Cambridge A Level Mathematics 9709 — 2024 May/June Paper 3 · Variant 1

21 Expand ( 3 + x)( 1 - 2x) 1 in ascending powers of x, up to and including the term in x2, simplifying the coefficients. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 1 State correct unsimplified first two terms of the expansion of   1 2 1 2 x , e.g.   1 1 2 2  x B1 Symbolic coefficients are not sufficient. 1 – x State correct unsimplified term in x2, e.g.   2 1 1 1 2 2 2 2!         x B1 Symbolic coefficients are not sufficient. 2 1 2  x Obtain sufficient terms of the product of (3 + x) and the expansion up to the term in x2 M1 Obtain final answer 3 – 2x – 2 5 2 x A1 4

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2 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. 2 7 5 e 0    x x (allow decimals) A1 Obtain answer x = 35.71 only A1 4

More questions on Logarithmic and exponential functions

3 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 lnx 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 ln ln ln y a b x Carry out a completely correct method for finding lna or lnb M1 E.g., from ln ln 0.336 a b   1.5ln ln 1.31. a b   Obtain value a = 7 A1 Obtain value b = 5 A1 4

More questions on Logarithmic and exponential functions

4 The complex number u is given by u = - 1 - i 3 . (a) Express u in the form r (cos i+ i sini) , 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 cos 16 1 r + i sin 6 r b l. v (b) Express the complex number in the form reii where r 2 0 and - r 1 i G r . [2] u ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a) State or imply r = 2 B1 State or imply θ = – 2 3 π B1 2 4(b) State or imply r = 5 2 B1FT FT 5 . 2 their State or imply θ = 5 6 π B1FT FT 1 π 6 – their – 2 π. 3 2

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5 The equation of a curve is y = 2 for 0 G x G 2 r. cos x dy Find and hence find the x-coordinates of the stationary points of the curve. 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Mark scheme: 5 Use correct quotient (or product) rule *M1 Obtain correct derivative   sin 3 sin 2 2 e cos ( 2e sin cos ) cos  x x x x x x or equivalent A1 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 sin 2lncos   y x x or equivalent. Obtain 1 d sin cos 2 d cos   y x x y x x (A1) Or equivalent. Equate d d y x to zero (DM1) Continue as for the original 7

More questions on Differentiation

= ex - 3 has exactly one6 (a) By sketching a suitable pair of graphs, show that the equation cosec 12 x root, denoted by a, in the interval 0 1 x 1 r . [2] (b) Verify by calculation that a lies between 1 and 2. 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(c) Show that if a sequence of values in the interval 0 1 x 1 r given by the iterative formula x n + 1 1 = ln ( cosec 2 x n + 3) converges, then it converges to a. 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(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. 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(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) Correct shape, correct vertical intercept B1 Sketch a second relevant graph, e.g. cosec 2  x y (correct shape, minimum above the axis) and justify the given statement. Need to mark intersection with a dot, a cross, or say root at points of intersection, or equivalent B1 2 6(b) Calculate the values of a relevant expression or pair of expressions at x = 1 and x = 2 M1 Use of degrees is M0. 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) State 1 ln cosec 3 2         x x and rearrange to the given equation cosec e 3 2   x x B1 AG. Or vice versa and obtain the iterative formula. 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 change in the interval (1.495, 1.505) A1 1.5156, 1.4940, 1.4978, 1.4971. 3 1 -2 y x O π Question Answer Marks Guidance 6(e) 4 B1 1

More questions on Numerical solution of equations

7 (a) On a single Argand diagram sketch the loci given by the equations z - 3 + 2i = 2 and w - 3 + 2 i = 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 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 Show the perpendicular bisector of the line joining (–3, 4) and centre of the circle FT is on the position of (–3, 4) and centre of the circle B1FT 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 Im Re -2 4 -3 3

