2012 HSC Mathematics 'Sample Answers' - Board Of Studies

2012 HSC Mathematics‘Sample Answers’When examination committees develop questions for the examination, they maywrite ‘sample answers’ or, in the case of some questions, ‘answers could include’.The committees do this to ensure that the questions will effectively assess students’knowledge and skills.This material is also provided to the Supervisor of Marking, to give some guidanceabout the nature and scope of the responses the committee expected students wouldproduce. How sample answers are used at marking centres varies. Sample answers maybe used extensively and even modified at the marking centre OR they may be consideredonly briefly at the beginning of marking. In a few cases, the sample answers may not beused at all at marking.The Board publishes this information to assist in understanding how the markingguidelines were implemented.The ‘sample answers’ or similar advice contained in this document are not intendedto be exemplary or even complete answers or responses. As they are part of theexamination committee’s ‘working document’, they may contain typographical errors,omissions, or only some of the possible correct answers.–1–

2012 HSCMathematicsSample AnswersSection IIQuestion 11 (a)2x 2 7x 3 ( 2x 1)( x 3)Question 11 (b)3x 1 2 2 3x 1 2 1 3x 3Hence 1 x 13Question 11 (c)y x2,dy 2xdxslope of tangent at x 3 is 2 3 6 6 y 32 y 9 x 3x 3Hence the equation of the tangent is y 6 ( x 3) 9 6x 9Question 11 (d)y′ 5 ( 3 e 2x ) 2e 2x4 10e 2x ( 3 e 2x )4–2–

2012 HSCMathematicsSample AnswersQuestion 11 (e)x 2 16 ( y 2 ) 4.4 ( y 2 ) , so a 4the vertex is at ( 0, 2 ) , so the focus is at ( 0, 2 4 ) ( 0, 6 )Question 11 (f)Area of the sector is given by A θ 2r2ie 50 θ 2r2 θ 2 62 θ 362 18θ θ 5018now, l rθ 6 length of arc 501850cm3Question 11 (g)π 2π xx 2 sec 2 dx 2 tan 22 0 0 2 tanπ 2 tan 04 2 1 0 2–3–

2012 HSCMathematicsSample AnswersQuestion 12 (a) (i)y′ loge x ( x 1) loge x 1 1x1xQuestion 12 (a) (ii)y′ x 2 sin x 2x cos xx4 ( x sin x 2 cos x )x3Question 12 (b) 4x 2xdx 2 2dx 2 x 6 x 6 2 loge ( x 2 6 ) CQuestion 12 (c) (i)Every row has two tiles more than the previous row and the first row has three tiles.It is an arithmetic sequence and T20 3 19 2 41ie There are 41 tiles in row 20.–4–

2012 HSCMathematicsSample AnswersQuestion 12 (c) (ii)The number of tiles for the 20 rows is(203 T202 10 ( 3 41) 440S20 )Question 12 (c) (iii)We want()n3 Tn 200,2where Tn 3 2 ( n 1) 2n 1.Hencen(3 2n 1) 2002ie n ( n 2 ) 200n 2 2n 200 0 n 2 4 8002 1 201 13.1774Hence Jay can make 13 complete rows.Question 12 (d) (i)()If ƒ x denotes depth at distance x from the river bank then, by Simpson’s rule, theapproximate area is:A 3( f ( 0 ) 4 f (3) 2 f (6) 4 f (9) f (12))3 1( 0.5 4 2.3 2 2.9 4 3.8 2.1) area 32.8 m 2–5–

2012 HSCMathematicsSample AnswersQuestion 12 (d) (ii)Volume through the cross-section in 10 seconds is(32.8 0.4 10 ) m 3 131.2 m 3Question 13 (a) (i)Coordinates of A : y 0, 2x 8, sox 4ie A ( 4, 0 )coordinates of B : x 0, y 8ie B ( 0, 8 )By Pythagoras’ theorem AB 4 2 82 80 4 5Question 13 (a) (ii)By the cosine ruleAC 2 AB 2 BC 2 2 AB BC cos ( ABC )25 80 65 2 4 5 65 cos ( ABC )40 13 cos ( ABC ) 120cos ( ABC ) Hence313 ABC 33.69 ,so the angle is 34 to the nearest degree.–6–

