Cambridge A Level Mathematics 9709 — 2019 Oct/Nov Paper 1 · Variant 2

4 A r cm O 21 rad T B The diagram shows a circle with centre O and radius r cm. Points A and B lie on the circle and angle AOB radians. The tangents to the circle at A and B meet at T. = 21 (i) Express the perimeter of the shaded region in terms of r and [3] 1. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) In the case where r 5 and 1.2, find the area of the shaded region. [4] = 1 = ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(i) Arc length AB = 2rθ B1 Tan θ = or AT BT r r → AT or BT = r tan θ B1 Accept or 2 2 cos r r θ    −           or sin sin 2 θ θ π   −     r NOT (90 – θ) P = 2rθ + 2r tan θ B1FT OE, FT for their arc length + 2 × their AT 3 Question Answer Marks Guidance 4(ii) Area ∆AOT = ½ × 5 × 5 tan 1.2 or Area AOBT = 2 × ½ × 5 × 5 tan 1.2 B1 Sector area = ½ × 25 × 2.4 (or 1.2) *M1 Use of ½r2θ with θ = 1.2 or 2.4. Shaded area = 2 triangles – sector DM1 Subtraction of sector, using 2.4 where appropriate, from 2 triangles Area = 34.3 (cm2) A1 AWRT Alternative method for question 4(ii) Area of ∆ ABT = ½ × (5 × tan 1.2)2 × sin(π – 2.4) (= 55.86) B1 Segment area = ½ × 25 × (2.4 – sin 2.4) (= 21.56) *M1 Use of ½r2 (θ – sin θ) with θ = 1.2 or 2.4 Shaded area = triangle – segment DM1 Subtraction of segment from ∆ ABT, using 2.4 where appropriate. Area = 34.3 (cm2) A1 AWRT 4

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7 G Q F D E P C B k j O i A The diagram shows a three-dimensional shape OABCDEFG. The base OABC and the upper surface DEFG are identical horizontal rectangles. The parallelograms OAED and CBFG both lie in vertical planes. Points P and Q are the mid-points of OD and GF respectively. Unit vectors i and j are parallel to −−→OA and −−→OC respectively and the unit vector k is vertically upwards. The position vectors of A, C and D are given by −−→OA 6i, −−→OC 8j and −−→OD 2i 10k. = = = + (i) Express each of the vectors −−→PB and −−→PQ in terms of i, j and k. 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(ii) Determine whether P is nearer to Q or to B. 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(iii) Use a scalar product to find angle BPQ. 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Mark scheme: 7(i) PB = 5i + 8j – 5k ( ) JJJG PQ = 4i + 8j + 5k B2,1,0 B2 all correct, B1 for two correct components. Accept column vectors. SC B1 for each vector if all components multiplied by –1. 4 7(ii) (Length of PB =) ( ) 2 2 2 5 8 5 + + = ( 114 ≈ 10.7) (Length of PQ =) ( ) 2 2 2 8 5 4 + + = ( 105 ≈ 10.2) M1 Evaluation of both lengths. Other valid complete comparisons can be accepted. P is nearer to Q. A1 WWW 2 7(iii) ( ) . JJJG JJJG PB PQ = 20 + 64 – 25 M1 Use of x1x2 + y1y2 + z1z2 on their JJJG PB and JJJG PQ ( 114)( 105)cos ( 59) = Their their BPQ their M1 All elements present and in correct places. BPQ = 57.4(°) or 1.00 (rad) A1 AWRT Calculating the obtuse angle and then subtracting gets A0. 3

