CAPE PURE MATHEMATICS UNIT 1
CAPE UNIT 1 2016 SOLUTIONS
Question 1
a.
i.
Given 𝑓(𝑥) = 2𝑥 3 − 𝑥 2 + 𝑝𝑥 + 𝑞. If 𝑥 + 3 is a factor then 𝑓(−3) = 0.
𝑓(−3) = 2(−3)3 − (−3)2 + 𝑝(−3) + 𝑞 = 0
−54 − 9 − 3𝑝 + 𝑞 = 0
−3𝑝 + 𝑞 = 63
----------
(1)
Also 𝑓(−1) = 10 then
𝑓(−1) = 2(−1)3 − (−1)2 + 𝑝(−1) + 𝑞 = 10
−2 − 1 − 𝑝 + 𝑞 = 10
−𝑝 + 𝑞 = 13
------
(2)
Subtracting (1) from (2) we have
2𝑝 = −50, 𝑝 = −25
Substitute 𝑝 = −25 into (2) we have
−(−25) + 𝑞 = 13, 𝑞 = −12
ii.
𝑓(𝑥) = 2𝑥 3 − 𝑥 2 − 25𝑥 − 12
𝑥 + 3 is a factor therefore using long division we have
2𝑥 2 − 7𝑥 − 4
𝑥+3
2𝑥 3 − 𝑥 2 − 25𝑥 − 12
2𝑥 3 + 6𝑥 2
−7𝑥 2 − 25𝑥
−7𝑥 2 − 21𝑥
−4𝑥 − 12
−4𝑥 − 12
0
𝑓(𝑥) = (𝑥 + 3)(2𝑥 2 − 7𝑥 − 4)
= (𝑥 + 3)(𝑥 − 4)(2𝑥 + 1)
1
When 𝑓(𝑥) = 0, 𝑥 = −3, − 2 , 4
1
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2016
b.
When 𝑛 = 1, 61 − 1 = 5 which is divisible by 5 therefore the statement is true for 𝑛 = 1
Assume the statement is true for 𝑛 = 𝑘. Therefore 6𝑘 − 1 = 5𝑚 where 𝑚 𝜖 𝐍.
When 𝑛 = 𝑘 + 1 we have
6𝑘+1 − 1 = 6(6𝑘 − 1) + 5
= 6(5𝑚) + 5
= 5(6𝑚 + 1) which is divisible by 5.
Therefore the statement is true for 𝑛 = 𝑘 + 1
Since the statement is true for 𝑛 = 1, 𝑘 and 𝑘 + 1. It is true for all natural numbers n.
c.
p
T
T
F
F
ii.
q
T
F
T
F
𝐩→𝐪
T
F
T
T
𝐩˅𝐪
T
T
T
F
𝐩∧𝐪
T
F
F
F
𝐩 → 𝐪 and (𝐩˅𝐪) → (𝐩 ∧ 𝐪) are logically equivalent because both have the same
truth values in their output column.
Question 2
a.
(𝐩˅𝐪) → (𝐩 ∧ 𝐪)
T
F
T
T
log 2 (10 − 𝑥) + log 2 𝑥 = 4
log 2 (10 − 𝑥)𝑥 = 4
𝑥(10 − 𝑥) = 24
10𝑥 − 𝑥 2 = 16
𝑥 2 − 10𝑥 + 16 = 0
(𝑥 − 2)(𝑥 − 8) = 0
𝑥 = 2, 8
2
b.
𝑥+3
Given the function 𝑓(𝑥) = 𝑥−1 , 𝑥 ≠ 1.
