add maths paper 1 penang malaysia spm 2011 trial paper

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Answer for Add Maths Paper 1 Penang Malaysia SPM 2011 Trial Paper. Look for the solutions to the problems provided by [email protected]

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NAMA TINGKATAN :

r,^rrulru

r Dl\\'r/r

UA Slrl(OLAH

MALAYSIA

CAWANGAN PULAU PINANG

MODUL PENILAIAN SPM 2OIIADDITIONAL MATHEMATICS KertasISeptember 2 iam

3472t1Dua jam

JANGAN BUKA KERTAS SOALAN INI SEHINGGADIBERITAHU l . Tuliskan nama dan tingkatan anda pada ruanganyangdisediakan. r Untuk Kesunaan PemeriksaSoalan Markah Penuha J

Markah Diperolehi

2 . Kertas soalan ini adalah dalam dwibahasa.a J.

I)

J

Soalan dalam bahasaInggeris mendahului soalan yang sepadandalam bahasaMelayu.

3 4f,

31 J 1 J

4 . Calon dibenarkan menjawab keseluruhan atau sebahagiansoalan sama ada dalam bahasaInggeris atau bahasaMelayu.

6 7 8 9

4

3Ja

4a J

10 5 . Calon dikehendakimembacamaklumat dihalaman belakang kertas soalan ini. 1l

4J

t2l3

t4l5 l6

4 4a J

2a a

t718

4

t9 202l

3a a

JJ

22 23 24 25

J Ja J

a

ruMLAH

80

Kertassoalanini mengandungi22halaman bercetak.

3472t1

[Lihat halamansebelah]

3472t1THE UPPER TAIL PROBABILITY Q(z) FOR THE NORMAL DISTRIBUTION N(0, 1) KEBARANGKALIAN HAJANG ATAS Q@)BAGI TABURAN NORMAL N(0, 1)z

0 0.4960 0.4562 0.4168

2 0.4920 0.4522 0.4t29 0.3745

J

5

6

7

8

9

2

3456 Minus / Tolak

E

9

0.0 0.s000 0 . 1 0.4602 0.2 0.4207 0.4 0.3446 06 0.2743

0.4880 0.4483 0.4090 0.3707

0.4840 0.4443 0.4052 0.3669

0.4801 0.4404 0.4013 0.3632 0.3264 0.2912 0.2578 0.2266 0.t977 0.1469

0.4761 0.4364 0.3974 0.3594 0.3228 0.2877 0.2546 0.2236 0.1949 0.I 685 0.1446

0.4721 0.4681 0.4641 0.4325 0.4286 0.424't 0.3936 0.3897 0.38s9 0.3557 0.3520 0.3483 0.3192 0.3156 0.3121 0.2843 0.2810 0.2776 0.25t4 0.2206 0.I 660 0.t423 0.2483 0.2t't7 0.I 635 0.2451 0.2148 0.t61 I 0.1 70 1

4 ,1 ,r 1 4

0.3 0 . 3 8 2 1 0.3783

0.3409 0.3372 0.3336 0.3300 0.2709 0.2090 0.2676 0.206t 0.2643 0.2033 0.155 l 0.26n 0.2005 0.1492

8 8 8 7 '7

12:116 20 12 20 116 l2lr5 t9 llll5 19 ,rlrr 18

2 4 28 ,1: 36 2 4 28 3: 36 23 -1 I J) 2 2 26 30 34 lQ l) 22 )\

0 . 5 0.3085 0.3050 0.301s 0.2981 0.2946 0 . 7 0.2420 0 . 8 0.2119 0.9 0.I 84r 1 . 0 0.I 587l.l

33 r 3 l :

i7 6 5 5 5

rolra nt0ll3 e lt2 8 11 I 8ll0 719 16 15 t4 13 t2

zole l8 16 ts t4

)t

1l

23 21 t9 t6 14 ll l0 8 789 678 566 455 344

16 21 22

29 27 25 23

0.2389 0.23s8 0.2327 0.2296 0.1814 0.1788 0.t762 0.1562 0.1 s39 0 . 13 l l r 0.095 0.0793 0.065s 0.0537 0.0436 0.0281 0.0222 0.1 12 1 0.0934 0.0778 0.0643 0.0526 0.0427 0.0214 0.0217

0.1922 0.1894 0.1867 0.1401 0.1379

0.1736 0.l7l 1

18 20

19 2l 16 l8 I7 13 1l 13 t4 ll t0

0.t357 0.0968

0.1335 0.1314 0.1292 0.t271 0.091 8 0.0764 0.0630 0.0s 16 0.0418 0.0268 0.0212 0.0207

