# Skema jawapan Kertas 1 Matematik Tambahan SPM Number Solution by act50979

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```									                               Skema jawapan Kertas 1 Matematik Tambahan SPM

Sub   Fu
Number                                  Solution and Marking Scheme
Marks Mar
1      7                                                                       2
2(1)  5                                                             B1     2

2      g ( x)  2 x  1                                                        3
g ( y )  2( x  3)  5                                                B2
3
y  x3                                                                B1
3             3               1
m  and k                                                            3
4               4
3             1                                                 B2
m  or k  
4             4
3 x
f 1 ( x )                                                           B1    3
4 4
4 (a)   2x  x  1  0
2                                                                    2
B1
2( x 2  2 x  1)  5 x  3

(b)    h9                                                                     2
2(4)  h(4)  4  0
2                                                                 B1    4
5      p 1                                                                    3
 4 p  4                                                             B2
3
(2) 2  4(1)( p )  0                                                 B1
6 (a)   k=2                                                                    1
(b)   x=2                                                                    1
(c)   (2, 3 )                                                               1     3

1                                                                     3
7      pq
1
B2
log a 5  log a 7
1
log a 35                                                               B1    3

9      x2 + y2 – 4x – 14y + 37 = 0.                                           3
(x – 2)2 + ( y – 7)2 = 42                                              B2
or equivalent x2 – 4x + 4 + y2 – 14y + 49 = 16

AP = 4     or       ( x  2) 2  ( y  7) 2  4                        B1
3

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Sub   Fu
Number                                      Solution and Marking Scheme
Marks Mar
10           3                                                            3
y=      x4
4
3
Gradient of line perpendicular to AB, m =                         B2
4
4
Gradient of AB:                                                  B1    3
3
11     244                                                               3

2 3
x         2

 72
B2
4
B1
x= 7                                                                    3
12
15
r    cm                                                          3

10
r
2                                                              B2

3
4    2
AOB  2   OR                                                B1
3
3    3
13      3x  1                                                           3
2x 1
x
2x 1                                                          B2
2x 1
1
1             
2 x  1  x( )(2)(2 x  1) 2                                    B1    3
2
14            1                                                          3
P(2, )
2
2(2 x  5)  2 or x  2                                          B2
dy
 2(2 x  5)                                                  B1    3
dx
15          27                                                           2
y 2
x
3
3    93                  3
y  x   or    or a  x and r                                 B1    2
x    x x                 x

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Sub   Fu
Number                                 Solution and Marking Scheme
Marks Mar
16     17                                                            3
x6                                                          B2
3
(4 x  1)  (3 x  1)  (6 x  3)  (4 x  1)                B1
17    47                                                           3
99
0.47
B2
1  0.01
0.47  0.0047  0.000047  ...                               B1    3
18     k  8, h  1                                                  3
k  1 6   2 or 3  1 h   2                             B2
y                                                                  3
 2 x                                                    B1
x2
19 (a)   5                                                           1

(b)   12.5                                                        3
4
3 
10   x 2                                                  B2
 2 1
4           4
2  g ( x)dx   3xdx                                        B1    4
1              1
20       10i   13 j                                                 3

269    269

3a  b  10 2  13 2  269                                   B2

10 
 
13                                                         B1    3
 
21       
        3                                                 3
PY = b  a
2

           1
PY   a  b  ( a )                                        B2
2

                1
PQ   a  b or QY   a                                     B1    3
2

22      210 , 330                                                  4
1
sin x =  , sin x =  2 ( both)                              B3
2
(2sin x  1)(sin x  2)  0
B2
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Sub   Fu
Number                                Solution and Marking Scheme
Marks Mar

cos 2 x  1  2sin 2 x                                      B1    4

23 (a)   924                                                         1

(b)   112                                                         3
5
C4 7 C2  5C5 7 C1
B2
5
C4 7 C2 or 5C5 7 C1
B1    4
24 (a)                                                                2
13
25                                                          B1
3 3 2 2
  
5 5 5 5
2
24
(b)                                                                B1    4
125
2
 2 3
2   
5 5
25 (a)   1.667                                                       2
54  50
Z                                                          B1
2.4

(b)    0.9505                                                      2
1  0.00177  0.04776
43  50      54  50
P(          Z          )
2.4          2.4                                      B1    4

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