# Menjawab Soalan SPM Matematik Tambahan

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```					MENJAWAB SOALAN SPM

12 MAC 2012
FORMAT KERTAS SOALAN
PAPER 1
   25 soalan – jawab semua soalan.
   80 markah.
   Markah setiap soalan antara 2 hingga 4.
   Masa 2 jam (120 minit).
   Purata masa menjawab 1 soalan = 4.8 minit.
   Condong ke arah format objektif.
   Menguji tahap pengetahuan dan kefahaman.
asas tajuk-tajuk yang diuji.
Pemarkahan Kertas 1
1. Diagram 1 shows the relation between set P
and Q.
State
(a) object of 5,
(b) type of relation.            [2 marks]

(b) …………….(1)
6. Diagram 2 shows the graph of function
y = (x – p) – 1, where p is constant.
Find,
(a) the value of p,
(b) the equation of axis of symmetry,
(c) the minimum point.                [3 marks]

(b) …………….(1)
(c) …………….(1)
17. Given cos θ = t, 00 < θ < 900,
express in terms of t.
(a) sec θ,
(b) cos (900 – θ)           [3 marks]

(b) …………….(2)
11. Diagram 3 shows a straight line graph
of 1/y against 1/x .
Variables x and y ere related by equation
y = …………..
Find the values of a and b.       [3 marks]

b = ………….}(both)
16. Given P(1, 2), Q(8, - 4), a = i + j and
b = 2i – 3j. If PQ = ma + nb, with m and
n as constant.
Find the values of m and n. [3 marks]

n = ………..} (both)
9. Given the first three terms of an
arithmetic progresssion are x, y – 1
and 5 + 2x. Express y in terms of x.
[3 marks]

10. The first term and the third term of a
geometric progression are 108 and 12
respectively. Find
(a) the common ration,
(b) the sum of infinity of the
progression.               [4 marks]

(b) …………….(2)
2. Given f(x) = 4x + 2 and g(x) = x2 – 3x – 1,
find
(a) f -1(x),
(b) gf(2)                            [4 marks]

(b) …………….(3)
8. Given 2 log2M = 2 + 4 log4N, express
M in terms of N.           [4 marks]

PAPER 2
   100 markah.
   3 Sections, A, B dan C.
   Masa 2 jam 30 minit (150 minit).
   Menguji tahap pengetahuan, kefahaman
dan aplikasi.
   Langkah kerja matematik yang sistematik
perlu ditunjukkan.
   Markah diberi kepada jawapan dengan
langkah kerja matematik yang betul.**
Markah diberi untuk jawapan yang betul sahaja

Markah diberi untuk kerja yang betul mengikut

K1   Markah diberi untuk tahap pengetahuan

M1   Markah diberi untuk kaedah kerja

A1   Markah diberi untuk nilai
1.   2.   3.   4.   5.
K1   K1        K1   M1

A1   M1
M1   A1   A1

A1   A1        A1   A1
CONTOH RANGKAIAN MARKAH

a

If a = 0, then b = 0 and c = 0
b
If a = 1 and b = 0, but c can be 1 or 0

c
a
If a = 0, then b = 0, c = 0

If d = 0, then c = 0
b
If d = 1, and a = 0, then c = 0
d

c
SECTION A:
 40 markah.
 6 soalan – jawab SEMUA soalan.
 Setiap soalan antara 5 hingga 8
markah.
 Purata masa setiap soalan
= 0.4 x 150 / 6 = 10 minit.
SECTION B:
 40 markah.
 5 soalan tetapi jawab mana-mana 4
soalan.
 Boleh jawab semua 5 soalan, markah
akan dikira dari 4 soalan dengan
markah tertinggi.
 Setiap soalan 10 markah.
 Purata masa setiap soalan
= 0.4 x 150 / 4 = 15 minit.
SECTION C:
 20 markah.
 4 soalan. Jawab mana-mana 2 soalan.
 Boleh jawab semua 4 soalan tetapi markah
dikira dari 2 soalan dengan markah tertinggi.
 Setiap soalan 10 markah.
 Purata masa setiap soalan
= 0.2 x 150 / 2 = 15 minit.
 Soalan digubal hanya dari 4 tajuk yang telah
ditetapkan, 2 tajuk dari tajuk Form 4 dan 2
lagi dari tajuk Form 5.
PAPER 1

Form 4:
   Simultaneous Equation
   Solution of Triangles
   Index Number

Form 5:
 Motion in Straight Line
 Linear Programming
PAPER 2
Section A: (40 marks)       Section B: (40 Marks)
Form 4:                     Form 4:
* Functions                  * Coordinate Geometry
* Quadratic Functions        * Circular Measures
* Simultanous Equations      * Differentiation
* Coordinate Geometry
* Statistics
* Circular Measures
* Differentiation

Form 5:                     Form 5:
* Progression                * Linear Law
* Integration               * Integration
* Vectors                   * Vectors
* Trigonometric Functions   * Probability Distribution
PAPER 2
Section C: (20 Marks)

