Cambridge A Level Mathematics 9709 — 2017 May/June Paper 1 · Variant 3

4 Relative to an origin O, the position vectors of points A and B are given by ` a ` a −−→ 5 −−→ 5 OA = 1 and OB = 4 . 3 −3 −−→ −−→ The point P lies on AB and is such that AP = 1 AB. 3 (i) Find the position vector of P. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the distance OP. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Determine whether OP is perpendicular to AB. Justify your answer. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(i) ( ) 5 5 0 4 1 3 3 3 6 OB OA AB           − = = − =           − −      uuur uuur uuur B1 5 0 5 1 1 3 2 3 3 6 1 OP         = + =         −     uuur M1 A1 If OP uuur not scored in (i) can score SR B1 if seen correct in (ii). Other equivalent methods possible Total: 3 4(ii) Distance OP = 2 2 2 5 2 1 + + = 30 or 5.48 B1 FT FT on their OP uuur from (i) Total: 1 4(iii) Attempt . . AB OP uuur uuur Can score as part of ( )( ) . cos AB OP AB OP θ = uuur uuur Rare ALT: Pythagoras 2 2 2 5 30 OP AP OA + = + = uuur uuur uuur M1 Allow any combination of . AB PO uuur uuur etc. and also if AP uuur or used PB uuur instead of AB uuur giving 2‒2 = 0 & 4‒4 = 0 respectively. Allow notation × instead of . . (0 + 6 ‒ 6) = 0 hence perpendicular. (Accept 90º) A1 FT If result not zero then 'Not perpendicular' can score A1FT if value is 'correct' for their values of , AB OP uuur uuur etc. from (i). Total: 2

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2 sin 1 + cos 1 5 (i) Show that the equation = 2 tan 1 may be expressed as cos21 = 2 sin21. [3] sin 1 + cos 1 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 2 sin 1 + cos 1 (ii) Hence solve the equation = 2 tan 1 for 0Å < 1 < 180Å. [3] sin 1 + cos 1 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(i) 2sin cos 2sin sin cos cos θ θ θ = + M1 Replace tan by sin / cos 2 2 2 2 2sin cos cos 2sin 2sin cos 2 c s θ θ θ θ θ θ + = + ⇒ = M1 A1 Mult by c(s + c) or making this a common denom.. For A1 simplification to AG without error or omission must be seen. Total: 3 5(ii) 2 2 2 tan 1/ 2 or cos 2 / 3 or sin 1/ 3 θ θ θ = = = B1 Use 2 2 tan / c or c s 1 s θ = + = and simplify to one of these results 35.3 or 1 44.7 θ = ° ° B1 B1 FT FT for 180 ‒ other solution. SR B1 for radians 0.615, 2.53 (0.196π, 0.804π) Extra solutions in range amongst solutions of which 2 are correct gets B1B0 Total: 3

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6 The line 3y + x = 25 is a normal to the curve y = x2 −5x + k. Find the value of the constant k. 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Mark scheme: 6 Gradient of normal is – 1/3 → gradient of tangent is 3 SOI B1 B1 FT FT from their gradient of normal. dy/dx = 2x – 5 = 3 M1 Differentiate and set = their 3 (numerical). x = 4 *A1 Sub x = 4 into line → y = 7 & sub their (4, 7) into curve DM1 OR sub x = 4 into curve → y = k ‒ 4 and sub their(4, k ‒ 4) into line OR other valid methods deriving a linear equation in k (e.g. equating curve with either normal or tangent and sub x = 4). k = 11 A1 Total: 6

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7 C 8 cm 10 cm A B D The diagram shows two circles with centres A and B having radii 8 cm and 10 cm respectively. The two circles intersect at C and D where CAD is a straight line and AB is perpendicular to CD. (i) Find angle ABC in radians. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the area of the shaded region. 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Mark scheme: 7(i) sin 8 /10 0.927 3 ABC ABC = → = B1 Total: 1 7(ii) ( ) 6 Pythagoras 8 6 48.0 AB BCD = → ∆ = × = M1A1 OR 8×10sin0.6435 or ½×10×10sin((2)×0.927)=48. 24or 40or80 gets M1A0 Area sector ( ) 2 ½ 10 2 0.9273 BCD their = × × × *M1 Expect 92.7(3). 46.4 gets M1 Area segment = 92.7(3) – 48 *A1 Expect 44.7(3). Might not appear until final calculation. Area semi-circle ‒ segment = ( ) 2 ½ 8 92.7 48 their π × × − − DM1 Dep. on previous M1A1 OR ( ) 2 2 8 ½ 8 44.7 . their π π × − × × + Shaded area = 55.8 – 56.0 A1 Total: 6

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10 (a) y y = h y = x2 −1 x O Fig. 1 Fig. 1 shows part of the curve y = x2 −1 and the line y = h, where h is a constant. (i) The shaded region is rotated through 360Å about the y-axis. Show that the volume of revolution, V, is given by V = 0 12h2 + h . [3] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ (ii) Find, showing all necessary working, the area of the shaded region when h = 3. 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(b) h Fig. 2 Fig. 2 shows a cross-section of a bowl containing water. When the height of the water level is h cm, the volume, V cm3, of water is given by V = 0 12h2 + h . Water is poured into the bowl at a constant rate of 2 cm3 s−1. Find the rate, in cm s−1, at which the height of the water level is increasing when the height of the water level is 3 cm. 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Mark scheme: 10(a)(i) Attempt to integrate 1 d V y y π = ∫ + M1 2 1 ½ h h h ∫ + = + is M0. Use of 2d y x ∫ is M0 ( ) 2 2 y y π   = +     A1 2 2 h h π   = +     A1 AG. Must be from clear use of limits 0→ h somewhere. Total: 3 10(ii) ( ) 1/2 1 d y y ∫ + ALT 6 ‒ ( ) 2 1 d x x ∫ − M1 Correct variable and attempt to integrate ( ) 3/2 1 y + ⅔ oe ALT 6 ‒ ( 3x x − ⅓ ) CAO *A1 Result of integration must be shown [ ] 8 1 − ⅔ ALT 8 1 6 [ 1 1 3 3     − − − −         ] DM1 Calculation seen with limits 0→3 for y. For ALT, limits are 1→2 and rectangle. 14/3 ALT 6 ‒ 4/3 = 14/3 A1 16/3 from 8 × ⅔ gets DM1A0 provided work is correct up to applying limits. Total: 4 Question Answer Marks Guidance 10(b) Clear attempt to differentiate wrt h M1 Expect ( ) d 1 d V h h π = + . Allow h + 1. Allow h. Derivative = 4π SOI *A1 2 derivative their . Can be in terms of h DM1 2 1 or or 0.159 4 2 π π A1 Total: 4

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