PROGRAM 2Q PPDHL 2011
JAWAPAN DAN PEMARKAHAN
Additional Mathematics
SPM
PAPER 2
Disediakan oleh:
Ng Kok Lye
( SMK Tinggi Kajang )
Zarinah bte Zakaria
( SMKA Maahad Hamidiah )
TOPIC:
SIMULTANEOUS EQUATIONS
1. 2005 / SECTION A / Q1
x + ½ y = 1 ….(1)
y2 – 10 = 2x …(2)
(1) x 2: 2x + y = 2 …(3)
(3): 2x = 2 – y OR y = 2 – 2x .…(4) √ K1
Eliminate x or y by substitute (4) into (2)
y2 – 10 = 2 – y √ M1 OR (2 – 2x)2 – 10 = 2x √ M1
factorise y + y – 12 = 0 4x2 – 10x + 6 = 0
2
(y + 4)(y – 3) = 0 √ M1 2x2 – 5x + 3 = 0
y = − 4, y = 3 (2x + 1)(x – 3) = 0
√ M1
√ A1
x = 3, x = − ½ x = 3, x = − ½ √ A1
√ A1 y = −4, y = 3
√ A1
OR using quadratic formula method:
y2 + y – 12 = 0 2x2 – 5x + 3 = 0
a = 1, b = 1, c = −12 a = 2, b = −5, c = 3 √ M1
√ M1
1 12 4(1)( 12)
(-5) (-5) 2 4(2)( 3)
y= x=
2(1) 2(2)
y = − 4, y = 3 √ A1 x = 3, x = −½ √ A1
x = 3, x = −½ y = − 4, y = 3
√ A1 √ A1
OR using completing the square method:
2. 2006 / SECTION A / Q1
2y + x = 1 ……(1)
1- x
x = 1 – 2y OR y=
2
……(3) √ K1
x2 + 2y2 + xy = 5 ……(2)
Eliminate x or y by substitute (3) into (2)
1- x 1- x
2y + (1 – 2y)
2 2+ y(1 – 2y) = 5 OR 2( )+x 2 + ( )x=5
2 2
4y2 – 3y – 4 = 0 √ M1 x2 + x – 8 = 0
( -3) ( -3) 4( 4) ( 4)
2
( -1) ( -1) 4( 1) ( 8)
2
y=
2( 4)
M1 x=
2( 1)
√
√
y = −0.693, y = 1.443 (3 d.p) A1 x = −1.886, x = 2.386 (3 d.p)
x = −1.886, x = 2.386 (3 d.p) y = −0.693, y = 1.443 (3 d.p)
√ A1
3. 2008 / SECTION A / Q1
y+4
y = 3x – 4 or x= ——— ………….√ K1
3
Eliminate x or y:
(3x – 4) 2 + x(3x – 4) – 40 = 0 …………√ M1
3x 2 -7x – 6 = 0
factorization (3x + 2) (x – 3) = 0
or using quadratic formula ………..√ M1
or completing the square method.
x = - 2/3 , x = 3. ……………………√ A1
y = - 6, y = 5 ……………………√ A1
K1 can also be obtained in the substitution
process, if not stated explicitly.
( 3x – 4 )2 + ( 3x – 4 )x – 40 = 0
K1
2 y 4
y +( 3 )y – 40 = 0
Obtained K0 but substituted correctly into the
correct equation – got first M1.
2 because straight
(y+ 3 )( y – 3 ) = 0 M0 – from calculator
M1 - if substituted correctly into
quadratic formula even the values of a,
b and c are wrong from the first M .
Copy wrongly the question, but correct
in solving the problem
– marks giving are K0M1M1A0A1.
