 − t −1 ρ relevant scalar product to zero, or equating the derivative of OP to zero, or e.g. −1 + 2t . 2 = 0 taking a specific point Q on m, e.g. (1, – 1, – 2), using Pythagoras in triangle −2 + 2t 2 OPQ 7 A1 Obtain t = 9 Carry out correct method to find OP DM1 5 A1 Obtain the given answer from full and correct working. Obtain 3 Alternative method for question 9(c) Take a specific point Q on m, e.g. (– 1, 3, 2) and use a scalar product to find *M1 QN, the projection of OQ on m 11 A1 Obtain QN = , or equivalent 3 Use Pythagoras to obtain ON DM1 Obtain the given answer correctly A1 4
Question
The complex number 1 + 2i is denoted by u. The polynomial 2x3 + ax2 + 4x + b, where a and b are real constants, is denoted by p x. It is given that u is a root of the equation p x = 0. (a) Find the values of a and b. [4] (b) State a second complex root of this equation. [1] (c) Find the real factors of p x. [2] (d) (i) On a sketch of an Argand diagram, shade the region whose points represent complex numbers z satisfying the inequalities z − u ≤ 5 and arg z ≤ 4 π. 1 [4] (ii) Find the least value of Im z for points in the shaded region. Give your answer in an exact form. [1] 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 in the polynomial and attempt expansions of x 2 and x3 M1 u 2 = −3 + 4i, u 3 = −11 − 2i Full substitution but need not simplify. Equate real and/or imaginary parts to zero M1 −18 − 3a + b = 0, 4 + 4a = 0 Obtain a = – 1 A1 Obtain b = 15 A1 4 10(b) State second root 1 – 2i B1 1 10(c) State the quadratic factor x 2 − 2 x + 5 B1 State the linear factor 2x +3 B1 2 10(d)(i) Show a circle with centre 1 + 2i B1 Show circle passing through the origin B1 4i Show the half line y = x in the first quadrant (accept chord of circle) B1 Shade the correct region on a correct diagram B1 2 4 10(d)(ii) State answer 2 − 5 B1 1
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Mathematics 9709/31 for October/November 2021 is out of 75 marks, with 1 hour 50 minutes to answer it.
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