Chung Cheng High . 2012 Prdlim Exam 1 . AM P1

o2r+lA curve has the equation y = ;, where x+ -3 6- Solvetheequation cos39=2sin30:2sin0+cos?.for 0Sd(180". t5l

t2)

tzl

Prove that cot (45'+ A) =:

cot 75".

tot A -1'. H.n"e find the exact value ofcor A +l

Find the range of values of .r for which I + x2 > -?J+ Hence or otherwise,3

findtheminimumvalueofyif y=1+x2 *'* a" . t41

(ii) Find the.r-coordinate ofeach ofthe stationarypoints of.the curve forwhich 0 <x<2n. t5j

Express f (x) -- 2x2 - 54 . 7 in the form a(x - b)2 + c, where a, D and c are

constants. Hence, sketch the graph of lf<xll=lZx' -Sx-ll for -l < x < 4

anddeterminethenumberofsolutionsforwhich lftrll=1. t6l

8

)

A polynomial, f(x), of degree 4 has a quadratic factgr of 12 - ,r + 2 and .a linear

factor of (x + 3). Given that f(x) leaves a remainder of 8 arird, -24 when divided

by (x - l) and (x + l) respectively, find

(i) theremaining factor off(;r), t3l

(ii) the number of real roors of the equation f(x) = 0, ;urligrin*

yowanswer, LZI

(iii) the remainder when f(x) is divided by x. tl1

4. Given that m' + n2 +12 = 0 and mn =B , find two quadratic equations whose .

roots are * *. * t5l

5. A curve has the equation y = 8 sin x cos3 x .

(i) Find an expression for the gradient of the curve. t21

9. A particle P moves in a straight line so that, , seconds after passing through a

fixed point O, its velocity v cm/s is given by v = 3t2 - 15r + 18 .

Calculate

(D the values of t for which P is instantaneously at rest, L2)

(ii) an expression, in terms of t, for the distance of P from O at time t, [2]

(iii) the total distance travelled by P in the first 4 seconds after passingthrough O. t3l

10. The equation of a circle is xz + y2 + 4x - 4y -17 = 0.

(i) Find the radius and the coordinates ofthe centre ofthe circle- t3l

(ii) Show that the circle touches the line y = l. " l2l

A line of equation L : x + 7 y + 13 = 0 intersects the circle at the points P and Q.

(iii) Find the coordinates of the points P and Q. t3l

(iv) Hence, or otherwise, find the shortest distance of the centre of the circlefrom/-. t3l

CCHY Preliminary Exam 1 (2012) Additional ivlathematics (Sec 4Ei 5 NA) pg1of6

t43CCHY Preliminary Exam 1 (2012) Additional Mathematics (Sec 4El 5 NA) pg 2 of 6

(i) nind 9.dx: ,:, :, ,.. ,. :.. , '.

(ii) Hence, stating your reasons clearly, determine if the graph ofe7x+l

) = -

is decreasingorincreasing forwhich x >O, x * -3.x+3tsl

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In the diagram, ABC is a straight line and IBAE = IDBE = ZBCD =9OoGiventhat 0 car,vary, where 0e <e<9O",BE=3m andBD=7m.

3m

(iv) ShowthattheareaoftrianeleDEFis 29 sin.(0+ a)

"2

Answer the whole of this question on a single sheet of graph paper.

The table below shows the experimental values of two variables x and y-

x 0.1 0.2 0.3 0.4 0.5 0.6

15.9 16.9 18.1 19.5 21.2 35.1

END OF PAPER

B

c F

(i) Show that AC = 7 sin 0 +3 cos 0 .

(ii) Express AC in the form R sin (d + a) where R is positive and

a is acute. Hence, find the value of dfor which A C = 6 m-

(iii) State

(a) the line in the diagram with length R m,

(b) the angle in the diagram with value a.

The point Fii on ttre line CD such that EF is parallel to AC.

tzt

t3l

tll

tl1

t2l

tzlFind the maximum value of the area of triangle DEF and statethe corresponding value of d.

(v)

7m

v

:

CCHY Preliminary Exam 1 (2012) Additional Mathematics (Sec 4El 5 NA) pg 3 ol 6CCHY Preliminary Exam 1 (2012) Additional Mathematics (Sec 49 5 NA) pg 4 91 6

ll..1)

v

It is known that x and y are related by the equation y = ADt', where A and b are

constants. One of the values ofy is due to an abnormally large error.

(D Plot lg y against x for the above data, using a scale of 2 cm to represent

0.1 units on the .r axis and 2 cm to represent 0.2 units on the lg y axis.