If the function is one-to-one then, 𝑓(𝑎) ≠ 𝑓(𝑏), 𝑎 ≠ 𝑏, (𝑎, 𝑏) ≠ 1
𝑎+3 𝑏+3
≠
𝑎−1 𝑏−1
(𝑎 + 3)(𝑏 − 1) ≠ (𝑏 + 3)(𝑎 − 1)
𝑎𝑏 − 𝑎 + 3𝑏 − 3 ≠ 𝑎𝑏 − 𝑏 + 3𝑎 − 3
4𝑏 ≠ 4𝑎,
𝑎≠𝑏
Therefore a and b are distinct and hence a maps to f(a), and b maps to f(b)
For any 𝑥 𝜖 𝐑, where 𝑥 ≠ 1 and 𝑓 −1 (𝑥) =
𝑓(𝑓
−1 (𝑥))
𝑥+3
𝑥−1
𝑥+3
+3
𝑥+3
= 𝑓(
)=𝑥−1
𝑥+3
𝑥−1
𝑥−1−1
𝑥 + 3 + 3𝑥 − 3
=(
)
𝑥+3−𝑥+1
=
4𝑥
=𝑥
4
Therefore the function is a one-to-one and onto because for
(𝑥, 𝑦) 𝜖 𝐑, where (𝑥, 𝑦) ≠ 1, 𝑥 = 𝑓 −1 (𝑦) <=> 𝑦 = 𝑓(𝑥)
c.
i.
Given the roots of the equation 2𝑥 3 − 5𝑥 2 + 4𝑥 + 6 = 0 are 𝛼, 𝛽, and 𝛾.
5
𝑥 3 − 𝑥 2 + 2𝑥 + 3 = 0
2
𝑥 3 − (𝛼 + 𝛽 + 𝛾)𝑥 2 + (𝛼𝛽 + 𝛽𝛾 + 𝛼𝛾)𝑥 − 𝛼𝛽𝛾 = 0
5
𝛼+𝛽+𝛾 = ,
2
𝛼𝛽 + 𝛽𝛾 + 𝛼𝛾 = 2,
3
and 𝛼𝛽𝛾 = −3
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2016
c.
ii.
An equation whose roots are
1
𝛼2
,
1
𝛽2
and
1
𝛾2
has
1
1
1
𝛽2 𝛾 2 + 𝛼 2 𝛾 2 + 𝛼 2 𝛽2
+
+
=
𝛼 2 𝛽2 𝛾 2
𝛼 2𝛽2𝛾 2
=
(𝛼𝛽 + 𝛽𝛾 + 𝛼𝛾)2 − 2𝛼𝛽𝛾(𝛼 + 𝛽 + 𝛾)
(𝛼𝛽𝛾)2
5
(2)2 − 2(−3) ( ) 19
2 =
=
(−3)2
9
1
1
1
𝛾 2 + 𝛼 2 + 𝛽2
+
+
=
𝛼 2𝛽2 𝛽2𝛾 2 𝛼 2 𝛾 2
𝛼 2𝛽2𝛾 2
(𝛼 + 𝛽 + 𝛾)2 − 2(𝛼𝛽 + 𝛼𝛾 + 𝛽𝛾)
(𝛼𝛽𝛾)2
=
5 2
(2) − 2(2)
=
1
𝛼 2𝛽2𝛾 2
𝑥3 −
(−3)2
=
1
(𝛼𝛽𝛾)2
=
1
1
=
(−3)2 9
19 2 1
1
𝑥 − 𝑥− =0
9
4
9
36𝑥 3 − 76𝑥 2 − 9𝑥 − 4 = 0
4
=−
1
4
Question 3
a.
i.
Prove sec 2 𝜃 =
LHS
=
cosec 𝜃
cosec 𝜃−sin 𝜃
1
sin 𝜃
1
sin 𝜃 − sin 𝜃
1
= sin 𝜃2
1 − sin 𝜃
sin 𝜃
=
1
1
=
2
1 − sin 𝜃 cos 2 𝜃
= sec 2 𝜃 𝐏𝐫𝐨𝐯𝐞𝐧.
ii.
Given
cosec 𝜃
cosec 𝜃−sin 𝜃
4
=
4
3
3
sec 2 𝜃 = 3 , cos2 𝜃 = 4
cos 𝜃 = ±
√3
2
this gives an acute angle
𝜋
6
𝜋 5𝜋 7𝜋 11𝜋
Therefore 𝜃 = 6 , 6 , 6 ,
6
b.
i.
