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Cambridge AS & A Level Mathematics (9709), Paper 3 Pure Mathematics 3, May/June 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_s24_qp_319709_s24_ms_319709/31/M/J/24
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 1 B1 Symbolic coefficients are not sufficient. State correct unsimplified first two terms of the expansion of (1 − 2 x ) 2 , 1–x 1 e.g. 1 + ( −2 x ) 2 11 B1 Symbolic coefficients are not sufficient. − 1 ( −2 x ) 2 2 2 1 State correct unsimplified term in x2, e.g. − x2 2! 2 Obtain sufficient terms of the product of (3 + x) and the expansion up to the term M1 in x2 5 2 A1 Obtain final answer 3 – 2x – x 2 4
Question
Solve the equation ln (x - 5) = 7 - ln x . Give your answer correct to 2 decimal places. [4]
Mark scheme
2 Use law of logarithm of a product (or quotient) on correct terms *M1 Use correct method to eliminate logarithm DM1 Obtain a correct quadratic in x, e.g. x2 − 5x − e7 = 0 (allow decimals) A1 Obtain answer x = 35.71 only A1 4
Question
y (1.31, 1.50) (0.336, 1.00) O ln x The variables x and y satisfy the equation a y = bx , where a and b are constants. The graph of y against ln x is a straight line passing through the points (0.336, 1.00) and (1.31, 1.50), as shown in the diagram. Find the values of a and b. Give each value correct to the nearest integer. [4]
Mark scheme
3 State or imply that y ln a = ln b + ln x B1 Carry out a completely correct method for finding ln a or ln b M1 E.g., from ln a = ln b + 0.336 1.5ln a = ln b + 1.31. Obtain value a = 7 A1 Obtain value b = 5 A1 4
Question
The complex number u is given by u = - 1 - i 3 . (a) Express u in the form r (cos i + i sin i) , where r 2 0 and - r 1 i G r . Give the exact values of r and i . [2] The complex number v is given by v = 5 bcos 16 r + i sin 16 rl. v (b) Express the complex number in the form re ii where r 2 0 and - r 1 i G r . [2] u e sinx
Mark scheme
4(a) State or imply r = 2 B1 State or imply θ = – 23 π B1 2 4(b) State or imply r = 52 B1FT 5 FT . their 2 State or imply θ = 56 π B1FT 1 2 FT π – their – π. 6 3 2
Question
The equation of a curve is y = for 0 G x G 2r . cos 2 x dy Find and hence find the x-coordinates of the stationary points of the curve. [7] dx
Mark scheme
5 Use correct quotient (or product) rule *M1 esin x cos3 x − ( −2esin x sin x cos x ) A1 Obtain correct derivative or equivalent ( cos x ) 2 2 Equate numerator to zero DM1 Obtain equation in one unknown DM1 E.g. sin2x – 2sinx – 1 = 0. Solve a 3 term quadratic in sin x to find a value for x M1 Obtain a correct solution to the quadratic equation, e.g. 3.57° A1 At least 3sf. Obtain a further correct solution, e.g. x = 5.86° and no others in the interval A1FT At least 3sf. FT 3π – their 3.57. Alternative Method for the first 3 marks: Take logarithms of both sides and simplify (*M1) ln y = sin x − 2ln cos x or equivalent. 1 dy sin x (A1) Or equivalent. Obtain = cos x + 2 y dx cos x dy (DM1) Equate to zero dx Continue as for the original 7
Question
(a) By sketching a suitable pair of graphs, show that the equation cosec 12 x = e x - 3 has exactly one root, denoted by a , in the interval 0 1 x 1 r . [2] (b) Verify by calculation that a lies between 1 and 2. [2] (c) Show that if a sequence of values in the interval 0 1 x 1 r given by the iterative formula xn + 1 = ln (cosec 12 xn + 3) converges, then it converges to a . [1] (d) Use this iterative formula with an initial value of 1.4 to determine a correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] (e) State the minimum number of calculated iterations needed with this initial value to determine a correct to 2 decimal places. [1]
Mark scheme
6(a) Sketch a relevant graph, e.g. y = ex – 3 B1 y Correct shape, correct vertical intercept x B1 Sketch a second relevant graph, e.g. y = cosec (correct shape, minimum above 1 2 O the axis) and justify the given statement. Need to mark intersection with a dot, a π x cross, or say root at points of intersection, or equivalent -2 2 6(b) Calculate the values of a relevant expression or pair of expressions at x = 1 and M1 Use of degrees is M0. x=2 Complete the argument correctly with correct calculated values A1 E.g. −0.282 < 2.086, 4.389 > 1.188 1 < 1.626, 2 > 1.432 2.36 > 0, –3.2 < 0 At least 2sf. Condone truncation. 2 6(c) 1 x B1 AG. State x = ln cosec x + 3 and rearrange to the given equation cosec = e x − 3 Or vice versa and obtain the iterative formula. 2 2 1 6(d) Use the iterative formula correctly at least twice M1 Use of degrees in M0 (might see 1.38….). Obtain final answer 1.50 A1 Show sufficient iterations to 4 dp to justify 1.50 to 4 dp or show there is a sign A1 1.5156, 1.4940, 1.4978, 1.4971. change in the interval (1.495, 1.505) 3 6(e) 4 B1 1
Question
(a) On a single Argand diagram sketch the loci given by the equations z - 3 + 2i = 2 and w - 3 + 2i = w + 3 - 4i where z and w are complex numbers. [4] (b) Hence find the least value of z - w for points on these loci. Give your answer in an exact form. [2]
Mark scheme