Draw a straight line graph to represent the equation y = AD*', with values

on the lg y axis being corrected to 2 decimal places. t3l

(ii) Identify the incorrect reading ofy and estimate its correct value. tzl

(iii) ' Use your graph to estimate the values ofA and b. 12)

(iv) By drawing a siritable straight line on your graph, solve the equation

l_.,Abt, _t1z -" . l2l

tts

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Answers

(i) ez'*t (2x + 5)

Forx>0,r+3, (x+3)2 >0

"2"1 70, (2x+5)>0;

Tho, 4 > 0 and y is an incieasing functiondx

(ii)

2. *r--l o, x(-3; -ll 24

3. (i) 2.x- | (ii) 2 real roots (iii) - 6

4. 8x2 -Zxi1=O and Bxz +2x+1=0

5 (i) 8cos2x(cos2x-3sin2x) (iDr3nnlln5nTz" 2'2'6' 6'6'6 radians

6. 0 = 0", 90', 1 16.6", 180o

7 (ii) 2-J36.

x.t = z(. =r|l-',4,' 3 sorutions

9. (i) t=2 or ,=3*(ii) .=(13-1912+18r)cm'2(iii) 17 cm

10. (i) Radius = 5 units Cenue = (-2,2)(iii) P(-6,-1) andQg,-2)

(iv) iJ7 units or 3.54 units (3 sf)2

11 (ii) AC= J56-sin(d +23.2"); 0=28.8o(iii) Line with length R m is DE; Angle with value a is ZBDE(v) Max area = 14.5 m2; 0=21-8"

12. (D

(ii)

(iiD(iv)

lgy =blgA+ xlgAFrom the graph, the incorrect reading is y = 35.1The correct reading should be y = 21.9.A=2.ll,b=3.78x = 0.065

CCHY Preliminary Exam 1 (2012) Additional Mathematics (Sec 4U 5 NA) pg5ol6

t4

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;,:rr

The populatiori, x, of a certain micro-organism, present at , hours after initially beingobsewed, is given by the formula x = 200 + 200eor'

(i) Calculate the initial popularion..

(ii) Calculate the tinie taken for the population to reach 100O. Give your answer tothe nearest hour.

after initially being observed.

(iv) Calculate the rat! of increase of the population on the first 24 hours.

i

2. (a) wh"n f++."]o, *h"r" n > Ois expanded fully, the first 3 terms in ascending\x' )'-144924poweror xare : _;-+p+F+...

Find the values of,k, n and c.

x2 -!-2x

where p is a positive constant, the term

independent ofx is 5376. Find the value ofp.

3. Solve the following equations.

(a) 82r+t - 3x-2

(b) logr(3:r - l)

4. (a)

6. (i) Express'. 3x+8

3x3 +7x2 -4in partial fractions.

(ii) Hence. evaluate f =3**l a*' Jz3xt +7x'-4

7. Given that lagp 2 = x and, logo 3 : y ; find in terms of x and y, :.,

(i) logp21^fpi

(ii) logu36./p

. f;nA curve C is defined as , = il= .

(i) show rhat 3v'(x-3)'4= -5'dx

t5l

t3l

32

tll

t2)

tllt21

12)

t3l

t3l

t4l

i3l

t4l

t3l

t3l

8.

(ii) Determine the value of .r in which y is undefined.

9. In the diagram, OR is a diameter of the circle, I is the midpoint of PR and

ZPQR = ZPBA . PK, BC atd QR are three lines parallel to each other. O1( is a tangenl

to the circle at 0. Prove that

(D APBA is similar to APQR

(ii) pA= PR'

2PQ(iiD PC:PQ=112.

(iv) ZKHQ=ZPRQ.

K

2 (" 9\- l.Etz, = loc4lx'+-J

Given that 9'+t x22r+3 = 4r-", evaluate I 2' without using a calculator. L-eave

your answer in surd form. - .

or+l nn-l

Giventhat o

;o =at-3,findthe valueof c.4',

t2l

t2t

t2t.

tzt

(b)

5 Solve the simultaneou! eqlations

logo x-log, y =l ,

t6l

."(n:-tvtZ

)

4

Chuhg Cheng High School (Yishun) . 2012 Prelim Exam 1 . AM P2

201 2 CCHY 4E5N Pretim I AM p2

,sl2012 CCHY 4E5N Prelim I AM P2':.:'

(b)

14)

tll

3

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':.,--'.' t'

10:

and C. .. .

(iii) Find the area of the shaded region bounded by y = ?,me xaxis, the y axis,x

-1d1l = sin

= , y = e and the vertical line through C giving your answer correct to' 2'

3 significant figures.

v2e,-

):e

-1)=sin

ltx

,

(i) Showthatf(-r)= Atan3x,whereA isaconstant.

(ii) State the period,p radians off(x).

(iii) sketchthe graptr y=l/(x)l for - p < x < p.

