Cambridge A Level Mathematics 9709 — 2023 May/June Paper 1 · Variant 1

1 Solve the equation 4 sin 1 + tan 1 = 0 for 0Å < 1 < 180Å. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 4sin tan    0 ⇒   sin 4sin 0 cos     sin tan . cos    BOD if  missing. ⇒   sin 4cos 1 [ 0    ⇒ 1 sin 0 or ] cos 4     M1 WWW Factorise, not divide by sin or tan. May see    tan 4cos 1 0    or    sin 4 sec 0     . θ = 104.5° A1 AWRT 1.82 rads A0. Ignore answers outside (0, 180°). If M1 M0, SC B1 for θ = 104.5° max 2/3. 3

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4 C D 1 B A 8 cm The diagram shows a sector ABC of a circle with centre A and radius 8cm. The area of the sector is 16 πcm2. The point D lies on the arc BC. 3 Find the perimeter of the segment BCD. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 4 2 1 16 8 2 3     ⇒ 6  B1 SOI OE e.g.   2 , 0.524 3 s.f. 12 Use of degrees acceptable throughout provided conversion used in formulae for sector area and arc length. Arc length = 8 6  their [= 4.1887…] M1 OE FT their θ. Look for 4 3 . [BC =] 1 2 8 sin 2 6          their [= 4.1411…] M1 Attempt to find BC or 2 BC (see alt. methods below) FT their θ. Look for 16sin12  or 4 6 4 2  . Perimeter = 8.33 A1 AWRT Must be combined into one term. Question Answer Marks Guidance 4 Alternative methods for Question 4: 2nd M1 mark (use normal scheme for the other marks) ALT 1   2 2 2 8 8 2 8 8cos 4.14 6               BC their BC ALT 2     2 2 2 8 4 3 4 4.14       BC BC ALT 3   8 4.14 5 sin sin 6 12                   BC BC ALT 1 Substitute into correct cosine rule. FT their θ Look for 128 64 3  ALT 2 Find lengths 4 and 4 3 then use Pythagoras in the left hand triangle. ALT 3 Substitute into correct sine rule. 4 8  / 6 4 4√3

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5 The line with equation y = kx −k, where k is a positive constant, is a tangent to the curve with equation y = −1 2x. Find, in either order, the value of k and the coordinates of the point where the tangent meets the curve. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 5 1 2   kx k x ⇒   2 2 2 1 0    kx kx OR quadratic in 2 1 : 2 2 0 2                y k y x y y ky k y k k k *M1 OE e.g.   2 1 0 2    kx kx ,   2 1 0 2    x x k Equate line and curve to form 3-term quadratic (all terms on one side).   2 4 0   b ac ⇒       2 2 4 2 1 0    k k or 2 4 8 [ 0   k k ⇒   4 2 0   k k ] OR using equation in : y    2 2 4 2 0   k k DM1 Use discriminant correctly with their , , a b c not in quadratic formula. DM0 if x still present. May see  2 1 4 0 2         k k or 1 1 4 0 2         k . k = 2 only A1 If DM0 then k = 2, award A0 XP then B0 B0 Allow A1 even if divides by k to solve. If 0  k also present but uses 2  k , award A1.   2 2 4 4 1 0 2 1 0          x x x ⇒ 1 2  x B1 1 2 2 1 2     y B1 Question Answer Marks Guidance 5 Alternative method for Q5 2 d 1 d 2  y x x or 2 1 2  x *M1 Differentiate 1 2 x M0 for 2 2  x . No errors.   2 2 1 1 1 2 2 2 y x x x x    or 2 2 1 1 2 0 2         x x x x DM1 Sub their d d y x into equation of line or set gradient = k to form equation in x. 1 2  x only A1 If DM0 then 1 2  x , award A0XP then B0 B0. 1 2 2 1 2           y B1 2  k B1 5

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p2 6 The first three terms of an arithmetic progression are , 2p −6 and p. 6 (a) Given that the common difference of the progression is not zero, find the value of p. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Using this value, find the sum to infinity of the geometric progression with first two terms p2 and 2p −6. [2] 6 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a)     2 2 2 2 6 3 12 0 6 6        p p p p p OR       2 2 2 6 2 6 3 12 0 6 6          p p p p p p OR   2 1 0 6   d d quadratic in p (all terms on one side) or 2-term quadratic in . d OE e.g.   2 18 72 0    p p ,   2 1 9 36 0 2    p p .   2 18 72 0    p p ⇒     6 12 0    p p or     2 18 18 4 1 72 2    OR   1 1 0 6 6           d d d DM1 Solve a 3-term quadratic in p by factorisation, formula or completing the square or solve a 2-term quadratic in d by factorisation. p = 12 only A1 Since p = 6 gives d = 0. If *M1 DM0 then p = 12 only, award SC B1, max 2/3 marks. A0 XP if error in either factor and 12  p only. 12  p only by trial and improvement 3/3. 3 Question Answer Marks Guidance 6(b) For GP r = 2 2 6 6              p p = 18 3 24 4        B1 OE SOI. Sum to infinity = 24 3 1 4  = 96 B1 FT FT their value of p if used correctly to find r (B0 if ‘ p ’ used) provided 1  r . e.g. 18  p ⇒  54 121.5 5 1 9    S . 2

