# ADDMA G.C.E P2 2021 SOLUTIONS ```SUGGESTED SOLUTIONS FOR ADDITIONAL
MATHEMATICS PAPER 2
4030/2
JULY/ AUGUST (G.C.E) EXAMINATIONS – 2021
PREPARED &amp; COMPILED
BY
KACHAMA DICKSON. C
0970 295655/0966 295655/0955 295655
&copy; 2021 – Copperbelt Mufulira
Why should you fail mathematics when success is guaranteed?
Page 2 of 12
QUESTION 1
2𝑥𝑥 + 3𝑦𝑦 + 𝑧𝑧 = 17 ……………. (i)
𝑥𝑥 − 3𝑦𝑦 + 2𝑧𝑧 = −8 …….………(ii)
5𝑥𝑥 − 2𝑦𝑦 + 3𝑧𝑧 = 5……….…… (iii)
Step 1: Eliminate 𝑦𝑦 from equations (i) and (ii) by adding the two equations.
3𝑥𝑥 + 3𝑧𝑧 = 9
𝑥𝑥 + 𝑧𝑧 = 3…………………..(iv)
Step 2: Eliminate 𝑦𝑦 from equations (i) and (iii) by multiplying (i) by 2 and (iii) by 3
4𝑥𝑥 + 6𝑦𝑦 + 2𝑧𝑧 = 34
+15𝑥𝑥 − 6𝑦𝑦 + 9𝑧𝑧 = 15
19𝑥𝑥 + 11𝑧𝑧 = 49 ……………(v)
Step 3: Solve equations (iv) and (v) simultaneously by substitution method.
𝑥𝑥 = 3 − 𝑧𝑧
19𝑥𝑥 + 11𝑧𝑧 = 49
19(3 − 𝑧𝑧) + 11𝑧𝑧 = 49
𝑥𝑥 = 3 − 1
𝑥𝑥 = 2
57 − 19𝑧𝑧 + 11𝑧𝑧 = 49
−8𝑧𝑧 = 49 − 57
−8𝑧𝑧 = −8
𝑧𝑧 = 1
Step 4: Substitute the values of 𝑥𝑥 and 𝑧𝑧 in equation (i) to find 𝑦𝑦.
2(2) + 3𝑦𝑦 + 1 = 17
4 + 3𝑦𝑦 + 1 = 12
3𝑦𝑦 = 12 − 5
3𝑦𝑦 = 12 ⟹ 𝑦𝑦 = 4
Therefore 𝒙𝒙 = 𝟐𝟐, 𝒚𝒚 = 𝟒𝟒 and 𝒛𝒛 = 𝟏𝟏 Ans
Prepared by Kachama Dickson C/High Standards in Mathematics/ Mufulira/2021
Page 3 of 12
QUESTION 2
(a) 𝟑𝟑𝒖𝒖𝟐𝟐 − 𝟖𝟖𝟖𝟖 &lt; 3
3𝑢𝑢2 − 8𝑢𝑢 − 3 &lt; 0
3𝑢𝑢2 + 𝑢𝑢 − 9𝑢𝑢 − 3 &lt; 0
𝑢𝑢(3𝑢𝑢 + 1) − 3(3𝑢𝑢 + 1) &lt; 0
(3𝑢𝑢 + 1)(𝑢𝑢 − 3) &lt; 0
The critical points are
(3𝑢𝑢 + 1) = 0 or 𝑢𝑢 − 3 = 0
1
𝑢𝑢 = − 3 or 𝑢𝑢 = 3
1
−3
Factors
(3𝑢𝑢 + 1)
𝑢𝑢 − 1
(3𝑢𝑢 + 1)(𝑢𝑢 − 3)
3
1
1
𝑥𝑥 &lt; − 3
− 3 &lt; 𝑢𝑢 &lt; 3
+
−
−
−
𝑥𝑥 &gt; 3
+
+
−
+
+
1
Therefore the range of 𝒖𝒖 values lies between − 3 &lt; 𝑢𝑢 &lt; 3