More questions on Complex numbers

8 Use the substitution u = 1 - sin x to find the exact value of 2 3 r sin 2x dx . yr 1 - sin x Give your answer in the form a + b 2 where a and b are rational numbers to be determined. 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Mark scheme: 8 State or imply du = – cosx dx B1 Use sin 2x = 2sinxcosx and write the integral in terms of u *M1 Obtain (1 ) 2 d   u u u or equivalent A1 Integrate correctly to obtain 1 3 2 2  au bu DM1 Obtain correct 1 3 2 2 4 4 3   u u A1 Correctly use limits u = 2 and 0 in an expression of the form 1 3 2 2  au bu OR limits 3 2 π x  and 1 2 π in an expression of the form 1 3 2 2 (1 sin ) (1 sin )    a x b x DM1 Obtain 8 4 3 3 2  A1 7

More questions on Integration

9 The equations of two straight lines l1 and l2 are n ( 3i - 2j - 2k ) , l1: r = i - 2j + 3k + m ( 2i - j + ak ) and l2: r =- i - j - k + where a is a constant. The lines l1 and l2 are perpendicular. (a) Show that a = 4 . 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The lines l1 and l2 also intersect. (b) Find the position vector of the point of intersection. 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The point A has position vector - 5i + j - 9k . (c) Show that A lies on l1. 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The point B is the image of A after a reflection in the line l2. (d) Find the position vector of B. 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Mark scheme: 9(a) Carry out correct process for evaluating the scalar product of direction vectors, equate the result to zero and obtain given value of a = 4 B1 E.g. 2(3) + (–1)(–2) + a(–2) = 0. 1 9(b) Express general point of at least one line correctly in component form, i.e. (1 + 2λ, –2 – λ, 3 + 4λ) or (–1 + 3µ, –1 – 2µ, –1 – 2µ) B1 The third component could be implied by a correct 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

More questions on Vectors

10 (a) Given that 2x = tan y , show that = 2 . 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Mark scheme: 10(a) Obtain 2 = sec2y d d y x or equivalent d x y = sec2y by differentiation with respect to y. Use sec2y = 1 + tan2 y M1 Replace tan y with 2x and rearrange to obtain given answer 2 d 2 d 1 4   y x x A1 3 10(b) Integrate by parts and reach 2 2 1 2 tan 2 d 1 4     x ax x b x x *M1 Obtain 2 2 1 1 2 2 tan 2 d 1 4 x x x x x     A1 OE Reduce integral to expression of the form 2 d 1 4    n m x x M1 Complete integration and reach 2 1 1 tan 2 tan 2     px x qx r x M1 Obtain 2 1 1 1 1 1 2 4 8 tan 2 tan 2 x x x x     A1 OE Use limits of 1 2 x  and 1 2 3 x  in the correct order, having integrated twice DM1 Obtain answer 5 1 1 48 8 8 π 3   or exact equivalent A1 7

More questions on Integration

11 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) . 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(b) Using partial fractions, solve the differential equation and obtain an expression for t in terms of a single logarithm involving x. 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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) State or imply equation of the form d (300 ) d   x kx x t and use d 0.2 and 1 d   x x t Obtain k = 1 1495 and rearrange to the given answer A1 d 1495 (300 ). d x x x t   2 Question Answer Marks Guidance 11(b) Separate variables correctly B1   1 1 d d 300 1495        x t x x Correct integration of t term B1 E.g. obtain t or . 1495 t State or imply partial fractions of the form 300   A B x x B1 Correct method to find A or B M1 A = 1 300 and B = 1 . 300 May see 1495 299. 300 60 A B    Obtain terms 1495 1495 ln ln(300 ) 300 300   x x A1 OE. May see   1 1 ln ln 300 . 300 300 x x   Use t = 0, x = 1 to evaluate a constant or as limits in a solution containing terms of the form ln , ln(300 ) x x  and t. M1 Obtain correct answer in any form A1 E.g.   1495 1495 ln ln(300 ) ln 299. 300 300 x x t    Use law of logarithms twice to obtain an expression for t M1 Obtain final answer 299 299 ln 60 300   x t x or equivalent single logarithm A1 9

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