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7 A curve has equation y = 2 + 3 sin 12x for 0 ≤x ≤4π. (a) State greatest and least values of y. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Sketch the curve. [2] y x 0 π 2π 3π 4π (c) State the number of solutions of the equation 2 + 3 sin 12x = 5 −2x for 0 ≤x ≤4π. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) [Greatest =] 5 B1 No inequality required. [Least =] ‒1 B1 No inequality required. Condone   1, 5  or equivalent. 2 Question Answer Marks Guidance 7(b) B1 One complete cycle starting and finishing at y = 2. Maximum and minimum in correct quadrants. Shape and curvature approximately correct. B1 FT Maximum and minimum (indicated on y -axis with numbers or lines, or labelled on graph). FT their greatest and least values. Award B1 for 5 and 1  even if their values were incorrect in (a). 2 7(c) 1 B1 WWW 1 π 2π 3π 4π −2 2 4 x y

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9 Water is poured into a tank at a constant rate of 500cm3 per second. The depth of water in the tank, t seconds after filling starts, is hcm. When the depth of water in the tank is hcm, the volume, V cm3, of water in the tank is given by the formula V = 4 25 + h 3 −62500 . 3 3 (a) Find the rate at which h is increasing at the instant when h = 10cm. 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(b) At another instant, the rate at which h is increasing is 0.075cm per second. Find the value of V at this instant. 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Mark scheme: 9(a)   2 d 4 3 25 d 3    V h h [= 4900 when h = 10] d V h . d d d d d d   V h V h t t   2 d d "4 25 10 " 500 d d       h h their t t 500 4900       M1 Use chain rule correctly to find a numerical expression for d d h t . Accept e.g. 500 2500 2000 400   . 1 d 0.102 cms d       h t A1 AWRT OE e.g. 5 49 ISW. 3 9(b) d d d 500 d d d     V V h t h t   2 "4 25 " 0.075   their h *M1 SOI Use chain rule correctly to form equation in h.   2 5000 25 [15.8 3           h h 248…] DM1 Solve quadratic to find h. Exact value of h is 5000 25 3  or 50 6 25 3  25 40.82    h 69900  V cm3 A1 AWRT ISW Look for 698(88.5) . 3

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10 y A 1, 4 4 y = 2 2x −1 1 B 32, 1 x O 1 4 The diagram shows part of the curve with equation y = and parts of the lines x = 1 and y = 1. 2x −1 2 The curve passes through the points A 1, 4 and B, 32, 1 . (a) Find the exact volume generated when the shaded region is rotated through 360Å about the x-axis. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) A triangle is formed from the tangent to the curve at B, the normal to the curve at B and the x-axis. Find the area of this triangle. [6] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 10(a)           4 4 3 16 16 π d π 16 2 1 d π 2 1 3 2 2 1                   x x x x x *M1 Integrate 2 y (power incr. by 1 or div by their new power). M0 if more than 1 error or   3 16 2 1 6    x x .    3 16 π 3 2 2 1            x A1 OE e.g.   3 8 2 1 3          x .  16 16 π 6 8 6 1             112 7 π π 48 3         DM1 Sub correct limits into their integral: F 3 2       F(1). Must see at least 1 8 . 3 3         Allow 1 sign error. Decimal: 2.33 π or 7.33 . Volume of cylinder 2 1 1 π 1 π 2 2           OR    1.5 1 1 π 1 d π 2   x B1 1 π 2 or 3 π 1 2         seen. Volume of revolution 7 1 π π 3 2         11π 6  A1 A0 for 5.76 (not exact). If DM0 for insufficient substitution, or B0, SC B1 for 11π 6 . 5 Question Answer Marks Guidance 10(b)      3 d 8 2 1 2 d            y x x B2, 1, 0 OE B1 for each correct element in {}. At B gradient = 2  B1 Eqn of tangent 3 1 " 2" 2          y their x OR Eqn of normal 1 3 1 " " 2 2         y their x M1 SOI Following differentiation OE e.g. 2 4   y x or 1 1 2 4   y x . (Must have 1  N T m m for M1). Tangent crosses x-axis at 2 or normal crosses x-axis at 1 2  A1 SOI For at least one intercept correct or correct integration. Area = 5 4 A1 From intercepts: 1 5 5 1 2 2 4   or 1 5 1 4 4   , from lengths: 1 5 5 5 2 2 4    or by integration. 6