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8 (a) Over a 21-day period an athlete prepares for a marathon by increasing the distance she runs each day by 1.2 km. On the first day she runs 13 km. (i) Find the distance she runs on the last day of the 21-day period. [1] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ (ii) Find the total distance she runs in the 21-day period. 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(b) The first, second and third terms of a geometric progression are x, x and x respectively. −3 −5 (i) Find the value of x. 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(ii) Find the fourth term of the progression. [2] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ (iii) Find the sum to infinity of the progression. 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Mark scheme: 8(a)(i) 21st term = 13 + 20 × 1.2 = 37 (km) B1 1 Question Answer Marks Guidance 8(a)(ii) S21= ½×21 × (26 + 20 × 1.2) or ½ × 21 × (13 + their 37) M1 A correct sum formula used with correct values for a, d and n. 525 (km) A1 2 8(b)(i) 3 5 3 − − = − x x x x oe (or use of a, ar and ar2) M1 Any valid method to obtain an equation in one variable. (a = or x =) 9 A1 2 8(b)(ii) r = 3 −       x x or 5 3 −     −   x x or 5 − x x = ⅔. Fourth term = 9 × (⅔)3 M1 Any valid method to find r and the fourth term with their a & r. 2⅔ or 2.67 A1 OE, AWRT 2 8(b)(iii) S∞ = 2 3 9 1 1 a r = − − M1 Correct formula and using their ‘r’ and ‘a’, with r <1, to obtain a numerical answer. 27 or 27.0 A1 AWRT 2

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10 y 4 y = 1 − 2 2x + 1 B x O A 4 The diagram shows part of the curve y 1 . The curve intersects the x-axis at A. The 2 = − 2x 1 + normal to the curve at A intersects the y-axis at B. dy (i) Obtain expressions for and y dx. [4] dx Ó ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the coordinates of B. 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(iii) Find, showing all necessary working, the area of the shaded region. 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Mark scheme: 10(i) [ ] 3 d 0 (2 1) d y x x −   = + +  × [+ 16] B2,1,0 OE. Full marks for 3 correct components. Withhold one mark for each error or omission. ∫ydx = [ ] [ ] 1 (2 1) 2 −   + + × +   x x (+c) B2,1,0 OE. Full marks for 3 correct components. Withhold one mark for each error or omission. 4 10(ii) At A, x = ½. B1 Ignore extra answer x = −1.5 d d y x = 2 → Gradient of normal ( ) ½ =− *M1 With their positive value of x at A and their dy dx , uses m₁m₂ = −1 Equation of normal: ( ) 0 ½ ½ − = − − y x or y − 0 = −½ (0 – ½) or 0 = −½×½ + c DM1 Use of their x at A and their normal gradient. B (0, ¼) A1 4 Question Answer Marks Guidance 10(iii) ( ) ( ) 1 2 2 0 4 1 d 2 1 − + ∫ x x *M1 d y x ∫ SOI with 0 and their positive x coordinate of A. [½ + 1] – [0 + 2] = (−½) DM1 Substitutes both 0 and their ½ into their ∫ydx and subtracts. Area of triangle above x-axis = ½ × ½ × ¼ 1 16   =     B1 Total area of shaded region = 9 16 A1 OE (including AWRT 0.563) Alternative method for question 10(iii) ( ) 0 1 3 2 1 1 d 2 (1 ) − − − ∫ y y *M1 d ∫x y SOI. Where x is of the form 1 2 1 ) −   − +       k y c with 0 and their negative y intercept of curve. [ ] 3 2 4 2   − −−+     = (½) DM1 Substitutes both 0 and their –3 into their ∫xdy and subtracts. Area of triangle above x-axis = ½ × ½ × ¼ 1 16   =     B1 Total area of shaded region = 9 16 A1 OE (including AWRT 0.563) Question Answer Marks Guidance Alternative method for question 10(iii) 1 2 0 1 1 d 2 4 − + − ∫ x y x *M1 ∫(their normal curve) with 0 and their positive x coordinate of A. Curve [½ + 1] – [0 + 2] = (−½) DM1 Substitutes both 0 and their ½ into their ∫ydx and subtracts. 1 2 0 1 1 d 2 4 − + ∫ x x = 2 4 4 − + x x = [ ] 1 1 – 0 16 8 −   +     1 16   =     B1 Substitutes both 0 and ½ into the correct integral and subtracts. Total area of shaded region = 9 16 A1 OE (including AWRT 0.563) 4

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