No working/steps to solve QE – minus
1 mark
4. 2009 / SECTION A / Q1
k – 3p = −1 …..(1)
p + pk – 2k = 0 …..(2)
1k
(1): k = 3p – 1 …..(3) OR p = ….. (3) √ K1
3
Eliminate p or k by substitute (3) into (2)
1k 1k
p + p(3p – 1) – 2(3p – 1) = 0 OR 3 +( 3 )k – 2k = 0
√ M1
3p2 – 6p + 2 = 0 k2 − 4k + 1 = 0
( -6) ( -6) 4( 3) ( 2) √
2
M1 ( -4) ( -4) 4( 1) ( 1)
2
p= k=
2( 3) 2( 1)
p = 0.423, p = 1.577 √ A1 k = 0.269, k = 3.731
k = 0.269, k = 3.731 √ A1 p = 0.423, p = 1.577
TOPIC:
TRIGONOMETRIC
FUNCTIONS
1. 2004 / SECTION A / Q3
y Shape of graph √ K1
(a) y = cos 2x Maximum 1 & Minimum – 1 √ K1
1 ½ Period √ K1
0 x
90o 180o
−1 √ M1
√ K1
(b) 2
2s in x 1 c os 2x 1 y 2
x x
y 1
180 180
X 0 180o
Number of solution = 2 √ A1 y −1 0
2. 2005 / SECTION A / Q5
(a) LHS = 1 + cot2x – 2 sin2x – cot2x
= 1 – 2 sin2x √ M1
= cos 2x √ A1
= RHS
1 2
or LHS = 2 – 2 sin2x – cos x
sin x 2
sin x √ M1
= 1 – 2 sin2x
= cos 2x √ A1
(b) (i)
y Shape of graph √ K1
1 Maximum 1 & Minimum – 1 √ K1
x 2 Periods √ K1
0 π 2π
−1
√ M1 ( Straight line - must pass through π )
x 1
(ii) Straight Line Equation: y √ A1
3π 3
Number of solution = 4 √ A1
3. 2006 / SECTION A / Q4 (Edited)
(a) y Π
Basic shape of graph √ K1
y
x √ M1 Reflection graph √ K1
2 x Maximum 2 & Minimum – 2 √ K1
x 2 Periods √ K1
x
0 x
π 2π
−2
y = - 2 sin 2x
To plot curve y = π/x
(b) Π
2 s i n2 x 0 y
Π x π/2 π 2π
√ K1 y 2 1 ½
x x
No. of solution = 4 √ A1
4. 2007 / SECTION A / Q3
(a) y
Basic shape of graph
Reflection of graph
√ K1
√ K1
y = | 3 cos 2x | Maximum 3 & Minimum –3 √ K1
2 Periods √ K1
3
x
x
0 x
π 2π
−3
x x x √ M1
(b) 2 3 c o s2 x 2y y 2
2Π 2Π 2Π √ K1
X 0 2π
No. of solution = 8 √ A1 y 2 1
5. 2010 / SECTION A / Q2 (Edited)
y
y = 1 + 3 sin x
4
(a) (b) 6 s i nx 4π 3 x
3
4π 3x
3 s i nx
1 2π 2π
0 x 3x
π
√ M1 1 3 s i nx 3
2π
2π
−2
( Straight line must pass through at y-intersect = 3 and 2π ) 3x
y 3 √ K1
2π
Basic shape of graph √ K1 X 0 2π
Y-intersect at 1 √ K1
Maximum 4 & Minimum – 2 √ K1
y 3 0 √ A1
1 Period √ K1 No. of solution = 3
TOPICS:
* FUNCTIONS,
* QUADRATIC EQUATIONS,
* QUADRATIC FUNCTIONS,
* COORDINATES GEOMETRY,
* DIFFERENTIATIONS
1. 2005 / SECTION A / Q2
dy
(a) dx
= px2 – 4x ; Line: y + x – 5 = 0
y=−x+5
Gradient tangent = −1
dy
√ K1
dx = −1 ==> px2 − 4x = −1
At (1, 3); p(1)2 – 4(1) = −1 √ M1
p = 3 √ A1
dy
(b) dx = 3x2 – 4x
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 = x3 – 2x2 + 4 √ A1
