View
586
Download
6
Category
Preview:
Citation preview
1 © Penerbitan Pelangi Sdn. Bhd.
Additional Mathematics Form 5 Answers
Question Bank
Form 4
Chapter 1 Functions
1 1. (a) a = –1, b = — 4
(b) (i) 2 (ii) 2
2. (a) 2 (b) 8
3. (a) g(x) = x 2 (b) 49
4. 2(x – 1)————
x – 2 2 – x 1 – x 2 5. (a) ——— (b) ——— 3 3
6. (a) (i) {a, c, d} (ii) Many-to-one relation
(b) {(1, c), (2, a), (3, a), (4, d)}
Chapter 2 Quadratic Equations
1. p , –1 or p . 2
2. p = 10, x = 3 or 7 49 3. (a) p = 6, q = –20 (b) – —– 2 4. p . 5
5. (a) α is a root of the equation,∴ α2 – 2α – 4 = 0 α2 = 2α + 4 α3 = 2α2 + 4α = 2(2α + 4) + 4α = 8(α + 1)
(b) α2 – 2α– 4 = 0 α(α – 2) = 4
α– 2———2
= 2—α
6. m = –1, n = 5 or m = 2, n = 2x 2 – 3x – 18 = 0
Chapter 3 Quadratic Functions
1. (a) –7 < x < 3(b) x , –2, x . 2
2. (a) a = 4, b = 2 (b) y = 9, x = –2(c) x = –2
1 3. (a) p = —, r = 1, q = –2 2
1(b) y = – —(x – 2)2 – 1 2
4. 4 < x < 6
5. (a) p = 8, q = –4(b) 16, 32
6. (a) p , – 1—2
(b) x + y = q y = q – x x 2 – 2x + 2y 2 = 3 x 2 – 2x + 2(q – x)2 = 3 3x 2 – (2 + 4q)x + 2q 2 – 3 = 0 b 2 – 4ac = 0 (2 + 4q)2 – 4(3)(2q 2 – 3) = 0 (1 + 4q + 4q 2) – (6q 2 – 9) = 0 –2q 2 + 4q + 10 = 0 q 2 = 2q + 5
Chapter 4 Simultaneous Equations
1. x = 13, y = 4 or x = – 4, y = –13
2. x = 5, y = 7 or x = 7, y = 5
3. x = 1—2
, y = 2 or x = –2, y = – 1—2
4. (7, 8), (–8, –7)
x + (4 + x) 5. ————— (y) = 18 2 xy + 2y = 18 x + y + (4 + x) + 5 = 20 2x + y = 11 1 x = – —, y = 12 or x = 4, y = 3 2
6. 30 cm2
Chapter 5 Indices and Logarithms
1. 4
2. 4 log10 3 + 4—5
log10 2
3. (a) 3.096 (b) 0.834 4. x = –7 or 3 5. x = –1, y = 1 6. (a) x = 1 or –2
(b) x = 3 or –2
Chapter 6 Coordinate Geometry
1. 2y = –x + 19, 4y = –x + 7, P(31, –6) 1 2. y = – —x + 9, 8 : 1 2 3. (a) –2
(b) y = –2x + 5(c) A(1, 3), B(5, 5), C(3, –1)(d) 10 unit2
4. (a) 7 (b) 5y = 2x + 10(c) y 2 + x 2 – 4y – 25 = 0
5. (a) A(0, 2 – 3m), B(3 + 2m, 0)(b) ±2
Answers
© Penerbitan Pelangi Sdn. Bhd. 2
Additional Mathematics Form 5 Answers
6. (a) (i) B1 4—3
, 4—3 2
(ii) 2 unit2
(b) h = 4(k – 1)
(c) x 2 – 8y + 16 = 0
Chapter 7 Statistics
1. (a) k = 5, m = 9 (b) 30.29
2. (a)
30
Time (hours)
Num
ber
of s
tude
nts
20
10
4.5 14.5 24.5 34.5 44.5 54.50
Mode = 30.0 hours(b) 30.8 hours
3. (a)
120
40
60
80
100
20
10 20 30 400
Volume (Litres)
Cum
ulat
ive
frequ
ency
(b) (i) 20.0 (ii) 12
4. (a) 10(b) 6
5. (a) 63.1 mm(b)
20
30
40
50
10
34.524.5 44.5 54.5 64.5 74.5
43
84.5 94.50
Length (mm)
Cum
ulat
ive
freq
uenc
y
14%
4x + 3 6. (a) ———– 2
(b) 2(c) 5.07
Chapter 8 Circular Measure
1. If ∠ODB = 90°,
then DB—–OB
= sin �—6
rad.
