3

4',13

201 8 4E AMath Prelim - Tanjong Katong Secondary School

2(D

3 (a) Giventhat P=ri"-'(-f) ,express d intemsof z.

Hence, find the exact value of sin 20 * tan 0.

(b)v

o 1'

y=atan(bx)

BP-415

18;-2r+tq

Exoress---------------- in oarlial fractions.' f l-.rx4 +.rr)

On the same axes sketch the curves y = -1, and y = -rfT, x3 .Find the r-coordinates of the points of intersection ofthe two cunr'es.

tsl

tzl

12)

t4l

(iD

x

The figure shows part ofthe graph of y -- a lan (Dx) and a point

malked. Find the value ofeach ofthe constants a and 6.

,(T'-,)12)

4 The equation ofa curve is y = 6' * U" '.

(i) Find the coordinates ofthe stationary point ofthe curve, leaving your answerin exact form. t4)

(iD Determine the nature of this point. t21

4047l\lsec 4 Prelims 2018

I

4144

5 (i) Sketch the graph ofy = l- -;l- f. indicating clearly the vertex and the

intercepts on the coordinate axes.

BP-416

(iD

(iiD

(iv)

State the range ofy.

Find the values of;r for 4 -7--6.2

t3l

trl

t2I

tr I

t4l

trl

t4l

t3l

t4l

6 (a)

(b)

The graphy: V -;l- 1 is reflected in rhey-axis.Write down the equation of the new graph.

Find the maximum and minimum values of (l - cos l)2 - 5 and

the corresponding value(s) of,4 where each occurs for 0o < A

4155

A curve Cis such thatf :8 cos 2randP Q,z",ll -z)is apointon C.

(D The normal to the curve at P crosses the.r-axis at Q.

Find the coordinates of Q.

(ii) Find the equation ofC.

Civen thaty: sin 4x, show that S x fl = - :z sin ar.

lf p and q are roots of the equation I + 2t - | = 0 and p > q,

""p..r, { in the form a - b,,12. where a and b ars integers.

BP - 417

8 (a)

(b)

l0

9 (a) Find the range ofvalues of* for which Lr(2x + k) + 6 - 0 has no real roots.[4]

C

t3l

t3l

t4l

tsl

trl

t2l

14)

(b)

rcm

E D

A hexagon ABCDEF has a fixed perimeter of 2l 0 cm.BCD and AFE are 2 equilateral triangles and ABDE is a rectangle.The length ofBC is represented as x cm.

BA

F

(i)

(iD

(iii)

Express,4B in terms ofx.

Show that the area ofthe hexagon, ll is given by,:(+-z)x2 +rosx.

Find the value of x for which ll is a maximum4047 h lsec 4 Prelims 2018

4',t65

Y=bt+l

BP-418

ll

-I

O

P (8, -8)

The diagram shows triangle PQR in which the point P is (8, -8) and angle PQR is 90".t3

The gradient of PR is - ; and the equation of QR produced is y -- Lr +1 .

The line PR makes an angle d with QR produced.

(D Find the coordinates ofQ. t4l

(ii) Find the value of d. t3l

4047l1lsec 4 Prelims 2018

,1

417

TANJONG KATONG SECONDARY SCHOOLPreliminary Examination 20'l 8Secondary 4

BP-41 9

CANDIDATENAME

CLASS II INDEX NUMBERADDITIONAL MATHEMATICSPapet 2

AdditionalMaterials: WritingPaperGraph Paper

4047t02Tuesday 28 August 2018

2 hours 30 minutes

READ THESE INSTRUCTIONS FIRST

Write your name, class and index number on the work you hand in.Write in dark blue or black pen.You may use an HB pencil for any diagrams or graphs.Do not use staples. paper clips, highlighters, glue or correction fluid

Answer all the questions.Write your answers on the writing paper provided.Give non-exact numerical answers correct to 3 significant figures, or 1 decimal in the case of angles indegree, unless a different level of accuracy is specified in the question.The use of a scientific calculator is expected, where appropriate.You are reminded of the need for clear presentalion in your answers.

At the end of the examination, fasten all your work securely together.The number of marks is given in brackets [ ] at the end of each question or part question.

The total number of marks for this paper is 100.

This document consists of 6 printed pages.