2. 2006 / SECTION A / Q2
x 2 (c) hg(x) = 2x + 6
(a) f −1(x) = 3
√ A1 Let g(x) = y
x
And g(x) = 1
(b) f −1g(x) =f −1( x 1 ) 5
5
x
x y= 1
( 1) 2 5
= 5 √ M1
3
x = 5y – 5
And: √ M1
x 1 5
= √ A1 hg(x) = h(y) = 2x + 6
15
M1 √ = 2(5y – 5) + 6
h(y) = 10y – 4
h(x) = 10x – 4 √ A1
3. 2007 / SECTION A / Q4
dy 2
(a) At turning point, = 2x
x2 = O √ K1
dx
2x3 – 2 = 0; ==> At(k, 8), 2k3 − 2 = 0
√ M1 k = 1 √ A1
d 2y 4 d 2y
(b) = 2 ==> At (1, 8), =6>0
dx 2 x3 dx 2
√ K1
(1, 8) is a minimum point. √ A1
(c) y = ( 2x 2x2 ) dx
2
8 1 c
2
y = x2 + 2x-1 + c ==> At (1, 8), 1
√ K1 c = 5 √ M1
y x2
2
x
5 √ A1
4. 2008 / SECTION A / Q2
(a) At y-axis, x = 0, f(x) = −0 + k(0) – 5 = 5
A (0, 5) √ A1
(b) f(x)= −[x2 − kx + (−k/2)2 – (−k/2)2] – 5
= −[(x – k/2)2 – (k2/4)] – 5
√ M1 = −(x – k/2)2 + (k2/4) – 5 √ K1
k/2 = 2 ==> k = 4 √ A1
p = (42/4) – 5 ==>p = −1 √ A1
(c) − x2 + 4x – 5 ≥ −5 x
2 − 4x ≤ 0 ==> x(x – 4) ≤ 0 0 4
x √ M1
0 ≤ x ≤ 4 √ A1
5. 2009 / SECTION A / Q3
dy
(a) = hx2 – kx
dx
dy
At turning point (3, −4); = 0; √ K1
dx
h(3)2 – k(3) = 0 9h – 3k = 0 ……(1)
dy
At x = −1; dx = 8 √ K1
h(−1)2 – k(−1) = 8 h + k = 8 ……(2)
(2) x 3: 3h + 3k = 24 …..(3)
(1) + (3): 12h = 24 √ M1
h = 2 √ A1
h = 2 (2): k = 6 √ A1
dy
(b) dx
= 2x2 – 6x
y= ( 2x 6x) d x √ K1
2
y = (⅔)x3 – 3x2 + c
At (3, −4); −4 = (⅔)(3)3 – 3(3)2 + c √ M1
−4 = 18 – 27 + c
c=5
y = (⅔)x3 – 3x2 + 5 √ A1
6. 2009 / SECTION A / Q2
(a)(i) x2 – 5x + 6 = 0 (b) Roots:
h+2=3+2=5
(x – 2)(x – 3) = 0 3k – 2 = 3(2) – 2 = 4
x = 2, x = 3 √ M1 (x – 5)(x – 4) = 0 M1
since h > k; x2 – 9x + 20 = 0
√
h = 3, k = 2 √ A1 OR either one
(ii) x2 – 5x + 6 > 0 s.o.r = 5 + 4 = 9
(x – 2)(x – 3) > 0 p.o.r = 5(4) = 20
√ M1
√ M1
Quadratic equation:
√ M1 2 3
x
x2 – 9x + 20 = 0
x 3 √ A1 √ A1
7. 2010 / SECTION B / Q8
(a) dy = 3x2 – 12x + 9 √ K1
dx √ M1
dy
At (2, 3); dx = 3(2)2 – 12(2) + 9 = − 3 √ A1
(b) m┴ = ⅓, Equation: y – 3 = ⅓ (x – 2)
√ K1 √ M1 y = ⅓x + 7/3 or 3y = x + 7
(c) 3x2 – 12x + 9 = 0 x2 – 4x + 3 = 0 √ A1
(x – 3)(x – 1) = 0
x = 3, x = 1
Select x = 1, y = 13 – 6(1)2 + 9(1) + 1 = 5 √ M1
Q (1, 5) √ A1
d 2y
2
= 6x − 12 6(1) – 12 = − 6 (− )( ) = −1 √ M1
2 6
k = 4 √ A1
3
(a)(ii) y=− 2
x ...…..(1)
6y = 4X + 26 …….(2)
(1) (2): 6 (− 3 x) = 4x + 26 √ M1
2
x = −2, y = 3
A (−2, 3) √ A1
(b)(i) 6y = 4x + 26 intersects at y-axis, x = 0
13 B(x, y)
6y = 26 ==> y =
3
AQ : QB = 1 : 2 2
13
Q (0, ) 13
3 1
