St Stithians College



Draft 4: 4 April 2018VCAAMATHEMATICS PAPER IIMARKING GUIDELINES SECTION AQUESTION 1(a)In the diagram below E(1;5) is the point of intersection of the diagonals of the parallelogram ABCD.783590187960ABCDExy00ABCDExyB 8;5and C(9;2) are given.(1)Determine the co-ordinates of points A and D.xA+92=1yA+22=5xA=-7yA=8A-7;8D-6;5(6)(2)Determine the length of BC.BC=12+9=10(3)(3)Calculate the size of the angle BCA.mAC=6-16=-38∠inc AC=159,4omAD=-3∠inc AD=108,4o∴DAE=BCA=51o alt l’s AD||BC(5)ORmBC=-3∠inc BC=108,4omAC=6-16=-38∠inc AC=159,4o∴BCA=51o (5)(b)In the diagram, a circle with centre M(-3;5) is drawn. 647065335915AM -3;5 .Oyx00AM -3;5 .OyxThe line with equation 5x-7y=24 is a tangent to the circle at A.(1)Determine the coordinates of A.Equation MA:y-5=-75x+35y-25=-7x-217x+5y=4 ……equn 15x-7y=24….. equn 2Rewriting:35x+25y=2035x-49y=168∴y=-2; x=2A2; -2(7)(2)Determine the equation of the circle.x+32+y-52=r2Sub A2; -22+32+-2-52=r225+49=r2x+32+y-52=74(3)[24]QUESTION 2(a)Use the diagram below to prove the theorem: The line drawn from the centre of a circle perpendicular to a chord, bisects that chord.1397635278765AOCB00AOCBConstruction:OB and OA radiiRTP:AC=CBProof:In ?OAC and ?OBCOA=OBRadiiOC is commonC1=C2 Right angles given∴ ?OAC ≡ ?OBCR,H,S∴AC=CB ?OAC ≡ ?OBC(4)282321025146000(b)In the diagram below, the radii of two concentric circles (having the same centre), are 18 cm and 30 cm.Tangent PR touches the smaller circle at Q.ST=RQ-4. Determine QR and hence the length SR.Construction: OQ (O is centre of circle)OV (V is midpoint of TS)OROSProof:In ?QORQ=90oRadius ? Tangent302=182+RQ2PythagorasQR=24∴ST=20 and VS=10In ?OVSV=90oChord ? Radius at midpoint182=102+OV2PythagorasOV2=224In ?OVR302=224+VR2PythagorasVR=26∴SR=16(6)[10]QUESTION 3(a)Without the use of a calculator, determine the value of cos (90o+x)346138524765000 if 6tanx+8=0; 180o<x<360o.tanx=-864th Quadrant (diag)r=10 (Pythag)cos (90o+x)=-sinx=--810=45(5)(b)Express sinx+sin2x1+cosx+cos2x as a single trigonometric ratio.sinx+sin2x1+cosx+cos2x=sinx+2sinxcosx1+cosx+2cos2x-1 =sinx1+2cosxcosx1+2cosx=tanx(5)(c)Determine the general solution of siny?cosy=-25 .siny?cosy=-252siny?cosy=-45sin2y=-452y=-53,1o+k360oor2y=233,1o+k360o k?Zy=-26,6o+k180oory=116,6o+k180o (5)(d)Determine the maximum and minimum value of 4sinx - 9Max value of sinx is 1, and min value -1∴ max value: 4-1 - 9=-25∴ min value: 41 - 9=-12(4)[19]QUESTION 4In the diagram below the functions p(x)=asinx+b and q(x)=coscx are sketched for the domain -100o<x<200o.879423674(a)Determine the values of a; b and c.a=2b=1c=360o120o=3(4)(b)How many points of intersection will px and qx have for the interval 0o<x<360o?3(1)[5]QUESTION 5What is parkrun?It is a 5km run - it's you against the clock.When is it?Every Saturday at 8:00am.How fast do I have to be?We all run for our own enjoyment. Please come along and join in whatever your pace!Bryanston is an affluent residential suburb in Sandton, Johannesburg.Delta Park is situated in Victory Park, a leafy suburb of Johannesburg. It is described as one of the oldest and more trendy suburbs.Ballito is a holiday town located on the Dolphin Coast in Kwa-Zulu Natal. As well as being a thriving community there are a number of retirement villages.