𝑓(𝜃) = sin 𝜃 + cos 𝜃
𝑟 sin(𝜃 + 𝛼) = 𝑟 sin 𝜃 cos 𝛼 + 𝑟 sin 𝛼 cos 𝜃
Therefore 𝑟 cos 𝛼 = 1 and 𝑟 sin 𝛼 = 1
𝜋
Hence, tan 𝛼 = 1, 𝛼 = 4 , and 𝑟 = √2
𝜋
sin 𝜃 + cos 𝜃 = √2 sin (𝜃 + 4 )
ii.
The max value of 𝑓(𝜃) is √2
𝜋
This occurs at (𝜃 + 4 ) =
𝜋
2
Therefore the smallest non-negative value of 𝜃 is
5
𝜋
4
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2016
c.
Prove
tan(𝐴 + 𝐵 + 𝐶) =
tan 𝐴 + tan 𝐵 + tan 𝐶 − tan 𝐴 tan 𝐵 tan 𝐶
1 − tan 𝐴 tan 𝐵 − tan 𝐴 tan 𝐶 − tan 𝐵 tan 𝐶
tan(𝐴 + 𝐵 + 𝐶) = tan(𝐴 + (𝐵 + 𝐶)) =
=
tan 𝐴 + tan(𝐵 + 𝐶)
1 − tan 𝐴 tan(𝐵 + 𝐶)
tan 𝐵 + tan 𝐶
tan 𝐴 + 1 − tan 𝐵 tan 𝐶
tan 𝐵 + tan 𝐶
1 − tan 𝐴 (1 − tan 𝐵 tan 𝐶 )
tan 𝐴 (1 − tan 𝐵 tan 𝐶) + tan 𝐵 + tan 𝐶
1 − tan 𝐵 tan 𝐶
=
(1 − tan 𝐵 tan 𝐶) − tan 𝐴 (tan 𝐵 + tan 𝐶)
1 − tan 𝐵 tan 𝐶
=
tan 𝐴 + tan 𝐵 + tan 𝐶 − tan 𝐴 tan 𝐵 tan 𝐶
1 − tan 𝐴 tan 𝐵 − tan 𝐴 tan 𝐶 − tan 𝐵 tan 𝐶
Proven
Question 4
a.
i.
Given sin 𝜃 = 𝑥, sin2 𝜃 = 𝑥 2
1 − sin2 𝜃 = cos 2 𝜃
cos 𝜃 = √1 − 𝑥 2
tan 𝜃 =
=
sin 𝜃
cos 𝜃
𝑥
√1 − 𝑥 2
𝐒𝐡𝐨𝐰𝐧
6
Given 𝑦 = tan 2𝑡 and 𝑥 = sin 𝑡
ii.
𝑦=
2 tan 𝑡
𝑥
, and tan 𝑡 =
2
1 − tan 𝑡
√1 − 𝑥 2
𝑥
2(
)
2
√1
−
𝑥
𝑦=
2
𝑥
1−(
)
√1 − 𝑥 2
2𝑥
2
= √1 − 𝑥2
𝑥
1−
1 − 𝑥2
2𝑥
𝑥2
= √1 −
1 − 𝑥2 − 𝑥2
1 − 𝑥2
=
2𝑥(1 − 𝑥 2 )
√1 − 𝑥 2 (1 − 2𝑥 2 )
2𝑥√1 − 𝑥 2
𝑦=
1 − 2𝑥 2
b.
1
2
Given 𝐮 = (−3) and 𝐯 = (1)
2
5
i.
|𝐮| = √1 + 9 + 4 = √14
|𝐯| = √4 + 1 + 25 = √30
𝐮∙𝐯
cos 𝜃 = |𝐮||𝐯|
ii.
=
c.
2 − 3 + 10
√14 × √30
= 0.439
At any time the point 𝑃(𝑥, 𝑦) is 2𝑎 from the origin and a from the x-axis. Therefore its
distance from the y-axis is given by √4𝑎2 − 𝑎2 = 𝑎√3 using Pythagoras Theorem.