(b) −3𝑥𝑥 2 + 24𝑥𝑥 − 38 = −3 �𝑥𝑥 2 − 8𝑥𝑥 −
38
3
�
= −3 �𝑥𝑥 2 − 8𝑥𝑥 + (−4)2 −
= −3 �(𝑥𝑥 − 4)2 −
38
3
86
38
3
− 16�
− (−4)2 �
= −3 �(𝑥𝑥 − 4)2 — 3 �
= −𝟑𝟑(𝒙𝒙 − 𝟒𝟒)𝟐𝟐 + 𝟖𝟖𝟖𝟖 as required
∴ the coordinates of the turning point are (𝟒𝟒, 𝟖𝟖𝟖𝟖) Ans
Prepared by Kachama Dickson C/High Standards in Mathematics/ Mufulira/2021
Page 4 of 12
QUESTION 3
1 −3
(a) 133𝑥𝑥 −4 = �4�
(b)
133𝑥𝑥−4 = (4−1 )−3
log 3 (𝑥𝑥 + 25) = 2 + log 3 (2𝑥𝑥 − 1)
log 3 (𝑥𝑥 + 25) − log 3 (2𝑥𝑥 − 1) = 2
𝑥𝑥+25
133𝑥𝑥−4 = 43
log 3 �2𝑥𝑥−1� = 2
(3𝑥𝑥 − 4) lg 13 = 3 lg 4
𝑥𝑥 + 25 = 9(2𝑥𝑥 − 1)
𝑥𝑥+25
log 133𝑥𝑥 −4 = lg 43
2𝑥𝑥−1
3𝑥𝑥 lg 13 − 4 lg 13 = 3lg 4
𝑥𝑥 + 25 = 18𝑥𝑥 − 9
3𝑥𝑥 lg 13 =3lg 4 + 4 lg 13
𝑥𝑥 =
= 32
25 + 9 = 18𝑥𝑥 − 𝑥𝑥
3lg 4+4 lg 13
34 = 17𝑥𝑥
3 lg 13
6.261953383
𝑥𝑥 = 1.873809642
𝒙𝒙 = 𝟐𝟐 Ans
𝒙𝒙 = 𝟏𝟏. 𝟖𝟖𝟖𝟖 Ans
QUESTION 4
(a) Let 𝑓𝑓(𝑥𝑥) = 𝑥𝑥 3 + 3𝑥𝑥 2 − 𝑞𝑞𝑞𝑞 − 24
If f(x) is divisible by 𝑥𝑥 − 2, then (𝑥𝑥 − 3) is the factor of 𝑓𝑓(𝑥𝑥).
𝑥𝑥 − 3 = 0 implies that 𝑥𝑥 = 3 and 𝑓𝑓(3) = 0.
f(3) = 33 + 3(3)2 − 𝑞𝑞(3) − 24 = 0
27 + 27 + 3𝑞𝑞 − 24 = 0
54 − 24 = −3𝑞𝑞
30 = −3𝑞𝑞
𝒌𝒌 = −𝟏𝟏𝟏𝟏Ans
(b) 𝑓𝑓(𝑥𝑥) = 𝑥𝑥 3 − 7𝑥𝑥 + 6 = 0
By trial and error, we have 𝑓𝑓(1) = 0, therefore (𝑥𝑥 − 1) is the factor of 𝑓𝑓(𝑥𝑥).
By synthetic division, we have
1
1
1
0
−7
6
1
1
−6
1
−6
0
Prepared by Kachama Dickson C/High Standards in Mathematics/ Mufulira/2021
Page 5 of 12
𝑥𝑥 2 + 𝑥𝑥 − 6
𝑥𝑥 2 − 2𝑥𝑥 + 3𝑥𝑥 − 6
(𝑥𝑥 − 2)(𝑥𝑥 + 3)
∴ (𝑥𝑥 − 1)(𝑥𝑥 − 2)(𝑥𝑥 + 3) = 0
𝑥𝑥 − 1 = 0 𝑜𝑜𝑜𝑜 𝑥𝑥 − 2 = 0 𝑜𝑜𝑜𝑜 𝑥𝑥 + 3 = 0
𝒙𝒙 = 𝟏𝟏, 𝒙𝒙 = 𝟐𝟐 𝒐𝒐𝒐𝒐 𝒙𝒙 = −𝟑𝟑 Ans
QUESTION 5
(a) (i)
Joseph
C4 &times; 4! &times; 1C1 &times; 1!
