# 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
2π₯π₯ + 3π¦π¦ + π§π§ = 17 ……………. (i)
π₯π₯ − 3π¦π¦ + 2π§π§ = −8 …….………(ii)
5π₯π₯ − 2π¦π¦ + 3π§π§ = 5……….…… (iii)
Step 1: Eliminate π¦π¦ from equations (i) and (ii) by adding the two equations.
3π₯π₯ + 3π§π§ = 9
π₯π₯ + π§π§ = 3…………………..(iv)
Step 2: Eliminate π¦π¦ from equations (i) and (iii) by multiplying (i) by 2 and (iii) by 3
4π₯π₯ + 6π¦π¦ + 2π§π§ = 34
+15π₯π₯ − 6π¦π¦ + 9π§π§ = 15
19π₯π₯ + 11π§π§ = 49 ……………(v)
Step 3: Solve equations (iv) and (v) simultaneously by substitution method.
π₯π₯ = 3 − π§π§
19π₯π₯ + 11π§π§ = 49
19(3 − π§π§) + 11π§π§ = 49
π₯π₯ = 3 − 1
π₯π₯ = 2
57 − 19π§π§ + 11π§π§ = 49
−8π§π§ = 49 − 57
−8π§π§ = −8
π§π§ = 1
Step 4: Substitute the values of π₯π₯ and π§π§ in equation (i) to find π¦π¦.
2(2) + 3π¦π¦ + 1 = 17
4 + 3π¦π¦ + 1 = 12
3π¦π¦ = 12 − 5
3π¦π¦ = 12 βΉ π¦π¦ = 4
Therefore ππ = ππ, ππ = ππ and ππ = ππ Ans
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Page 3 of 12
QUESTION 2
(a) ππππππ − ππππ &lt; 3
3π’π’2 − 8π’π’ − 3 &lt; 0
3π’π’2 + π’π’ − 9π’π’ − 3 &lt; 0
π’π’(3π’π’ + 1) − 3(3π’π’ + 1) &lt; 0
(3π’π’ + 1)(π’π’ − 3) &lt; 0
The critical points are
(3π’π’ + 1) = 0 or π’π’ − 3 = 0
1
π’π’ = − 3 or π’π’ = 3
1
−3
Factors
(3π’π’ + 1)
π’π’ − 1
(3π’π’ + 1)(π’π’ − 3)
3
1
1
π₯π₯ &lt; − 3
− 3 &lt; π’π’ &lt; 3
+
−
−
−
π₯π₯ &gt; 3
+
+
−
+
+
1
Therefore the range of ππ values lies between − 3 &lt; π’π’ &lt; 3
(b) −3π₯π₯ 2 + 24π₯π₯ − 38 = −3 οΏ½π₯π₯ 2 − 8π₯π₯ −
38
3
οΏ½
= −3 οΏ½π₯π₯ 2 − 8π₯π₯ + (−4)2 −
= −3 οΏ½(π₯π₯ − 4)2 −
38
3
86
38
3
− 16οΏ½
− (−4)2 οΏ½
= −3 οΏ½(π₯π₯ − 4)2 — 3 οΏ½
= −ππ(ππ − ππ)ππ + ππππ as required
∴ the coordinates of the turning point are (ππ, ππππ) Ans
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Page 4 of 12
QUESTION 3
1 −3
(a) 133π₯π₯ −4 = οΏ½4οΏ½
(b)
133π₯π₯−4 = (4−1 )−3
log 3 (π₯π₯ + 25) = 2 + log 3 (2π₯π₯ − 1)
log 3 (π₯π₯ + 25) − log 3 (2π₯π₯ − 1) = 2
π₯π₯+25
133π₯π₯−4 = 43
log 3 οΏ½2π₯π₯−1οΏ½ = 2
(3π₯π₯ − 4) lg 13 = 3 lg 4
π₯π₯ + 25 = 9(2π₯π₯ − 1)
π₯π₯+25
log 133π₯π₯ −4 = lg 43
2π₯π₯−1
3π₯π₯ lg 13 − 4 lg 13 = 3lg 4
π₯π₯ + 25 = 18π₯π₯ − 9
3π₯π₯ lg 13 =3lg 4 + 4 lg 13
π₯π₯ =
= 32
25 + 9 = 18π₯π₯ − π₯π₯
3lg 4+4 lg 13
34 = 17π₯π₯
3 lg 13
6.261953383
π₯π₯ = 1.873809642
ππ = ππ Ans
ππ = ππ. ππππ Ans
QUESTION 4
(a) Let ππ(π₯π₯) = π₯π₯ 3 + 3π₯π₯ 2 − ππππ − 24
If f(x) is divisible by π₯π₯ − 2, then (π₯π₯ − 3) is the factor of ππ(π₯π₯).