Hence, 𝑥 = 𝑎√3 and 𝑦 = 𝑎.
𝑥 = 𝑦√3,
𝑦=
𝑥
√3
7
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2016
d.
2𝑥 + 𝑦 + 3 = 0
------
𝑥2 + 𝑦2 = 9
(1)
-------
From (1), 𝑦 = −(2𝑥 + 3)
(2)
-----
(3)
Substituting (3) into (2) we have
2
𝑥 2 + (−(2𝑥 + 3)) = 9
𝑥 2 + 4𝑥 2 + 12𝑥 + 9 = 9
5𝑥 2 + 12𝑥 = 0
𝑥(5𝑥 + 12) = 0
𝑥 = 0, −
12
5
When 𝑥 = 0, 𝑦 = −3
When 𝑥 = −
12
5
12
9
, 𝑦 = −2 (− 5 ) − 3 = 5
Therefore the points of intersection are (0, −3) and (−
12 9
5
, 5)
Question 5
a.
Given ∫(𝑥 + 1)1/3 𝑑𝑥, Using the substitution 𝑢 = 𝑥 + 1 we have
𝑑𝑢
= 1,
𝑑𝑥
𝑑𝑢 = 𝑑𝑥.
∫(𝑥 + 1)1/3 𝑑𝑥 = ∫(𝑢)1/3 𝑑𝑢
=
(𝑢)4/3
+𝐶
4
3
=
3 4/3
𝑢 +𝐶
4
=
3
(𝑥 + 1)4/3 + 𝐶
4
8
b.
0
V = 𝜋 ∫−1 𝑥 2 𝑑𝑦
𝑦 =3 − 1,
𝑥 = (𝑦 + 1)1/3
𝑥 2 = (𝑦 + 1)2/3
0
𝑉 = 𝜋 ∫(𝑦 + 1)2/3 𝑑𝑦
−1
3
0
= 𝜋 [ (𝑦 + 1)5/3 ]
−1
5
c.
𝑎
=
3𝜋
cubic units
5
𝑎
Given ∫0 𝑓(𝑥) 𝑑𝑥 = ∫0 𝑓(𝑎 − 𝑥) 𝑑𝑥 𝑎 > 0
1
1
𝑒𝑥
𝑒 1−𝑥
∫ 𝑥
𝑑𝑥 = ∫
𝑑𝑥
𝑒 + 𝑒 1−𝑥
𝑒 1−𝑥 + 𝑒 (1−(1−𝑥))
0
0
1
𝑒 1−𝑥
= ∫ 1−𝑥
𝑑𝑥
𝑒
+ 𝑒𝑥
0
Dividing both numerator and denominator by 𝑒 𝑥 we have
1
𝑒 1−2𝑥
= ∫ 1−2𝑥
𝑑𝑥
𝑒
+1
0
1
1
= [− ln(𝑒 1−2𝑥 + 1)]
0
2
1
= − [(ln(𝑒 −1 + 1)) − (ln(𝑒 + 1))]
2
1
𝑒+1
= − [ln |
| − ln|𝑒 + 1|]
2
𝑒
1
= − [ln|𝑒 + 1| − ln 𝑒 − ln|𝑒 + 1|]
2
1
1
= − (−1) =
2
2
9
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2016
d.
i.
Bacteria grow exponentially at a rate of 2% per hour where 𝑦 = 𝑓(𝑡) is the
number of bacteria present t hours later is given by the differential equation
𝑑𝑦
= 0.02𝑦
𝑑𝑡
Separating variables and integrating both sides we have
∫
𝑑𝑦
= ∫ 0.02 𝑑𝑡
𝑦
ln 𝑦 = 0.02𝑡 + 𝐶
𝑦 = 𝑒 0.02𝑡+𝐶
𝑦 = 𝑒 𝐶 𝑒 0.02𝑡
When 𝑡 = 0, 𝑦 = 1000, therefore
1000 = 𝑒 𝐶 , 𝑦 = 1000𝑒 0.02𝑡
ii.