-3453411425100-34534114251004826586211553123126542545Bryanston Ballito Delta00Bryanston Ballito DeltaThe above box and whisker plots represent the times (in minutes) that participants completed the Parkrun in. The following additional information was calculated.BryanstonBallitoDeltan11733241224Q2 (median)38,0342,4843,25x39,0042,4243,46σ10,2311,6311,55(a)What percentage of people participating in the Bryanston parkrun,run longer than 30 minutes?75%(1)(b)How many people ran faster than 43,25 mins at Delta?0,5×1224=612(2)(c)Suggest a reason for the large range (Max – Min) at the Ballito Parkrun.Young members of community runningOlder members (lots of them) walking(2)(d)A runner has a personal best of 20,00 minutes. At which run would heget the best place finish? Justify your answer.Ballito because closest to the minimum value/fewer entries/fewer fast runners(2)A measure of skewness can be calculated using the following formula:Skewness=3(x-Q2)σ(e)Using the formula given above, determine if the data at the Delta Parkrunis more skew than the data at the Bryanston Parkrun.Suggest a reason for your findings.Delta:Skewness = 0,0545Bryanston:Skewness = 0,28445Bryanston’s data is more skewed to the rightAffluent area – afford gym fees and time out to get fit.(3)[10]QUESTION 6During ‘The Championship’ held at Wimbledon in 2017 the following statistics were recorded for the Women’s competition:RoundR1R2R3R4Quarter FinalSemi FinalNumber of games643216842Number of players (x)12864321684Average number of unforced errors per player (y)22,521,921,321,017,911,0An ‘unforced error’ is a missed shot or lost point (as in tennis) that is entirely a result of the player's own blunder and not because of the opponent's skill or effort(a)Write down the equation of the regression line of ‘Average number of unforced errors per player (y)’ on ‘Number of players (x)’ correct to 3 decimal places.y=0,056x+16,911(3)(b)Interpret the gradient of the regression line obtained in (a).The gradient is positive – as the number of players in the tournament decreases, the number of unforced errors decrease The better players are left; they make fewer errors.(1)(c)Use your equation to determine the expected number of unforced errorseach player will make in the final game of the tournament, correct to 3 decimal places.y=0,0562+16,911=17,023OR(1)y=17,024(d)The correlation coefficient for the above data is 0,611. Comment on thereliability of your prediction in (c) above. Not a very strong correlation, not very reliable/fairly reliable.(1)(e) The actual average number of unforced errors per player in the finalwas 18. How would it influence or change the regression line?Make it less steep and increase the y-intercept(2)(f)With justification state which game you would rather have watched. The semi-final would have been a thrilling game with tacticsand ‘winners’ being the deciding factors.(1)[9]5400675889077 marks0077 marksSECTION BQUESTION 7(a)If sin22o=a, express the following in terms of a: cos(-26o)?sin318o+cos402o?sin(-154o)=cos26o?-sin42o+cos42o?