Cambridge A Level Mathematics 9709 — 2021 Oct/Nov Paper 1 · Variant 2

3 1 Solve the equation 2 cos 1 = 7 − for −90Å < 1 < 90Å. [4] cos 1 ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 [ ] 2 2cos 7cos 3 0 θ θ − + = M1 Forming a 3-term quadratic expression with all terms on the same side or correctly set up prior to completing the square. Allow ± sign errors. ( )( ) 2cos 1 cos 3 0 θ θ − − = DM1 Solving their 3-term quadratic using factorisation, formula or completing the square. 1 [cos or cos 3 2 θ θ = = leading to] 60 or 60 θ θ = − ° = ° A1 60 60 θ θ = − ° = ° and A1 FT FT for ± same answer between 0° and 90° or 0 and π 2 . π or 1.05 3 ± ± AWRT scores maximum M1M1A0A1FT. Special case: If M1 DM0 scored then SC B1 for θ 60 60 θ = − ° = ° or , and SC B1 FT can be awarded for ( ± their 60 )° . 4

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5 The first, third and fifth terms of an arithmetic progression are 2 cos x, −6 3 sin x and 10 cos x respectively, where 2π1 < x < π. (a) Find the exact value of x. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Hence find the exact sum of the first 25 terms of the progression. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) 6 3sin 2cos 10cos 6 3sin x x x x − − = + leading to 12 3sin 12cos x x   − =   OR [(1st term + 5th term) = 2 × 3rd term leading to…] 12cos 12 3sin x x = − *M1 OE. From the given terms, obtain 2 expressions relating to the common difference of the arithmetic progression, attempt to solve them simultaneously and achieve an equation just involving sinx and cosx. Elimination of sinx and cosx to give an expression in tanx 1 tan 3 x   = −     DM1 For use of sin tan cos x x x = [ ]5π 6 x = only A1 CAO. Must be exact. 3 5(b) d = 2cosx or d = 2cos(their x) B1 FT Or an equivalent expression involving sinx and cosx e.g. ( ) ( ) 3 3sin cos their x their x − − 3   = −   FT for their x from (a) only. If not ± 3 , must see unevaluated form. S25 ( ) ( ) ( ) ( ) ( ) 25 2 2cos 25 1 2 their x their d = × + − × ( ) ( ) ( ) 12.5 2 3 24 3   = × − + −     M1 Using the correct sum formula with 25 2 , (25 — 1) and with a replaced by either 2(cos(their x)) or ± 3 and d replaced by either 2(cos(their x)) or ± 3 . 325 3 − A1 Must be exact. 3

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6 The second term of a geometric progression is 54 and the sum to infinity of the progression is 243. The common ratio is greater than 2.1 Find the tenth term, giving your answer in exact form. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 6 or 54 and 243 1 a their a ar r = = − B1 SOI ( ) 54 243 1 r r = − leading to [ ] 2 243 243 54 0 r r − + = [ 2 9 9 2 0] r r − + = OR [ ] 2 243 13122 0 a a − + = *M1 Forming a 3-term quadratic expression in r or a using their 2nd term and S∞. Allow ± sign errors. ( )( )[ ] 3 2 3 1 0 k r r − − = OR ( )( )[ ] 81 162 0 a a − − = DM1 Solving their 3-term quadratic using factorisation, formula or completing the square. If factorising, factors must expand to give ±their coefficient of 2 r . 2 54 3 their a   ÷ =     OR ( ) 54 81 their r ÷ = DM1 May be implied by final answer. Tenth term = 9 8 512 2 2 OR 81 OR 54 243 3 3       × ×               A1 OE. Must be exact. Special case: If B1M1DM0DM1 scored then SC B1 can be awarded for the correct final answer. 5