When the bacteria population is double 𝑦 = 2000,
2000 = 1000𝑒 0.02𝑡
2 = 𝑒 0.02𝑡
ln 2 = 0.02𝑡,
𝑡=
ln 2
= 34.66 hrs
0.02
Question 6
a.
Given 𝑓(𝑥) = 2𝑥 3 + 5𝑥 2 − 𝑥 + 12
𝑓 ′ (𝑥) = 6𝑥 2 + 10𝑥 − 1
The gradient of the tangent at the point where 𝑥 = 3, is given by
𝑓 ′ (3) = 6(3)2 + 10(3) − 1 = 83
When 𝑥 = 3, 𝑓(3) = 2(3)3 + 5(3)2 − (3) + 12 = 112
Therefore the equation of the tangent at the point where 𝑥 = 3 is given by
𝑦 − 112 = 83(𝑥 − 3)
𝑦 = 83𝑥 − 137
10
2
b.
i.
Given 𝑓(𝑥) = {𝑥 + 2𝑥 + 3
𝑎𝑥 + 𝑏
𝑥≤0
𝑥>0
lim 𝑓(𝑥) = 02 + 2(0) + 3 = 3
𝑥→0−
lim 𝑓(𝑥) = 𝑎(0) + 𝑏 = 𝑏
𝑥→0+
ii.
For 𝑓(𝑥) to be continuous at 𝑥 = 0, lim− 𝑓(𝑥) = lim+ 𝑓(𝑥)
𝑥→0
𝑥→0
Therefore 𝑏 = 3, and 𝑎 𝜖 𝐑.
iii.
Given 𝑓 ′ (0) = lim
𝑓(0+𝑡)−𝑓(0)
𝑡→0
𝑡
For 𝑥 ≤ 0
𝑎(0 + 𝑡) + 3 − (𝑎(0) + 3)
𝑡→0
𝑡
= lim
𝑎𝑡 + 3 − 3
𝑡→0
𝑡
𝑎𝑡
= lim = 𝑎
𝑡→0 𝑡
= lim
For 𝑥 > 0
(0 + 𝑡)2 + 2(0 + 𝑡) + 3 − (02 + 2(0) + 3)
𝑡→0
𝑡
𝑓 ′ (0) = lim
(𝑡)2 + 2(𝑡) + 3 − 3
= lim
𝑡→0
𝑡
𝑡 2 + 2𝑡
𝑡→0
𝑡
= lim
= lim 𝑡 + 2 = 2
𝑡→0
If the 𝑓(𝑥) is differentiable at 𝑥 = 0 then 𝑓′(0) = 2
Therefore 𝑎 = 2
11
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2016
c.
Given 𝑓(𝑥) = √𝑥 therefore 𝑓(𝑥 + ℎ) = √𝑥 + ℎ
√𝑥 + ℎ − √𝑥 √𝑥 + ℎ + √𝑥
×
ℎ→0
ℎ
√𝑥 + ℎ + √𝑥
𝑓 ′ (𝑥) = lim
= lim
(𝑥 + ℎ) − 𝑥
ℎ→0 ℎ(√𝑥
= lim
+ ℎ + √𝑥)
ℎ
ℎ→0 ℎ(√𝑥
= lim
ℎ→0 (√𝑥
=
+ ℎ + √𝑥)
1
+ ℎ + √𝑥)
1
2√2
12
CAPE PURE MATHEMATICS UNIT 1
SOLUTIONS FOR 2015 EXAM
1.
a.
i.
The inverse ~𝐩 → ~𝐪 and the contrapositive ~𝐪 → ~𝐩
ii.
iii.
p
q
~𝐩
~𝐪
𝐩→𝐪
~𝐪 → ~𝐩
T
T
F
F
T
T
T
F
F
T
F
F
F
T
T
F
T
T
F
F
T
T
T
T
𝐩 → 𝐪 and ~𝐪 → ~𝐩 are logically equivalent because both final columns
are the exactly same.
b.