Form 4:
• Solution of Triangles
• Index Numbers

Form 5:
• Motion in Straight Line
• Linear Programming

• Indices & Logarithms
• Permutations & Combinations
• Probabilty
CONTOH
JAWAPAN DAN PEMARKAHAN
SPM

PAPER 1
in this blog
Q.1:

(a) -1, 1/ -1 and 1 / {-1, 1} ..√(1)
reject: -1 or 1 / (-1, 1) / [-1, 1]

(b) many-to-one /                     ….√(1)
many with one /
many → one
Q.2(a):
(a)    f(x) = 4x + 2
f -1(x) = y
f(y) = x
= 4y + 2   √ -----------(Z1)
x = 4y + 2

x-2
y = ———         √------------(2)
4
Q.2(b):
(b) gf(2) = g[f(2)]
= g[4(2) + 2]     √………..………..(Z1)
= g(10)
= 102 – 3(10) – 1
= 69              √…………………..(2)

f(2) = 4(2) + 2
= 10             √ ………………….(Z1)
Q.3:
   gh(x) = g[h(x)]
= mx2 + n – 3 ……..(Z1)
= 3x2 + 7
Bandingkan:
m=3            ……...(Z2) - either one
n–3=7             ………(Z2)
n = 10        ………(3) (both m and n correct)
Q.4:
   2x2 + p + 2 = 2px + x2
x2 – 2px + p + 2 = 0           √ ………..(Z1)
    a = 1, b = - 2p, c = p +2
    b2 - 4ac > 0
   (-2p)2 – 4(1)(p + 2) > 0
4p2 – 4p – 8 > 0              √ ………..(Z2)
p2 – p – 2 > 0
          Let: p2 – p – 2 = 0                       x
(p – 2)(p+ 1) = 0     -1         2

p = 2 atau p = - 1

   Therefore: p > 2 atau p < -1 √ ..…….(3)
Q.5:
   (x – ⅔)(x + 4) = 0      ….…….(Z1)
x2 – ⅔ x – 4x – 8/3 = 0
3x2 + 10x – 8 = 0       …………(2)

OR:
x2 – (⅔ - 4)x + (⅔)(-4) = 0 …………(Z1)
3x2 + 10x – 8 = 0           …………(2)
Q.6:

 (a) p = 3 (axis of symmetry)√.….(1)
 (b) x = 3                   √….(1)
 (c) (3, -1)                 √….(1)

   Note: Minimum value = -1
Q.7:
   32x + 1 = 4x
   (2x + 1)log10 3 = xlog10 4       √ …..…….(Z1)
   2xlog10 3 + log10 3 = xlog10 4

   2x(0.4771) – x(0.6021) = - 0.4771 √.………..(Z2)
   0.9542x – 0.6021x = - 0.4771
   0.3521x = - 0.4771
- 0.4771
          x = ————
0.3521

             = - 1.355              √ ………...(3)
Q.8:
   2log2M = 2 + 4log4N
4log2N
2log2M = 2 + ———          √……….(Z1)
log24

log2M2 = log24 + 2log2N   √..……..(Z2)

log2M2 = log24N2          √..….…(Z3)
M2 = 4N2
M = √4N2
M = 2N                 √..…....(4)
Q.9:
 (y – 1) – x = (5 + 2x) – (y – 1) √....(Z1)
 y – 1 – x = 5 + 2x – y + 1
         2y = 3x + 7             √….(Z2)
3x + 7
          y = ———                √….(3)
2
Q.10(a):

    a = 108

ar2 = 12

ar2 = 12    √ ……….(Z1)
a     108

r2 = ¹/9

r = ¹/3    √..……….(2)
Q.10(b):
108
S∞ = ——— √…….….(Z1)
1 – ¹/3

108
= ———
⅔

=   162   √…..…….(2)
Q.11:

x
   y =     ————
a + bx
a + bx
¹/y = ————
x

¹/y =   a/       +b   √ ……….(Z1)
x

= - 6/4 = - 3/2    √ ……...(Z2) – either a or b correct

b = y-intercept
= 6                √……….(3) – both a and b are correct
Q.12:

 y/2 - y/4 = 1
4x - 2y =        8      √..……(Z1)
2y =      4x - 8
y =    4/ x – 4
2
y–7=       2(x – 2)
y =   2x + 3 √.……(3)
Q.13:
              4m + 9s                        2m + 6t √…. (Z1)
S   =                      or      t=
5                          5
P(2m, m)                Q(s, t)       R(3s, 2t)

