add. maths pahang 2011 - tutor mansor · pdf file3 . 347211. the following formulae may be...
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3472/,1ADDITIONAL MATHEMATICS I PKertas 1
September 2011Dua jam
347211
NAMA
TINGKATAN
AD DITIO NAL MATH EMATICSTINGKATAN 5
KERTAS 1
2 JAM
JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU
1 . Kertas soa lan ini adalah dalamdwibahasa.
2. Soa/a n dalam bahasa lnggerismendahului soa/a n yang sep adandalam bahasa Melayu.
3. Calon dibenarkan menjawabkeseluruhan atau sebahagian soalandalam bahasa Inggeris atau bahasaMelayu.
4. Calon dikehendaki membacamaklumat di halaman belakangkertas soa/a n ini.
U ntuk Kesunaan PemeriksaKod Peme riksa:
SoalanMarkahPenuh
MarkahDiperoleh
1 32 33 24 35 36 37 3I 4ov 410 311 412 313 214 315 316 317 418 419 420 321 222 423 324 425 3
Jumlah 80
Kertas soalan ini mengandungi 24 halaman bercetak.
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Example I Contoh:
34721r
rf x,- N(0, 1), then
Jika X -- N(0, l), makn
PW> k)= Q&)
P(X> 2.1) = 8(2.1) = 0.0179
z
Qk)=€
Ir@a,k Q@)
THE UPPER TA|L PROBABILTil Q(z) FOR THE NORMAL DISTR|BUTION ir(0, i)KEBARAN G I<ALIAN HUJ U NG A TAS BAGI TABURAN NORMAI w(0, I
a 0 I 32 654 98734567
Minus / Tolak
II27
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.1
2,5
2.6
2.7
2.8
2.9
3.0
0.5000
0.4602
a.42A7
0.3821
0.3446
0.3085
0,2743
o.zqio
0.2119
0.1M,|
0.1587
0.1357
0.1151
0.0968
0.080E
0.0668
0.0548
0.0446
0.0359
0.0287
0.0228
0.0179
0.0r39
0.0107
0,00820
0.0002r
0.00466
0.00347
0.00256
0,00187
0.00135
0.4960 0.4920 0.4880
0.4562 0.4s22 0.4483
0.4168 0.4129 0.4090
0.3783 0.3745 0.3707
0.3409 0.3372 0.3336
0.3050 0.3015 0.2981
0.2709 0.2676 0.2643
0.2389 0,2358 0.2327
0.2090 0.2061 0.2033
0.1814 0,1788 0.1762
0.1s62 0.1539 0.1515
0.1335 0.1314 0,1292
0.1131 0.1112 0.1093
0.0951 0.0934 0.0918
0.0793 0.0778 0.0764
0.0655 0.0643 0.0630
0.0537 0.0526 0.0516
0.0436 0.M27 0,0418
0.0351 0.0344 0.0336
0.0281 0,0274 0.0268
0.02n 0.0217 0.0212
0.0174 0.0170 0.0166
0.0136 0.0132 0.0129
0.0104 0.0102
0.00990
0.00798
0.00604
0.00453
0.00336
0.00248
0.00427
0.00317
0.00233
0.00181 0.00175 0.00169
0.00131 0.00126 0.00122
0.4840 0.4801 0.4761
0.4443 0.4404 0.4364
0.4052 0.4013 0.3s74
0.3669 0.3632 0.3594
0.3300 0,3264 0.3228
0.2946 0,2912 0,2877
0.2611 0.2578 0.2546
0.2296 0.2266 0.2236
0.2005 0.1977 0.1949
0.1736 0.1711 0.1685
0.1492 0.1469 0.1446
0.1271 0.1251 0.1230
0.1075 0.1056 0.1038
0.0901 0.0885 0.0869
0.0749 0.0735 4.4721
0.0618 0.0606
0.0505 0.0495
0.0409 0.0401
0.0329 0.0322
0.0262 0.02s6
0.0594
0.04E5
0.0392
0.0314
0.0250
0.0207 0.0202 0.0197
0.0162 0.0158 0.0154
0.0125 0.0122 0.0119
0.00964 0,00939 0.00914