Q( 0, )
3
2(2) 1( x) A(-2, 3)
0= ==> x = 4
12
either one
√ M1
13
3 = 2 ( 3 ) 1 ( y )
12
==> y = 7 B (4, 7) √ A1
3
b(ii) Gradient, mBC = mOA = − 2
3
Equation BC: y – 7 = – 2
(x – 4) √ M1
2y – 14 = − 3x + 12
2y = − 3x + 26 √ A1
(c) 2PA = PB, Let P(x, y),
2√ (x + 2)2 + (y – 3)2 = √ (x – 4)2 + (y – 7)2
4[(x + 2)2 + (y – 3)2] = [(x – 4)2 + (y – 7)2] √ M1
3x2 + 3y2 + 24x – 10y – 13 = 0 √ A1
4. 2005 / SECTION B / Q9
(a)(i) Equation BC: 2y + x + 6 = 0
y = −½ x – 3
Gradient, mBC = −½ ==> mAB = 2 √ M1
Equation AB: y – 9 = 2(x + 4) √ K1
y = 2x + 17 √ A1
(a)(ii) 2y + x + 6 = 0 ….(1)
y = 2x + 17 ….(2)
(2) (1): 2(2x + 17) + x + 6 = 0 √ M1
x = −8; y = 1
B (−8, 1) √ A1
A( − 4, 9)
(b) 2 2( x) 3( -4)
−8= ==> x = −14
23
B( − 8, 1)
2 ( y ) 3 ( 9 )
either one √ M1
3 1= 23
==> y = −11
D(x, y) D ( −14, −11) √ A1
(c) PA = 5, Let P(x, y),
√ (x + 4)2 + (y – 9)2 = 5 √ M1 P(x, y)
A(−4, 9)
x2 + 8x + 16 + y2 – 18y + 81 = 25
√ M1
x2 + y2 + 8x – 18y + 72 = 0
√ A1
5. 2006 / SECTION B / Q9
0 6 3 0
(a) Area Δ AOB = ½ | |
0 2 4 0
= ½ | 24 – 6 | √ M1
=9 √ A1
2( -3) 3( 6) 2( 4) 3(2)
(b) C = ( , )
23 23 √ M1
1 2 2 either one
=( , )
5 5 √ A1
(c)(i) PA = 2PB, Let P(x, y)
√ (x + 3)2 + (y – 4)2 = 2√ (x – 6)2 + (y + 2)2 √ M1
[(x + 3)2 + (y – 4)2] = 4[(x – 6)2 + (y + 2)2]
3x2 + 3y2 − 54x + 24y + 135 = 0 √ M1
x2 + y2 − 18x + 8y + 45 = 0 √ A1
(c)(ii) Locus P: x2 + y2 − 18x + 8y + 45 = 0
Intersect at y-axis, x = 0;
y2 + 8y + 45 = 0 √ K1
a = 1, b = 8, c = 45
b2 – 4ac = 82 – 4(1) 45
= − 116 1.6)
= 1 – 0.0548
= 0.945 / 0.9452 √ A1
(b) (ii) P(X > m) = 0.6 0.6
0.4
√ M1
m 3.2
P(Z > 0.5
) = 0.6 z
z 0
m 3.2
1 – P(Z > − 0.5
) = 0.4 OR P(Z > z) = 0.6
1 – P(Z > −z) = 0.4
m 3.2
P(Z 7) = P( 7 5 .8 )
z √ M1
1 .5
= P(z > 0.8) 0.2119
= 0.2119
√ A1 7 5 .8
z
1 .5
(b) P(5 0.8)
= 0.491 √ A1
z
- 0.533 0 0.8
Number of grade B lemons = 0.491 x 500
= 246
√ A1
(c) P(X > k) = 0.857
√ M1
= 1 – P(X 2) = 1 – P(X ≤ 2)
√ M1 = 1 – [ P(X = 0) + P( x = 1) + P(X = 2) ]
= 1 – [ 8C0 (0.4)0 (0.6)8 + 8C1 (0.4)1 (0.6)7 +
8C (0.4)2 (0.6)6 ]
2 √ M1
= 0.6846 √ A1
(b) µ = 130, σ = 16 0.7357
√ M1
(i) P(114 1) – P(z > 1.25) −1 0 1.25
= 1 – 0.1587 – 0.1056 √ M1
= 0.7357 √ A1
(ii) P(X > 150) = P(z > 1.25) √ M1
= 0.1056
132
Total number of workers = = 1250 √ A1
0.1056
4. 2008 / SECTION B / Q11
µ = 300, σ = 80 √ M1
1 6 8 3 0 0 0.9505
(a) P(X > 168) = P(z > 80 ) 0.0495
= P(z > −1.65) z
−1.65 0
