CSSA 2001 Ext 1 THSC With Solutions

8/8/2019 CSSA 2001 Ext 1 THSC With Solutions

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MarksQuestion 2

(a ) Find

Begin a new page

d z 1 7(,d-x'

(b) a(-t ,+) and B( r, .V) are wo points. hepoint 414 ,-O) aivirteshc ntervalABcxtcrnally n the ratio 5 : 3. Find thecoordinates f B.

Find henumberol'ways n which he etters f thcword EXTENSIONcanbc arrangcdna straight inc so thatno two vowclsarcncxt to eachother.

\Z t ,f ) is a variable oint whichmoves n theparabolax' -- 4y. The angento th cparabola l P cuts thc x axis at T. M is themidpointof Pf.(i ) Show hat he angent P'f hasequation I x - y - tz - O.

(ii) Show hat M hascoordinatcs(+ , {)\ 2 2 )(iii) Hence lnd the Cartesian quation [ the ocusof M as P moveson theparab


3/16

Question 3 Begin a new page

(a) (i) By cxpandingcc.,sZe + A), show that(ii) Hcnccshow hat l' 2 cosA- x + fx

c o s 3 4 4 c o s tA 3 c o s A .then 2cos3A:x3 * 4x

23

(b) Thc unction (.r) is givenby /(t) - . f ,r+6 lor x > -6(i) Fincl he nverse 'unction -'(r) and ind its domain.(i i) On thcsame iagram, ketch hegraphs f y : I(r) and y - f-'("), showingclearlythe ntcrceptson thecoordinatcaxes. Draw in the ine y -- x .(iii) Show hat hex coordinatcs f any pointsol'intersection f thegraphsy - ,f(x) and

y- f - t ( r ) sadsfyheequat ionx '- tc-6=0. Hence indanypointsofntersect iono[ the tlvo graphs.

Question 4 Begin a new page(a) UseMathemat icalnduct iontoshowthat '+ Z ( f f ) isamul t ip leof 3 foral l

positiventegersn.

(b)

Two concentric ircleswith centre O haveradti 2 cm and 4 cm . The points P and alie on the argercirclc zurd /. POQ = y , where 0


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MarksQuestion 5 Begin a new page

(a) Evaluate fo' I

J , f f i * u s i n g t h e s u b s t i t u t i o n u 2 : x , t t > o . G i v e t h e a n s r v c rin simple.sl xact orm,

(b) At anypoint on thecurve y - IG), the gradientunction is givenby # = sin' r ' 4Find he atue f f (+) - f (?)

(c ) A partictes pcrfonningSimpleHarmonicMotion n a straight inc.At timc I secondstha i velocity v metrespcr seiond, anddisplacement metres rom a ixcd point (J onthe inc, where x - 5,ns #'

(i ) Find thepenodof themotion. I

(ii) Findan expressionor v in terms of t, andhence how hat v' = * (tS- "t ) . 3Find rhespeedof the particlc when it is 4 rnetres o thenght of O.


5/16

Marks

Question(a)

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A personon horiz.ongal round s looking at an aeroplaneA througha telescoPeT'The aeroplanes approachingat a speed.ol go ms-'at a constant ltitudeof 200 rnetresabove he erescope. hen the honiontal distance f theaeroplanerom thetelescopesr metres,he anile of elevationof the aeroplanes 0 radians'

(i) Showhat I =d0(ii)Show hat d,

(i i i) Find he at eatwhich e is changingwhen 0 : La, giving heanswern degreespersecond orrect o thenearest egrce'

(b) A particlemoves n a straight ine.At time / secondsts displacement s x metres roma fixetl point O on the ine, its accelerations a ms

-"and ts velocity s v ms

-t

w h e r e Y i s g i v e n b Y ' - 2 ;(i ) Find an exPressionor a rntermsof x'

(ii) Show hat , = [ # dx , andhence how hatx' =& - 6Oe- '

(iii) sketch hegraphof r' against r anddescribehe imiting behaviour [ theparticle'

-, 200tan x16000

x 2+4O000


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MarksQuestion 7 Begin I new Page

(a) Four farr diceare rolled. Any die showing 6 is left alone,while the rematningdice arerolledagain.(i) Find rhe probability (cone ctto 2 decimalplaces) hatafterthe first roll of the dice, Iexactlyone of the four dice is showing 6:(ii) Find theprobabitity (correct ro 2 decimalplaces) thatafter thesecond oll of the dice 4exactly wo of the four dice are showing 6.