0,00734
0.00714 0.00695
0,00554 0.00539 0.00523
0,00415 0.00402 0.003e1
0.00307 0.00298 0.00289
0,00226 0,00219 0.00212
0.00164 0.00159 0,00154
0.00118 0,00114 0,0011.|
0.4721 0.4681 0.4641
0.4325 0.4286 0,4247
0.3936 0.3897 0.3859
0.3557 0.3520 0.3483
0.3192 0.3156 0.3121
0.2843 0.2810 0.2776
0.2514 0.2483 0.2451
0.2205 0,2177 0.2148
0.1922 0.1894 0.1867
0.1660 0.1635 0.1611
0.1423 0.1401 0.1379
0.1210 0.1190 0.1170
0.1020 0.1003 0.0995
0.0853 0.0838 0.0823
0.0708 0,0694 0.0681
0.05s2 0,0571 0.0559
0..047 0.0465 0.0455
5 0.0375 0.0367
0.0384 0.0301 0.0294
0.0307 0.0239 0.0233
0.0244
0,0192 0,0188 0,0183
0,0150 0,0146 0,0143
0,0116 0.0113 0.0110
0,00889
0.00676
0.00508 0.00494 0.00480
0.00379 0.00368 0.00357
0.00280 0,0a272 0.00264
0.00m5 0,00199 0.00193
0,00149 0.00144 0.00139
0.00107 0.00104 0.00100
4812481248124 V 11
4711
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124123123112112
0
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I10
16 20 24
16 20 24
15 19 23
15 19 22
15 18 22
14 17 2A
13 16 19
12 15 18
11 14 16
10 13 15
91214810127 9" 11
6810678
2232221221.l210 13 15
91214
8 11 13
7911
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222
28 3;,2 36
28 32 36
27 31 35
26 30 34
25 29 32
24 27 31
23 26 29
21 24 27
19 22 25
18 20 23
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14 16 18
13 15 17
11 13 14
10 11 13
81011789678566455
3443s423322718 20 23
16 16 21
15 17 19
13 15 17
1t
I7
5
3
3
t2 14
910896644
43
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3 347211
The following formulae may be helpful in answering the questions. The symbolsgiven are the ones commonly used.
Rumus-rumus beikut boteh membantu anda menjawab soalan. Simbot-simbolyang dibei adalah yang biasa digunakan.
-b!1. )(=2a
2. a^ x en = am+n
3. e* + ao = e*-n
ALGEBRA
8. logo fi =log,b
log, a
9. Tn = Q+ (n-I)d
10. ,Sn
11. Tn
12, ,sn
4.
5.
(a*). = a*n
logo mn= log am+logon
logo ln -logo m-logonn
7.lo& m' =nlogom
=llua+(n-l)dl2'= arn-l
_ a(r' :,!) = a(\- r')
,r *!r -1, L-r
13. s. =*,lrl .tt'\
o.
CALCULUS / I<ALKULUS
1' !=uv )
dv dv duL-u + v"dx dx dx
4. Area under a curveLuas di bawah lengkung
bb
= Irtu or (atau) = I*ay
5. Volume of revolutionlsi padu kisaranb
= y Y'dx or (afau)
b
= [o *'ay
v't, u dY-z. y-- ,vdx
.\ dy dy .. du\). :-71-
dx du dx
du dv-udx dx
- 4ac
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4
STATTSTTCS / SilArrSTtK
34721r
Fr1. x=u
N,f.z. x-fr
11, 4X - rY' C,p' q*', p*8 =L
12. Mean lMin , F=nP
,
13. o=lnpqar-X-1tt
=-
14' tHI
o
GEOMETRY IGEOMETRI
5. lrl =,tW
3. g=
4.6- w(!N-F')
5.m=L+L?y
6. [ -0, xlooQo
l.Distance I Jarak
=
2, Midpoint I Titik tengah( xt * )cz /r + /z') 6.