A water dispensef in the shape of a cylinder with radius t0 cm is filled with water.The water is dispensed at a steady rate into an empty cup in the form of an invertedcone oFheight 8 cm and base mdius 4 cm.(a) After , seconds, the depth of water in the conical cup is x cm, show that the

volume of water in the conical "ro

ia ft' "rn'-12

(b) Ifthe dapth of water in the cylihder ddcreased by 0.0025.cm s=', find the rate,ofchange of the volume of water in the conical cup

(c) Hence, calculate, at the instant when the volume ofwater in the conical cup is

?r cm3 therateof increase of,3

(D the depth ofthe liquid in the conical cup,

(ii) the area of the horizontal surface of t}le tiquid in the conical cup.

12.

x(D Find the coordinates ofA. tll

t3l

tzl

t2l

L2)

t2)

x

, ",,I

8cm

Conical Cup

1 3. The diagram shows a quadrilatenl ABCD where AC is parallel to the line3y = 2x + t8 and A is on the x-axis. The coordinates of P and D are (-3, 2) and

(-5, 7) respectively. PC = UP, ZBAC =gO" , AB :2.18 unirs and APC is a

saaight line. Find

B

(a) the equation ofAC,

(b) the coordinates ofA,

(c) the equation ofA8,

(d) the coordinates of C and B,

(e) the area of the quadrilateral ABCD.

v t2).

tl I

Lzl

t5l

tzl

D (-5,1)

t3l

t11

t2)

C

x

5

B

Pe3,2)

201 2 CCHY 4E5N Prelim I AM P2

r6l

2012 CCHY 4E5N Prelim I AM P2

,.t'

I

I

ll. A functionf(.t) is given by /(r)="or3r(l-*r-"*")

Water Dispenser

6

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14' ' The diagram berow shows a sorid body rvhich consists of a cyrinder fixed, with nooverlap, to a prism. The prism has a triangular base of sides 3i cm, 4.r cm and 5.r cmand a height ofr cm. The cylinder has a radius ofx cm and a height ofycm

(i) Given that the total volume of the solid is 96 cm3, express y in terms of x..:..(ii) Hence show th:it rhe lotal surface areaofthe solid is, S 192 +1i,*r.

'

xGiven that x can vary, find

(iii) rhe value of-r for which S has a stationary value,(iv) the stationary value ofS.(v) Determine whether the stationary value of S is a maximum or a minimum.

(i)' .m(ii) r =7 hours (nearesthour)(iii) 24500 (nearest hundred)

(iv) ff = oil*micro-organismftour

'2. (a) n=22,k=1,c=2 (b)

3. (a) -1.40 (3 s0 (b)

(a) (b)

r=4. v=l2

(i) -l 91+-+.=-4(3x-2) 4(x+2)

121

(21

Answers

(D

8. (iD

1q.- (i)

72.

13.

14.

P=4.'':

x=2

63

(iD 0.278

t2l

tll12)

4.

5.

6.

7.

,"'-2J',J

x+1

3y-5x+2!'2x=3

A=(2, e)

I 1. (ii) Period = 4radians

(ii)

(ii) B=(2,0) c=(4,0) (iiD 7.93 units2

'2+ t"hlv

x

3x

6't

(b)

(a)

(d)

(i)

lo" '1"4

v =?r* q o)'3

3=1-2,-6)' C=(3,6)

(cXi)I-cnl/s4

a = (-6 ,0)

(e)

(iiD ? (iu) tM

1ii1 =]rcm2ls

(c) y=-1x-92

units212

96-6x3J--

1d'(v) minimum

2OI2 CCHY 4E5N Priliri I AM P2

r63

2012 CCHY 4E5N Prelim I AM P2 87

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2012 CCHY Prelim AMaths P'l

CCHY Preliminary Exam 1 (2012) Additional Mathematics (Sec 4U 5 NA) pg 1 ot 10

1(i)

(ii) Forx > 0, x* 3, (x+3)2 > 0

"2x+1 ,O , (Zx+5) > 0,

fnus 4 > 0 and y is an increasing function *ax

e2'*1(2x+s)= (r*rf u

6+31(2e2'*1)_ r2x+1dy ,_

"2x+1' x+3

drc (.r+3)2

(b)

2. t*f >-7*-*3

3+3x2 +7x-x2 >O

2x2 +7x+3>O(2x+1)( x+3) >'0

1x>-e or x<-3 *

3

4Min value ol y occurs when .r =

N/in value ol y = 2

-3 r

12

I #24

-3 + (-;)

I)' .,(-i).'I

(

3(i) Let /(.r) =(x2 -x+2)(x+3)(ax+b).f(1)= 8a+8b -$=1 a+b=1

"f (1) = 2(a)(a + b)8 =8(a+b)

3 a+b=1 --------(1)

/(-1)= 4(21(-a+b)-24=8(-a+b)=-a+b=-3------(2)(1) +(2), 2b = -2

b=-1substb=-1 into (1), a-1 = 1

d=2

t+1

:r.'