-sin26o=-sin42o?cos26o-sin26o?cos42o=-sin42o?cos26o+sin26o?cos42o=-sin68o=-cos22o=-1-a2(5) (b)In the diagram below ?RST is an isosceles triangle with RS=RT.291846061595R=36o and TP=TS=1 unit.(1)Prove that RT=2cos36o.RTsin108o=TPsin36oRT=sin108osin36oRT=sin72osin36oRT=2sin36o ?cos36osin36o∴RT=2cos36o(4)(2)Given that: ?RTS///?TSP: prove that 4cos236o-2cos36o=1RTTS=RSTP=TSSPRT1=RS1=1SP∴RT.SP=1RTRS-RP=1 RTRT-RP=1RS=RT isos triangleRT2-RT.RP=12cos36o2-2cos36o=1(6)[15]QUESTION 8394335475615 A man standing on top of a tower sees a car coming directly toward the tower. The car is traveling at a constant speed. It takes 20 minutes for the angle of depression to change from 30o to 60o. 4648200146685T00TDetermine the time remaining for the car to reach the tower.Inserting angles and time onto diagramQPsin30o=20sin30o∴QP=20or QP=QT Isos trianglecos60o=RQPQ=RQ20∴RQ=10 10 minutes remaining.[6]QUESTION 9(a)Two straight line with equations px-5y=-4 and 4x-2y=-5 are given. Determine the value for p in each of the following cases:(1)The lines are opposite sides of a square.y=p5x+45andy=2x+52p5=2∴p=10(3)(2)The lines are the diagonals of a square.p5=-12p=-52(2)(3)The acute angle between the two lines is 550, and p is positive.165163512700arctan2=63,430d=8,430tan8,430=0,148p5=0,148p=0,7(5)(b) Tangents to a circle with centre the origin, touch the circle atE(4;7) and F(8; 1).228028576200OxyF(8;1)E(4;7)GG00OxyF(8;1)E(4;7)GGThe tangents intersect at G.Slope OE=74Slope OF=18∴slope EG=-47∴slope FG=-18Equation EGEquation FGy=-47x+cy=-8x+c7=-47(4)+c1=-88+c∴c=927∴c=65y=-47x+927y=-8x+65Equating -47x+927=-8x+65x=7,5(8)[18]QUESTION 10(a)In the diagram below, a circle with centre U is given with RT a diameter parallel1532890217805to chord QP.194691013398500R=35o. 385191025209500Determine the size of SConstruct TP and RSRPT=90oL in a semi-circleT=55oL’s in triangle supplementaryRSP=55o L’s in same seg∵P1=R=35oAlt L’s RT//QP∴RSQ=35oL’s in same seg∴S=20o(7)(b)In the sketch below, DA and DB are tangents to the circle at A and B.AB produced cuts the line through D, which is parallel to FB, at C.222885196215AF produced meets DC at E.AF=FB. DAE=x(1)Find, giving appropriate reasons, 5 angles which are equal to x.B1=xTan ChordC=xCorresp L’s, CD//BFA=xIsos triangle, BF=AFB2=xIsos triangle, ?ADBD1=xAlt L’s, CD//BF (5)(2)Prove that ABED is a cyclic quadrelateral.A2=D1=x∴ABED is cyclicline segm subtends equal angles(2)(3)Prove that ABE=3A1B3=A1=xL’s in same segment∴B1+B2+B3=3x=3A1(3)(4)Prove that AD=BC.AD=BDGivenBD=BCSides opp equal L’s D1=C∴AD=BC(3)[20]QUESTION 11In the figure below ABCD is a rectangle.413385139065(a)Find, giving reasons, the size of each of the following in terms of x.(1)D3D3=xAlt L’s, opp sides rectangle parallel(2)(2)D1D1=90o-xadj L’s on str line.(2)(b)(1)Prove that ADB ||| DEAIn ADB and DEAD=ERight angles, given, prop of rectangleA2=xL’s in triangle∴B=A2∴ ADB ||| DEAA, A, A(5)(2) Hence complete: AD2= ADDE=ABAD=BDAEresult of similarityAD2=AB?DE(1)(c) It is further given that ADB ||| ?BCD. Write down an expression for:BD2= BD2=AB?CD(1)(d)If CDDE=9 find the value of BDADBD2AD2=AB?CDAB?DEBD2AD2=CDDE∵CDDE=9∴BDAD=3505206049911073 marks0073 marks(3)[14]TOTAL: 150 marks ................
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