418 BP-4202

l.Quadratic Equation

For the equation al + bx + c : 0,

Mathetnatical Fomulae

ALGEBRA

-b+ b2 - 4ac2a

Binomial Theorem

(a+b)'=tr+

Identities

Formulae for AABC

r

where z is a positive integer and

n

I )o-'u*(;)" n. .(:)" ,u+ +u,

(;)_ nt _ n(n 1).......(n-r +l'S

(n - r)lr! rl.

2. TRIGONOMETRY

sinzA+cos2A:7sec2A:l+tan2A

cosec2,{=l+coeAsin(A+B)= sinl cosB+ cos,4 sin.B

cos (l + B) : cos AcosBI sinl sin -Btanl + tan B

tanU + Bl=--' 1+tan/tanBsin2A=2 sinl cos I

cos 2A : cos2 A - sin2 A : 2 cos2 A - | : | - 2 sin2 AZlan A

t^n 1A = _1-tan2 A

0 b c

sin A sin B sin C

a2: b2 + c2 -2bc cos A

t: !bcsinA2

4047 l2lSec4PrcLm2917

BP-4214193

Answer all questions.

I The amount ofenergy, E erg, generated in an earthquake is given by the equationE: l}a+ bM. where rz and b are constants and Mis the magnitude of the earthquake.

The table below shows some corresponding values of M and E.

M I 2 4 5

E (erg) 2.0 x l0r3 6.3 x l0la 2.0 x l016 6.3 x l0t1 2.0 x lOre

(i) Plot lg E against M.

(ii) Using your graph, find an estimate for the value of a and of b.

(iii) Using your answers from (ii), find the amount of energy generated, in erg, by anearthquake of magnitude 7.

2 (i) Write down the expansion of (3 - .r)r in ascending powers of -r.

(ii) Expand (3 + 2tf, in ascending powers ofx. up to the term in .tr.

(iii) Write down the expansion of (3 - r)3 (3 + 2r-)8 in ascending powers of x, up to l.

(iv) By letting -r = 0.01 and your expansion in (iii), find tbe value of 2.993 x 3.028,giving your answer correct to 3 significant figures.Show your work ings clearll.

(v) Explain clearly why the expansion in (iii) is not suitable for finding the value of23 x 58.

(i) Bywriting3das(20+ 0), show that sin (3d):3 sin0-4sin3 0.

(ii) Solve sin (30):3 sin dcos Ofor 0'< 0

4204

In the diagram,,4 . B, C and D are points on the circle centre O.AP and BP are tangents to the circle at,4 and B respectively.DQ and CQ are tangents to the circle at D and C respectively.POQ is a straight line.

D

A

P

C

(i) Prove that an gle COD:Z x angle CDQ.

(ii) Make a similar deduction about angle AOB.

(iii) Prove that 2 x angle OAD = angle CDQ + angls 3trp

BP-422

5

o

t3l

trl

t4l

6 (i) Differentiate y = 2d- 0 - 2-r) *'ith respect to x.

(ii) Find the range ofvalues of.x for which y is decreasing.

(iii) Given that x is decreasing at a rate of 5 units per second, find the rate of changeofy at the instant when x - *1.5.

t3l

trl

t3l

(i) By using an appropriate substitution, express 23'-r - 22a+2 + 2a as a cubic function. [3]

(ii) Solvetheequation23o*l-22a+2+2a=0. I5l

(iii)Findtherangeofvaluesofftforwhich23o*1-22a+2+k(za)=0hasatleastonereal solution. t3l

7

4047I2ISec4PteLm201Iffurn over

421

5

8 The diagram shows the graphs ofy : f(x) and y = f'(;s).

BP-423

v

-1, : f'(.r

l6-v = f(r)

The function f (x) = at + b* + 24x + l6 has stationary points at -r :p and .r : 4.

(i) Find an expression for f'(r), in terms ofa and D.(ii) Find the value ofa and ofb.

(iii) Find the value ofp.State the range of values of /., where [ > 0 and y : f(r) - /r has only one real root.

(iv) Find the mininrum value ofthe gradient off(x).

9 The diagram shows the graph of

24

C

tr l

t3l

t3l

12)

tll

t3lt6l

t:-){x-z)o+to.

AB and AC are tangents to thecurve at B and C respectively.

-B f ies on the y-axis and AB : AC.

(i) Find the gradient function ofthe curve.(ii) Find the equation of the tangent at B.

Hence. state the coordinates ofl.(iii) Find the area of the shaded region.