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7 15cm P B Q A 9 cm 15 cm C In the diagram the lengths of AB and AC are both 15 cm. The point P is the foot of the perpendicular from C to AB. The length CP = 9 cm. An arc of a circle with centre B passes through C and meets AB at Q. (a) Show that angle ABC = 1.25 radians, correct to 3 significant figures. 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(b) Calculate the area of the shaded region which is bounded by the arc CQ and the lines CP and PQ. 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Mark scheme: 7(a) EITHER By using trigonometry: ˆ BAC = 0.6435… and ˆ ABC = π 0.6435 2 − OR By Pythagoras: AP = 12 ⇒ BP = 3 so tan ˆ ABC = 9 3 OR Using ∆PBC and either the sine or cosine rule 3 sin 10 ˆ ABC = or 10 cos 1 ˆ 0 ABC = M1 3 0.9486 10 = … 10 0.3162 10 = … ˆ ABC = π 0.6435 2 − or tan-1 9 3 or 1 3 sin 10 − or 1 10 cos or 10 − ( ) 1 .249 04 or 71.56 … ° = 1.25 radians (3 sf) A1 AG. Final answer must be 1.25, more accurate value 1.24904… with no rounding to 3sf seen as the final answer gets M1A0. If decimals are used all values must be given to at least 4sf for A1. 2 7(b) BC = ( ) 2 2 3 9 their + or 9 sin1.25 [= 90 , 3 10 or 9.48697…] M1 Using correct method(s) to find BC. Area of sector = ( ) [ ] 2 1 1 tan 3 56.207 56.25 2 their BC or − × × = M1 Using 1 tan 3 or 1.25 − and their BC, but not 9 or 15, in correct area of sector formula. Area of triangle PBC = 13.4 to 13.6 or 1 9 3 2 × × B1 [Area = (56.207 or 56.25) – their 13.5 =] 42.7 or 42.8 A1 AWRT 4

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9 The volume V m3 of a large circular mound of iron ore of radius r m is modelled by the equation V = 3 r −1 3 −1 for r ≥2. Iron ore is added to the mound at a constant rate of 1.5 m3 per second. 2 2 (a) Find the rate at which the radius of the mound is increasing at the instant when the radius is 5.5 m. 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(b) Find the volume of the mound at the instant when the radius is increasing at 0.1 m per second. 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Mark scheme: 9(a) d d V r   =     2 9 1 2 2 r   −     B1 OE. Accept unsimplified. 2 d d d 1.5 1.5 1.5 d d d d 112.5 9 1 5.5 d 2 2 r r V V t V t their r       = × = = =     −         M1 Correct use of chain rule with 1.5, their differentiated expression for d d V r and using 5.5 r = . 0.0133 or 3 225 or 1 [ 75 metres per second] A1 3 9(b) d d 1.5 or or 1 5 d d 0.1 V V their r r = OR 1.5 0.1 d d V their r = 2 2 1.5 OE 1 9 2 r     ×   =     −         B1 FT Correct statement involving d d V r or their d d V r , 1.5 and 0.1. 2 9 1 15 2 2 r     − =            r = 1 10 2 3 + B1 OE e.g. AWRT 2.3 Can be implied by correct volume. [Volume =] 8.13 AWRT B1 OE e.g. 3 5 30 3 −+ . CAO. 3

More questions on Differentiation

11 y 1 7 1 y = 2x + 10 − 1 3 x −2 A 3, 65 x O 5 2 1 1 and the normal to the curve The diagram shows the line x = 52, part of the curve y = 12x + 107 − x −2 3 at the point A 3, 6 . 5 (a) Find the x-coordinate of the point where the normal to the curve meets the x-axis. 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(b) Find the area of the shaded region, giving your answer correct to 2 decimal places. 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Mark scheme: 11(a) ( ) 4 3 d 1 1 d 2 3 2 y x x = + − B1 OE. Allow unsimplified. Attempt at evaluating their d d y x at x = 3 ( ) 4 3 1 1 5 2 6 3 3 2     + =   −     *M1 Substituting x = 3 into their differentiated expression – defined by one of 3 original terms with correct power of x. Gradient of normal = 1 dy their dx − 6 5   = −     *DM1 Negative reciprocal of their evaluated d d y x . Equation of normal ( )( ) 6 normal gradient 3 5 y their x − = − 6 4.8 5 6 24 5 y x y x   = − +  = − +     DM1 Using their normal gradient and A in the equation of a straight line. Dependent on *M1 and *DM1. [When y = 0,] x = 4 A1 or (4, 0) 5 Question Answer Marks Guidance 11(b) Area under curve = ( ) [ ] 1 3 1 7 1 d 2 10 2 x x x      + −     −   M1 For intention to integrate the curve (no need for limits). Condone inclusion of π for this mark. ( ) 2 3 2 3 2 1 7 4 10 2 x x x − + − A1 For correct integral. Allow unsimplified. Condone inclusion of π for this mark. 2 3 9 3 6.25 3 0.5 2.1 1.75 4 2 4 2   ×     + − − + −           M1 Clear substitution of 3 and 2.5 into their integrated expression (with at least one correct term) and subtracting. 0.48[24] A1 If M1A1M0 scored then SC B1 can be awarded for correct answer. [Area of triangle =] 0.6 B1 OE [Total area =] 1.08 A1 Dependent on the first M1 and WWW. 6