2012 HSCMathematicsSample AnswersQuestion 13 (a) (iii)Slope of AB is 2, so the slope of CN isEquation of CN is1.21 y 4 ,2 x 7so y 1( x 7) 42x 1 .2 2The coordinates of N are obtained by the intersection of AB and CN:y 2x 8 11x 22 4x 16 x 115 5x3 xSince y 2x 8y 6 8 2Hence N has coordinates ( 3, 2 )Question 13 (b) (i)For the x-coordinate at the intersection of the two parabolasx 2 3x 5x x 22x 2 8x 02x ( x 4 ) 0x 0,4so x 4 is the x-coordinate of point A.–7–

2012 HSCMathematicsSample AnswersQuestion 13 (b) (ii)Area is given by4 22 ( 5x x ) ( x 3x ) dx 04 8x 2x 2 dx 04 2 4x 2 x 3 3 0 64 1283 192 1283area 64units23Question 13 (c) (i)3 39 5 7 35Question 13 (c) (ii)Complement of (i) : 1 926 35 35Question 13 (c) (iii)Probability of 2 red probability of 2 white:9 2 4 17 35 5 7 35–8–

2012 HSCMathematicsSample AnswersQuestion 14 (a) (i)ƒ ′ ( x ) 12x 3 12x 2 24x 12x ( x 2 x 2 ) 12x ( x 2 )( x 1)Hence the stationary points are at x 0, x 1, x 2Now ƒ ( 0 ) 0, ƒ (1) 3 4 12 and ƒ ( 2 ) 316 48 124 5 32Hence the statinary points are( 2, 32) , ( 0, 0 ) , (1, 5)Now ƒ ′′ ( x ) 12 ( 3x 2 2x 2 )ƒ ′′ ( 0 ) 24 0, so ( 0, 0 ) is a maximumƒ ′′ (1) 12 3 0, so (1, 5) is a minimumƒ ′′ ( 2 ) 12 6 0, so ( 2, 32 ) is a minimum–9–

2012 HSCMathematicsSample AnswersQuestion 14 (a) (ii)Question 14 (a) (iii)ƒ ( x ) is increasing for 2 x 0 or for x 1Question 14 (a) (iv)k is the vertical shift of the graph of f .To make sure the equation has no solution (ie the new graph should not cut the x-axis)move the graph up by the smallest minimum, so k 32.– 10 –

2012 HSCMathematicsSample AnswersQuestion 14 (b) Volume is given by V π y 2 dx 1 9V π dx4 0 ( x 2 )1 4 9π ( x 2 ) dx 01 9π 3 ( x 2) 3 0 3π ( 3 3 2 3 ) 1 1 3π 27 8 Volume 19πunits372Question 14 (c) (i)N ( 20 ) 1000e 20k 2000e 20k 220k ln 2ln 2k 0.034720Question 14 (c) (ii)N (120 ) 1000e120k 1000 e120 0.0347Number of bacteria 64 328– 11 –

2012 HSCMathematicsSample AnswersQuestion 14 (c) (iii)dN kN , so from (ii)dtdN 0.0347 64 328 2232 when t 120dtrate of change 2232 bacteria/minuteQuestion 14 (c) (iv)At t 0N 1000Find t so thatHence100 000 1000e kt100 e ktln100 ktln100t kln100 0.0347time 132.7 minutes– 12 –

2012 HSCMathematicsSample AnswersQuestion 15 (a) (i)Length in cm is10 10 0.96 10 0.962 10 0.969 10 (1 0.96 10 0.969 ) 1 0.9610 10 1 0.96 83.79Question 15 (a) (ii)()Since 0.96 1 the limiting sum 10 1 0.96 0.962 exists. 1 10The limiting sum is 10 250 1 0.96 0.04As 250 cm 300 cm, a strip of length 3 m is sufficient.Question 15 (b) (i)Initial velocity is x ( 0 ) 1 2 cos 0 1 m/sQuestion 15 (b) (ii)x 2sin t 0 if t 0, π , 2π , The first maximum velocity is at t πx (π ) 1 2 cos π 3 m/s– 13 –