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l Edu

catio

n

(ii)Hence, the remaining factor a-r + b = 2x - 1 *When .f(x) = (* - * + 2\(x+ 3)(2x- 1) = 0,

or ;+3=0 or 2x-1=0b2_ 4ac'

(iii)

= (-1)2-4(1X2)--7<0

= *' - x + 2 =0 has no solutionHence,/(.r) = 0 only has 2 real roots *

/(.r) is divided by x, let x = 0

"f (0) = 2(3X- 1)

--6f

4.sum ol roots: 1* -1 - m+n -------- (1)

nxnmnProduct of roots: 1

m1n)

1

mn

1

=8rno=8 ------(2)

Given: m2+n2+12=O(m+n)2-2mn+12=0(m+n)2-2(8) +12=0

(m+n)2=4m + n =x2 ---------- (3)

subst (2) and (3) into (1),1

sum of roots = t 4The two quadratic equations are:

,'-1 **1 =o and ,21.1r..1=g4848or 8x2 -2x+1=O and 8-r2 + 2x+1=0 *

5(i). y=Ssinxcos3x

(ii) ff = arin r(3.o.2 r|-rin r; + ("or3 rla.osr;

= -24 sinz x coSzx + 8 cosax= 8 cos2x ( cos2.r - 3 sin2x)

For stationary points,8 cos2 x ( cos2x- 3 sin2x) = 08cos2x=0

CCHY Preliminary Exam 1 eA12) Additional Mathematics (Sec aU 5 NA) pg 2 of 10

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catio

n

COSx=011 3tr

2'2or CoS2x-3sin2x:0

tan x=t;|tr llx 5n 7n

Y=-

-6' 6'6

lta

3nzT, 6,

lln 5n 7rradians

6 6 6

6. cos 30-2 sin 30 =2 sin 0 + cos 0 0"<e<180ocos 30- cos 0 = 2 (sin 0 + sin 30)

-2 sin 30+0

s'n 3e-e

= z( zrin9199"o, e-ge')2 2 \ 2 2)-2 sin20 sin0 = 4 sin20 cos(- 0)

-2 sin20 sin0 = 4 sin20 cos02 sin20 (2 cos0 + sin0 ) = g

2 sin20 = 0sin20 = 020 = 0", 180",360o

0 = 0o, 90o, 180o

or 2cos0+sinO=0tan0=-2

Cr = 63'4349"0=180o-63'4349'

: 1 16'5651"- 116.6.

Hence, 0 = 0o, 90", 1 16.6o, 180o *

7(i).

(ii )

cot (45'+A) =1

;ilr";;)=1+

tarr45' + tan A

1- tan 45" tan A

. 1+tanA=l=-

1- tan Al-tanA +tanA

1+tanA +tanAcorA -1cotA+1

(proven)

trl

CCHY Preliminary Exam 1 (2012) Additional Mathematics (Sec 4U 5 NA) pg 3 of 10

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catio

n

cot 30" + 1

Js-r Js-r=:X:Js+t Js-r_ 3-2J3+1

2

= 2-J3

cot 75" = cot (45"+30o)cot 30o - 1

f 5x 7

=2(x2-2.-L,

=,[(,-;l;-(;)']

='('-;)'-+

"('-'|)'-'4When '.*'2 -S*-7 =O

(%-7)(x+1)=Q1*=$, or -r=-1

when x = 4, f(*) = 2(16) - 5{4) -75

1Turnins noint is (r 4,01'l

8)

lf(x)l=lZx'?-5x-71

7

flx)=g

o x

For there are 3 solutions.

ccHY Preliminary Exam 1 (2012) Additionat Mathematics (Sec 4u 5 NA) pg 4 of 10

8.

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catio

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9.(i)

(ii)

(iii)

v=gtz -15r+18When P is instantaneously at rest, v = 0

3r2-15t+18=0+3, t'-5r+6=0(t-2Xr-3)=o ' '

t=2 or /=3*

s-J3t2-15r+18 dt

= 13-1512+18r+c2

Whenr=0,s =0, ... c=0. = (,t -if+18r) cm s

When r = 2, .=8-lq(a)+18(2)=14

whenr=3, s=27-f fnl + 18(3):---

13-5

When t = 4, S = 64-f tr6) + 18( )_ 1C=lO

Totaldistanc€ = 14 + (14 - 13.S) + (16 - 13'5) cm_ 1-r,= lr CITI*

10. (i)

(ii)

(ii i)

*2 + y2 + 4x-4y -17 =O(x+2)2+lv-2)2=17+4+4

Radius=5units =25

Centre ={-2,2)

(x +2)2 +(y - 2)'= 25. -------- (1)

! = T'-----------(Z)subst (2) into (1),

(x +2)z +(7 - 2)2 = 25(x +2)2 = g

x = -2 (repeated roots)Since the quadratic equation has repeated roots,3 ) = 7 is a tangent to the circle