= 1 – P(z m) = 435 = 0.87 √ K1
500 0.13
0.87
m 300
P(z > ) = 0.87 √ M1
80
z
−1.127 0
m 300
P(z 3 inequalities
II : y ≤ 4x √ K1
or
III : y – x ≥ 5 √ K1 No equal sign on all the
inequalities
- minus 1 mark
(b) Lukis garis-garis lurus graf
(Guna kaedah dua titik)
I : x + y = 100 II : y = 4x III : y – x = 5
X 0 100 X 0 20 X 0 20
y 100 0 y 0 80 y 5 25
y Sekurang-kurang satu garis lurus dilukis
dengan betul
√ K1
20
x
√ A1
Tiga garis lurus dilukis dengan betul
x
16
70
60 Lorek dengan label R √ A1
R
Not using given scales or
40 Wrong label of axes or
35 Not using graph paper
– minus 1 mark
x
20
x
0 x x x
0 20 30 40 60 80 100
(C)(i) From graph drawn (shown by
extrapolation method on the feasible
region): √ A1
Range = 35 < y < 70
(ii) Objective function:
√ M1
50x + 60y = k
Maximum point = (20, 80) √ A1
Maximum fees:
50(20) + 60(80) = RM5,800 √ A1
2. 2006 / SECTION C / Q14
(a) I : 60x + 20y ≤ 720
3x + y ≤ 36 √ K1
II : 30x + 40y ≥ 360
3x + 4y ≥ 36 √ K1
III : x/y ≥ 1/3
√ K1
3x ≥ y or y ≤ 3x
(b) Lukis garis-garis lurus graf
(Guna kaedah dua titik)
I : 3x + y = 36 II : 3x + 4y = 36 III : y = 3x
X 10 12 X 0 12 X 0 6
y 6 0 y 9 0 y 0 18
y Sekurang-kurang satu garis lurus
dilukis dengan betul
√ K1
20 Tiga garis lurus dilukis dengan
(6, 18) betul
x
√ A1
16
12 x
√ A1
x R
8
x
4
0 x xx x
0 2 4 6 8 10 12
C(i) When x = 7, y = 4
Minimum number of racks type Q per
day = 4 √ A1
√ M1
(ii) Objective function: 24x + 32y = k
Maximum point = (6, 18) √ A1
Maximum profit = 24(6) + 32(18)
= RM720.00 √ A1
3. 2007 / SECTION C / Q14
(a) I : x + y ≤ 500 √ K1
II : y ≥ ⅓ x
x ≤ 3y √ K1
III : 15x + 12y ≥ 4 200
5x + 4y ≥ 1 400
√ K1
(b) Lukis garis-garis lurus graf
(Guna kaedah dua titik)
I : x + y = 500 II : x = 3y III : 5x + 4y = 1 400
X 0 500 X 0 150 X 0 280
y 500 0 y 0 50 y 350 0
y
Sekurang-kurang satu garis lurus dilukis
dengan betul
√ K1
500
√ A1
Tiga garis lurus dilukis dengan betul
400
300
R √ A1
225
200
(375, 125)
100
X
0 x
0 100 200 300 400 500
(c)(i) When x = 100, y = 225
The minimum number of Q = 225
√ A1
(ii) Objective function: 15x + 12y = k
√ M1
Maximum point: (375, 125) √ A1
Maximum profit = 15(375) + 12(125)
= RM 7 125.00 √ A1
4. 2008 / SECTION C / Q15
(a) I : x + y ≤ 8 √ K1
II : x ≤ 2y or y ≥ ½ x √ K1
III : 800x + 300y ≤ 4 000
8x + 3y ≤ 40 √ K1
(b) Lukis garis-garis lurus graf
(Guna kaedah dua titik)
I: x+y=8 II : x = 2y III : 8x + 3y = 40
