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

1 Solve the equation 8 3 - 6 x = 4 # 5 -2 x . Give your answer correct to 3 decimal places. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 1 Use law of the logarithm of product or quotient on each side *B1 Allow logs to any base, as well as decimals, throughout. ln 83 + ln 8−6x and ln 4 + ln 5−2x. Allow for ln 3 8 4 and ln 86x – ln 52x. (3 − 6x) ln 8 and ln 4 + ln 5−2x gains next DB1 as well. Use law of logarithm of a power involving x on ONE side, e.g. ln 83 + (−)6x ln 8 or (3 − 6x) ln 8 or   9 18x  ln 2 or ln 4 − 2x ln 5 DB1 SC If *B0 DB0, then allow B1 (1/4) for a correct logarithm law seen anywhere. Obtain a correct linear equation in x, e.g.     3 6 ln8 9 18 ln2 ln 4 2 ln5 x x x      B1 If in decimals, allow small errors in 2nd and 3rd dp. Obtain answer x = 0.524 B1 3dp required. No working scores 0/4 marks. After *B1 DB1 to correct answer with no more log working seen, then SC B1 for x = 0.524. Maximum 3/4 possible. Alternative Method for Question 1 Use laws of indices to get to a = b ±2x or c ± x in a correct form so now only ONE log power law required (B2) (83/4) and (5/83)−2x or (52/86)−x opposite sides or (4/83) and (83/5)–2x or (86/52)–x opposite sides Obtain a correct linear equation in x, e.g. 3 3 8 8 ln 2 ln 4 5 x  (B1) −2x ln (5/83) or 2x ln (83/5) or x ln (86/52) or – x ln (52/86). SC: If B0 then allow B1 (1/4) for a correct term seen anywhere. If in decimals, allow small errors in 2nd and 3rd dp. Obtain answer x = 0.524 (B1) 3dp required. No working scores 0/4 marks. From the first line to correct answer with no log working seen, then B2 and SC B1 for x = 0.524. Maximum 3/4 possible. Question Answer Marks Guidance 1 Alternative Method 2 for Question 1 Use laws of indices to get to any correct form with indices combined so now TWO log power laws are required (*B1) Allow 27 – 18x and 5–2x on opposite sides or 29 – 18x and 22 – 4.64x on opposite sides. Use law of logarithm of a power involving x on ONE side, e.g. (7 – 18x) ln 2 = ln 5–2x or ln 27 – 18x = − 2x ln 5 or … Allow 7 – 18x ln 2 or 9 18 ln2 x  (DB1) e.g. (7 – 18x) ln 2 or   9 18 ln2 x  or –2x ln 5 or (2 – 4.64x) ln 2 SC: If *B0 DB0 then allow B1 (1/4) for a correct term seen anywhere. E.g. any term in *B1 shown above. Obtain a correct linear equation in x, e.g. (7 – 18x) ln 2 = − 2x ln 5 or (9 – 18x) ln 2 = (2 – 4.64x) ln 2 (B1) If in decimals, allow small errors in 2nd and 3rd dp. Obtain answer x = 0.524 (B1) 3dp required. No working scores 0/4 marks. From the first line to correct answer with no log working seen, then *B1 and SC B1 for x = 0.524. Maximum 2/4 possible. 4

More questions on Logarithmic and exponential functions

2 Find the exact coordinates of the stationary point of the curve y = e 2 x sin 2x for 0 G x G 1 r . [5] 2 .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 Use correct product rule cos 2x may be 1 – 2 sin2x or … Allow M1 if only error is ex instead of e2x in one of terms, then maximum 1/5. Obtain correct derivative 2 2 2e sin2 2e cos2 x x x x  A1 OE, e.g.   2 2 2 2 4e sin cos 2e cos sin . x x x x x x   Equate derivative of the form ae2xsin2x + e2xbcos2x to 0 and solve for 2x or x using a correct method Note may have substituted for sin2x and/or cos2x M1 Obtain 2x = tan−1(− their b/their a) OE. Allow one slip in rearranging. Allow degrees. Variety of other methods available, such as solving quadratic equation in sin x or tan x e.g. tan² x – 2tan x – 1 = 0 leading to x = tan-1(1 + √2). Obtain x = 3 8 π only or exact equivalent A1 CWO 67.5° gets A0. Ignore any answers outside interval 0 ⩽ x ⩽ π . 2 Obtain y = 3π 4 1 2e 2 only or exact simplified equivalent A1 CWO, ISW. Not 3π 4 3 sin πe 4         . Ignore any answers using x outside interval 0 ⩽ x ⩽ π . 2 5