3                  2

5s – 9s = 4m,           or 5t – 6t = 2m
- 4s = 4m,                 - t = 2m √……(Z2)
- s = m,
- 2s = - t
s = t/2             √……………………...(3)
Q.14:
   PA : PB = 2 : 1                       √..…(Z1)
reject:   PA/      = 2 /1
PB
   PA = 2 √(x – 5)2 + (y – 0) 2 √…..(Z1) or
PB = √(x + 2)2 + (y + 3)2               √….(Z1)
   PA2 = 4[(x – 5)2 + y 2 ] =
PB2 = (x + 2)2 + (y + 3)2                √....(Z2) or
   4[(x – 5)2 + y 2 ] = (x + 2)2 + (y + 3)2 √…..(Z2)
   3x2 + 3y2 - 44x + 6y + 87 = 0 √....(3)
Q.15
D                   C
AC = AB + BC
= AB - CB                 .........(Z1)
A                  B                 = (i + 2j) – (-5i – 6j)

= 6i + 8j                 ….…(2)

6i + 8j
Unit vector AC = ————                        ……..….(Z1)
√ 6 2 + 82

6i + 8j 3i + 4j
= ———— or ————                …..……(2)
10       5
Q.16
PQ = PO + OQ
P(1, 2)
= - OP + OQ
O                          = - (i + j) + (8i + 4j)
= 7i + 3j               ….(1)
Q(8, -4)

7i + 3j = ma + nb
= m(i + j) + n(2i – 3j)
= (m + 2n)i + (m – 3n)j   ……………….…..(Z1)

Compare:   m + 2n = 7              5n = 4        .….(Z1)
m – 3n = 3              n = 4/5 )..(Z2 either)
m = 27/5 )..(3 - both)
Q.17:
   sec θ = 1/cos θ
1
1 – t2

         = 1/t   ………(1)
t

cos (90 – θ) = sin θ       ……….(Z1)

= √ 1 – t ……….(2)
2
Q.18:
   2 sec2θ – 3tanθ = 4
   2(1 + tan2θ) – 3tanθ = 4                 …..…..(Z1)
2 + 2 tan2θ – 3tanθ = 4
2tan2θ - 3tanθ – 2 = 0
   (2tanθ + 1)(tanθ – 2) = 0                ………..(Z2)
   tanθ = -½ atau tanθ = 2
θ = (360o – 26o 34’), (180o - 26o 34’)
   θ = 333o 26’, 153o 26’                   ………...(Z3)
θ = 63o 26’, (180o + 63o 26’)
   θ = 630 26’, 2430 26’                 ….....….(Z3)
   θ = 630 26’, 153o 26’, 2430 26’, 333o 26’. …….….(4)
Q.19:
   ̸̱ BOC = π – 1
= 3.142 – 1
= 2.142       ………..(Z1)

s = jθ
BC = 15 x 2.142    ………..(Z2)
= 32.13         ………..(3)
Q.20:
2
dy     (x – 4)(4x) – (2x + 3)(1)
— = ———————————                         ….…(Z2)
dx             (x - 4)2
4x or 1   …….(Z1)

2
2x – 16x - 3
= ———————                          ..……(3)
(x - 4)2
Q.21:
   dy/dx = 4x – 3                 ………..(Z1)
      δx = 0.01                   ………..(Z1)
       x = 2


δy∕      ≈ dy∕dx
δx

          δy ≈ dy∕dx (δx)

             = [4(2) – 3](0.01)   …………(Z2)
             = 5(0.01)
             = 0.05               …………(3)
Q.22:
3           3
   ∫ f(x) + ∫ (kx)dx = 20 .......(Z1)
1           1
3
       8 + [ kx2 ∕ 2 ]        = 20    …..(Z2)
1

3
            [ kx2 ∕ 2 ]       = 12 …..(Z2)
1

9k/2 – k/2 = 12,         8k/2 = 12   …..(Z3)
k = 3     …..(4)
Q.23:

6                      4
P           or    P             ….(Z1)
4                  3

6               4
P x P = (6 x 5 x 4 x 3) x (4 x 3 x 2) ..(Z2)
4           3

= 360 x 24
= 86400      ……...(3)
Q.24:
9               5
   (a)        C       x       C                                           ……….(Z1)
6               4

atau 9 x 8 x 7
3x2
     = 84                                                                 …………(2)

5           9               5       9           5       9
   (b)        C x C               +   C x C           +   C x C           …...….(Z1)
5              5           4       6           3       7

= 948                                                            …………(2)
Q.25:

   0.5 – 0.225   ……….(Z1)
 = 0.275         ……….(2)
PENUTUP
SEBAB-SEBAB CALON TAK CAPAI
KEPUTUSAN YANG DIHARAPKAN

1.    Kurang Latihan
2.    Kurang sistematik
3.    Asas matematik lemah
4.    Kurang kefahaman tajuk yang dipelajari
5.    Tidak beri tumpuan semasa P&P
6.    Tak hadir sekolah / ponteng
7.    Suka buat ‘short cut’ utk berjaya
MENJAWAB SOALAN
SPM
THE ENDPAPER 1
MENJAWAB SOALAN
SPM
PAPER 1

THANK YOU

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