= the circle touches the line y = 7 (shown) o

(x +2)z +(y - 2)2: 25 ------- (1)x+7y+13=0

ccHY Preliminary Exam 1 (2012) Additionat Mathematics (Sec 4FJ 5 NA) pg 5 of 10

,s3

't

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catio

n

x = -7y - 13 --** (2)subst (2) into (1),(-7y - 13 +2)2 +lry - 2\2 = 25

(7y+11)2+(y-2\2=2549yz + 154y + 121 + y2 - 4y + 4 = 25

50y2+150y+100=0+50, y2 +3y +2=O

(y+1)(y+2)=0!=-1 or !=-2

subst ), = -1 into (2), x = -7(-1 )- 13--6

subst ! = -2 into (2), a = -7(2 )- 13

=1Points P and Q are (-6, -1) and (1, -2) respectively 6

ttilid-point of PQ =

Shortest distance =

(-

+b

2

+(-2)2

1 1

5

,(-2) - ez)tz +12 -(-|lt2 units

= 1J, units or 3.54 units (3sf) n2

:)

1 1 (i).

(rr I Tsin 0 + 3 cos0 = R sin (0 + cx)

R = ^17 *s'

='683

tanc[=7

o = 23'1986o- 23.2

AC = -.68 sin(o + 23.2) *

and ,ino=BC7

BC = Tsin 0

AC=BC+AB= Tsin 0 + 3 cosO (shown) *

.ABcos0-_3

=AB=3cOS0,

ccHY Preliminary Exam 1 (2012) Additional Mathematics (Sec 4u 5 NA) pg 6 of 10

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catio

n

G8 sin(e + 23.1986.) = 6

sin(0+23'1986')= +' Jsa

- 28-8o #

Line with length R m is DEAngle with value a is ZBDE

Area of LDEF

xDFxEF

(7 cosg -3sine)[ sin 0 + 3 cosO)

x Jsg cos(o + a) Jse sin(o + cr)

1

-21

-21

-2

=*(*+rq)29 sin 2 (0+ cr)

2(shown)

Max value of L,DEF occurs when sin 2(0 + o) ='l

Max area = 29

^"2= 14'5 m2

sin2(0+c)=12(0+o)=$Q"

0+(x=45"0+23'1986o=45"

0 = 21'8014"- 21.9 t

CCHY Preliminary Exam 1 (2012) Additional Mathematics (Sec 4U 5 NA) pg 7 of 10

tss

l: t..,:

.. .'

:'j''.

(iii)(a)(b)

(iv)

(v)

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catio

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Question 12

,1.6

1.4

1.2

,t

0.8

0.6

0.4

0.2

'0

0.2 0.40x

0.6

: 6 points

correctly

: Best fit line

0.3.26

M1

vxlg

(i)

0.11.2

0.21.23 'l

0.41.29

0.51.33

f = Abn'

lgy = blg A+ xlgA

(ii) From the graph, the incorrect reading is y = gg-1. The correct reading shouldbe Y = P'l'9.

lgy=1'34t0'olY = 101's

= 21'8776- 21'9 accept 21.4 - 22-4

A2

..... toA_1.33-1.20(lr) " 0.5_0.14=2j1 (10.1) A1

CCHY Preliminary Exam 1 (ZO|Z) Additional Mathematics (Sec 4U 5 NA) pg 8 of 10

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catio

n

blgA=1.18b=3.78 (t0.1)

Ab*- = 10

3

A1

(iv)

Y = 102--5' A1

algy =i-S*

From the graph, x = 0.065 A1

CCHY Preliminary Exam 1 (2012) Additional N/athematics (Sec 4U 5 NA) pg 9 of 10

rYf

t

r: I

Danya

l Edu

catio

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so\u*r o n L.-\ - ?rdrn, I Arn )

l. The population, -r, of a certain micro-organiSm, present at , hours after initially beingobserved, is given by the formula x = 200 + 20}eo2' .

(i) Calculate the initial population.(ii) calculate the time taken for the population ro reach 1000- Give your atswer to

the nearest hour. '

(iii) Find, to the nearest hundred, the population of the micro-organism, one dayafter initially being observed.