4
= 4P4 &times; 1
= 4&times;3&times;2&times;1
= 24 Ways
(ii) MJ, S, Jos, Jam
4! &times; 2!
24 &times; 2
𝟒𝟒𝟒𝟒 Ways
(b) Possibilities
Women(5)
Men(8)
0
6
1
5
2
4
3
3
4
2
5
1
Number of ways = 5C0 &times; 6C5 + 5C1&times; 8C5
Number of ways = 6 + 280
Number of ways = 286 ways Ans
Prepared by Kachama Dickson C/High Standards in Mathematics/ Mufulira/2021
Page 6 of 12
QUESTION 6
(a) 3 sin(60&deg; − 𝑥𝑥) = sin 𝑥𝑥
3 sin 60&deg; cos 𝑥𝑥 − 3 cos 60&deg; sin 𝑥𝑥 = sin 𝑥𝑥
1
√3
3 � 2 � cos 𝑥𝑥 − 3 �2� sin 𝑥𝑥 = sin 𝑥𝑥
3√3
2
3√3
2
3√3
2
3
cos 𝑥𝑥 − 2 sin 𝑥𝑥 = sin 𝑥𝑥 (Divide throughout by sinx
3
cot 𝑥𝑥 − 2 = 1
1
6√3
𝑦𝑦 = tan−1 � 10 �
3
&times; tan 𝑥𝑥 = 1 + 2
3√3
2 tan 𝑥𝑥
𝑦𝑦 = 46.1
5
=2
Now tan is positive in 1st and 3rd quadrants
10tan 𝑥𝑥 = 6√3
tan 𝑥𝑥 =
tan 𝑦𝑦 =
6√3
𝑥𝑥 = 𝑦𝑦 = 46.1
In the 3rd quadrant: 𝑥𝑥 = 180&deg; + 𝑦𝑦 = 180&deg; + 46.1
10
6√3
10
𝒙𝒙 = 𝟒𝟒𝟒𝟒. 𝟏𝟏, 𝟐𝟐𝟐𝟐𝟐𝟐. 𝟏𝟏 Ans
5
(b) (𝐢𝐢)R cos(𝑥𝑥+∝) = 4 cos 𝑥𝑥 − 3 sin 𝑥𝑥
Now R cos(𝑥𝑥+∝) = R cos 𝑥𝑥 cos ∝ − Rsin 𝑥𝑥 sin ∝
5
R cos 𝑥𝑥 cos ∝ − Rsin 𝑥𝑥 sin ∝ = 4 cos 𝑥𝑥 − 3 sin 𝑥𝑥
R cos 𝑥𝑥 cos ∝ = 4 cos 𝑥𝑥
5
R cos ∝ = 4 and Rsin ∝ = 3
5
R2 cos2 ∝ + R2 sin2 ∝ = 42 + �3 �
R2 (cos2 ∝ + sin2 ∝) = 16 +
R2 (1) =
5
and − Rsin 𝑥𝑥 sin ∝ = − 3 sin 𝑥𝑥
144 + 25
9
25
9
2
∝= 22.6&deg;
5
∴ 4 cos 𝑥𝑥 − 3 sin 𝑥𝑥 =
13
3
cos(𝑥𝑥 + 22.6&deg;)
(ii) Therefore the maximum value is 13.
169
R2 = �
R=
13
3
9
since R &gt; 0
Rsin ∝
5
5
tan ∝ = Rcos ∝ = 3 &divide; 4 = 12
5
∝= tan−1 �12 �
Prepared by Kachama Dickson C/High Standards in Mathematics/ Mufulira/2021
Page 7 of 12
QUESTION 7
𝑛𝑛
(a) (i) Sn = 2 [2𝑎𝑎 + 𝑛𝑛 − 1)𝑑𝑑]
5
Comment: Multiply both sides
by 2 and divide both sides by 5.