X 0 8 X 0 10 X 2 5
y 8 0 y 0 5 y 8 0
y
Sekurang-kurang satu garis lurus dilukis
dengan betul √ K1
10 √ A1
Tiga garis lurus dilukis dengan betul
8 x
6
x
4 R √ A1
2
(4, 2)
0 x x
0 2 4 6 8 10
(c)(i) When x = 3, the minimum number of
vans rented, y = 2 √ A1
(ii) Objective function: 48x + 12y = k √ M1
Maximum point: (4, 2) √ A1
Maximum number of members
= 48(4) + 12(2) = 216 √ A1
5. 2009 / SECTION C / Q14
(a) I : x + y ≤ 100 √ K1
II : y ≥ ⅓ x √ K1
x ≤ 3y
III : y ≤ x + 30 √ K1
(b) Lukis garis-garis lurus graf
(Guna kaedah dua titik)
I : x + y = 100 II : x = 3y III : y = x + 30
X 0 100 X 0 60 X 0 20
y 100 0 y 0 20 y 30 50
y
Sekurang-kurang satu garis lurus dilukis
dengan betul
100 √ K1
Tiga garis lurus dilukis dengan betul √ A1
80
(35, 65)
60
x
40
√ A1
x R
20 x
0 x
75
0 20 40 60 80 100
(c)(i) The maximum number of small tiles
can be used, x = 75 √ A1
(ii) Objective function: 1.5x + 3y = k √ M1
Maximum point: (35, 65) √ A1
Maximum total cost =1.5(35) + 3(65)
= RM 247.50 √ A1
TOPIC: INDEX NUMBERS
1. 2004 / SECTION C / Q12
(a)(i) 3 7 . 7 0
x1 0 0 1 3 0 √ M1
PS / 9 3
PS/93 = 29.00 √ A1
PP / 9 5 P P
(ii) x1 00 P / 9 5x P / 9 3x1 00 √ M1 PP / 9 5 1 3 5
PP / 9 1 PP / 9 3 PP / 9 1 PP / 9 3 1 0 0
135 120 PP / 9 3 1 2 0
x x100 √ M1 P
100 100 P /9 1 100
= 162 √ A1
(b)(i) Item I W IW
P 135 40 5440
Q x 30 30x
√ K1
R 105 10 1050
S 133 20 2660
∑ 100 9150 + 30x
9 1 5 0 3 0 x
1 2 8 √ M1
100
x = 121.67 √ A1
(ii) P9 5 100
x100 128 P9 3 x 32 √ M1
P9 3 128
P93 = 25 √ A1
2. 2005 / SECTION C / Q13
(a) 1. 00
x x100 1 2 5 √ A1
0. 80
y 140
x 100 140 y x 2. 00 2. 80 √ A1
2. 00 100
0. 40 0. 40
x100 80 z x 100 2. 80 √ A1
z 80
(b)(i) Item I W IW
4660 0
P 125 80 10 000 I √ M1
Q 140 120 16 800 360
R 150 100 15 000 = 129.44 √ A1
S 80 60 4 800
∑ 360 46 600
√ K1
(b)(ii) 2 9 8 5
x1 0 0 1 2 9 . 4 4 √ M1
Q2 0 0 1
2 985
Q2 0 0 1 x 100 = 2 306 09 √ A1
129.44
(c) Q2 0 0 7 1 5 0 Q2 0 0 4
and x 1 0 0 1 2 9 . 4 4
Q2 0 0 4 1 0 0 Q2 0 0 1
Q2 0 0 7 Q2 0 0 7 Q2 0 0 4
x 100 x x 100
Q2 0 0 1 Q2 0 0 4 Q2 0 0 1
150
x 1 2 9 . 4 4 = 194.16 √ A1
100
√ M1
3. 2006 / SECTION C / Q15
w 120
(a) x 100 120 w x 5. 0 0 6. 0 0
5. 0 0 100
y y 125 M1 √ A1
(b) x 100 125
x x 100
4y = 5x ……... (1)
M1
y = x + 2.00 …..(2)
(2) (1) 4(x + 2.00) = 5x
x = 8.00 √ A1
y = 10.00 √ A1
30.60 30.60
(c)(i) x 1 0 0 1 2 7 . 5 Q 2 0 0 4 x 1 0 0 M1
Q2 0 0 4 127.5
Q2004 = 24.00 √ A1