More questions on Differentiation

3 The square roots of 24 - 7i can be expressed in the Cartesian form x + yi , where x and y are real and exact. By first forming a quartic equation in x or y, find the square roots of 24 - 7i in exact Cartesian form. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 3 Square x + iy obtaining three terms when simplified and equate real and imaginary parts to 24 and −7 respectively Obtain equations x2 – y2 = 24 and 2xy = –7 A1 Allow 2xyi = –7i. Eliminate one variable by correct method and find a horizontal equation in the other M1 All powers of x or y are positive and are in the numerator. Obtain 4x4 – 96x2 – 49 = 0 or 4y4 + 96y2 – 49 = 0 or 3-term equivalents A1 Obtain answers 7 2 2 i 2 2  and 7 2 2 i 2 2   or exact equivalents and no others A1 E.g. 7 2 2 i , 2 2           but not 7 2 2 i 2 2          or 7 2 2 i . 2 2           Allow coordinates or x =…, y =… paired correctly. ISW converting to different form. Must simplify 49. 5

More questions on Complex numbers

4 1n y (5.10, 2.21) (2.80, 0.372) O x The variables x and y satisfy the equation ky = ecx , where k and c are constants. The graph of ln y against x is a straight line passing through the points (2.80, 0.372) and (5.10, 2.21), as shown in the diagram. Find the values of k and c. Give each value correct to 2 significant figures. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 4 State or imply that ln k + ln y = cx or ln y = cx + ln 1 k etc. B1 Allow ln k + ln y = cx lne Carry out a completely correct method for finding ln k or c M1 Equations must have been formulated correctly. Obtain value c = 0.80 A1 AWRT Allow 0.8 for 0.80. Not a fraction. Accept in the equation ky = ecx. Obtain value k = 6.5 A1 AWRT Not a fraction. Accept in the equation ky = ecx.

More questions on Logarithmic and exponential functions

6 x 2 - 2x + 2 5 Express in partial fractions. [5] ( x - 1)( 2x + 1) .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 5 State or imply the form 1 2 1 B C A x x     B1 Use a correct method for finding a constant M1 Correct appropriate method. Obtain one of A = 3, B = 2 and C = –3 A1 Obtain a second value A1 Obtain a third value A1 Alternative Method for Question 5 Divide numerator by denominator to reach A = 3 (M1) May be implied by 3 [+]    1 2 1 ax b x x    with a and b not both 0. Obtain 3 +    5 1 2 1 x x x    (A1) State or imply the form 1 2 1 D E x x    (B1) Obtain one of D = 2 and E = − 3 (A1) Obtain a second value (A1) 5

More questions on Algebra

6 (a) On an Argand diagram shade the region whose points represent complex numbers z which satisfy both the inequalities z - 4 - 3i G 2 and arg ( z - 2 - i) H 1 r . [5] 3 (b) Calculate the greatest value of argz for points in this region. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 6(a) Show a circle centre (4, 3) Allow dashes for coordinates on axes B1 Note full circle is not required but must show centre and include relevant arc. Show a circle with radius 2. Can be implied by at least two of the points (2, 3), (6, 3), (4, 1) and (4, 5) being correct B1FT FT centre not at the origin. Point representing (2, 1) B1 Half-line or ‘correct’ full line extending into the third quadrant implies point (2, 1). Show a half-line at their (2, 1) at an angle of 1 3 , cutting top of circle between x = 3 and x = 5 B1FT FT the point (±2, ±1) or (±1, ±2). Shade the correct region Needs correct half-line or “correct” full line extending into the third quadrant AND correct circle B1 5 6(b) Carry out a correct method for finding the greatest value of arg z in the correct region in (a) M1 E.g. sin−1(2/√(25)) + tan−1(3/4) or sin−1(2/√(25)) + sin−1(3/5). Or, e.g., substitute y = kx in circle equation, solve when discriminant = 0, to get tan−1 6 21 6          . Obtain answer 1.06, or 1.05 or 1.055 or 1.056 or 60.4° or 60.5° A1 The marks in (b) are available even if errors in (a). No working seen scores 0/2 marks. 2