(b) A particlesprojectedrom apoint O with speecl 0ms r atanangle f elevation ,andmovesreelyunder ravity,where 8 = l0 ms ''

(i) Wntedownexpressionsor thehorizontal ndvertical isplacementsf theparticle t Itime secondselerred o axes OxandOy.(ii) Hence how hat heequarion f thepathof theprojectile, ivenasaquadratic quation 2

in tanO, is . r t an '0 - 500,ranO (" t +5ooy)-0 '(iii) Hence how hat here re wo values f 0 , 0 < 0 .7, forwhich heprojectile asses 2

rhroughagivcnpointX,f) providedthat00Y < 62500 X2-(iv) If theprojectileasseshroughhepoint (X, X) whosecoordinatesatisfyhis nequaliry, 2

and he wovalues f 0 are a and P , findexpressionsn termso[ X forr a n s + t a n p a n d l 4 t a t a n B , a n d h e n c e s h o w t h a t+ P = + .


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MATHEMATICS EXTENSIONTRIAL EXAMINATION 2OO1

Page4 QUESTION 2(c)

Page QUESTION6(b)

Find the numberof ways in which the lettersof theword EXTENSION canbe arranged n a straight ine sothat no two vowelsand no two consonantsare next toeachother.

A particlemoves n a straight ine. At time / secondstsdisplacements x metres toq a fixed point O on thelini, its accelerations d -s-2, and ts velocity is v ms-lwherev is givenby u --Z - + The particlex zstarts2 metres o theright of O

A


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Mathematics Extension I CSSA HSC Trial Examination 2001Marking GuidelinesQuestion I(a) Outcomes Assessed:H5, H9

Markin

A n s w e r :4

I ( - r ) ' * t = - l !+2t -31 4t '= 19(b) i;'r.",,,., Assessed:4

Guidel ines

Markin Guidelines

Answer:AB hasgradientm, = J Ix+2y*1= 0 has radi ntmr= - d + tan0

. - .Q -81052 '

(c ) OutcomesAssessed:i) PS (ii) PE3Markin

Answer:(i ){-r) = (-r)' + c (-.r)' + b (-x)

= - r 5 - a x 3 - b x= { r t + a x t + b . r )= -{r) for all .r

.' . t) is odd.

(ii) WhenP(") isdividedbY { +2) ,remainders y'1-Z)=- 42) sinceP(t) is odd=-5 since42)=S

Guidelines

2603 2

or simplification. onemark lor valueof sum

. onern"arkor valuesof gradtents. one mark for value of tan 0I one mark for sizeof angle =1,:-#l'(i) . on" mark or showing r) is odd(ii) . onemark or showing emainders -(z)oone mark for value of remainder


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br copying dtagram(ii) 'one mark or reason(iii) . onemark for reason(iv) ' onemark for showing -/- MAB= I ABD. onemark for showing I ACB - /. ACD

(d) Outcomes Assessed: t) ( i i ) PE3 ( i i i )PE3 (iv)H5' PEz,PE3Marki Guidel ines

Answer :( i) (ijr) z ACD = I ABD becausehe anglessubtendednthesamesegmentat B and C by the arc AD areequal.(iv)ZMAB = ^t ABD (equal alternateangles, W ll BD)/.AcB- z.ACDaln = ./.ACB,ABD = /ACD)

AC biSECtS BCDrL)Z ACB= Z MAB becausehe anglebetween hetangentMA and the chord AB through the point ofcontactA is equal to the angleACB in the alternatesegment.