B

o -t

40 47 I 2 I Se c4P r eli n2O 1 A ffurn over

,l

4226

BP-424

10 A particle, P, travels along a straight line so that, / seconds after passing a fixed point O,its velocity, v m/s is given by

u : 112e t' + 18), where t is a constant.

(i) Find the initial velocity ofthe particle.

Two seconds later, its velocity is 40 m/s.(ii) Show that t = 0.3031, correct to 4 significant figures.

(iii) Sketch the graph ofv= l2ek'+ I8, for0

BP-425

4232018 Add Math Prelim Paper 1 Mark kheme

an Working Marks1 8* -2x+19 A Bx*C*--4*xt(l-r[a+/) 1-x

8l-2rr19 : A(4+fr +(Bx + Q(l - r)Subx:1,8-2+19:5A A=5Sub .r: 0, 19 =4{, + C C : -lCompare coeff of f ,8:A - B B:-3

s* -x+tg s 3x + |(l-xX4+l) t-x 4+*

B I correct PF

MI

A'2 For all 3conectA1 For 2 correct

{et Ontyifelawarded

Total 5 nrarks2(1) v

,

x

: -rlj

G1

G1

2(n)

1-xZ : .132 x3x : 32x6*\1 -32;)=0r:0or1

2

MI

A1Total 4 rnarks

3(a) ^fi

2sin O cos O + tar 0: 2 (-9 (, + (-{o3/=-!3

B1Bl value of cos IBl value of tan 0BI

3(b) a-2Period: Zn: i b

1BIB1

Total 6 nmrks4{r) !: ., -2e-r = odx

x = h.12

y = era\C+ ze'tn\l:.'tz+ftxft:za Pourt is 0n rl2,2r/Z)

deIr4l :: o

dxB I DfferentiateAl value of:

Bl o.e

4(ii) fl:e +z*',:^&,#:2+*>oMinimurn point

Ml Knowing testCorrect conclbased on test

{ar

Total 6 marksI

r

BP-426

4242018 Add Math Prelim Paper 1 Mark Scheme

Working5(1)

x

(8,- 4

G1 vertexGl :r irtsGly int

5(ii) y>- I

5(iri)l+ -:,1- r: aln-il:,+-i:t w +- ,

2

t:- 6 or 22M 1 or by countiagA1

5('v) r:l++r)-t B1

Total 7 marks6(") (l cosl)2 - 5

Max valle: (l_(_t))r_ 5 = 1When cosl : -1,1 : 1800

Minvalue:(1*1)'-s:-sWhen cosl : l,l : 00,360o

B1B1

B1B1

6(bX1) 90o

BP-427

4252018 Add Math Prelim Paper 1 Mark Scheme

Qn Working Marks7(aXD

d /rx1 4x (r1) - +rn,i"-V=)-@4-4Ln-x

1.6x2l-lnr - .=;;t (shown)

Ml Quotent ruleMI diffln xBI workrrg seen

7(a)(ii)J

1-lm-d , lnldl:- + cr

l f lnx+J7d*=

lL lnx| -x-zdx--+c1J4 4xr 1- tnrr__-+cr-4 4,

t lm I lnrd)c:-_-_r+c

81 use integn asreverse of diffIgnore if +c ismissrng81 rearrange termsBI ! x-2 aj1

81 must have +c

1b) (r(x) dx + "sJ, [1'@) +u)ax: Ii rcoa* + li f o);a* + f uax:r.F,];=8+

79t)

BP-428

4262018 Add Math Prelirn Paper 1 Mark Scheme

an Working Marks9(a) be\2x+k)+6:0

4*+2b+a-oDiscriminant < 0

Qk)2 - 4(4)(q q

M 1 rationalise

Ml simplify

A1Total 9 marks

100 4x + zUB):210AB=105-2t B1

10(ii) H:2 60+(105-2r>:E**fi5x-2.*

2/-Bl| - z) *' + 105x (shown)

Bl Area oldBl sub & working

ro(iii) dil ^l.tz& \2!!:nx.: 46.3

2)x + 10s

-=.!j _4

BP-429

2018 Prelim Exam427

Add Maths P2 Marking Scheme

Ar Key Steps Marks / RemartsI (i)