(iv) Calculate the rate of increase of the population on the first 24 hours.(i) x= 200+200eo'h

When , = 0,x=200+200

= 400

(ii) 1000 = 200+ 2O0eo'2'

200e0'2' = 800

eO.2, = 40.2t =ln4

ln4o.2

= 6.9314.'. [ =7 hours (nearest hour)

liii; wtren t =24,x =20O+20oea'8

=24502.0835= 24500 (nearest hundred)

(iv) 4 = 4Oeo2''dtWhen t =24,lL = 40"0sdt

= 4860.416dx.'.A = 4860 micro-organism/hour

2. (a) When (++rr) , *f,"r" n > Ois expanded fully, the first 3 terms in ascending

-144924power Of x are : 66 + 62 + --;- + ..xxxFind the values of t, n and c.

o,(j'*",)"= (;X*lkn nckn-l nh -'l[zkn-z

= -

r

-

r _-l-----a-r- ,3, ' *3n-4 ' 2r3n-ABy comparing coeffi cients,

2012 CCHY 4E5N Prelim I AM P2

k ),.[lxi)'-'.,,,.(;x*l 'Gi, *

tu

t21

tllt2)

t4)

3

165-

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catio

n

3n = 66 k22 =7 Z2(l)(c) = AA,

n=22 k =l c=2

(b) In the expansion or [r' -*)'*n"* p is a positive constant, the term t3l

independent ofx is 53?6. Find the vaJue ofp.

(b) 4+r =(l)t".r'l*)'

=(l)6'*'{-f)'(,-'

=1;;h"*'(-*)'l8-3r=0

r=6

)

[3X *)'=5376

r/d)= 5376[64J

p6 = 4096

P=4

3. Solve the following equations.-2x+l ax-2(a) o =r

o 2r+l ; -t-26 =J

8"18;=3'111

o+'(8) =:'ti)

{fi1' =tz

,rr*=ts7?le'l2

ls:"&

t3l

(a)

= -1.3974

= _r.40 (3s./)

(b) rogr(3x -tl-ffi-,"*.( ,9x- +-)

2012 CCHY 4E5N Prelim I AM P2

4 t4l

4

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logr(3.r-l) -2logrJi =log,

Iogz

logr(3x- l) - logr, = lrrer1; *1>

logr(3;r- l)2 -log, q =bgr$' +1')

los, **= toe,(r'+i)

(3x-l)2 =(rr+2)44'

9x2 -6x +l = 4x2 +9

.5x2 -6x-8=0(5x+4)(x- 2)=O

x=2 or t=-: eejectecl)

2)4

+(x'

4

4. (a) Given that 9t+t x22'+3 = 4t-: , evaluate l2' without using a calculator. lraveyour answer in surd form. '

gx+t X22r+3 _ 43-.r

32x+2 X22t+3 = 26-2..

32' 791x22' 18) = )612-z,t

62'(72)=*(+)

,nz, -&72

lrt= Ellqo-E

tax_ o\ o

J

Given that *"*t -,*n-'

= a 2n-!,find the value of a.4',

t3l

(b)

gtr+l _ g"-l

t3l

2n-34n

52012 CCHY 4E5N Prelim I AM P2

r(t

i:

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^3r+3 ^3r-3t =t = a2"-32"

!%L=ez.-j

7!8

I

= sr(r

2" (8 _ 1) = ah" (*)

a=

8

=63

Solve the simultaneous equationsIogox-log, y=2,

- -3v2.'

5t6l

-xJ

logox-logry=)los^ x

-los. v=2logr4I-logrx-log, y = 4

logrx-logry2 -4

tog,\ = 4v'

x

7 ='u

x = l6y'Sub (l) into (2)

t6y' -7 -6yl6y2 +6y-7 =O(8y +7)(2y - l) = 0

7l or y=-8-2

(rejected)

.'.x=4,r=1'2

J= 9

?:-3 v2'

3' = 3o(3.u.').r a7-6rJ =J

x=7 -6y

tr[

v-

2012 CCHY 4E5N prelim I AM p26

:

.. . .:-,-.

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6' (i) Express ## *in patial fractions.

(ii) Hence, evaruare l,##_*t5l

t3l

(i) Let f(x) = 3x3 + 7x2 -4Tryx+1, Ietx+l=0,fCl)=3(-l)+7(t)-4

-0.'. (x + l) is a factor.

f(x) = (x+1)(3x3 + 4xz - 47

= (x + l)(3x - 2)(x + 2)

3x+8

-1 3

3

44

o

0

-4

-4

7

-3

4

3x3 +7x2 -4

2=8cI

L -- 4

ABC=- T-x*7 3x-2 x+2

5=-5AA=-1

3x + 8 = A(3x - 2)(x + 2) + B(x + lXx + 2) + C(x + l)(3x - 2)

When x = -2 when x = -l2

whenx= -J

rc =49Poo

B=14

3x+B -1 9 1'iV;iFJ= r+1 +

4(sr-a * a1**43x+B

7. Giventhat log12=.r and logr3=y,find in termsofxandy,

27 ^[7(i) toe,t,

(i) log, += lo*p 27 +togr(ps1) -,,g,,sz

$D JT+7x2 -4

=t4-1 * 9 * 1 d,'' x+1 4(3x-2) 4(x+2)

= [-

rnr,+ ri 4[*{.2].i,,o. o1'

=t-rn 5 +]ln r 0+ 1tn 6) -(-ln 3+1,n+ * 1,n

+i

= 0.27775

= 0.278 (3s.0

t2)

7

tq

2012CCHY 4E5N Prelim I AM P2

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= llogp,s+f, roe, p -5logr2

= 3v-5x+21'2

(ii) bgu36,! p = lo96 36 +lo96 p'

= r*lroru o

=r*t[ ' ]2l. log, 6.) t3l

1 1-1t

2 log o2+log13

8.

