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```					 MENJAWAB SOALAN
MATEMATIK TAMBAHAN

SMK SUBANG JAYA
Jalan SS14/6, Subang Jaya, Selangor

19 JULAI 2012
oleh:

Ng K.L.
SMK Tinggi Kajang
PAPER 1
( Jawab Semua Soalan )
Question 1:
(a) { (1,p), (2,r), (3,s), (4,p) } √ 2

Any one pair correct √ A1

Common mistakes:
* No bracket and wrong type of bracket
* No comma
* Wrong pairing
* Wrong position of object-image
Question 1:

(b) many – to – one √ 1

Common mistakes:
* one – to – many
* many – to – one, one – to – many
* fancy ways of writing the type of relation
* mistaken as notation of relation
Question 2:
(a)   7 √1

(b) f(x) = x + 2
or
f:x      x+2 √1
Common mistakes:
• Mistaken as the type of relation.
• Weakness in deducing the f(x) from
the diagram of the relation.
Question 3:
(a) 2, 4 √ 1

(b) { 1, 2, 3, 4 } √ 1

Common mistakes:
* No bracket and wrong type of bracket
* No comma
* Listed objects as the answer or
mistaken objects as the range
Question 4:
2
(a)   3   √1   from: 3m – 2 = 0, substitute x as m

(b) 0 ≤ f(x) ≤ 13 √ 2    Common mistakes:

13 (seen) √ A1       * 2 ≤ f(x) ≤ 13
* x = 0, f(x) = 2
*0 ≤ x ≤5
* Do not understand the
meaning of range
Question 5:
(a) p = 8, q = −3       √ 3 (both correct)

4x + 8x – 3 √ A2
2                      gf(x) = 4x 2 + px + q

(2x + 1)2 + 2(2x + 1) – 6 √ A1       gf(x) = g(2x + 1)

Weaknesses:
* Not knowing that the question is about
finding the composite function or do not
understand in finding the composite function.
* Not realise to compare the composite
functions to get the value of p and q.
Question 6:
• Majority candidates did
not include this term or
did not know how to find.
(a)   3/2 √ A1

–1           3x - 5       1
(b)   f        (x) =          , x     √2
1  2x       2

y5
x        √ A1              • Show the process of
3  2y
finding the inverse
Weakness:                     function
* Do not know how to find the inverse
function.
Question 7:
5x 2 – 3x – 4 = 0 √ A1
(convert into general form of the quadratic equation)

 ( 3)  ( 3)2  4(5)( 4)
x
2(5)             √ A2

x = 1.243, x = − 0.6434 √ 3 (both correct –
in 4 significnt
figures)
Question 8:
Common mistakes:
k<2    √3                  • 36 - 4k -4(3) > 0

• 36 – 12k + 12 > 0
6 – 4(k + 1)(3) > 0 √ A2
2

• −12k > − 48
k > - 48/-12
6 – 4(k + 1)(3)
2
therefore: k > 2
or b2 – 4ac > 0
or a = (k+1), b = 6, c = 3 √ A1
Question 9:                 Common mistakes:
•α+β =−5
2
x – 5x + 6 = 0 √ 3          •α+β =5
• α + β = − 5/2
• αβ = 3
4αβ = 6                     • x 2 + 5x + 6 = 0

2(α + β) = 5 √ A2 (both correct)

α + β = 5/2 √ A1 (either one)
αβ = 3/2
Question 10:
2                  2
b – 4ac > 0 or 4 – 4(2)(m) √ A1
42 – 4(2)(m) > 0 √ A2
16 – 8m > 0
16 > 8m
8m < 16               Common mistakes:
* − 8m > −16/−8
m<2 √3                     m>2