2012 HSCMathematicsSample AnswersQuestion 15 (b) (iii) x x dt 1 2 cost dt t 2sin t CWe are given that x ( 0 ) 3, sox ( 0 ) 2sin 0 C 3Hence C 3 and the displacement isx t 2sin t 3Question 15 (b) (iv)The particle is at rest if x 0, sox 1 2 cost 01ie cost 2πHence t 3 π π πThe displacement is x 2sin 3 metres 3 3 3 π 3 3 metres 3 Question 15 (c) (i)A2 360 000 (1 0.005) M (1 0.005) M 360 000 (1.005) M (1 1.005)2– 14 –

2012 HSCMathematicsSample AnswersQuestion 15 (c) (ii)Generalising from (i)An 360 000 (1.005) M (1 1.005 1.005n 1 )n(1.005n 1)n() 360 000 1.005 M(1.005 1)We require A300 0, so360 000 (1.005)300 M360 000 (1.005)300M (1.005300(1.005300 1)(1.005 1)0.005 1) 2319.50Question 15 (c) (iii)We want to find the smallest n so that An 180 000360 000 (1.005) Mn(1.005n 1)0.005 180 000360 000 (1.005) 463 900 (1.005n 1) 180 000n103 900 (1.005) 283 900n(1.005)n 2.7324Hence n log 2.7324 201.5log1.005After 202 months An will be less than 180 000 for the first time.– 15 –

2012 HSCMathematicsSample AnswersQuestion 16 (a) (i)EFCD since CDEF is a rhombusEDFC since CDEF is a rhombus( corresponding angles, EF CA)( corresponding angles, ED BC ) FEB DAE FBE DEAHence EBF is similar to AED since two (and therefore all) angles are equal.Question 16 (a) (ii)Using that EBF is similar to AED,xb x ( corresponding sides of similar triangles)a xxx 2 ( b x )( a x )x 2 ba ax bx x 20 ba x ( a b )abx a bQuestion 16 (b) (i)T has coorindates ( cosθ , sin θ )The line OT has slopesin θcosθHence the line PT perpendicular to OT has slope Hence the equation of PT is: cosθy sin θ sin θx cosθ x cosθ cos2 θ ysin θ sin 2 θx cosθ ysin θ cos2 θ sin 2 θ 1– 16 –cosθand passes through T.sin θ

2012 HSCMathematicsSample AnswersQuestion 16 (b) (ii)Q is the point of intersection of the line y 1 with the line from (i).Hence the x-coordinates of Q satisfiesx cosθ 1sin θ 11 sin θx cosθ1 sin θThe length of BQ iscosθQuestion 16 (b) (iii)Area of trapezium is given byA 1OB (OP BQ )2P is on the line x cosθ ysin θ 1 with y 0,so x 1cosθie OP 1cosθOB 1 and from (ii) BQ A 1 sin θcosθ1 11 sin θ 2 cosθcosθ 1 2 sin θ 2 cosθ 2 sin θ2 cosθ– 17 –

2012 HSCMathematicsSample AnswersQuestion 16 (b) (iv)Differentiate area with respect to θ : 2 sin θ 2 cosθ dAd dθ dθ cos2 θ ( 2 sin θ ) sin θ2 cos2 θ2sin θ ( cos2 θ sin 2 θ )2 cos2 θ2sin θ 12 cos2 θNeed to solve 2sin θ 1 0ie sin θ 12Hence θ If θ πis a critical point6πthen the area of the trapezium becomes very large: A 2If θ 0, thendA1 0, so the area is decreasing.dθ2As there is only one stationary point it must be minimum.Hence θ πgives the minimum area.6– 18 –

2012 HSCMathematicsSample AnswersQuestion 16 (c) (i)Find the points of intersection of the parabola y x 2 and a circle x 2 ( y c ) r 2 :2y ( y c) r 22y y 2 2cy c 2 r 2y 2 (1 2c ) y c 2 r 2 0The circle is tangent if there is precisely one solution, so the discriminant has to vanish.(1 2c )2 4 (c2 r 2 ) 0(1 2c )2 4 (c2 r 2 )1 4c 4c 2 4c 2 4r 24c 1 4r 2 as requiredQuestion 16 (c) (ii)y must be positive to be a solution since the circle is inside the parabola.As the discriminant is zero1(1 2c ) 02so 1 2c 01 c211If c there is only one point, so c .22y – 19 –

'Sample Answers' When examination committees develop questions for the examination, they may write 'sample answers' or, in the case of some questions, 'answers could include'. The committees do this to ensure that the questions will effectively assess students' knowledge and skills.