Question2(a) OutcomesAssessed: 7, PEs

Markin Guidelines

Answer:L , " = Z x e "dr

(b ) Outcomes Assessed:P 4

Answer:s x -3 x ( -1 )5 -3l Y - 3 x ( 4 )5 - 3

( z r r " = zk" )+ (z . t )z r r " )=z ( r+2x ' )e "dxd z , r- e =dx'

Markin

= 1 4 + 5 r + 3 = 2 8 . ' . = 5

- - 6 = 5 Y - 1 2 = - 1 2 . ' . ) = 0

ti

a(s, )

oone mark for lirst denvafveoone mark for secondderivative

u eunesCrlteria Marksoonemark for equation n yoonemark for coordinatesof B

3


10/16

. one rnark or numberof arrangements I vowels. onemark for numberof arrangements f consonanG. onemark for total numbcr of arrqnSgments

,comesAssessed:PE3Markin Guidel ines

Answer:The vowels( E, E, I, O) can be arranged npositions 2, 4, 6, 8 in ; -- L2 waYS'Hence he total number of arrangements s 12 x 60 =72O .

The consonants N,N, S, T, X ) canbe arrangedn5!positions, 3, 5, 7, 9 in 2 = 6O lvays'

(d) OutcomesAssessed: i) PE3' PE4Markin

Answer:

(ii)PE3 (iii)PE3Guidelines

( i ) y i x '.' . angent tandequation

d y r+ _ 2 . = i Xfuc4,,r ') has radienty - t ' - t ( x - Z t )

t x - y - t / = O

l(zt) - t

( i i ) A t r , ) = 0 + t x - 0 - t 2 = 0 - x = lHence I hascoordinates(t, O), andM isthemidPointof \Zt ,t'\ and T (l '0 ) 'withmrdinatesW, 4, ) = (+, A

3t 2x( i i i ) A t M , x : 7 - ' = ar ? , (2x \ ': . Y - t ' = t [ ; J\ J . ,

Hence he ocushaseqution2xz

9? xz - 9 y .

@equationof tangent(ii) 'one mark lorcoordinatesf T. onemark or coordinates f M(iii) . onemark for

tr,,,')


11/16

Quest ion 3(a) Outcomes Assessed: ( i )

Answer:(i )

P 4 ( i i )PE 3Marking G idel i

=cos2Acr lsA-s in2^4sinA= (Zcos A-1)ms A - (2sinAos. )sinA=2os tA - cosA 2s inzAcosA- 2 c o s ' A - c o s A Z ( l - o o s 2 a ) m s e=4cos3A 3cosA

(b)OutcomesAssessed:i)P5, HE4 (ii)pS, HE4 (i i i)p4Markine Guidel in

Answer :( i )

Y -,[ff i Inbrchanging ,r and yy ' = x * 6 g i v e s y = x ' - 6x = ! ' -6 . - . I - t ( r ) = x , -6

Range f /(r) is Domain d f-'(x) is{ y : v> o } = ) { . r : x> o }(i i i)Where =;(x), y -.f - '(rI != x intersecr,f - t ( . t ) - r = + . r t - 6 = r

!r - - J - 6 = 0( . r + 2 ) ( . r - 3 ) = oBut r*-2 (oursideomain). -. r= 3

(ii2cos3A=8cost -6cosA

= (2cos )' -3 (2msA)/ l \ 3 _ / l \= [. ' *=) -3[ . r* ; )

=13*3 . r+ * f1 ) ' - r r - 1x \ J / x= r t + 1 -*3JI

cos 4 - rcs(2A+ A)

( i i )

n ul( le l tnesCriteria Marks(i ) ' onemark or expansion nd expressionsor cos2.4' sin24' onemark for simplification to obtain final expression lorcos3,4(ii) . onemark for expressing cos34 in termsof ( ** !' )\ x J

oonemark for binomial expansion f f r *1') '\ x Joonemark for simplification to obtain final expression or cos3A

in termso[ cosA

in terms of x

5

u esCriteria Marks(i ) . onemarkforfinding hffi. onemark or thedomainof the nverseunction(ii) . onemark or thegraphof y = f(t) and ntercepts. onemark or thegraphof y - f -t(r) and ntercepts. onemark or the ine y = J passinghroughhepointof intersection(iii) . onemark or theequationoonemark or thecoordinatesf thepointof intersection

7

l = xv- f(x)

Hcncc nterscct ionointol ' thccur! 'cs s (3,3).