(t,y) = i
-\-2'2)
3. A point dividing a segment of a lineTitik yang membahagi suatu tembereng garis
(x,y)-( n*,***,,ry\\ m+n m*n )
4. Area of triangle I Luas segitiga"
=:l(", lz * xz!g* rry, ) -(*,ry, * *ryz + xrlrl
I"' -,Hv
NIt" - *)'
Zf @- *)' -Zr
x'+y'
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3472/r
TRIGONOMETRY I TRI,G.ONOMETRI
l.Arclength, s=r0 8. sin(l+B)=sinlcosBtcos AsinBPanjang lengkok, s = j0 sin(,{tB) =sin,4kos B*kos AsinB
2. Area of sec.tor, tr=lr,o 9' cos(l t 8) = cos 'r4cos 'B T sin Asn B
2' - kos (l t B) = kosl kosB T sin lsin.B
luas secfor, [' = *r',3. sin2 A+co$2 A=t
sin2 A+kos'A=1
4. sec2 A=1+ tanz A
sek'A-1+tan2 A
S.cosec'A=l+cot2A 1., a b c
kosek'A=l+kotzA ''' rjuif.- titB - titc
a. sin2A=ZsinAcosA 19. az =b2 +c'-ZbccosAsin2A-ZsinAkoM az =b2 +c' -2bckosA
7. cos 2A = cos2 A - sin2 A 14. Area of triangle / Luas segifiga
= 2cos' A-1 - labsinC=l-Zsinz A 2
koslA=kos2A-sin2 A
= 2 kos'A-l=1- 2sin2 A
10. tan(ArB) =ffi
11. tan 2A=.2.t*!' 1-tanz A
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ForExaminer'sUse
6 347Ut
Answer All questions.Jawabsemua soalan.
1. Diagram 1 shows part of the mapping under function f followed by
function g.
Rajah 1 menunjukkan sebahagian pemetaan di bawah tungsi f diikutidengan fungsig.
Diagram IRdah I
StateNyatakan
(a) s (4),
(b) the value of x where g f (x) = x,
nilai x dengan keadaan g f (x) = x,
(c) codomain of function f.kodomain bagi fungsi f.
Answer I Jawapan :
(a)
(b)
(c)
13 rnarksJ
13 markahl
4,
2
-2
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7 3472n
2. Given the composite function fg : x + x' -x + 5, function g : x + L - lac
and f :x+!*' *nr, where m andkare constants .'4Find the value of k and of m.
ForExaminer'sUse
Diberifungsi gubahan fg:x->x2 -x+5, fungsi g:x-+1-lec^1f : x-; ^. x2 + m, dengan keadaan k dan m ialah pemalar..r4
Cari nilai k dan nilai m.
Answer I Jawapan:
dan
3. Diagram 3 shows the graph of function f where f(x) = 2v11.
Function g is the reflection of f about the line y = x.
Rajah 3 menunjukkan grcf bagi fungsi f dengan keadaan f(x)=l2a+1.Fungsi g merupakan pantulan fungsi f pada gais ypx.
Dagram 3
Rajah 3
Find function g.
Cari fungsi g.
Answer / Jawapani
13 marksJ
13 markahl
12 marksJ
12 markahl
3
HO
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ForExaminer'sUse
8 347Ar
4. Find the value ofp where the curve pf- p, +3 = 4x-.f touches the
x-axis at only one point.
Cartkan nilai p dengan keadaan lengkung pi - p, + 3 = 4x - fmenyentuh paksi-x pada satu titik sahaja.
13 marksJ
13 markahl
Find the range of values of x for which (x - 1)'> 4.
Cari julat nitai x bagi (x - 1)' > 4.
13 marksl
13 markahlAnswer I Jawapan :
Answer I Jawapan :
5
HO
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g 3472/l
6. Solve the equatio^ T- 7" : 0.
Se/esaikan persamaan + '7* :0.4
[3 marks]13 markahl
Answer / Jawapan:
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10 3474r
7. Diagram 7 shows a curve f(x) = - (x- 1)'-2which intersects the straight
line Y = - 3 at Point P and Point Q.
Rajah 7 menunjukkan suatu lengkung f(x) = -( x- 1 )' -2 yang bersitang
dengan garis lurus y = -g pada titik P dan titik Q.
StateNyatakan
(a) the equation of the axis of symmetry,persamaan paksi simeti,
(b) the coordinates of point Q,
koordinat titik Q,
(c) the type of roots when f(x)= Q.
jenis punca bila f(x) = Q.
13 marksJ
13 markahlAnswer I Jawapan
(a)
(b)
(c)
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ll 347211
8. Given that lag23 = p and lagzS: g, €XpleSshog47-Z in terms of p and g.
Diberi log2l--pd,tm logp2S - q,ungkpkanlag$'2 dalam sebutan p dan q.
14 marksl14 markahl
Answer / Jawapan:
9. Given an arithmetic progression p + I , 2p + 10 , 7p -1, ".wherepisaconstant.Diberi suatuianiang aritmetik p + I , 2p+ 10, 7P-1,...dengan keadaan p ialah pemalar.