7(a) Show a circle centre (3, –2) B1 Im 4 Show a circle with radius 2 FT centre not at the origin B1FT Show the point representing (–3, 4) or the midpoint (0, 1) B1 3 Show the perpendicular bisector of the line joining (–3, 4) and centre of the circle B1FT -3 Re -2 FT is on the position of (–3, 4) and centre of the circle 4 7(b) Carry out a correct method for finding the least value of |z – w| M1 (Distance (3, –2) to (0, 1)) – 2. Obtain answer 18 − 2 or 3 2 − 2 A1 2
Question
Use the substitution u = 1 - sin x to find the exact value of 3 2r sin 2x y r 1 - sin x dx . Give your answer in the form a + b 2 where a and b are rational numbers to be determined. [7]
Mark scheme
8 State or imply du = – cosx dx B1 Use sin 2x = 2sinxcosx and write the integral in terms of u *M1 (1 − u ) A1 Obtain ±2 u du or equivalent 1 3 DM1 Integrate correctly to obtain au 2 + bu 2 4 1 3 A1 Obtain correct −4u 2 + u 2 3 1 3 DM1 Correctly use limits u = 2 and 0 in an expression of the form au 2 + bu 2 OR limits x = 32 π and 12 π in an expression of the form 1 3 a(1 − sin x) + b(1 − sin x) 2 2 Obtain 83 − 43 2 A1 7
Question
The equations of two straight lines l1 and l2 are l1 : r = i - 2j + 3k + m (2i - j + ak) and l2 : r =- i - j - k + n (3i - 2j - 2k) , where a is a constant. The lines l1 and l2 are perpendicular. (a) Show that a = 4 . [1] The lines l1 and l2 also intersect. (b) Find the position vector of the point of intersection. [4] The point A has position vector - 5i + j - 9k . (c) Show that A lies on l1 . [2] The point B is the image of A after a reflection in the line l2 . (d) Find the position vector of B. [2] dy 2
Mark scheme
9(a) Carry out correct process for evaluating the scalar product of direction vectors, B1 E.g. 2(3) + (–1)(–2) + a(–2) = 0. equate the result to zero and obtain given value of a = 4 1 9(b) Express general point of at least one line correctly in component form, B1 The third component could be implied by a correct i.e. (1 + 2λ, –2 – λ, 3 + 4λ) or (–1 + 3µ, –1 – 2µ, –1 – 2µ) final answer. Equate at least two pairs of corresponding components and solve for λ or for µ M1 Obtain λ = –1 or µ = 0 A1 Obtain position vector of point of intersection is – i – j – k A1 4 9(c) Equate one component of l1 to matching component of A and solve to find λ M1 Use λ = –3 in equation of l1 and show this gives position vector of A A1 AG Or show λ = –3 for all three components equated. 2 9(d) Method to find position vector of B M1 E.g. ± 2×their (– i – j – k) ± (–5i + j – 9k) Obtain position vector of B is 3i – 3j + 7k A1 2
Question
(a) Given that 2x = tan y , show that = . [3] dx 1 + 4x 2 3 (b) Hence find the exact value of y x tan (2x) dx . 1 2 -1 [7] 2
Mark scheme
10(a) dy B1 dx Obtain 2 = sec2y or equivalent E.g. 2 = sec2y by differentiation with respect to y. dx dy Use sec2y = 1 + tan2 y M1 dy 2 A1 Replace tan y with 2x and rearrange to obtain given answer = dx 1 + 4 x 2 3 10(b) x2 *M1 Integrate by parts and reach ax 2 tan −1 2 x + b 1 + 4x2 dx x2 A1 OE Obtain 12 x 2 tan −1 2 x − 1 + 4 x2 dx n M1 Reduce integral to expression of the form m + 1 + 4x2 dx Complete integration and reach px 2 tan −1 2 x + qx + r tan −1 2 x M1 Obtain 12 x 2 tan −1 2 x − 14 x + 81 tan −1 2 x A1 OE Use limits of x = 12 and x = 12 3 in the correct order, having integrated twice DM1 Obtain answer 485 π − 18 3 + 18 or exact equivalent A1 7
Question
In a field there are 300 plants of a certain species, all of which can be infected by a particular disease. At time t after the first plant is infected there are x infected plants. The rate of change of x is proportional to the product of the number of plants infected and the number of plants that are not yet infected. The dx variables x and t are treated as continuous, and it is given that = 0.2 and x = 1 when t = 0 . dt (a) Show that x and t satisfy the differential equation dx 1495 = x (300 - x) . [2] dt (b) Using partial fractions, solve the differential equation and obtain an expression for t in terms of a single logarithm involving x. [9] Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown.
Mark scheme
11(a) dx dx M1 M0 for verification. State or imply equation of the form = kx (300 − x ) and use = 0.2 and x = 1 dt dt 1 A1 dx Obtain k = and rearrange to the given answer 1495 = x (300 − x ). 1495 dt 2 11(b) Separate variables correctly B1 1 1 dx = dt x ( 300 − x ) 1495 Correct integration of t term B1 t E.g. obtain t or . 1495 A B B1 State or imply partial fractions of the form + x 300 − x Correct method to find A or B M1 1 1 A= and B = . 300 300 1495 299 May see A = B = = . 300 60 1495 1495 A1 1 1 Obtain terms ln x − ln(300 − x ) OE. May see lnx − ln ( 300 − x ) . 300 300 300 300 Use t = 0, x = 1 to evaluate a constant or as limits in a solution containing terms M1 of the form ln x, ln(300 − x) and t. Obtain correct answer in any form A1 1495 1495 E.g. [ln x − ln(300 − x)] = t − ln 299. 300 300 Use law of logarithms twice to obtain an expression for t M1 299 299 x A1 Obtain final answer t = ln or equivalent single logarithm 60 300 − x 9
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Mathematics 9709/31 for May/June 2024 is out of 75 marks, with 1 hour 50 minutes to answer it.
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