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dy 11 The equation of a curve is such that = 6x2 −30x + 6a, where a is a positive constant. The curve dx has a stationary point at a, −15 . (a) Find the value of a. 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(b) Determine the nature of this stationary point. 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(c) Find the equation of the curve. 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(d) Find the coordinates of any other stationary points on the curve. 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Mark scheme: 11(a) 2 6 30 6 0    a a a [ ⇒ 6 4 0]   a a B1 Sub  x a into d 0 d  y x . May see 2 5 0    a a a . a = 4 only B1 2 Question Answer Marks Guidance 11(b) 2 2 d 12 30 d   y x x or correct values of d d y x either side of 4  x M1 Differentiate d d y x (mult. by power or dec. power by 1) M0 if no values of d d y x , only signs. At 2 2 2 2 d d 4, 0 minimum or 18 minimum d d y y x x x     or concludes minimum from d d y x values A1 WWW A0 XP if 4  a obtained incorrectly in (a) Must see ‘minimum’. If M0, SC B1 for ‘minimum’ from d d y x sign diagram. 2 11(c)    y    3 2 6 30 6 3 2    x x their a x c B1 FT Expect   3 2 2 15 24    x x x c . B1 poss. even if uses ‘ a ’ – no value in (a) – max 1/3.       3 2 2 15 2 "4" 15 "4" 6 "4"      their their their c M1 Sub x = their"4", y = –15 into integral (must incl +c ) Look for –15 = 128 – 240 + 96 + c [⇒ c = 1]. 3 2 2 15 24 1     y x x x A1 Coefficients must be correct and simplified. Need to see ‘  y ’ or ‘  f  x ’ in the working. 3 11(d)    2 d 6 30 6 "4" 0 d     y x x their x If correct,     6 1 4 0    x x or     2 30 30 4 6 24 12    M1 OE Forming a 3-term quadratic using the given d d y x and solving by factorisation, formula or completing the square. Check for working in (b). Coordinates   1,1 2 A1 Allow 1, 12   x y (ignore 4  x if present). If M0, award SC B1 for   1,1 2 . 2

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12 y Q A P x O The diagram shows a circle P with centre 0, 2 and radius 10 and the tangent to the circle at the point A with coordinates 6, 10 . It also s ows a second circle Q with centre at the point where this tangent meets the y-axis and with radius 5 5. 2 (a) Write down the equation of circle P. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the equation of the tangent to the circle P at A. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Find the equation of circle Q and hence verify that the y-coordinates of both of the points of intersection of the two circles are 11. 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(d) Find the coordinates of the points of intersection of the tangent and circle Q, giving the answers in surd form. 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Mark scheme: 12(a) 2 2 2 100    x y 2 2 2 0 2 10     x y ISW. 1 12(b) Gradient of radius = 10 2 4 6 0 3          or gradient of tangent 3 4   M1 OE SOI Use coordinates to find gradient of radius or differentiate to find T m e.g. 2  d d 3 2 2 0 d d 4      y y x y x x at (6, 10)     1 2 2 2 d 1 3 2 100 100 2 d 2 4          y y x x x x . Equation of tangent is   3 3 29 10 6 4 4 2             y x y x A1 OE ISW Allow e.g. 58 4 . 2 12(c) Coordinates of centre of circle Q are 29 0, 2       their M1 SOI From a linear equation in (b). Equation of circle Q is 2 2 2 29 5 5 125 2 2 4                         x y their A1FT OE e.g.     2 2 0 14.5 31.25     x y ISW.   2 2 11 2 100     x 2 19  x and 2 2 29 125 11 2 4          x  2 19  x OR e.g.   2 2 125 29 2 100 25 275 11 4 2                y y y y B1 OE e.g.  2 2 19, 19 19     x x x Correct argument to verify both y -coords are 11 ISW. 3 Question Answer Marks Guidance 12(d) 2 2 2 2 3 29 29 125 25 125 20 4 2 2 4 16 4                     x x x x or   2 29 199 0    y y M1 Substitute equation of their tangent into equation of their circle. May see 2 31.25 14.5    y x . 2 5  x or 29 3 5 2   y A1 OE e.g. 20  x For 2 x-values or 2 y -values or correct   ,x y pair. 3 29 29 3 5 20 4 2 2                  y A1 OE e.g. 58 3 20 4 4  , 58 3 20 4 4  Correct   ,x y pairs. 3

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