12/16

Question 4(a ) Outcomes Assessed:HE2 ar l n ur tnesCriteria Marks. onemark or establishing he ruthof S(l)

oon emark or s(ft) true + 5* + z(tt ') -l t t for some ntegeroonemark fo r t * '+z ( t t ' . ' ) = s (s * )+zz ( t t - )oone mark for deducing s(k) tme + s(k + I) true. onemark or deducing s(") true for al l integers n> |

M.5

Answer :D e f i n e t h c s e q u e n c e o f s t a t e m e n t s( n ) : 5 n + Z ( f f " ) i s a m u l t i p l e o f 3 , n = 1 , 2 , 3 , . . .

ConsiderS(l): 5' + Z(tt') =Z l = 3x 9 .'. S(t) is rue.I f S(t) is true, then5t +Z(t l r)=Zttt forsomeintegerM. **considers( t+t) : t ' r+z(t t*" )=s(s^)+zz(t t - )t f i t ' ) *z( t t ' ) l * rz(rr*). ' .st* ' z 01'. ' ) 5QM)+ tz0r-)=3{sM + rr ')} i f s(t) s rue, singx

But M andk integral+ {su+ +(rr -} is an nteger.S(ft) rue=r S(ft l) true k =1,2,3,...Hences(t) is rrue and f S(/


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stion 5Outcomes Assessed:HE6 lVlarkins Guidel i

(b) OutcomesAssessed:H5 Ma kt Guidel i

Answer:dY - sin z. rdx

:+ ( t -cos2 ; )y =+(r -+sin2.r)+, c coratant

t sinf + .)c)

(c) OutcomesAssessed:i) HE3Mat

(ii)H5,kine G

HE3ideli

Answer :(i ) Penod is Zrc + t = 4 seconds( i )

x = S g ( > s f tdxv = - = 5 ( - f , s i n * r )dt

v 2 = ( + ) . 2 5 s i n 2 * r

vz= (+) :s(r cost,)= ?(25 -25cos ' t , ), ' =T ( : s_ . r ' )x = 4 : + ' - ? ( : S - 1 6 )+Spcecls ? *. -'

a uloe i lnesCriteria Marks. onemark or change l-limits. one mark for changeo[ variable. onemark or integration. one mark for evaluation4

I __1,,=1,=1 [ rn

1- j ( l n

+t=- dxa (, r+ Jx)A n s w e r

r 1 9L e t I - |J ,

I ^TVqZuduItE.'J ctu

(a )l l8 - ln2)-1 ln4=In2

Then

t t z x , u > Ozr=L a dxdux = l g r = 1 ,

=2u dux = 4 9 = - u - 7

n uloettnesCriteria Marl(s. onemark or expressingin? r in termsof cos2-roonemark or integration,ncluding onslant [ integration. onemark or evaluation t f (+), t (T). onemark or valueof difference

4

arKtn u nesCriteria Marks(i) . one mark for finding the periodof the motion(ii) . onemark for expressingv t in termsof I

. one mark for expressingv t in termsof x. onemark for the value of the speed,4


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(i) ' one mark for expression or 0(ii; . on" mark for expressionfo, {d-x

. one mark for exprcssion fo, {d t(iii) onemarkorvalue #

. onemark for valueof 0

Question 6(a) Outcomes Assessed:

Answer:(i)

(i )P4, HE4 (i i )HE4, HEs (i i i )H5Markins Guidel ines

ITAP =0(alt. ./ s, parallel ines)

200tan9 =- x r^00 - tan'r "Lx

(ii)P

200d0 | ( 200\

- t - G t :dr l+(+) ' \ x"d0 de dx - 200- =d t - a * a t - f + 4 o o o o

de 16000dt x '+4om

32 64 - .uzAnswer :( i )

, (3? r\ t lo24V - = | - - '\ . r i l = ' F 3 r -o - ! 8 , , ) = + ! ( r y - 3 3d r * - ' " d r \ . r -- 1024 ,ru = d T l

-2C0-x'+ 40000(-80)

( i i i ) W h e n 0 = 3 , T P = A P : + r = 2 N , a n d # = m % = o . 2 r a d i a n s p e r s e c o n d .Hence e is increasingat I 1o s -r (conect to thenearest egree)

(b) OutcomesAssessed:i) HEs (ii) H3' H5, HE4 (i i i) HE3' HE 7Marklne Guidelinesg (' uCriteria Marks