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12 y B P A x O R Q The diagram shows the circle with equation x2 + y2 −6x + 4y −27 = 0 and the tangent to the circle at the point P 5, 4 . (a) The tangent to the circle at P meets the x-axis at A and the y-axis at B. Find the area of triangle OAB, where O is the origin. 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(b) Points Q and R also lie on the circle, such that PQR is an equilateral triangle. Find the exact area of triangle PQR. 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Mark scheme: 12(a) Centre is (3, – 2) B1 Gradient of radius = ( ) ( ) 2 4[ 3 5 their their − − = − 3] *M1 Finding gradient using their centre (not (0, 0)) and P (5,4). Equation of tangent ( ) 1 4 5 3 y x − = − − DM1 Using P and the negative reciprocal of their gradient to find the equation of AB. Sight of [x =]17 and [y =] 17 3 A1 1 17 289 Area 17 2 3 6   = × × =     A1 Or 48 1 6 or AWRT 48.2. Alternative method for question 12(a) d d 2 2 6 4 0 d d y y x y x x + − + = B1 At P: d d d 1 10 8 6 4 0 d d d 3 y y y x x x   + − + =  =−     *M1 Find the gradient using P (5,4) in their implicit differential (with at least one correctly differentiated y term). Equation of tangent ( ) 1 4 5 3 y x − = − − DM1 Using P and their value for the gradient to find the equation of AB. Sight of [x =]17 and [y =] 17 3 A1 1 17 289 Area 17 2 3 6   = × × =     A1 Or 48 1 6 or AWRT 48.2. Question Answer Marks Guidance 12(a) cont’d Alternative method for question 12(a) ( ) ( ) ( )( ) 1 1 2 2 2 2 d 2 40 3 OE leading to 3 31 6 d y y x x x x x −   = −± − − = − + −       B1 OE. Correct differentiation of rearranged equation. ( ) ( ) ( ) ( ) 1 2 2 d d 1 3 5 31 6 5 5 d d 3 y y x x −  = − + −  =−     *M1 Find the gradient using x = 5 in their differential (with clear use of chain rule). Equation of tangent ( ) 1 4 5 3 y x − = − − DM1 Using P and their value for the gradient to find the equation of AB. Sight of [x =]17 and [y =] 17 3 A1 1 17 289 Area 17 2 3 6   = × × =     A1 Or 48 1 6 or AWRT 48.2. 5 Question Answer Marks Guidance 12(b) Radius of circle = 40 , B1 Or 2 10 or 6.32 AWRT or 2 40 r = . Area of ∆CRQ = 2 1 1 3 ( ) sin120 40 2 2 2 their r   × = × ×     OR Area of ∆CQX = 1 2 40cos30 40cos60 × × OE 1 30 10 2   = × ×     OR Area of circle ‒ 3× Area of segment = 40π ‒ 3 × (40 π 3 ‒ 10 3) OR 120 or 2 30 QR = and area = 2 1 sin60 2 QR M1 Using 2 1 sin 2 r θ with their r and 120 or 60 [ 3 × ] Using 1 2 base height × × in a correct right-angled triangle [ 6 × ]. Use of cosine rule and area of large triangle 30 3 A1 AWRT 52[.0] implies B1M1A0. 3 See diagram for points stated in ‘Answer’ column. C X

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