More questions on Complex numbers

7 Let f ( )x = 8x 3 + 54x 2 - 17x - 21. (a) Show that x + 7 is a factor of f ( )x . 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(b) Find the quotient when f ( )x is divided by x + 7 . 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(c) Hence solve the equation 8 cos 3 i+ 54 cos 2i - 17 cos i - 21 = 0 , for 0° G i G 360° . 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Mark scheme: 7(a) This is sufficient if no errors seen. [ – 2744 + 2646 +119 – 21 = 0] Or complete division of 8x3 + 54x2 – 17x – 21 by x + 7 to get quotient 8x2 – 2x – 3 and remainder of 0 Or state (x + 7)(8x2 – 2x – 3) is sufficient Factors must be stated again in (b) to collect marks there Correct division: 8x2 −2x −3 . x + 7 8x3 + 54x2 − 17x − 21 8x3 + 56x2 . − 2x2 −17x − 2x2 −14x . − 3x − 21 − 3x − 21 1 7(b) Commence division and reach partial quotient of the form 8x2 ± 2x or 8x3 + 54x2 – 17x – 21 = (x + 7)(Ax2 + Bx + C) and reach A = 8 and B = ± 2 or C = –3 M1 Condone no visible working. Obtain quotient 8x2 – 2x – 3 with no errors seen Stating (x + 7)(8x2 – 2x – 3) is sufficient A1 Division can terminate with 0 or −3x – 21 stated once or twice. The working of division and finding quotient may be seen in (a) but results required here to collect marks. 2 Question Answer Marks Guidance 7(c) Solve quadratic from (b) to obtain a value for 𝜃 = 1 1 cos 2         or 1 3 cos 4       M1 (x + 7) (8x2 − 2x – 3) = (x + 7)(4x – 3)(2x + 1) = 0 x = cos 2 4 96 1 3 and . 16 2 4     Obtain one answer, e.g. 𝜃 = 120° A1 Obtain three further answers, e.g. 𝜃 = 240°, 41.4° and 318.6° (condone 319°) and no others in the interval A1 Accept more accurate answers. Answers in radians, maximum 2/3. 3

More questions on Quadratics

8 (a) Express 3 cos 2x - 3 sin 2 x in the form R cos ( 2x + a) , where R 2 0 and 0 1 a 1 1 r . Give the 2 exact values of R and a. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 1 r 12 3 (b) Hence find the exact value of dx , simplifying your answer. [5] y 2 0 `3 cos 2x - 3 sin 2xj ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 8(a) State R = 12 or exact equivalent B1 ISW Use trig formula to find α M1 Allow 1 3 30 or tan 3             or cos−1 3 2          or sin−1 1 2        Allow M1 if – 1 3 tan 3         etc. NB: If cos = 3 and sin = 3 seen then M0 A0. Obtain α = 1 π 6 A1 CWO, so A0 if from 1 3 tan . 3          3 Question Answer Marks Guidance 8(b) Express integral in the form A   2 sec 2 ... d x x   or A   2 sec 2 ... d x x   B1FT FT α from (a). Integrate and reach B   tan 2 ... x  or B   tan 2 ... x  B1FT FT α from (a). Where B = A or 2A or 0.5A. Obtain   1 tan 2 ... 8 x  B1FT OE FT α from (a). Allow 1 8 as 1 1. 4 2  Coefficient must be correct. Use limits of 0 x  and 1 12 π x  in the correct order in expression of form B   tan 2 ... x  so B tan ... 6         −B  tan ... or B tan ... 6         −B   tan ...  M1 Allow with tan still present. FT α from (a). SC: B1 3 12 OE after 1 1 tan 2 π 8 6 x        with no working. Obtain answer 1 12 3 or 1 4 3 or 1 48 or single term exact equivalent A1 1 8 ( 3 – 1 3 ) = 1 8 3 1 3        needs simplifying. 5 Note: allow all marks in (b) even if α = 1 π 6 found by an incorrect method in (a).