M 2 3 1 5lnE 133 14.8 16.3 17.8 19.3

lnE

M

B1 TOV

Bl Lrne passes tlrough pts

1

(ii) lgE:a+bMa : vertical irtercept : 11,86 : gradtent (their rise/run): 1.5

B1M1AI

11.7to 11.9workiag fm gradientl-49 to 1-51

(o lgE:11 8+ 1.5(7)-22.3E :2.0 x l0zz Erg

M1AI l34xlazz to 2.95 x 102

2(i) (3- x)3:27 -27r+ 9i - I B1

10

(n) (3 + Zr)8

:,'. [f Ji,r'r,r. (l)ou,,'-' [i) or,,,

: 6 561 + 34 992f +81 6,t8y''+ 108 86413 +. .B3 1m for each telm (2nal to ,lfth)

-lm if I st tefm missingB0 is all not evalualid

(rr) (3 - ,)r (3 + 2x)8= their (i) " thei(ii)= 177 147 + 767 637x + 2 854 035x'?+...

MiAI

choosing correct pairs

(r") 2.9f 1.A2"- r77 147+ 767 637(0.01)+2 854 035(0.01)z:185108.7735: 185 000

81 Subn mustbe seen

Bl reject i84 956

(v) For 23 x 59, needto use r: 1

Since 1 is Large in comparison to 0.01, the value isinaccurate because a sigaifieantly large value isremoved after the 3rd term

Bl -r: 1 seen

81 o.e. "big" or "large" seen

1

BP-430

2018 Prelim Exam428

Add Maths P2 Marklng Scheme

Qn Key Steps Marks I Remarks3(i) s,J.(o+ 20)

: sin d cos. 20+ oos 0 sin 2d=sinA(1 -2 sinz6)+cos €Qsin9cosA= sin A (1 - 2 s1n'z 0) + 2 sin 6 cos2 0=sin0(I -2sinz0)+2sinA(1 -- sirf 0)=3sind-4sirfd

BiBI

Use courpound angle

Any double angle seen

B1 Use identity

8

(n) sin (3Q: 3 sin 0 cos d3 sind- 4 sirf 0 :3 sindcosdsin (3 - 4 strf 0 -3cos r) : 0sind-0 .'. A= 18tror3 - 4 strl 0- 3 cos d:03-4(l - cos2 9) - 3 cos d:04cosz 0- 3 cos 6- l:0(4 cos 0 + l)(cos d - 1) -- 0

"o. a: - I or cos g: 1 (NA)

4flence, d: 104,5", 255.5"

Bl d: l8f seen

B2 -lm for extra alswer

4(' a+ p:-b afr:c 81 Both correct

7

(ii) d+ 92 :(8+ p)'-ZaP:bz-2c

d2g'.:"',EEr: *-(F-zc)x+ cz:o

B1B1B1

(in) For 2 distiact rootr,(b?-2c\2-4d>Ab2 (rr -{c)>0Snce b2 > 0,hare ,2 -,k > 0

B1 Ccrrect Dak if HDz- 2c)l? tr (&2 - !;)2o.e.B1

(iv) b:5,c:2 Bl o.e.

Ml Solve a Quadradc81 Use identity

Correct sumCorrect prodrctEquation seen

I

BP-431

2018 Prelim Exam429

Add Maths P2 Markjng schenie

Qn Key Steps Marks / Remarks5(' Let ZCDQ: a

ZODQ:90" (tan-L raQ:. loDC- 900 - a:. ZCOD = 18tr - 2(90p - a) (Zsurl1 LCOD)

B1

B1

B1

wrth reason

with reason

8

(ii) /AOB=2xZBAP B1

(iii) From (i) and (ii),z(zcDQ+ 1_BAP): ICOD+ AOB

ICDQ+ 13AP : + (/CoD+ ZA2B)

: ZAOP + lmQ (1 prcp of chorQ:t80'-lAoD

:ZLOAD

81 anempt to use (i) and (ii)

B18 1 lm forreason

B1

6(' y:2e3' (l - 2x)

ff :ra, gr1* 6ntx (t -2x): Ze3, (l _ 6a)