Given 𝑓(𝑥) = 𝑥 3 + 𝑝𝑥 2 − 𝑥 + 𝑞
i.
If (𝑥 − 5) is a factor then 𝑓(5) = 53 + 𝑝(52 ) − (5) + 𝑞 = 0
125 + 25𝑝 − 5 + 𝑞 = 0
25𝑝 + 𝑞 = −120 … … …. (1)
When divided by (𝑥 − 1) the remainder is 24 therefore
𝑓(1) = 13 + 𝑝(12 ) − (1) + 𝑞 = 24
1 + 𝑝 − 1 + 𝑞 = 24
𝑝 + 𝑞 = 24 … … …. (2)
Subtract (2) from (1) we have
24𝑝 = −144, 𝑝 = −6
−6 + 𝑞 = 24, 𝑞 = 30
13
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2015
ii.
𝑓(𝑥) = 𝑥 3 − 6𝑥 2 − 𝑥 + 30
𝑥2 − 𝑥 − 6
𝑥−5
𝑥 3 − 6𝑥 2 − 𝑥 + 30
𝑥 3 − 5𝑥 2
−𝑥 2 − 𝑥
−𝑥 2 + 5𝑥
−6𝑥 + 30
−6𝑥 + 30
0
𝑥 3 − 6𝑥 2 − 𝑥 + 30 = (𝑥 − 5)(𝑥 2 − 𝑥 − 6)
= (𝑥 − 5)(𝑥 − 3)(𝑥 + 2)
c.
Given 𝑆(𝑛) = 5 + 52 + 53 + 54 + ⋯ + 5𝑛 (𝐿. 𝐻. 𝑆) and 4𝑆(𝑛) = 5𝑛+1 − 5 (𝑅. 𝐻. 𝑆)
When 𝑛 = 1, 𝐿. 𝐻. 𝑆 = 4𝑆(1) = 4 × 5 = 20, and R. H. S = 52 − 5 = 20
Therefore result is true for 𝑛 = 1.
Assume result is true for 𝑛 = 𝑘, therefore
4𝑆(𝑘) = 4(5 + 52 + 53 + 54 + ⋯ + 5𝑘 ) = 5𝑘+1 − 5
When 𝑛 = 𝑘 + 1, we have
R.H.S = 4𝑆(𝑘 + 1) = 5𝑘+2 − 5
L.H.S = 4(5 + 52 + 53 + 54 + ⋯ + 5𝑘 + 5𝑘+1 )
= 4(𝑆(𝑘) + 5𝑘+1 )
= 4(𝑆(𝑘)) + 4(5𝑘+1 )
= 5𝑘+1 − 5 + 4(5𝑘+1 )
= 5 × 5𝑘+1 − 5
= 5𝑘+2 − 5
R.H.S = L.H.S therefore result is true for 𝑛 = 𝑘 + 1.
Since the result is true for 𝑛 = 1, 𝑘 and 𝑘 + 1, it is true for all positive integer n.
14
2.
a.
i.
A function is one-to-one if each element in the domain maps to one and
only one image in the co-domain and each element in the range is the
image of only one element in the domain. Therefore given that 𝑓: 𝐴 → 𝐵
and 𝑔: 𝐵 → 𝐶 are one-to-one functions, (𝑔 ° 𝑓) is a one-to-one function
because the co-domain of f is used as the domain for g and this makes
(𝑔 ° 𝑓) = 𝑔: 𝐵 → 𝐶 a one-to-one function.
𝑓(𝐴)
𝑎1
𝑎2 b
𝑎3
ii.
𝑔(𝐵)
𝑏1
𝑏2
𝑏3
𝑔°𝑓(𝐶)
𝑐1
𝑐2
𝑐3
A function is onto if each element in the co-domain is mapped unto at
least one element in the domain. Therefore given that 𝑓: 𝐴 → 𝐵
and 𝑔: 𝐵 → 𝐶 are onto functions, (𝑔 ° 𝑓) is a onto function because the
co-domain of f is used as the domain for g, and this makes
(𝑔 ° 𝑓) = 𝑔: 𝐵 → 𝐶 an onto function.
b.
i.