0.1251 0.1230 0.1210 0.1 90 1

1.2 0.1 51 1l .J

0 . 1 0 9 3 0 . 1 0 7 5 0 . 1 0 5 6 0 . 1 0 3 8 0.1020 0.1003 0.0985 0.0901 0.0885 0.0869 0.0853 0.0838 0.0823 0.0749 0.0505 0.0409 0.0735 0.0495 0.0401 0.0322 0.0202 0.0721 0.0485 0.0392 0.0314 0.0708 0..M75 0.0307 0.0244 0.0694 0.0465 0.0301 0.0239 0.0681 0.0455 0.0294 0.0233 0.0618 0.0606 0.0594 0.0582 0.0571 0.0559 0.0384 0.0375 0.0367

t . 4 0.0808 1 . 5 0.0668 1 . 6 0.0548 1 . 7 0.0446 1 . 9 0.0287 2.0 0.0228 2.2 0.0 139z,J

2 : r r r

2 .1 u r It 3 i r : 617 slo olu I ol, ,lo

ro t2e 8 i 6 5 ll 10 8 7 6

13 15

rrI

:1l

,l.,1,:12

443

sI4

1 . 8 0.0359 0.0351 0.03M

0.0336 0.0329

2 . 1 0 . 0 1 7 9 0.0t74

0.0197 0.0t92 0.0188 0.0183 0.0170 0.0166 0.0162 0.0158 0.0154 0.01s0 0.0146 0.0143 0.0136 0.0132 0.0129 0 . 0 1 2 5 0 . 0 1 2 2 0 . 0 1 1 9 0 . 0 1 1 6 0 . 0 1 1 3 0 . 0 1 1 0 0.00990 0.00964 0.00939 0.00914 0.00889

0 0ll 0li

I

li2

2

2

JJl

122 l12 222

0.0107 0.0104 0.0102

1A

0.00820 0.00798

0.00755 0.00734

3 0.00842 2 2

5 s 4

0.00714 0.00695 0.00676 0.00657 0.00639 0.00604 0.00587 0.00570 0.00554 0.00s39 0.00s23 0.00508 0.00494 0.00480 2.6 0.00466 0.00453 0.00440 0.00421 0.00415 0.00402 0.00391 0.00379 0.00368 0.00357 2 . 5 0.0062 I 2 . 1 0.00347 0.00336 0.00326 0.0037 I

z2 | |

4 ul,3 2 2l

8lt0 , ln u * I

1 3 l s l8 20 23 t2 14 16 16 2l l l 1 3 15 t'/ 19

e n8 6 s4

13 l5 ll 789 566 344 334 9910

t7

0.00307 0.00298 0.00289 0.00280 0.00272 0.00264 2 . 8 0.00256 0.00248 0.00240 0.00233 0.00226 0.00219 0.00212 0.00205 0.00199 0.00193 l 0.001 87 0 . 0 0 1 8 1 0 . 0 0 1 7 5 0 . 0 0 1 6 9 0.00164 0.00159 0.00154 0.00149 0.00144 0.00139 o 3.0 0.0013 0.0013 5 I 0.00126 0.00122 0 . 0 0 1 1 8 0 . 0 0 i l 4 0 . 0 0 l l l 0.00107 0.00104 0.00100 0

rlu rlr 3l4zlt

e 7 64

12 t4

.I,I

I tlz

2 32 2

/(z)

Example /Contoh: ffx- N(0, l), then Jika X -N(0, l), maka

O(z\=l f(z\dz

Q@\

P(x> k): Q&)P(X> 2 . r \ : Q Q . r ) : 0 . 0 1 7 e

z3472t1

3

34721r

The following formulae may be helpful in answering the questions. The symbols given are the ones commonly used. Rumus-rumusberikut boleh membantuanda menjawab soalan. Simbol-simbolyang diberi adalah yang biasa digunakan. ALGEBRA I '" x:-b!\lb'- 4* 2aat xa' = e^*n

g

log"b l o g ,b - ' - o , log" a

2

g l0

T,: a+(n-lV g, = !1Zo+ ( n - l) dl LT,=ar'-l a(r'-l)_a(t-r'),r*l l-r r -l

3

a- + an : (J--'

45

(a^l =a^'ll mn:log"m+log"n 1og, m log" - Logom-log,, ,. l o g " m '= n l o g o n 4 t2

s^:

o

7

13

s* =

*-, l

f

ltl't

CALCULUS KALKULUS

I

dv dv V:LtV. --!-='y-+Vdx dx dx

du

4

Areaunder a curveLuas di bawah lengkung u,

du ,L

dv dxv'