=z*l 1 I[2(x+ y)J

A curve C is defined oS y =

(i) Show rhar 3y'(x-3)2

(i)y=fu+(x -:).

E+,1t-1x-3 '

dv_-_1 - -\dr

t4l

dy

dx

(,-:)i (i)u. rl : - (,. r): (1)t,-rl{(,-:).1

(, -:)l[])t,-

rl : G + z\-1 g 4 - x -z)

- -s(x-r)-i (x+z) i{,-a)i

2

4

3(x-3)l

2012 CCHY 4E5N Prelirn I AM P2 8

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I

,ln4l,,Lt, -:;; ,l

_ -15(x+2) 3

4

3(x-l1t. (shown)

2

-3\' 4

3(x - llr--5

3y'

*rr'r:*(*)_ (x-sXtl-(x+zXo

(, - 3)2

-5= G-#

t;i3t-'V x-3x+2f x-3

4tdrc

Multiplying both sides by (,r-3)2, 3 y'(* - r*= - 5 (shown)

(ii) Determine the value of x in which y is undefined.

x=3

9- In the diagram, QRis a diameterof the circle, B is the midpoint of pR andIPQR = ZPBA. PK, BC and pR are rhree lines parallel ro each other. eK is atangent to the circle at Q. Prove

tll

I2l

t2l

12)

t2)

K

(i)

(ii)

(iii)(rv)

APBA is similar to LPQR

PA_ PR,

2PQPC: PQ=l:2.IKHQ = ZPRQ -

QR PR

PB

... ZAPB = ZRPQ ftommon Z)(i)' ' ZPBA = IPQR (given)

Since there are 2 pairs ofequal corresponding

PB BA PA

is similar to A,PQR.

(ii)PQ

PA

PR PQSince B is the midpoint of PR,

2012 CCHY 4E5N Prelim I AM P2 9

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pn =Lpn2

,o =L'oPR PQ

p1--!l'1-hown\2P8'

(iii) Since BCIIRQ and PB : PR = I :2Thus, PC : PQ = I : 2 (MidpointTheorem)

OR CB//QR,PB:BR=l:lBy Intercept Theorem

PC -

PB' cQ -BR

PC:PQ=l:2

I=T

(iv) ZPHQ= 180' - IKHQ (angles on a st line)

= 180' - IPRQ (angles in opp segment)

:.ZKHQ= ZPRQ

2e(r) y=-_r

When y=e, x=2,... 4 =(2, e)

'7d(tt) Y=51n-

When Y = Q,

sinII=o2

ct=0

E =0 .n.Zn2

x=0,2,4.'.8=(2,0) C=(4,0)

2012 CCHY 4E5N Prelim I AM P2

Find the coordinates of A.

The curve v=-sin4 cuts thexaxis atB and C. Find the .r-coordinates of B'2and C.

Find the area of the shaded region bounded by y =?9,th"x axis, the y axis,

y = sin 9, ! = e and the vertical line through C giving your answer correct to'23 significant figures. v

2e

r

tllt3l

t3l

)=sin1A

2

Y=e'I

xB

-l

r0

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(iii) Area of Shaded Region =

=2e

=2e

= = z"-[- ?"o"n*?"*o)]. fzetn+-2etn2l

: 7.93166

= 7.93 (3s.,f)

I t. A function/(x) is given by /(x) = "o.:-*fl-t'1\l-sin3x l+sin3x./(i) Show that/( x) = Atan3x, where A is a consranr.(ii) State the period, p radians of f(x).

(iii; Sketch thegraph y=fitril for - p < x < p.

"o.:*(^ (t+sin3x-l+sin3x\toti'(.

r -rin'3, .J

.'. the area = 7.93 units2

I _-- I

l-sin3r I +sin3x

t3ltllt2)

(i) =.",r,(#*)2sin 3x

cos3x

=2tan3x (shown)

(ii) Period = 4radians

v

(iii) 2

0

xtan/=3

TL

iI

IIII!