• 4 – 4(2)(m) > 0
(careless)
Question 11:
2
y = a(x + p) + q            minimum or maximum value

axis of symmetry

q=−1√1
p = − 2 √ 1 from: x + p = 0  2 + p = 0
2
At (0,3): 3 = a(0 – 2) + (− 2)
5 = 4a
a = 5/4 √ 1
Question 12:
2
5x – x – 4 ≤ 0 √ A1
x2 – 5x + 4 ≥ 0
2
Let x – 5x + 4 = 0
−     +

(x – 1)(x – 4) = 0
+
−     −      +
x                    x
1   4       + 1
1     −   4 4+

x = 1, x = 4 √ A2              √ A2
Then: x ≤ 1          or (x – 1)(x – 4) ≥ 0
x≥4√3           x ≥ 1, x ≥ 4 ?
(based on the graph)    Then: x ≤ 1, x ≥ 4 √ 3
Question 13:
2
f (x) = 5 + 4x – x
5 + 4x – x 2 ≤ 8
2
− x + 4x – 3 ≤ 0 √ A1        √ A2
x2 – 4x + 3 ≥ 0         1          3
x

2
Let x – 4x + 3 = 0
(x – 1) (x – 3) = 0
x = 1 , x = 3 ? √ A2
Then: x ≤ 1, x ≥ 3 √ 3
Question 14:                          Common mistakes:
* = lg2 x lg2 x lg5 ÷ lg2
lg20                        = 0.3 x 0.3 x 0.7 ÷ 0.3
log2 20 =    lg2
√ A1            * = 0.3 + 0.3 + 0.7 – 0.3

lg (2 x 2 x 5)               lg 2 2  lg 5
=                         or    =     lg 2
lg 2
lg 2  lg 2  lg 5              2 lg 2  lg 5
=                                =                   √ A2
lg 2                         lg 2
√ A2
0.3  0.3  0.7             2(0.3)  0.7
=                         √ A3 =              √ A3
0.3                          0.3
13
=         √4
3
Question 15:
Wrong Concept:
log5 125x 2
logx125x2   =               √ A1
log5 x                    log5 5 3 x 2
log5 x
log5 5 3  log5 x 2
=                     √ A2
log5 x                 3log5 x 2
=
3log5 5  2log5 x             log5 x
=                   √ A3
log5 x
3  2k
=        √4
k
Question 16:
 2n   
2 n .2 3  2 n  5       
 2     √ A1
      

n     5
2  8 - 1   √ A2
        2
19 
n
2     
 2 
n -1
2          (19) √ 3
Question 17:
(a) / AOC = 2π – 2.675 = 3.614
r=   7.2 = 1.992 ==> Use s = rθ
3.614
√2
√ A1

(b) Area Sector COB = ½ (1.992)2 x 2.675 √ A1
= 5.307 √ 2
Use A = ½ r2θ
Question 18: Coordinates Geometry
(Area of polygon)

3 1 5 3       √ A1
12 = ½
h 0 −h h

24 =     (0 – h + 5h) – (−3h + 0 + h) √ A2

6h = 24 , 6h = − 24
h=4       h = − 4 √ 3 (Both correct)
Question 19:
Line 1: 6x + 3y – 5 = 0  y = −2x + 5/3
m1 = −2 √ A1
Line 2: m2 = ½
x-intercept = a; y-intercept = −3

Then: −(−3) = ½ √ A1
a

a=6 √3
Question 20:
dy                du
(a)         = 10u or         =3     √ A1 (either one)
du                dx
dy
= 10(3x + 2) x 3   or    90x + 60 √ 2
dx

(b) δx = 0.01 √ A1
δy dy
δy = 30(3x2 + 2) x 3 x 0.01        Use     =
δx dx
= 7.2   √2                           x=2
Question 21: Differentiation
2
y = 3x – 2x
dy                                       dy
= 6x – 2 √ A1      At turning point    =0
dx                                       dx
At turning point; 6x – 2 = 0

x = ⅓ √ A2
y = 3(⅓)2 – 2(⅓)