=

lldxa

or (atau)

"

,,-uy-

dYv' dx

r

V--U

dx

b

= ^dvdvdu -dr' t d, A=

Ir

l*oY

5

Volume revolution ofIsi padu kisaranb fr

= | ry' dx or (atau) J;

: lm_d!I

3472t1

[Lihat halamansebelah]

4 STATISTICS STATISTIK x:

3472tr

>N

;l.= - LIy, t,,WI

ZI'"

'P- = '

''' (n- r)lnl (n - r)lrr.

Zr

I{ N

n(^ . = -'

l0

r(.tw a): 4e)* 4a)- \.e a n)p(x =rl'C,o'n"-', p + q = |MeanlMin, p:np o = {npq

n

[1' -.) m=L.lt , It[" )

t2 t3t4

7=x-Po

t =9x100QoGEOMETRY GEOMETRIDistance lJarakJ

{

l r |l |

t-

xt+y'

= J(x, - xr)' + (y, - yr)'Midpoint I Titik tengah

6

r=

lxA point dividing a segment a line of garis Titikyang membahagi suatu tembereng ( nx, + mx. nv, + mv.\\,)/I-r-;

xi+yi r- +y'

\

m+n

m+n

)

Area of triangleI Luas segi tiga =Ir.

;l(x,!z

* xzlt * xsh,) - (xryt * \lz + x'% )l

.l

3472/r

5

34721r

TRIGONOMETRY TRIGONOMETRI

1

s:r0 Ar c l e n g th , s:i lengkok, 0 Panjang sectorA: Lr'T Areaof , 2 sektor t: Luas , ! i'e 2-

8

AsnB sin(l i-B):sin,4cosBtcos + + B sin(,n B) : titt AkosB kosAsn t cos(,,4B) = cosI cosB T sin,4sinB sin,B kos(l+ A): kosAkosBTsin,'4

2

9

3

s i n ' A+ co s'A : IA stn2 +kos2A=l

l0

tan(lx n)=

tanA+ tanB l+_tanAtanB

4

sec2A:l+tanzA s e k 2 A = r + t a nA z A A: cosec2 | + cotz =l+ kotzA A kosekz sin 2A :2 siM cosl sn2A = 2stnAkosA c o s 2 A = c o sA - s n z A z =2cos2 -l A =l-2stn2 A kos2A = kos'A-sin' A = 2 k o sA - l 2 =l-2srnz A

II

tan2A

2tan A l-tan2 A

5

6

t2

abc

sin I

sin I

sin I

7

13

A a' : b' + c' -2bc cos a' =b' +c' -2bc kosA

14

Area of triangleI Luas segi tiga =!ab srr.C 2

347211

[Lihat halamansebelah]

6For Examiner's Use

3472t1

Answerall questions. Jawabsemuasoalan. DiagramI showsthe graphof the function -f (x):l .r + l , for the domain -3 -

For Examiner's Use

t8 t---l

11.,1of 19 Given I ftrr dr = 5 and [,'[yG)- *]ax= 8, findthevalue fr.

nilaik. dr=5 do, l,'Vrf(i-zx]dx=8,cari Diberi J'-/t..)[3 marks] 13 markahlAnswer I Jawapan:

r34721r[Lihat halaman sebelah]

19

o

3472t1Diagram 6 showsthe graph of a quadraticfunctionf (x) : (x + l)' - 16. The straightline PQwhich is parallelto thex-axisis a tangent the curve. to graf fungsi kuadratikf (x\: (x + l)2 - 16. Garis lurus PQ yang Rajah 6 menunjukknn selari denganpaksi-x ialah tangenkepadalengkung itu.For Examiner's Use

'

flx)

Diagram6 Rajah 6 (a) Statethe coordinates the minimum point of the curve. of Nyataknnkoordinat titik minimumbagi lengkungitu. (b) Write the equationof the straightline PQ. Tulispersamaan garii lurus PQ. (c) Find the rootsof the equation/(.r):0. Cari punca-puncabagipersamaanf(x): 0. [4 marks] [4 markah) Answer Jawapon'. I (a)

(b)

(c) 6

ffi

@3472tr[Lihat halamansebelahl

10For Examiner's Use

3472tr

Solve the equation 2zx14tx-21:l6'*3. ') = persamaan 2" (4t" l6'nt Selesaikan

[3 marl