# Program Permata Harapan SMK Jalan 4

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```					    PROGRAM
PERMATA HARAPAN
MATEMATIK TAMBAHAN
(TINGKATAN 4)

SMK JALAN EMPAT
BANDAR BARU BANGI
25 & 26 JUN 2012
oleh:
nklpunya
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         5  3x      1
(b)   f                     , x
2 √2
(x) =
2x  1

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) √ A2 or
x                                   use completing
2(5)                  the square method

x = − 0.6434, x = 1 .243 √ 3 (both correct)
Common mistakes:
1.                                 2. Answer not in four
 3   3 2  4 x 5 x  4      significant figures
x                                 or not round-off
2x 5                   correctly
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:
2
x – 5x + 6 = 0 √ 3         Common mistakes:
•α+β =−5
•α+β =5
4αβ = 6                    • α + β = − 5/2
• αβ = 3
2(α + β) = 5 √ A2 (both)   • x 2 + 5x + 6 = 0

α + β = 5/2 √ A1 (both correct)
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   Concept Mistakes:
              2n + 3 − 2n + 2n + 5(2n – 1)
n     5
2  8 - 1   √ A2                 * 2n + 3 − 2n + 2n + 10n – 1
        2
19 
n                             * 2n + 23 − 2n + 10n + 5−1
2     
 2 
n -1
2          (19) √ 3
Question 17:
34 or 35 √ A1       Wrong Concept:
34x – 4 = 35 + x    (3 4 ) x - 1
or                                  35
3x
33x – 4 = 35 √ A2
3 4x - 1
4x – 4 = 5 + x                      35
or                     3x
3x – 4 = 5 √ A3     3x − 1 = 5
x=2
x=3 √4
Question 17 - Wrong Concept: (continue)

4x – x – 1 – 5
*=3
= 33x – 6
* log3 4x – 1 – x = log353

( 34 ) x - 1 5
*    x        =3
3
4x – 4
x     =5
Question 18:
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 19:
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)
Question 20:
N = 9, x = 12   Use the formula to find
Σx          mean of ungroup 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
Question 1:
y = 2x – 1 √ K1
x 2 + 2(2x – 1) + x(2x – 1) = 5 √ M1
3x 2 + 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 √ A1         or did not mention, then leave
the answer in 4 significant figures
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(a):
(a)     x = y 2 - 2 √ M1          Let f -1(x) = y
y2 = x + 2                       f (y) = x
f -1 (x) = √ x + 2 √ A1
Let x = an integer value,
f -1 (x) = the value after substitution, √ M1
 an object has two images or
it’s a one-to-many relation or
the function undefined.
Therefore the invest function does not exist.
√ A1
Question 3(b):
Let the first function or
the ‘inner’ function
(b) Let h(x) = y
g(y) = 6x + 7     From: g[h(x)] = 3x + 4

g(y) = 3y + 4     From: g(x) = 3x + 4

3y + 4 = 6x + 7 √ M1
y = 2x + 1
h(x) = 2x + 1 √ A1
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                                   √ N1
• minimum point = (2, −9)
x
0      2

√ N1 −5
−9
(2, −9)   √ N1
Question 5:
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:
(a)     y–3 6–3
=       √ K1
0 – 6 14 – 6
y=¾
√ A1
then; A = (0, ¾)
12 – ¾   15
Gradient AE, mAE =        =    √ M1
3–0      4
Eqn. AE: y – 12 = 15 ( x – 3 )
4

y= 15 x + ¾ √ A1
4
Question 6: (continue)
2(x) + 3(3)           E (3, 12)
(b) 6 =
2+3                 2    B (6, 3)

√ M1 (either one)          3
D (x, y)
2(y) + 3(12)
3=
2+3

x= 21 , y = − 21 √ A1 (either one)
2         2

D( 21 , − 21 ) √ A1
2     2
Question 6: (continue)
6 14 21/2 6    √ K1
(C) Area ∆BCD = ½
3 6 −21/2 3

= ½ (36 + 147 + 63/2) – (−63 + 63 + 42)
√ M1
= 86.25 √ A1
TAMAT
SELAMAT BERJAYA
TERIMA KASIH

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