S5 = 2 [2𝑎𝑎 + (5 − 1)𝑑𝑑]
50 =
5
(2𝑎𝑎 + 4𝑑𝑑)
2
10(2𝑎𝑎 + 4𝑑𝑑) = 100
2𝑎𝑎 + 4𝑑𝑑 = 20
𝑎𝑎 + 2𝑑𝑑 = 10 ………………(i)
Next 10 terms implies that 5 + 10 = 15
S15 = 50 + 325 = 375
S15 =
15
375 =
2
[2𝑎𝑎 + (15 − 1)𝑑𝑑]
15
2
Comment: Multiply both sides by
2 and divide both sides by 15.
(2𝑎𝑎 + 14𝑑𝑑)
750 = 15(2𝑎𝑎 + 14𝑑𝑑)
2𝑎𝑎 + 14𝑑𝑑 = 50
𝑎𝑎 + 7𝑑𝑑 = 25…………..(ii)
Solving equations (i) and (ii) simultaneously gives 𝑎𝑎 = 4 and 𝑑𝑑 = 3
∴ the first term is 𝟒𝟒 and the common difference is 3. Ans
(ii) S13 =
13
2
[(2(4) + (13 − 1)3]
S13 = 6.5(8 + 12 &times; 3)
S13 = 6.5(44)
S13 =286 Ans
(b) (i) Since the annual salary increases by a ratio of 10% = 0.1, this is a compound
interest with p = 4000 and r = 0.1
𝐒𝐒 = 𝒑𝒑(𝟏𝟏 + 𝒓𝒓)𝒕𝒕
S = 4000(1 + 0.1)6
S = 4000(1.1)6
S = 7086.244
S = 708
Therefore the sum of the Doctor’s annual salary is K708.00
Prepared by Kachama Dickson C/High Standards in Mathematics/ Mufulira/2021
Page 8 of 12
(ii)
𝑝𝑝(1 + 𝑟𝑟)𝑡𝑡 &gt; 1 000 000
4000(1.1)𝑡𝑡 &gt; 1 000 000
(1.1)𝑡𝑡 &gt; 250
Taking log both sides we have
tlog1.1 &gt; log250
𝑡𝑡 &gt;
log 250
= 57.93149199
log 1.1
𝒕𝒕 ≈ 58 years Ans
QUESTION 8
𝒙𝒙
𝒙𝒙𝟐𝟐
𝒇𝒇𝒇𝒇
𝒇𝒇𝒙𝒙𝟐𝟐
cf
164 −168
166
27 556
𝒇𝒇
8
1 328
2 20 448
8
169−173
171
29 241
18
3 078
5 26 338
26
174 −178
178
31 684
28
4 928
8 67 328
54
181
32 761
21
3 801
6 87 981
75
184 −188
186
34 596
5
930
1 72 980
80
Scores
179 −183
∑ 𝑓𝑓 = 80
Totals
(a) Median =
𝑛𝑛
=
2
80
∑ 𝑓𝑓𝑓𝑓 = 14065
∑ 𝑓𝑓𝑥𝑥 2 = 2 475 075
= 40. The first number to reach 40 in the cumulative frequency
2
column corresponds to the median class. i.e. 54. Therefore, the median class is
174 −178.
(b) (i)
(ii)
Mean (𝑥𝑥̅ ) =
=
∑ 𝑓𝑓𝑥𝑥 2
SD = �
∑ 𝑓𝑓𝑓𝑓
∑ 𝑓𝑓
∑ 𝑓𝑓
14065
80
− (𝑥𝑥̅ )2
2 475 075
SD = �
80
=175.8125 Ans
− (175.8125)2
SD = √30 938.4375 − 30 910.03516
SD = √28.40234375
SD = 5.329384932
SD = 5.33 Ans
Prepared by Kachama Dickson C/High Standards in Mathematics/ Mufulira/2021
Page 9 of 12
QUESTION 9
(a) At the point of intersection A, the curve 𝑦𝑦 = 𝑥𝑥 2 and 𝑦𝑦 = 2𝑥𝑥 are equal.