=−⅓
Coordinate of turning point = (   ⅓, −⅓ )       √3
Question 22:
N = 9, x = 12   Use the formula to find
Σx          mean of ungrouped data:
(a) 12 =                       Σx
9                 x=
N
Σx = 108 √ 1
(b) N = 8, x = 11
Σx = 108 – p
108 − p
11 =          √ B1 p = −30 √ 2
8
PAPER 2
SECTION A, B and C
SECTION A & B
Question 1:
y = 2x – 1 √ K1
2
x + 2(2x – 1) + x(2x – 1) = 5 √ M1
2
3x + 3x – 7 = 0
 3  3 2  4(3)( 7)
x                       √ M1
2(3)

x = 1.107, − 2.107 √ A1         whether in 3 decimal places,
y = 1.214, − 5.214              or did not mention, then leave
the answer in 4 significant figures
√ A1
Question 2
x + 3y + 10 = 0    ...(1)     (y + 4) (y + 10) = 0
y = − 4,     √ M1
4    12
+     + 1 = 0 ...(2)         y = −10 √ A1
x     y
Sub. y (4):
4y + 12x + xy = 0 ...(3)        y=−4
x=2
x = − 3y – 10     ...(4) √ K1   y = −10
Sub. (4) into (3)               x = 20 √ A1

4y + 12(− 3y – 10) + (−3y – 10)y = 0 √ M1
−3y 2 – 42y – 120 = 0
3y 2 + 42y + 120 = 0
Question 3:
dy
(a)    = px2 – 4x ;
dx

Line: y + x – 5 = 0
y=−x+5
Gradient tangent = −1 √ K1

dy
dx   = −1, ==> px − 4x = −1 √ M1
2

At (1, 3);   p(1)2 – 4(1) = −1
p = 3 √ A1
dy
(b)    = 3x2 – 4x
dx

y=  ( 3 x2  4 x )dx

y = x3 – 2x2 + C        √ K1

At (1, 3); 3 = (1)3 – 2(1)2 + C
C=4       √ M1

y=x        3
– 2x + 4 √ A1
2
Question 4:
Axis of symmetry, from x value
of the coordinate given, (2, k)
(a) x = 2 √ A1
(b) f(x) = x2 + 2hx + h2 – h2 – 5 √ M1
= (x + h) – h – 5
2    2
√ K1
h = −2 √ A1           From: x + h = 0
x = 2; 2 + h = 0
k = −h – 5
2

= −4 – 5           minimum value = k
From the y value of
= −9 √ A1          coordinate given, (2, k)
Question 4(c):
From: f(x) = x2 + 2hx – 5

• a = 1 > 0; a minimum graph
f(x)
• y-intercept = 5
• axis of symmetry, x = 2                x=2
• minimum value = −9                                    √ A1
• minimum point = (2, −9)
x
0      2

√ A1    −5
−9
(2, −9)   √ A1
Question 5: Statistics
f              x              fx               fx2
( midpoint )

7            133             931            123823
8            138             1104         152352
13            143             1859         265837
10            148             1480         219040
8            153             1224         187272
4            158              632          99856
Σ f = 50          -           Σfx = 5 890 Σfx2= 1 048 180
√ M1            √ M1
Based on the formula of mean and variance, prepare the above
table for a much easier procedure of solving the problem
Question 5: (continue)
√ M1
5 890                                  Σfx
(a) mean =           = 117.8 √ A1              x=
Σf
50
1 048 180
(b) variance =                     (117.8)2 √ M1
50
= 7 086.76 √ A1                  2  Σfx2
σ =      − x2
Σf

(c) Std. dev. = 84.18       √ A1
σ = √ σ2
Question 6: Coordinate Geometry
y–3 6–3
(a)         =       √ K1
0 – 6 15 – 6

y = 1 √ A1
then; A = (0, 1)
7–1
Gradient AE, mAE =     =2      √ M1
3–0

Eqn. AE: y – 7 = 2( x – 3 )
y = 2x + 1 √ A1
Question 6: (continue)
2(x) + 3(3)          E (3, 7)
(b) 5 =
2+3                2      B (5, 3)