Find
Cart
(a) value of p,
nilai p,
(b) the sum of the next five terms.
hasil tambah lima sebutan berikutnya.
J4 marksl
14 markahlAnswer I Jawapan(a)
(b)
ForExaminer'sUse
9
HO
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t2 34721r
13 marksJ
| 3 markahl
ForExaminer'sUse
10. The sum to infinity of a geometric progression with common ratio ! is 42.
Hasil tambah hingga sebutan ketaktehinggaan bagi suatu janjang geometi
dengan nisbah sepunyaladatah 42
Find
Cari
(a) the first term,sebutan peftama,
(b) the sum of the first four terms of this progression.
hasiltambah empat sebutan pertama ianiang ini.
Answer / Jawapan:(a)
(b)
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3472/t13
11. The vertices of a parallelogram are A (2, 21, B ( S, 3 ),C ( 4, _1) andD ( p, -2).Bucu-bucu suatu segi empat selai atdalah A (2,2 ), B ( S, 3 ), C( 4, _11dan D ( p,-21.
FindCari
(a) the value of p,nilai p,
(b) the area of the parallelogram.luas segf empaf sela ri itu.
Answer I Jawapan
ForExaminer'sUse
14 marksJ
14 markahl
(a)
(b)
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14 34721r
12. Diagram 12 shows points P, Qand Ronthe line2y-x =4 where
PQ;QR=1:4
Rajah '12 menuniukkan titik P , Q dan R pada garis2y- x = 4 dengan
keadaan PQ;QR=1:4
Diagram 12Rqiah 12
Find the coordinates of point P and of point R.
Cari koordinat bagi titik P dan titik R,
Answer I Jawapan
l3 marksl
13 markahl
13. Given g = (Sh)I + (ti -2I is a non zero vector and is parallel to x-axis.
Find the value of h.
Diberi !: (shli + th- zy
paksix. Cari nilai bagi h, [2 marksJ
[2 markahl
Answer I Jawapan
adalah bukan vektor sifar dan selari dengan
Q{2,3}
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15 34721r
14. Diagram 14 shows the position of .the points A, B and C relative to theorigin, O.
Rajah 14 menunjukkan kedudukan titik-titik A, B dan C relatif terhadap titikasalan, O.
Diagram 1,4
Rajah 14
Given B ( 1, I ) , oE - 7r+ ?j and 6t:;m "
DiberiB ( 1,8 ), m - 7i+ 2i dan od - lffi
Find in terms of i and j,Cari dalam sebutan i dan j,
(a) ot(b) ffi
Answer I Jawapan(a)
(b)
[3 rnarksJ
13 markahl
.B (l, g)
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O
16 3474t
15. The variables xand yare related bythe equation y=Xf -4x. A straight
line graph is obtained by plotting r against.f as shown in Diagram 15.
Pembolehubah x dan y dihubungkan oleh persamaan y =lrf '4x. Grcf
garis lurus dipercleh dengan memplotkan E melawan.f, seperti pada
Rajah 15.
Diagftm 15Rajah 15
Find the value af h and of k.
Cari nilai h dan nilai k.
Answer I Jawapan
13 ma*sJ
13 maftahl
vr
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t7 3472/t
16. Diagram 16 shows a unit circle. T (-p, g) is a point on the rotating ray of anangle 0.
Rajah 16 menunjukkan sebuah bulatan unit. T(-p, q) adatah satu titik padasinarputamn suatu suduf 0.
v
T'{-n nl
(1,S
o
Diagrarn 16Rajah 16
Find in terms of p, the value ofCari dalarn sebufa fr p, nitai bagi
(a) cos e,
kos 0,
(b) sin 20 .