Item I W IW
(ii)
6.00
P x100 120 7 840
5.00
4.00
Q x100 160 3 480
2.50
R 125 m 125m
4.40
S x100 110 2 220
4.00
∑ - K1 12 + m 1540 + 125m
154 0 125 m
12 m
1 2 7 . 5 M1 m = 4 √ A1
4. 2008 / SECTION C / Q13:
P2005
(a) Use I = —— x 100
P2004
1.80
h = —— x 100 M1
1.50
= 120 A1
0.90
112.5 = —— x 100
k
k = 0.8 or 80 sen √ A1
(b) Item I W IW
Fish 150 30 4 500
Flour 120 45 5 400
Salt 112.5 15 1 687.5
Values of w : 30, 45, 15, 10
Sugar 105 10 1050
√ K1
∑ 100 12 637.5
_
I = 12 637.5 x 100 √ M1
100
= 126.38
√ A1
_ 150
(c)(i) I = —— x 126.38 √ M1
100
= 189.6 √ A1
P2009
(c)(ii) —— x 100 = 189.6 √ M1
25
P2009 = 47.39 √ A1
5. 2009 / SECTION C / Q13
2.10
(a)(i) x x1 00 = 75 √ K1
2.80
y 2.00x 130
(ii) x100 130 y = 2.60 √ K1
2.00 100
5.80 5.80
(iii) x100 116 z x 100 = 5.00 √ K1
z 116
(b) Item I W IW
√ M1
P 75 * 4 300
930 + 116m
Q 120 2 240 108.4 =
9+m
R 130 3 390
√ K1 m = 6 √ A1
S 116 m 116m
∑ - 9+m 930 + 116m
P2009
(c) x 100 = 108.4
525
108.4 x 525
= √ M1
100
P2009 = 569.10 √ A1
Q2 0 0 9 Q2 0 0 9 Q2 0 0 7 P2 0 0 9 1 3 2
(d) x 100 x x 100
Q2 0 0 7 Q2 0 0 7 Q2 0 0 8 P2 0 0 7 1 0 0
Q2 0 0 9
x 100
132 100
x x 100√ M1 P2 0 0 8 1 2 0
Q2 0 0 7 100 120 P2 0 0 7 1 0 0
= 110 √ A1
TOPIC: TRIANGLE SOLUTIONS
1. 2005 / SECTION C / Q12
(a) AC2 = 202 +152 – 2(20)(15) cos 65o √ M1
AC = 19.27 √ A1
A
D
(b) sin ADC sin 40
0
19.27 16
√ M1 B
65o
40o
D’
C
19.27sin 400
sin ADC
16
ADC = 50o 44’
A DC = 180 − 50 44 = 129 16 √ A1
o o ’ o ’
'
(c)(i) C AD = 180o - 40o - 50o 44’
= 89o 16‘ √ M1 A
CD 16
o '
89o 16’
si n 89 16 si n 40o √ M1 D
1 6 si n 8 9o 1 6' 40o
CD C
si n 4 0o
CD = 24.89 √ A1 (or use cos rule)
(ii) Area ABCD = Area ABC + Area ACD √ M1
= ½(20)(15)sin 65o + ½(19.27)(16)sin 89o16’
= 290.1 M1 √ √ M1 √ M1
√ A1
either one
2. 2006 / SECTION C / Q13
(a) Area ABCD = ½(5)(6)sin BCD = 13
BCD = 60o 4‘ √ A1 √ M1
(b) BD2 = 52 + 62 − 2(5)(6) cos 60o 4’
D
BD = 5.573 √ A1
√ M1 40O
si n BAD si n 40o √ M1 C
(c)
5. 573 9
A B
BAD = 23 27 √ A1
o ’
ABD = 180 − 40 − 23
o o o
17’
= 116o 33’ √ A1
(d) Area ABCD = Area BCD + Area BAD √ M1
= 13 + ½(9)(5.573)sin116o 33’
√ M1 √ M1
= 35.43 √ A1
3. 2007 / SECTION C / Q15
(a)(i) AC2 = 62 +16.42 −2(6)(16.4)cos50o
AC = 13.36 √ A1 M1
si n ADC si n 400
(ii)
19. 27
16
ACB = 23o 53’
A
√ A1
(b)(i) √ M1
B
A’
√ K1
50o
23o 53‘
C
(ii) BA’A = BAC
= 180o − 105o − 23o 53’
= 51o 7’ √ K1
BA’C = 180o − 51o 7’
= 128o 53’ √ K1