B1B1

B1

Product RuleDiff exponential firSimplify, ok if not factorised

7

(ir) dvFor decreasine firnctio[ -i- < 0

da.'. 1-5;r -6 B1

(iii)Given tlnt 4 =- S *lol,

dxd.y _ d,y .dxdt dr dt

:Zerx (l _ 6rX:5): 2"t(-t s) (l + 6 x 1.5)(_5):-1.11units/sec

81 with negahve seen

Bl with subs seen

B1

I

BP-432

2018 Prelim Exam430

Add Maths P2 Marking Scheme

Qn Key Steps Maks / Remarks

2!r+i -222+2+Za:2x23o.--4x2b+2a: ?:1'- 4*+ x

Let2:t B1B1

B1

lJ5gqf )P+e:2? y14Use of: (2$e: 2re

11

(ii) 2l-+*+r:o4*-4x+ r):o

otz*-4x+l:a-4+ 6-4xZxl

)c4

:7.7fr7 or0.2929

2:1.7A7 or0.2929le1.7O7 1s0.2929a:- oI-192 1g2

:0.7771 or -1.77

81 -r:0seen

M1

A1

Solving quad with wuting seen

Both;r

M1

AI

Using log (any base)

Both a

(iii) 2'd+ 1 - 22a+2 + (!02a :0 has at least one root: . 2i - 4( + & = 0 has at least one root.'. 16-4x Zxk>0

ks2

MIB1A1

Using quad part of eqrConect D with subs

8(r) f (x): ax3 + bl + 24x + 16f ' (x):1uz + 2bx + 2+ B1

9

(ii) Sub (4. 0) irto f' (.r) = 034t6)+ 2b(4)+% = 0.'. 48a+ 86 + 24 :0 (1)

Sub (4, 0) into f (r)a(64)+t6b+%{4)+16=0.'.64a+ 166+96+ 16=0 . (2)

a= 2, b= -15

81 Sub irfo their f ' (, md q:,r)

M1

A1

Solve simul eqn

Both

(ii') f'{*) * exz-3o,r+2*:6(x'/ -5x + 4):6(: - lXr- a)

:.p: LAtr:1, (r):2(1) - 15(1)+24(1)+ 16:27Herrc.t, k> 27

B1M1A1

Using thei p

(iv) :6(2.5)2 -3A

BP-4332018 Prelim Exam

431Add Maths P2 Marking scheme

Qn Key Steps Marks / Remarkse(i)

t= -ltx -z)o + ro,,.ff = -21* -21' B1 o.e

l0

(ii) Grad of AB : -2(-8) = t0AtB, x:0,.'. y:8Eqn AB: _y : 16, + 8

.. Ais(2,40)

B I Grad l.B seen

B1B1

Eqn l8 seen

(iii ) ATeaOBACD =(E+40)x2= 96 units2

Area bounded by curve and axes

= li(!t*-zt" * rc)a,= (-*{r-zf * ror)'=(-S x32+6a)-($ x 32)=57.6

.. shaded area - 96 - 57.6 = 38.4 units2

M1A1

Using composite figures

B1 Knowing lo use inleglal for arEa

Correct integration

Subs seen

BI

BI

BI

l0(i) vo: l2e klo) + 18 = 30 m/s 81 Sub need notbe seen

ll

(i i) w: 40 ... 40 = l\e k(2) + 18-2k - 1rE -T2k = tn (i)/r: 0.303 I

81 Sub into eqn

B1BI

Using logarithm

(iii )v (m/s)

(4, s8 3)

30

/ (s)

B1 Shape

Label /-intercept

Label (4, 58.3)

B1

B1

(i") Area under curve < Area oflrapeziumArea of Trapezium = 0.5 ( 30+60)

* 4= 180

60

30

4.'. distance tavelled < 180 m

B1 Find relevant distarcetravelled using any suitablemethod

B1 Making conclusion(v) Max accn occuls at , = 4 where the gradient is most steep

Max accn : 0.3031 x 12 e o lo3l (a)= 12.23 mls2

M1A1

Knowing to differentiate

BP-434

2018 Prelim Ex: rn432

Add Matha P2 Marking Scheme

a1 Key Steps11(' i+l-t*-+y-s=a

A b (4,2)

Radius = 42 +21 +5:5 (mits)BIMIAl

t2

(ii) l'z+f'?-8(1)-4(6)-5:oHence, (1, 6) lies on the circle. Bl Subs seen and statement

{ii') Gradient ofline joinine (+, 2) and (L O)_!3

Eqn oftangen at (1, 6) is4y-6:--(x-1)'3

4y -3x:21

B1 I grad seen

Find eqnB1

Bl o.e.

(iv) LtB,y:g

.'. 3 is (-7. 0)

M1 Finding.t

Ordered pair seenA1

(v) Distance between cenires:.hfi= Jns.'. radiue of Cr = ,6E - s

=s6 -s

M1 Find dist between centres

l\,I I

A1

Using sum radii : distance

Marks / Remarks