π₯π₯ − 3 = 0 implies that π₯π₯ = 3 and ππ(3) = 0.
f(3) = 33 + 3(3)2 − ππ(3) − 24 = 0
27 + 27 + 3ππ − 24 = 0
54 − 24 = −3ππ
30 = −3ππ
ππ = −ππππAns
(b) ππ(π₯π₯) = π₯π₯ 3 − 7π₯π₯ + 6 = 0
By trial and error, we have ππ(1) = 0, therefore (π₯π₯ − 1) is the factor of ππ(π₯π₯).
By synthetic division, we have
1
1
1
0
−7
6
1
1
−6
1
−6
0
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Page 5 of 12
π₯π₯ 2 + π₯π₯ − 6
π₯π₯ 2 − 2π₯π₯ + 3π₯π₯ − 6
(π₯π₯ − 2)(π₯π₯ + 3)
∴ (π₯π₯ − 1)(π₯π₯ − 2)(π₯π₯ + 3) = 0
π₯π₯ − 1 = 0 ππππ π₯π₯ − 2 = 0 ππππ π₯π₯ + 3 = 0
ππ = ππ, ππ = ππ ππππ ππ = −ππ 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
ππππ 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
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Page 6 of 12
QUESTION 6
(a) 3 sin(60&deg; − π₯π₯) = sin π₯π₯
3 sin 60&deg; cos π₯π₯ − 3 cos 60&deg; sin π₯π₯ = sin π₯π₯
1
√3
3 οΏ½ 2 οΏ½ cos π₯π₯ − 3 οΏ½2οΏ½ sin π₯π₯ = sin π₯π₯
3√3
2
3√3
2
3√3
2
3
cos π₯π₯ − 2 sin π₯π₯ = sin π₯π₯ (Divide throughout by sinx
3
cot π₯π₯ − 2 = 1
1
6√3
π¦π¦ = tan−1 οΏ½ 10 οΏ½
3
&times; tan π₯π₯ = 1 + 2
3√3
2 tan π₯π₯
π¦π¦ = 46.1
5
=2
Now tan is positive in 1st and 3rd quadrants
10tan π₯π₯ = 6√3
tan π₯π₯ =
tan π¦π¦ =
6√3
π₯π₯ = π¦π¦ = 46.1
In the 3rd quadrant: π₯π₯ = 180&deg; + π¦π¦ = 180&deg; + 46.1
10
6√3
10
ππ = ππππ. ππ, ππππππ. ππ Ans
5
(b) (π’π’)R cos(π₯π₯+∝) = 4 cos π₯π₯ − 3 sin π₯π₯
Now R cos(π₯π₯+∝) = R cos π₯π₯ cos ∝ − Rsin π₯π₯ sin ∝
5
R cos π₯π₯ cos ∝ − Rsin π₯π₯ sin ∝ = 4 cos π₯π₯ − 3 sin π₯π₯
R cos π₯π₯ cos ∝ = 4 cos π₯π₯
5
R cos ∝ = 4 and Rsin ∝ = 3
5
R2 cos2 ∝ + R2 sin2 ∝ = 42 + οΏ½3 οΏ½
R2 (cos2 ∝ + sin2 ∝) = 16 +
R2 (1) =
5
and − Rsin π₯π₯ sin ∝ = − 3 sin π₯π₯
144 + 25
9
25
9
2
∝= 22.6&deg;
5
∴ 4 cos π₯π₯ − 3 sin π₯π₯ =
13
3
cos(π₯π₯ + 22.6&deg;)
(ii) Therefore the maximum value is 13.