4
4
3 − (9)𝑥 − (81)𝑥 = 0
3−
4
4
− 2𝑥 = 0
𝑥
9
9
3(92𝑥 ) − 4(9𝑥 ) − 4 = 0 multiplying both sides by 92𝑥
(3(9𝑥 ) + 2)(9𝑥 − 2) = 0
(3(9𝑥 ) + 2) = 0,
2
9𝑥 = − ,
3
not possible
(9𝑥 − 2) = 0
9𝑥 = 2
𝑥=
log 2
= 0.315
log 9
15
factorising the equation
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2015
ii.
|5𝑥 − 6| = 𝑥 + 5
when 5𝑥 − 6 > 0 we have
5𝑥 − 6 = 𝑥 + 5
4𝑥 = 11
𝑥=
11
4
When 5𝑥 − 6 < 0 we have
−(5𝑥 − 6) = 𝑥 + 5
−5𝑥 + 6 = 𝑥 + 5
1 = 6𝑥
1
𝑥=6
c.
Given 𝑁 = 300 + 5𝑡
i.
When 𝑡 = 0, 𝑁 = 300 + 1 = 301
ii.
When 𝑁 = 3(301) we have
903 = 300 + 5𝑡
5𝑡 = 603
𝑡=
3.
a.
i.
log 603
= 3.98 hours
log 5
cos 3𝑥 = cos(2𝑥 + 𝑥)
= cos 2𝑥 cos 𝑥 − sin 2𝑥 sin 𝑥
= (2 cos2 𝑥 − 1) cos 𝑥 − 2 sin 𝑥 cos 𝑥 sin 𝑥
= 2 cos3 𝑥 − cos 𝑥 − 2 sin2 𝑥 cos 𝑥
= 2 cos3 𝑥 − cos 𝑥 − 2(1 − cos2 𝑥) cos 𝑥
= 2 cos3 𝑥 − cos 𝑥 − 2 cos 𝑥 + 2 cos3 𝑥
= 4 cos3 𝑥 − 3 cos 𝑥
16
ii.
cos 6𝑥 − cos 2𝑥 = 0
cos 6𝑥 = 4 cos3 2𝑥 − 3 cos 2𝑥
4 cos 3 2𝑥 − 3 cos 2𝑥 − cos 2𝑥 = 0
4 cos 3 2𝑥 − 4 cos 2𝑥 = 0
4 cos 2𝑥 (cos2 2𝑥 − 1) = 0
4 cos 2𝑥 = 0
2𝑥 =
𝑥=
𝜋 3𝜋 5𝜋 7𝜋
,
, ,
2 2 2 2
𝜋 3𝜋 5𝜋 7𝜋
, ,
,
4 4 4 4
cos 2 2𝑥 − 1 = 0
cos 2 2𝑥 = 1
cos 2𝑥 = ±1
2𝑥 = 0, 𝜋 2𝜋, 3𝜋 4𝜋
𝜋
3𝜋
𝑥 = 0, , 𝜋, , 2𝜋
2
2
b.
i.
𝑓(2𝜃) = 3 sin 2𝜃 + 4 cos 2𝜃
𝑟 sin(2𝜃 + 𝛼) = 𝑟 sin 2𝜃 cos 𝛼 + 𝑟 sin 𝛼 cos 2𝜃
𝑟 cos 𝛼 = 3 and 𝑟 sin 𝛼 = 4
4
tan 𝛼 = 3 , and 𝑟 = √32 + 42
4
𝛼 = tan−1 (3) = 0.927 and 𝑟 = 5
3 sin 2𝜃 + 4 cos 2𝜃 = 5 sin(2𝜃 + 0.927)
ii.