More questions on Integration

9 10- x x x A container in the shape of a cuboid has a square base of side x and a height of ( 10- )x . It is given that x varies with time, t, where t 2 0 . The container decreases in volume at a rate which is inversely proportional to t. When t = 1 , x = 1 and the rate of decrease of x is 20. 10 2 37 (a) Show that x and t satisfy the differential equation dx - 1 = . [5] td 2t `20 x - 3x 2j ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Solve the differential equation, obtaining an expression for t in terms of x. 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Mark scheme: 9(a) Obtain  d d V k t t  or  d 1 d V t kt  B1 Obtain 2 d 20 3 d V x x x   B1 Correct use of chain rule involving k M1 Use d d V t = d d . d d V x x t  Expressions for d d V t and d d V x must be seen to get M1. Obtain   2 d d 20 3 x k t t x x   or equivalent, A1 If this expression is first seen with numerical values, allow A1 when their value of k is substituted back into the general expression. Use 1 10 t  , 1 2 x  and d 20 d 37 x t  to obtain given answer which must be stated d 20 d 37 x t  needed to score final A1 A1   2 d 1 d 2 20 3 x t t x x    AG Need to at least see 20 37  = 1 3 10 10 4 k        if k t or 20 37  = 1 3 10 10 4 k         if k t  in working for correct k. d 20 d 37 x t  seen anywhere, then A0. 5 Question Answer Marks Guidance 9(b) Separate variables correctly & integrate at least one side correctly B1 Obtain terms 10x2 – x3 B1 May see –10x2 + x3 if negative sign moved across or e.g. 20x2 – 2x3 if 2 moved across. Allow 2 3 20 3 . 2 3 x x  Obtain term ln t with ‘correct’ coefficient from their separation of variables, for example a ln t for a t . B1FT FT sign and position of 2 from their separation but B0 if error from later manipulation. Use 1 10 t  , 1 2 x  to evaluate a constant or as limits in a solution containing terms of the form x2, x3 and ln t (or ln 2t) M1 Allow numerical and sign errors and decimals. Allow if exponentiate before substitution, even if exponentiation done incorrectly, allow for c or ec. Obtain correct answer in any form, for example 2 3 ln 19 ln0.1 10 2 8 2 t x x     A1 2 3 ln2 19 ln0.2 10 2 8 2 t x x     or 2 3 ln 10 2.5 0.125 1.15 2 t x x       Allow 1.14 to 1.16 for 1.15 and allow 2.44 to 2.46 for 2.45 Obtain answer 3 2 19 2 20 1 4 10 e x x t    or equivalent A1 ISW Need t =……… E.g. 3 3 2 2 2 3 19 2 19 4 2 20 4 19 20 20 2 4 0 . .1 e 1 , , e e 10 10e e x x x x x x     Allow decimals, allow 2.44 to 2.46 for 2.45, e.g. 3 2 2 20 2.45 e . x x   A0 if 1 ln10 e present in final answer. 6

More questions on Integration

10 The equations of two straight lines are r = i + j + 2ak + m ( 3i + 4j + ak ) and r =- 3i - j + 4k + n ( - i + 2 j + 2k ) , where a is a constant. (a) Given that the acute angle between the directions of these lines is 1 r, find the possible values 4 of .a [6] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Given instead that the lines intersect, find the value of a and the position vector of the point of intersection. 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Mark scheme: 10(a) Carry out correct process for evaluating the scalar product of direction vectors *M1 3 1 3 1 4 . 2 or 4 . 2 2 2 a a                                  3(–1) + 4(2) + 2a or –3 + 8 + 2a or 5 + 2a. Allow one slip in unsimplified form. Using the correct process for the moduli, divide the scalar product by the product of the moduli and equate to 2 , 2  or equate the scalar product to the product of the moduli and 2 2  *M1 *M1 marks independent of each other, so *M0 *M1 for failure to use both direction vectors, but must be using scalar product and same 2 vectors throughout. 2 Allow 2 or − 2 2 throughout question. State a correct equation in any form, e.g.  2 5 2 2 2 3 25 a a    Allow unsimplified as in guidance A1  2 5 2 2 2 9 16 1 4 4 a a       OE E.g. 5 + 2a =  2 2 9 16 1 4 4 2 a      If moduli initially correct but later has errors, award A1 when using 2 2 or 2 2  or − 2 . 2 Form a quadratic equation in a with 3 or more terms all on one side and solve for a. DM1 depends on BOTH *M1 DM1 Must square (5 + 2a) to get 3 terms and must remove square roots from both terms on other side. 25 + 20a + 4a2 = 9 2 (25 + a2) a2 − 40a + 175 = 0 hence (a – 5)(a – 35) = 0. 10(a) Obtain a = 5 and a = 35 A2 A1 for each, working not needed if quadratic correct. 6 Question Answer Marks Guidance 10(b) Express general point of at least one line correctly in component form, i.e. 1 3 1 4 2 a a                 or –3 – –1 2 4 2 µ µ µ             B1 Often the third point on the line occurs after M1 A1 is gained. Equate at least two pairs of corresponding components and solve for λ or µ or a M1 If solve for a first, they must have a complete method to eliminate both λ and µ. If using a to solve for λ or for µ, a must have been found from a valid method. Obtain λ = –1 or µ = –1 A1 Obtain a = 2 A1 Obtain position vector of the point of intersection is –2i – 3j + 2k Two different answers for point of intersection scores A0 even if one is correct A1 Accept coordinates, row or column, but not (–2i,– 3j,+ 2k) or 2 –3 2            i j k but ISW after correct form seen. 5

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