Answer I Jawapan(a)
ForExaminer'sUse
x
13 marksJ
13 markahl
(b)
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ForExaminer'sUse
18
17. The function f is defined by t (xl = 3 cos 2x lor 0" 5r 5 360'.
Fungsi f ditakrifkan sebagai f (x) = 3 kos 2x untuk 0' 5 r g 3600
(a) State
Nyatakan
(i) the amplitude of f,
amptitud bagif,
(ii) the number of solutions of (x) = 0 for 0" 5 r S 350' .
bitangan penyelesaian bagi f(x) = 0 untuk ll" ( r < 360'
(b) Solve 3 cos 2x= 2for 0'sr s 360'
Selesatkan 3 kos 2x =2 untuk 0'gr s 360'
Answer I Jawapan
(aXi)
(ii)
(b)
347211
14 marksl
14 markahl
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t9 34721r
18. Diagram 18 shows a circle, centre o with radius 6 cm. The tangent from ctouches the circle at B and CB = 10 cm.
Rajah 18 menunjukkan sebuah bulatan, pusat o dan berjejai 6 cm. Tangen
dart C menyentuh bulatan pada B dan CB = 10 cm.
Rajah 18
Usingn=3"142,find
Dengan menggunakan n = 3.1 42, carikan
(a) 4BOA in radians,
€fiOA dalam radians,
(b) the perimeter of the shaded region.
peri mete r kawasa n be rlore k.
ForExaminer'sUse
Answer I Jawapan
(a)
(b)
14 marksJ
14 markahl
Diagram 18
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20 347211
14 marksJ
14 markahl
ForExaminer'sUse
19. Diagram 19 shows the graph ol y = F +X*2 -& with its maximum point
alA (h, k). A straight line PQ passes through A and is parallelto the x-axis.
Rajah 19 menunJu0kan gnf .Y = 13+ 1"'- 6x dengan titik maksimum nw
di A( h,k). Garis lurus PQ melalaititikA dan selari dengan paksi-x.
Diagram 19Rajah L9
y=o'.+*o-*
FindCari
(a) value of h,
nilai h,
(b) the equation of PQ.persamaan PQ.
Answer I Jawapan(a)
(b)
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z}.Given that f#dx - f , nnd the value of k.
Dibei f #dx:; , cari nitai k.
Answer I Jawapan
21. The mean of seven numbers is 15. When three more numbers are added, the
new mean is 12. Find the mean of the added three numbers.
Min bagitujuh nombor adalah '15. Apabita tiga nombor di'tambah, min baru
adalah 12. Cart min bagi tiga nombor yang ditambah itu.
V marksJ
12 markahlAnswer I Jawapan
22. Find the total number of ways to anange all the letters P,A,R,D,O,N when
Cari bilangan cata menyusun semua huruf P,A,R,D,O,N apabila
(a) the letter D and O are to be adjacent to one another,
huruf D dan O adalah bersehelahan,
(b)thefirst letter is P, Aor R and the last letter is D,O or N.
huruf pertama adalah P, A atau R dan huruf tenkhir adalah D, Q atau N.
14 marksl
14 markahlAnswer lJawapan
(a)
(b)
ForExaminer'sUse
2l 34721t
13 marksl
13 markahl
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ForExaminer'sUse
22 34721r
23. The probability that Ronald does not score any goal in a football match is l.
Kebarangkalian Ronald tidak menjaring sebarang goldalam satu perlawanan
bota sepak adalah +
(a) Find the probability in the first two matches of a competition, he does not
score any goal in both matches.
Cart kebanngkalian dalam dua perlawanan yang peftama bagi suatu
pertandingan, dia tidak meniartng sebarang gol dalam kedua-dua
pedawanan itu.
(b) The probability that Ronald scores more than one goal in a football
match is a. Find the probability that he scores exactly one goal in a
match of the competition.
Kebarangkalian Ronatd menjaring lebih daripada satu gol dalam sesuafu
perlawanan ialah i:.Ctn kebanngka;lian bahawa dia meniaring tepat
satu gotdalam satu perlawanan bagi pertandingan itu.
[3 ma*s]
l3 markahl
Answer I Jawapan
(a)
(b)
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23 3472/t
24. The probability distribution of a random variable X is shown below:
Tabunn kebanngkalian suatu pembolehubah nwak X ditunjukkan di bawah:
X=x 2 3 4 5
P(X=x) 4.2 0.1 0.3 a
ForExaminer'sUse
Find
Cari
(a) the value of anilai a
(b) P(x>3)
Answer I Jawapan
(a)
(b)
14 marksJ
14 markahl
25. A continuous random variable X is distributed normally with a mean of fland a variance of 36. Given P(X>24)= 0.8413, find the value of p.