(ii) . onemark for expressing asan ntegral with respecto x. onemark for intcgrationt,o ind f in termso[ rI one mark for expression or x' in termsof r

(iii) . onemark or graphof x' as a functionof I. one mark or limiung valuesot 'x, v, Q. one mark or descnptionof limiting behavioulluygrdl -

7

( i i ) gd rt.r4. r t)4 )

-- lt -

d t(Y

r 22xG( 2-r) m

2r

[ =


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(i i i)( i i ) Cont.

| -- ln(o+ r') *r , ( a q - * ' \- . - r l { - I\ 6 0 )

. ' . x '=

= c = 1 n 6 0

_ f - r '60

l = 0 1>x = Z )e

& - 6 O e - 'A s t + o o , . r - + 8 - , - + t a - t = O ' , a - # f + * = O -Hence he particle.ism_ov_ingight andslowingdownas t approachests limiting position8 metres o the right of O.

Questlon 7(a) Outcomes Assessed: ( i )HE3 (i i )HE3

Markine Guldeli

Answer:(i) P(one6 on first oil) =aCr +)'(f)3 = 0.39 (to 2 decimal laces)(ii) P(rwoC s on irstroil andno 6 s onsecondoil)=aCz({)'(*)' *

\one 6 on firsr ot l and one6 onsecond ou) = .C,(*)t(*)' xftno Gs on"first olt and wo 6 s on second ott)- aco(t)'(*)' ". ' . 4mo 6s overa l t ) 0 .0804+0.1340+ 0.055g=0.27

'c.(*)'(f) = e.o8o4'c'(*)'(t)' = e'Bq'c'(*)'(*)' = Q'oss8

( to 2 decimalplaces)

(b ) outcomes Assessed: i) HE3 (ii)HE3 (iii)p4 , H2 (iv) p 4 , HzMarklns Guideli

ar u ne surtterla Marks(l) oonemartcor valueof probabiliry(ii) 'one mark or expressibnor pr6bability f two6'son firstroll andno 6'son second' onemark or expressionor probabilitiof one6 on irstroll andone6 onsecond' onemark or expressionor probabilify [ no 6'son irst roll and wo 6'son second. one nark or.value f probability

5

u nesurlterta Marks(i) oone mark for expressionsor ; and y in te.msof 0 und t(ii) . one rnark florexpression or y in termsof .roonemark for rearrangement squadratic n tan0(iii) onemark ordiscnrninant n terms f X .nd y' onemark for usingdiscnminant> 0 to giverequired nequality.(iv) ' onemark for the valuesof thesumandproduciof ran ', ran'p. one mark flor he valueof a + p in termso[

7


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Answer :( i ) x =5O cos0(i )

a n d ) = 5 0 t s i n g - 5 t 'x sin0- - _ - \ : = f -" 50cos0 - msO

500y= 500,r tan0 - x' sect= 5 0 0 . r t a n- x ' ( r * * ' 0 )= 50O r tan0 - ,'- x'tan t 0

;. x ' t ;rn, - 500.r an0 *(tt +5ooy)= o

(iv) tf theprojectilepasseshrough hepoint (X, X)x ' tan 'e 5ooxtano(x '+5oox)=otilt a+ tanp - ffi=X = 500 and tand tanF =x 2 x

. '. mn(a F) = tand + tanB -(-?) -,1- tana t:u,;.LBS i n c e< s+ F < n , a + B = + .

(iii) Projectilepasseshrough he point (X ' f) iftan0 satisfieshequadraticequation

x' tan'e - 5oox tan * (x' + 5oo )- oThis cquationhasnvo distinctsolutions lor tan0 ,andhence or I , provided ts discriminant A > 0.4 - (-5oo )' - 4 x2 x' * soo )

= 4x' (ozmo x'- 5oo). '. A> 0 Provided 5OOf < 63500 X'where 500X < 62500- X' , then heequation

has wo distinctreal roots tan d , tarlp whcreX2+ 500X

5 x '2500 cor t 0

x ' 500= 1 + X500=x

CSSA 2001 Ext 1 THSC With Solutions

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Transcript of CSSA 2001 Ext 1 THSC With Solutions