I

_1t -4J

T

12

.l

6IIIIII

I

,I!ttItII

IIrII

;

II

i

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12. A water dispenser in the shape of a cylinder with radius l0 cm is filled with water.The water is dispensed at a steady rate into an empty cup in the form of an invertedcone of height 8 cm and base radius 4 cm.(a) After r seconds, the depth of water in the conical cup is x cm, show that the

volume of water in the conical cuo is *' "^'.'t2

(a) Using similar triangle,

r4Vol of Water

t2)

-=-x8t (t-n1 -3\2 ,) ,

1 ta'r=-x' 2'- =a-cmj (shown)

(b) If the depth of water in the cylinder decreasecl by 0.0025 cm s-1, find the rateofchange of the volume of water in the conical cup

(b) Vol of cylinder = 100trh

4= 4.0025dtdv dv dh

dt dh dt

=l00ax({).0025)I

- -:-T c*' I s4

Therefore the rate of change of volume in the conical cup is ln r*' ls4

(c) Hence, calculate, at the instant when the volume of water in the conical cup isai r cmt the rate of increase of,J(i) the depth of the liquid in the conical cup,

)(ci)whenV=altJ

dv dv d.x

dt dx dt

t2)

t2)

ia3 2

t23x=2cm

i"=(i^'X#)when x=2dxl_-_cmlsdt4

(ii) the area of the horizontal surface of the liquid in the conical cup. t2)

2012 CCHY 4E5N Prelim I AM P2 l2

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(cii) =4i.)'nc'=-

4dA1-=-]ftrdx2dA dA dx

dr dx dt

-2 rE(2) x 14 cm2ls

=f,n"*'t "

8cm

, ""{

Conical Cup

13. The diagram shows a quadrilateral ABCD where AC is parallel to rhe line3y =2* +18 andA is on thex-axis. The coordinates of p andD are (-3, 2) and

(-5, 7) respectively. PC = ZAP, ZBAC = 90' , AB = ZJE units and ApC is astraight line. Find(a) the equation of AC,

3Y =2x+18",

v=-.r+6J

)Grad of AC =a

-1

2y=--r+cJ

At G3,2)2=1+c^_ AL -+

)equationof AC = y= |x+4

J(b) the coordinates of A,Wheny-0

I

v 12)

tlI

De5,7)

C

rA

B

l32012CCHY 4E5N Prelim I AM P2

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Water Dispenser

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2-x:4J

x=4A = (_6, 0)

(.i the equation of AB, :

Grad of AB = -a2

Jy=--r+c"2Ar (-6,0 )0=9+cc=-9-.

equationof AB>y- -1*-g2

(d) the coordinates of C and B,

(,,-1,_,

tzl

tsl

/ q 12(-6-x)'.(.i--rJ =2$3

^q36+ l2x + x' + I x' + 2'l x +81 = 52

4

9*'*39x+65=04

l3x2 +156x+260;0(l3x+26)(x+10) =9x=-2 or x=-10 (rejected)

l*,t B =

l=-6:.8=(_2,_6)lrngth of AP= -0)'+(-3+6)'?

=J4+g

=.11i units

Length of AC =3JE units

I,er c be U,!i++l

20I2CCHY 4E5N Plelim I AM P2 t4

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x2 + l2x + 36 *! r, *E * + 16 = | lj9391z a2.*-65=o.9 ' J: .'

l3x2 +156x-585=0(l3x-39)(.x+15)=g

x=3 or x=-15 (rejected)

!=6...9 =(3,6\

(e) the area of the quadrilateral ABCD.

=il_6-23-50-66 7. fl

12)

t2)

tzl

Area or AB cD = f,loe _ r, *2 r ) _ (_ r 8 _ 30 _ 42)l units2

-_l . u

= O I -untts-2

14. The diagram below shows a solid body which consists of a cylinder fixed, with nooverlap, to a prism. The prism has a triangular base of sides 3x cm, 4x cm and 5.r cm,and a height of x cm. The cylinder has a radius of x cm and a height of y cm.

(i) Given that the total volume of the solid is 96 cml, express y in terms of _r.

t-;(ax)(3x)(x)+ mt y =96'z

6x2 + ta2y =gg96-6x3

'- 1e(2

(ii) Hence show that the total surface area of the solid is, S =192 +lZ*'-x

s = 3x2 + 4x2 +5x2+ (4x)(3x)+ r^{go-o*' 1\m')

=24x2 *192 -12*'-x

-197 +l2r'(shown)x

Given that -x can vary, find20I2CCHY 4E5N Prelim I AM P2 l5

r1t

f----- -,-{(x+ol'

+(!x+4)2 =3JG,. ,,..:,,-,..

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(iii) the value of x for which S has a stationary value,

$ =-ry*24x=odx x'-192+24x3 =024x3 =192x3 =8x=2(iv) the stationary value ofS.

s =192 +121412"=144(v) Determine whether the stationary value of S is a maximum or a minimum. 121

tzl

tll

(v)

,1

2012 CCHY 4E5N Prelinr I AM P2

el?S

-=dt:z \+z+xWhenx=2d2s

a '-0x

since #>0, s = 144 is a minimum value

v

3x

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