A’BC = 180o −128o 53’ − 23o 53’
= 27o 14’√ K1
BC 5.6
BC = 10.77 √ K1
si n 1 2 8o5 3' si n 2 3o 5 3'
Area Δ A’BC = ½ (5.6)(10.77)sin27o14’
= 13.80 √ A1 M1
√
4. 2009 / SECTION C / Q12
(a) In KLN:
√ M1
LN 12.5
√ A1
sin 80O
sin 32O LN = 23.23
(b) KNL = 180 o
– 80o – 32o = 68o
NLM = KNL = 68o √ K1
√ M1
MN2 = 23.232 + 5.62 – 2(23.23)(5.6)cos68o
MN = 21.76 √ A1
(c) In NLM:
√ M1
si n LMN si n 68o
23. 23
21. 76
LMN = 81o 49’
LMN = 180o − 81o49’ √ K1
= 98o11’ √ A1
(because it’s an obtuse angle)
(d) Area LMN
=½(23.23)(5.6) sin68o √ M1
= 60.31 √ A1
TOPIC: STATISTICS
2005 / SECTION A / Q4
1(a) Marks f F LB
1- 10 6 6 K1
11- 20 9 15 = F
Median
class 21- 30 14 = fm 29 20.5 = Lm
31- 40 7 36
41- 50 4 40
Position of median class = 40 th = 20 th N
Median = L m + ( 2 F )c
2
fm
40 15 M1
Median = 20.5 + ( 2 )10
14
= 24.07 A1
If median class 31- 40, then Lm = 30.5, F = 29, fm = 7 substituted
correctly into the formula – M1
4(b) Marks f M.P ( x ) fx f x2
1-10 6 5.5 33.0 181.50
11-20 9 15.5 139.5 2162.25
21-30 14 25.5 357.0 9,103.50
31-40 7 35.5 248.5 8,821.75
41-50 4 45.5 182.0 8,281.00
∑ 40 960.0 28, 550.00
√ M1
x
fx 960
σ fx x
2
2
f 40 f
= 24 √ M1
28550
σ (24) 2 √ M1
40
Jawapn betul tetapi tiada jalan = 11.74 √ A1
kerja - minus 1 - either in (a) or (b)
TOPIC: PROGRESSIONS
1. 2008 / SECTION A / Q3
(a) r = 1.05 * √ K1 * follow through
Use T 6 = ar 5 √ M1
6 5
T = 18000 x (1.05*)
= RM 22 973.00 √ A1
(b) Use ar n-1 ≥ 36 000
18000 x (1.05*)n-1 ≥ 36 000 √ M1
n ≥ 16 (use logarithm) √ A1
* No working minus 1 mark if answer is correct
a(r n − 1)
(c) Use S6 = ———
r−1
18000(1.05*6 − 1)
S6 = √ M1
1.05* − 1
= RM 122 434 √ A1
TOPIC: LINEAR LAW
1. 2008 / SECTION B / Q8
log10 y 0.40 0.51 0.64 0.76 0.89 1.00
(a)
x 1.5 3.0 4.5 6.0 7.5 9.0
√ 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
(b) Plot graph log y against x
log y
√ 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
Not using the given scales – minus 1
Scales not uniform – minus 1
Not using graph paper – minus 1
(b) Plot log10 y against x √ M1
(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,
x = 5.6 ~ 5.7 √ A1
log10 y = log10 h + 2x log10 k √ K1
(ii) c = log10 h √ M1
h = 1.78 √ A1
(iii) m = 2 log10 k √ M1
k = 1.09 to 1.12 √ A1
PROGRAM 2Q PPDHL 2011
ADDITIONAL MATHEMATICS
SPM
PAPER 2
THE END
PROGRAM 2Q PPDHL 2011
ADDITIONAL MATHEMATICS
SPM
PAPER 2
THANK YOU