169
R2 = οΏ½
R=
13
3
9
since R &gt; 0
Rsin ∝
5
5
tan ∝ = Rcos ∝ = 3 &divide; 4 = 12
5
∝= tan−1 οΏ½12 οΏ½
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Page 7 of 12
QUESTION 7
ππ
(a) (i) Sn = 2 [2ππ + ππ − 1)ππ]
5
Comment: Multiply both sides
by 2 and divide both sides by 5.
S5 = 2 [2ππ + (5 − 1)ππ]
50 =
5
(2ππ + 4ππ)
2
10(2ππ + 4ππ) = 100
2ππ + 4ππ = 20
ππ + 2ππ = 10 ………………(i)
Next 10 terms implies that 5 + 10 = 15
S15 = 50 + 325 = 375
S15 =
15
375 =
2
[2ππ + (15 − 1)ππ]
15
2
Comment: Multiply both sides by
2 and divide both sides by 15.
(2ππ + 14ππ)
750 = 15(2ππ + 14ππ)
2ππ + 14ππ = 50
ππ + 7ππ = 25…………..(ii)
Solving equations (i) and (ii) simultaneously gives ππ = 4 and ππ = 3
∴ the first term is ππ 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
ππ = ππ(ππ + ππ)ππ
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
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Page 8 of 12
(ii)
ππ(1 + ππ)π‘π‘ &gt; 1 000 000
4000(1.1)π‘π‘ &gt; 1 000 000
(1.1)π‘π‘ &gt; 250
Taking log both sides we have
tlog1.1 &gt; log250
π‘π‘ &gt;
log 250
= 57.93149199
log 1.1
ππ ≈ 58 years Ans
QUESTION 8
ππ
ππππ
ππππ
ππππππ
cf
164 −168
166
27 556
ππ
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
∑ ππ = 80
Totals
(a) Median =
ππ
=
2
80
∑ ππππ = 14065
∑ πππ₯π₯ 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 (π₯π₯Μ ) =
=
∑ πππ₯π₯ 2
SD = οΏ½
∑ ππππ
∑ ππ
∑ ππ
14065
80
− (π₯π₯Μ )2
2 475 075
SD = οΏ½
80
=175.8125 Ans
− (175.8125)2
SD = √30 938.4375 − 30 910.03516
SD = √28.40234375
SD = 5.329384932
SD = 5.33 Ans
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Page 9 of 12
QUESTION 9
(a) At the point of intersection A, the curve π¦π¦ = π₯π₯ 2 and π¦π¦ = 2π₯π₯ are equal.
π¦π¦ = π¦π¦
When π₯π₯ = 0
π₯π₯ 2 = 2π₯π₯
π¦π¦ = 2(0) = 0
π₯π₯ 2 − 2π₯π₯ = 0
When π₯π₯ = 2
π₯π₯(π₯π₯ − 2) = 0
π¦π¦ = 2(2)
π₯π₯ = 0 or π₯π₯ − 2 = 0
π¦π¦ = 4
π₯π₯ = 0 or π₯π₯ = 2
Therefore, the coordinates of the point A are (ππ, ππ)
Gradient of each curve is found as follows:
ππ
(b) V = ππ ∫ππ (π¦π¦12 − π¦π¦22 ) ππππ
4
V = ππ ∫2 (2π₯π₯)2 − (π₯π₯ 2 )2 ππππ
4
V = ππ ∫2 (4π₯π₯ 2 − π₯π₯ 4 )ππππ
V = ππ οΏ½
4π₯π₯ 3
3
+
4(2)3
V = ππ οΏ½
3
32
π₯π₯ 5
+
V = ππ οΏ½οΏ½ 3 +
64
V = ππ οΏ½15 οΏ½
5
15π₯π₯οΏ½
(2)5
32
5
5
οΏ½οΏ½
4
2
οΏ½ − (0)
ππ
V = ππ ππππ ππ cubic units Ans
QUESTION 10
(a) V = 3π‘π‘ 2 − 2π‘π‘ + 2
ππ =
ππππ
ππππ
,
ππ
ππ = ππππ (3π‘π‘ 2 − 2π‘π‘ + 2)
ππ = 6π‘π‘ − 2
ππ = 6(1.5) − 2
ππ = ππ − ππ = ππππ/ππππ
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Page 10 of 12
(b) To find the distance, we integrate the velocity function.