DB = 141 1—2 2
= 7 cm
= Radius of sector CBD
25.66 cm2
1 2� � 2. ACB = —1—–2 = — 2 3 3 AC = 2 × r cos 30° 2r3 = ——— 2 = 3 r
�Length of arc APB = AC × — 3 3 = —— �r 3 2�Length of arc AQB = r × —– 3 2 3Total length = —�r + ——�r 3 3 � = —r (2 + 3 ) 3
3. (a) 4 cm (b) 9.26 cm (c) 2.9 cm2
4. (a) 2.35 rad. (b) 11.75 cm (c) 30.63 cm2
5. (a) 0.58 rad. (b) 6.13 cm2
6. Shaded area= Area of ∆OAC – Area of sector OAB 1 1= —r × r tan θ – —r 2θ 2 2 1= —r 2(tan θ – θ) 2
Chapter 9 Differentiation
1. (a) 4 (b) 1
2. (a) 2(3x – 2)(6x 2 – 2x – 3)
(b) 2(6 – x)————(x + 1)3
(c) 2—x 3
(d) –10(4x + 1)——————3(2x 2 + x)3
3. 1, y = –x + 1
�— rad. 6
3 © Penerbitan Pelangi Sdn. Bhd.
Additional Mathematics Form 5 Answers
4. –0.24 y 5 – y 5. (a) ——— = ——— 12 – x x xy = 60 – 12y – 5x + xy 60 – 5x y = ———— 12
x(60 – 5x) (b) ————— 12
(c) x = 6, y = 2 1—2
r 18 – h 6. (a) —– = ——— 12 18 3(12 – r) h = ———— 2
(b) 8
Chapter 10 Solution of Triangles
1. (a) 62°5 (b) 6.26 cm(c) 145.8 cm2 (d) 13.25 cm
2. (a) 8.38 cm (b) 10.3 cm(c) 34°23 (d) 25.88 cm2
3. (a) 1.63 m (b) 35°8(c) 84°53
4. (a) 5.13 cm (b) 28°16(c) 20.84 cm2
5. (a) 7.81 cm (b) 34°56(c) 75°15
6. (a) 10.88 cm (b) 72°25
Chapter 11 Index Number
1. x = 179, y = 140, z = 288
2. x = 1, y = 2(a) RM0.65 (b) RM10.45
3. 115, 120, 106, 113
4. (a) 120 (b) RM1080
5. x = 130, y = 3
6. (a) 129 (b) RM806.25
Form 5
Chapter 1 Progressions
1. (a) a = 3, d = 2 (b) 12(c) 25
2. 3 cm
3. (a) 3—2
, 3 (b) –1, 2
(c) 1—4
4. (a) 9 11—16
(b) 5
(c) 5—16
(d) 1—2
5. 1—2
6. (a) Areas of squares are
x 2, 1 x—2
22, 1 x—
4 22, …
\ x 2, x 2—4
, x 2—16
, … forms a geometric
progression with r = 1—4
.
(b) 4—3
x 2 cm2
Chapter 2 Linear Law
1. a = 2, b = – 1—2
2.
4
6
8
10
2
1
5 10 15 20 250
x + y Graph (x + y) against x 2
x2
a = 2.5, b = 1
3. p = 2, q = –4
4. (a) yx = ax + b
(b)
10
–5
20
30
40
01 2 3 4 5
x
Graph against xy x√ y x√
a = 7.80, b = –5
➤ Straight line Y = mX + c.
© Penerbitan Pelangi Sdn. Bhd. 4
Additional Mathematics Form 5 Answers
1 b 1 1 5. — = —1—2 + — y a x a
0.4
0.2
0.6
0.8
1.0
0.16
0.1 0.2 0.3 0.4 0.5 0.60
Graph against1y––
1y––
1x––
1x––
a = 6.25, b = 8.5
1 6. (a) log10 y = (–b log10 3)1—2 + log10 A x
(b)
1.0
1.5
2.01.75
0.5
0.02 0.04 0.06 0.080
log10 y
1x
––
Graph log10 y against 1x—
(c) A = 56.2, b = 31.44
Chapter 3 Integration
3 1. a = —, n = –1 2
2. (a) 11 (b) 14
1 1 1 3. (a) A1—, —2 (b) —– unit2
2 2 12 3(c) —–� unit3
80 1 4. (a) A(2, 3), B(5, 0) (b) 11— unit2
6
2 1 5. (a) k = –2, y = 5 — (b) –4 — 3 2
3 2 6. (a) A(1, 0), C12, —2 (b) 3—� unit3
2 3
7. (a) y = –x + 2, B(2, 0) (b) �—3
unit3
Chapter 4 Vectors
9 1. (a) 1 2 (b) 9i + 11j 11 ~ ~
2. (a) – 4i~ + j~
(b) 313 units
(c) m = 4, n = 7 (d) 1——W34
(5i~ + 3j~
)
→ 3. PS = 2a + b ~ ~→
XY = ma + nb ~ ~ → →Since XY // PS, then n 1 — = — m 2 m : n = 2 : 1
→ → → 4. (a) OC = 2i + 2j, OD = 4i + 5j, OE = 5i + 7j ~ ~ ~ ~ ~ ~
4 2(b) m = —, n = —, 4 : 1 5 5
1 1 2 5. (a) (i) —pa + —pb (ii) —qa + (1 – q)b 2 ~ 2 ~ 3 ~ ~
4 3(b) p = —, q = — 5 5(c) 4 : 1, 3 : 2
6. (a) A→C = –5i~ + 5j
~ ; B
→C = – 6i~ + 2j
~
(b) 5(c) 1 : 1
Chapter 5 Trigonometric Functions
1. (a) – 1—–3
(b) – 2—–3
(c) – 1—2
(d) –2 (e) 1
2.