√ M1 (either one)           3
D (x, y)
2(y) + 3(7)
3=
2+3

x=8,       y = −3 √ A1         (either one)

D ( 8, – 3 )   √ A1
Question 6: (continue)
5 15    8    5   √ K1
(C) Area ∆BCD = ½
3   6   −3   3

= ½ (30 – 45 + 24) – (−15 + 48 + 45) √ M1
= 34.5 √ A1
Question 7: Linear Law
(a)
x     1.5     3.0 4.5        6.0   7.5 9.0
log10 y 0.40     0.51 0.64      0.76 0.89 1.00

√ A1

•The value of log y must be ≥ 2 decimal places

• If table not shown, the A1 mark can be given on
all the points plotted correctly on the graph
Plot graph log y against x log y
log y
Not using the given scales – minus 1
Scales not uniform – minus 1
Not using graph paper – minus 1

√ M1
1.0                                                                 +
+
0.8                                                 +
√ A1
+
0.6           √ A1        +
+
0.4
√ M1
√ M1                                            √ A1
0.2 √ A1
√ A1
0.0                                                                     x
0      1     2        3      4         5     6       7       8   9
√ M1
(b) Plot log10 y against x
(correct axes and uniform scales)
6 points plotted correctly            √ A1

Lines of best fit √ A1
(c) (i) y = 4.8; log 4.8 = 0.68,
√ A1
x = 5.6 ~ 5.7
log10 y = log10 h + 2x log10 k √ K1
(ii) c = log10 h √ M1
√ A1
h = 1.78
(iii) m = 2 log10 k √ M1
k = 1.09 to 1.12     √ A1
Question 8. Circular Measures
o       o            o
(a) θ = 60 + 60 = 120               √ K1

B

60 o            π
= 120     = 2.095               √ A1
C       A              180

D

(b) Area ABCD = 2 ( ½ ) (4)2 sin60o                    √ M1

= 13.86 √ A1
Use Solution of Triangles Principles
(c) Area shaded region       √ M1

= Area Sector ABC – Area ∆ ABC
2
= ½ (4) (2.095) – ½ (Area ABCD)
2
= ½ (4) (2.095) – ½ (13.86)
√ M1 √ M1
= 9.832               √ M1

√ A1
Question 9: Index Number
Q2010
(a) Use               x 100 = 140
Q2008
√ P1
140
(i)    Q2010 =           x 8 = 112 √ A1
100

(ii)           I        W           IW                Σ IW
Use        = 132
P   140          30         4200                ΣW
Q       m        20         20m           9890 + 20m
R   110          15         1650                     = 132
100
S   104          10         1040         √ M1
m = 165
T   120          25         3000
√ A1
ΣW =100 ΣIW = 9890 + 20m
Question 9 (continue…)
(b)   Q2012
x 100 = 110 √ P1
Q2010
√ P1
110
Q2012 =         x 60 = 66       √ A1
100

Q2008       125         Q2010       132
(c)           =         and           =
Q2006       100         Q2008       100

Q2010         132 125
x 100 =    x    x 100 = 165 √ A1
Q2006         100 100
√M1   √M1
Question 10: Solution of Triangle
A
(a)            30o                Use Cos Rule.                √ M1
9               7
BC2 = 9 2 + 72 – 2(9)(7) cos 30o
B                       C
= 20.881 √ A1
BC = 4.570 √ A1
(b)
A
Use Sin Rule.
7            25 o       D    Sin / CAD       Sin 25
=            √ M1
11                     11              7
C
sin / CAD = 0.6641 √ A1

/ CAD = 41.61o √ A1
Question 10: (continue)

(c) Area ABCD √ M1
= ½(9)(7) sin30o + ½(7)(11) sin113.39o
√ M1     √ M1
= 39.84   √ A1
TAMAT
SELAMAT BERJAYA
TERIMA KASIH

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