ππ = ∫ π£π£ ππππ
S = ∫(3π‘π‘ 2 − 2π‘π‘ + 2) ππππ
S = π‘π‘ 3 − π‘π‘ 2 + 2π‘π‘ + ππ
Now S = 3 and t = 0
3=0+c
ππ = 3
S = π‘π‘ 3 − π‘π‘ 2 + 2π‘π‘ + 3 m
2
(c) S = ∫1 (3π‘π‘ 2 − 2π‘π‘ + 2)ππππ
S=οΏ½
3π‘π‘ 3
−
3
2π‘π‘ 2
2
+ 2π‘π‘οΏ½
2
1
S = οΏ½23 − 22 + 2(2)οΏ½ − οΏ½13 − 12 + 2(1)οΏ½
S= 8–2
S = 6m Ans
QUESTION 11
(a) V = ππππ 2 β
500 = πππ₯π₯ 2 β Divide both sides by ππππ 2
500
β = ππππ 2
(b) .A = 2ππππ 2 + 2ππππβ
500
A = 2πππ₯π₯ 2 + 2ππππ &times; ππππ 2
(c)
A = ππππππππ +
ππππ
ππππ
= ππππππ −
4ππππ −
ππππππππ
ππππ
ππππππππ
ππ
ππππππππ
ππππ
=0
Hence shown Ans
4πππ₯π₯ 3 − 1000 = 0
4πππ₯π₯ 3 = 1000
π₯π₯ 3 =
1000
4ππ
ππππ
, at the stationary, ππππ = ππ
π₯π₯ 3 = 79.57747155
A = 2ππ(4.30)2 +
1000
4.30
A = 36.98ππ + 232.5581395
A = 348.7
d2A
dx 2
d2A
dr 2
= 4ππ +
2000
π₯π₯ 3
at ππ = 4.30
2000
= 4ππ + (4.30)3 &gt; 0
this is a minimum point
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Page 11 of 12
3
π₯π₯ = √79.57747155
π₯π₯ = 4.301270069
ππ ≈ 4.30cm Ans
QUESTION 12
(a) π¦π¦ = ππ −2π₯π₯ − 3, at Q; π₯π₯ = 0
At P; π¦π¦ = 0
π¦π¦ = ππ −2(0) − 3
ππ −2π₯π₯ − 3 = 0
π¦π¦ = ππ 0 − 3
ππ −2π₯π₯ = 3
ln ππ −2π₯π₯ = ln 3
π¦π¦ = −2
∴ Q (0, −2) Ans
−2π₯π₯ ln ππ = ln 3
−2π₯π₯ = ln 3
π₯π₯ = −
(b)
∴ the coordinates of P and Q are οΏ½−
π¦π¦ = ππ 2π₯π₯ − 3
π₯π₯
π¦π¦
−1
4.4
−0.5
−0.3
0
0.5
ln 3
2
ln 3
2
and so P οΏ½−
ln 3
2
, 0οΏ½
, 0οΏ½ and (0, −2) respectively.
1
−2 −2.6 −2.9
π¦π¦ = ππ −2π₯π₯ − 3
-2
-1
0
1
−2
(c) π¦π¦ = ππ 2π₯π₯ − 3
π¦π¦ + 3 = ππ 2π₯π₯
π₯π₯ = ln √5 + 2π₯π₯
Now ππ 2π₯π₯ = π¦π¦ + 3
5+2π₯π₯ = π¦π¦ + 3
Prepared by Kachama Dickson C/High Standards in Mathematics/ Mufulira/2021
Page 12 of 12
1
π₯π₯ = ln(5 + 2π₯π₯)2
1
π₯π₯ = 2 lnβ‘
(5 + 2π₯π₯)
2π₯π₯ = lnβ‘
(5 + 2π₯π₯)
2π₯π₯
ln (5+2π₯π₯)
ππ = ππ
5 + 2π₯π₯ = ππ 2π₯π₯
5 − 3 + 2π₯π₯ = π¦π¦
ππ = ππππ + ππ Ans
Prepared by Kachama Dickson C/High Standards in Mathematics/ Mufulira/2021
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