𝑦𝑦 = 𝑦𝑦
When 𝑥𝑥 = 0
𝑥𝑥 2 = 2𝑥𝑥
𝑦𝑦 = 2(0) = 0
𝑥𝑥 2 − 2𝑥𝑥 = 0
When 𝑥𝑥 = 2
𝑥𝑥(𝑥𝑥 − 2) = 0
𝑦𝑦 = 2(2)
𝑥𝑥 = 0 or 𝑥𝑥 − 2 = 0
𝑦𝑦 = 4
𝑥𝑥 = 0 or 𝑥𝑥 = 2
Therefore, the coordinates of the point A are (𝟐𝟐, 𝟒𝟒)
Gradient of each curve is found as follows:
𝑏𝑏
(b) V = 𝜋𝜋 ∫𝑎𝑎 (𝑦𝑦12 − 𝑦𝑦22 ) 𝑑𝑑𝑑𝑑
4
V = 𝜋𝜋 ∫2 (2𝑥𝑥)2 − (𝑥𝑥 2 )2 𝑑𝑑𝑑𝑑
4
V = 𝜋𝜋 ∫2 (4𝑥𝑥 2 − 𝑥𝑥 4 )𝑑𝑑𝑑𝑑
V = 𝜋𝜋 �
4𝑥𝑥 3
3
+
4(2)3
V = 𝜋𝜋 �
3
32
𝑥𝑥 5
+
V = 𝜋𝜋 �� 3 +
64
V = 𝜋𝜋 �15 �
5
15𝑥𝑥�
(2)5
32
5
5
��
4
2
� − (0)
𝟒𝟒
V = 𝟒𝟒 𝟏𝟏𝟏𝟏 𝝅𝝅 cubic units Ans
QUESTION 10
(a) V = 3𝑡𝑡 2 − 2𝑡𝑡 + 2
𝑎𝑎 =
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
,
𝑑𝑑
𝑎𝑎 = 𝑑𝑑𝑑𝑑 (3𝑡𝑡 2 − 2𝑡𝑡 + 2)
𝑎𝑎 = 6𝑡𝑡 − 2
𝑎𝑎 = 6(1.5) − 2
𝒂𝒂 = 𝟗𝟗 − 𝟐𝟐 = 𝟕𝟕𝟕𝟕/𝒔𝒔𝟐𝟐
Prepared by Kachama Dickson C/High Standards in Mathematics/ Mufulira/2021
Page 10 of 12
(b) To find the distance, we integrate the velocity function.
𝑆𝑆 = ∫ 𝑣𝑣 𝑑𝑑𝑑𝑑
S = ∫(3𝑡𝑡 2 − 2𝑡𝑡 + 2) 𝑑𝑑𝑑𝑑
S = 𝑡𝑡 3 − 𝑡𝑡 2 + 2𝑡𝑡 + 𝑐𝑐
Now S = 3 and t = 0
3=0+c
𝑐𝑐 = 3
S = 𝑡𝑡 3 − 𝑡𝑡 2 + 2𝑡𝑡 + 3 m
2
(c) S = ∫1 (3𝑡𝑡 2 − 2𝑡𝑡 + 2)𝑑𝑑𝑑𝑑
S=�
3𝑡𝑡 3
−
3
2𝑡𝑡 2
2
+ 2𝑡𝑡�
2
1
S = �23 − 22 + 2(2)� − �13 − 12 + 2(1)�
S= 8–2
S = 6m Ans
QUESTION 11
(a) V = 𝜋𝜋𝑟𝑟 2 ℎ
500 = 𝜋𝜋𝑥𝑥 2 ℎ Divide both sides by 𝜋𝜋𝑟𝑟 2
500
ℎ = 𝜋𝜋𝜋𝜋 2
(b) .A = 2𝜋𝜋𝑟𝑟 2 + 2𝜋𝜋𝜋𝜋ℎ
500
A = 2𝜋𝜋𝑥𝑥 2 + 2𝜋𝜋𝜋𝜋 &times; 𝜋𝜋𝜋𝜋 2
(c)
A = 𝟐𝟐𝟐𝟐𝒙𝒙𝟐𝟐 +
𝐝𝐝𝐝𝐝
𝐝𝐝𝐝𝐝
= 𝟒𝟒𝟒𝟒𝟒𝟒 −
4𝜋𝜋𝜋𝜋 −
𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
𝒙𝒙𝟐𝟐
𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
𝒙𝒙
𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
𝒙𝒙𝟐𝟐
=0
Hence shown Ans
4𝜋𝜋𝑥𝑥 3 − 1000 = 0
4𝜋𝜋𝑥𝑥 3 = 1000
𝑥𝑥 3 =
1000
4𝜋𝜋
𝐝𝐝𝐝𝐝
, at the stationary, 𝐝𝐝𝐝𝐝 = 𝟎𝟎
𝑥𝑥 3 = 79.57747155
A = 2𝜋𝜋(4.30)2 +
1000
4.30
A = 36.98𝜋𝜋 + 232.5581395
A = 348.7
d2A
dx 2
d2A
dr 2
= 4𝜋𝜋 +
2000
𝑥𝑥 3
at 𝑟𝑟 = 4.30
2000
= 4𝜋𝜋 + (4.30)3 &gt; 0
this is a minimum point
Prepared by Kachama Dickson C/High Standards in Mathematics/ Mufulira/2021
Page 11 of 12
3
𝑥𝑥 = √79.57747155
𝑥𝑥 = 4.301270069
𝒙𝒙 ≈ 4.30cm Ans
QUESTION 12
(a) 𝑦𝑦 = 𝑒𝑒 −2𝑥𝑥 − 3, at Q; 𝑥𝑥 = 0
At P; 𝑦𝑦 = 0
𝑦𝑦 = 𝑒𝑒 −2(0) − 3
𝑒𝑒 −2𝑥𝑥 − 3 = 0
𝑦𝑦 = 𝑒𝑒 0 − 3
𝑒𝑒 −2𝑥𝑥 = 3
ln 𝑒𝑒 −2𝑥𝑥 = ln 3
𝑦𝑦 = −2
∴ Q (0, −2) Ans
−2𝑥𝑥 ln 𝑒𝑒 = ln 3
−2𝑥𝑥 = ln 3
𝑥𝑥 = −
(b)
∴ the coordinates of P and Q are �−
𝑦𝑦 = 𝑒𝑒 2𝑥𝑥 − 3
𝑥𝑥
𝑦𝑦
−1
4.4
−0.5
−0.3
0
0.5
ln 3
2
ln 3
2
and so P �−
ln 3
2
, 0�
, 0� and (0, −2) respectively.
1
−2 −2.6 −2.9
𝑦𝑦 = 𝑒𝑒 −2𝑥𝑥 − 3
-2
-1
0
1
−2
(c) 𝑦𝑦 = 𝑒𝑒 2𝑥𝑥 − 3
𝑦𝑦 + 3 = 𝑒𝑒 2𝑥𝑥
𝑥𝑥 = ln √5 + 2𝑥𝑥
Now 𝑒𝑒 2𝑥𝑥 = 𝑦𝑦 + 3
5+2𝑥𝑥 = 𝑦𝑦 + 3
Prepared by Kachama Dickson C/High Standards in Mathematics/ Mufulira/2021
Page 12 of 12
1
𝑥𝑥 = ln(5 + 2𝑥𝑥)2
1
𝑥𝑥 = 2 ln⁡
(5 + 2𝑥𝑥)
2𝑥𝑥 = ln⁡
(5 + 2𝑥𝑥)
2𝑥𝑥
ln (5+2𝑥𝑥)
𝑒𝑒 = 𝑒𝑒
5 + 2𝑥𝑥 = 𝑒𝑒 2𝑥𝑥
5 − 3 + 2𝑥𝑥 = 𝑦𝑦
𝒚𝒚 = 𝟐𝟐𝟐𝟐 + 𝟐𝟐 Ans
Prepared by Kachama Dickson C/High Standards in Mathematics/ Mufulira/2021
```