1
0
2
90° 180° 270° 360°
y = 2
y
x
y = 2 – cos x
k = 2
3. (a) – 1 – p 2
(b) 1———–1 – 2p2
(c) p
4. LHS = (cosec x + sec x)(cosec x – sec x) = cosec2 x – sec2 x 1 1 = ——— – ——— sin2 x cos2 x
cos2 x – sin2 x = ——————— sin2 x cos2 x 4 cos 2x = ——————— 4 sin2 x cos2 x 4 cos 2x = ———— sin2 2x = 4 cot 2x cosec 2x = RHS
5 © Penerbitan Pelangi Sdn. Bhd.
Additional Mathematics Form 5 Answers
sin 2θ 5. (a) LHS = —————– 1 + cos 2θ 2 sin θ cos θ = ————————– 1 + (2 cos2 θ – 1) = tan θ = RHS
(b) (i) x = 45°, 75°58, 225°, 255°58 (ii) x = 180°, 80°24, 279°36
6. (a) x = 19°28, 30°, 150°, 160°32
(b) (i) 5—–3
(ii) 5—6
1 (iii) —– 5
Chapter 6 Permutations and Combinations
1. 120
2. (a) 2880 (b) 1152
3. (a) 1440 (b) 720
4. (a) 10 (b) 10
5. (a) 31 (b) 35
6. (a) 84 (b) 1260
Chapter 7 Probability
1. (a) 3—5
(b) 3—5
(c) 9—20
2. (a) 3—5
(b) 7—10
9 3. —– 25
4. (a) 5—12
(b) 1—84
(c) 1—84
5. (a) 1—14
(b) 1—14
6. x = 4, y = 3
Chapter 8 Probability Distributions
1. (a) 0.2605 (b) 0.3953 (c) 0.8652
2. (a) 0.1536 (b) 0.0256 (c) 0.9744
3. µ = 3, σ = 1.64
4. (a) 0.7692 (b) 0.484
5. (a) 0.1360 (b) 0.6977
6. µ = 53.9, σ = 16.48
Chapter 9 Motion Along a Straight Line
1. (a) v = 2 m s–1, a = –2 m s–2
(b) 1.5 s(c) 2.25 m
2. (a) t = 3 s, s = –69 m (b) –36.75 m s–1
3. (a) –1.125 m s–1
(b)
20
v
t
23–
(c) 1 7—12
m
4. (a) 4 seconds (b) 38 m(c) 11 m s–1
5. (a) 80 m (b) 50 m s–1
(c) 4.5 seconds, 78.75 m from O
6. (a) v = 2t 2 + 4t – 6 (b)
0–6
v
t12
(c) 8 m
Chapter 10 Linear Programming
5 3 1. x > 3, y . – —x + 5, y , —x + 3 4 8
2. (a) x + 3y > 10 (b) x < y(c) y – x > 5
3. (a) x + 2y < 3000, 2x + y > 2400, y – x < 300(b)
1000750
1500
2000
2500
500
500 1000 1500 2000 2500 30000
y
x
R
y – x = 300
2x + y = 2400
x + 2y = 3000
3.2x + 2y = c
(3000, 0)
(c) (i) RM9600(ii) 750
© Penerbitan Pelangi Sdn. Bhd. 6
Additional Mathematics Form 5 Answers
4. (a) x + 2y 12, x + 4y 20, 5x + 2y 40(b)
10
15
20
(7, 2.5)5
5 10 15 200
y
xR
5x + 2y = 405x + 2.5y = c
x + 4y = 20
x + 2y = 12
(c) Maximumprofit =RM41.25
5. (a) 12x + 13y 2000, 20x + 17y 500, 2x y(b)
100
150
(167, 0)
200
50
50 100 150 2000
y
x
12x + 13y = 2000
y = 2x
80x + 40y = c
20x + 17y = 500R
(c) Optimum function=80x + 40y Total profit =RM13360
6. (a) 2x + y 600, 2x + 3y 900, x + y 350(b)
200
300
(250, 100)
(175, 175)
400
500
600
100
100 200 300 400 5000
y
x
2x + 3y = 900
2x + y = 600
2x + 1.5y = c
x + y = 350
R
(c) (i) 175 (ii) RM650
Recommended