Pembolehubah rcwa( selanjarx beftaburan secam normaldengan min Fdan varians 36. Diben P(X > 24 ) =0.8413, cari nilai y.
t3 marksl
l3 markahl
Answer lJawapan
END OF QUESTION PAPER
KERTAS SOALAN TAMAT
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Mark Scheme Additional Mathematics Trial Paper 1 2011
1
PEPERIKSAAN PERCUBAAN SPM 2011
ADDITIONAL MATHEMATICSTingkatan 5
KERTAS 1
PERATURAN PEMARKAHAN
UNTUK KEGUNAAN PEMERIKSA SAHAJA
3472/1(PP)Tingkatan LimaAdditional MathematicsKertas 1PeraturanPemarkahanSeptember2011
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Mark Scheme Additional Mathematics Trial Paper 1 2011
JAWAPAN PERCUBAAN SPM 2011PAHANG SECONDARY SCHOOLS
FORM 5 ADDITIONAL MATHEMATICS PAPER 1Question Solution and Mark Scheme Mark
1(a) 2 1 (b) 2 1 (c) { 4, 8 } 1
3
2 k = 2 , m = (both) 3
B2 k = 2 or m =
B1 fg(x) =
3
3 g(x) = or g(x) = 2
1
2
1 −x 2
B1 g(x) = or m = 2
1
24 p = 2 3
B2 or equivalentB1 (1+p) in general form or
a = p + 1, b = –p – 4 and c = 3 (all)3
5 3B2 ( x - 3)(x + 1)≥ 0 or 21,21 −≤−≥− xx (both)B1 or 21 ≥−x or 21 −≤−x
(do not accept 21 ±≥−x )3
6 1.677 3
B2 or equivalent
B1 (4x-2) lg 2= x lg 7 or equivalent (log both sides of the equation)3
7(a) x = 1 1 (b) ( 2, - 3 ) 1 (c) No (real) roots 1
3
Question Solution and Mark Scheme Mark
8 or equivalent 3
B2 4log
5log3log24log
2
222 −+ or equivalent
B1 or or
or equivalent3
9(a) 3 2B1 ( 2p +10 ) - ( p + 9 ) = ( 7p – 1 ) - ( 2p + 10 )
(b) 160 2
2
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Mark Scheme Additional Mathematics Trial Paper 1 2011
B1 - or
24+28+32+36+40 or
* follow through the value of a and d4
10(a) 28 1 (b) 41 2
B1
3
11
3
1128*
4
−
−
or 333
28*
33
28*
3
28*28*
××+
×++
* follow through the value of a from part (a) 3
11 (a) 1
B1 or or 2
42
2
5 +=+p
or p – 2 = 4 – 5 or p – 4 = 2 – 5
2
(b) 11 2B1
or equivalent3
Question Solution and Mark Scheme Mark12 P ( 0, 2) 1
R ( 10, 7 ) 2
B1 2 = or 3 =
or x = 2 + 4(2 – 0), y = 3 + 4(3 – 2) (both) 3
13 2 2B1 h – 2 = 0
214 (a) -9i +9j 2
B1 ( -7 i -2 j + i +8 j )
(b) 10 i – j 13
15 h = 12 , k = - 3 ( both) 3B2 h = 12 or k = - 3
B1
316(a) -p 1
(b) -2 2
B1 -2pq 3
3
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Mark Scheme Additional Mathematics Trial Paper 1 2011
17(a) (i) 3 1 (ii) 4 1 (b) 24.100 , 155.91o , 204.10o , 335.91o 2
B1 48.19 O
418(a) 1.031 radians 1 (b) 21.85 cm 2
B2 perimeter = 10+B1 Arc AB = 6 * (1.031) * follow through from part (a)
319(a) -2 2
B1 3x2 +3x -6 =0
Question Solution and Mark Scheme Mark19 (b) y =10 2
B1 *( -2)3+ * follow through
420 3 3
B2 or equivalent
B1 )3(1
)53( 1
−− −x
321 5 2
B1 15 7 + 3x = 10 12 or equivalent 2
22(a) 240 2
B1 5! 2! or (b) 216 2
B1 or 3 x 4! x 3
423(a)
9
4 1
(b) 2
B1 12
1
3
21)1( −−==XP or equivalent
324(a) 0.4 2
B1 0.2 + 0.1 + 0.3 + a =1 (b) 0.8 2
B1 1 – P(X = 2) or 0.1 + 0.3 + *0.4 * follow through4
25 30 3
B2 -1 =
B1 z = -13
4
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