Maximum value of occurs when 𝑓(𝜃) = 5
1
1
=
7 − 𝑓(𝜃) 7 − 5
=
1
2
Minimum value of occurs when 𝑓(𝜃) = −5
1
1
1
=
=
7 − 𝑓(𝜃) 7 + 5 12
17
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2015
4.
a.
i.
Given 𝐶1 : 𝑥 = √10 cos 𝜃 − 3;
𝐶2 : 𝑥 = 4 cos 𝜃 + 3;
From 𝐶1 : cos 𝜃 =
𝑦 = √10 sin 𝜃 + 2
𝑦 = 4 sin 𝜃 + 2
𝑥+3
;
√10
sin 𝜃 =
𝑦−2
√10
cos2 𝜃 + sin2 𝜃 = 1
𝑥+3 2
𝑦−2 2
(
) +(
) =1
√10
√10
(𝑥 + 3)2 + (𝑦 − 2)2 = (√10)
From 𝐶2 :
cos 𝜃 =
𝑥−3
4
;
2
sin 𝜃 =
𝑦−2
4
𝑥−3 2
𝑦−2 2
(
) +(
) =1
4
4
(𝑥 − 3)2 + (𝑦 − 2)2 = 42
2
(𝑥 + 3)2 + (𝑦 − 2)2 = (√10) ……….
ii.
From (1)
(1)
2
(𝑦 − 2)2 = (√10) − (𝑥 + 3)2 ………..
(𝑥 − 3)2 + (𝑦 − 2)2 = 42
(2)
…….. (3)
(𝑦 − 2)2 = 42 − (𝑥 − 3)2 …………..
(4)
Equating (2) and (4) we have
2
(√10) − (𝑥 + 3)2 = 42 − (𝑥 − 3)2
10 − (𝑥 2 + 6𝑥 + 9) = 16 − (𝑥 2 − 6𝑥 + 9)
10 − 𝑥 2 − 6𝑥 − 9 = 16 − 𝑥 2 + 6𝑥 − 9
10 − 16 = 6𝑥 + 6𝑥
−6 = 12𝑥
𝑥=−
1
2
18
1
Substituting 𝑥 = − 2 into (4) we have
1
(𝑦 − 2)2 = 42 − ((− ) − 3)
2
𝑦 2 − 4𝑦 + 4 = 16 −
𝑦 2 − 4𝑦 +
2
49
4
1
=0
4
4𝑦 2 − 16𝑦 + 1 = 0
𝑦=
16 ± √162 − 16
8
𝑦=
16 ± 4√15
8
𝑦 = 3.94,
0.0635
1
1
Points of intersection are (− 2 , 3.94) and (− 2 , 0.0635 )
b.
If the point 𝑃(𝑥, 𝑦) moves so that its distance from a fixed point (0, 3) is two
times the distance from the fixed point (5, 2) then;
𝑥 2 + (𝑦 − 3)2 = 4[(𝑥 − 5)2 + (𝑦 − 2)2 ]
𝑥 2 + 𝑦 2 − 6𝑦 + 9 = 4[𝑥 2 − 10𝑥 + 25 + 𝑦 2 − 4𝑦 + 4]
𝑥 2 + 𝑦 2 − 6𝑦 + 9 = 4𝑥 2 − 40𝑥 + 100 + 4𝑦 2 − 16𝑦 + 16
3𝑥 2 + 3𝑦 2 − 40𝑥 − 10𝑦 + 107 = 0
𝑥2 + 𝑦2 −
40
10
107
𝑥− 𝑦+
=0
3
3
3
20 2 400
5 2 25 107
(𝑥 − ) −
+ (𝑦 − ) −
+
=0
3
9
3
9
3
20 2
5 2 104
(𝑥 − ) + (𝑦 − ) =
3
3
9
20 5
This is the equation of a circle with centre ( 3 , 3) and radius
19
√104
.
3
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2015
5.
a.
Given
𝑓(𝑥) = {
sin(𝑎𝑥)
if 𝑥 ≠ 0, 𝑎 ≠ 0
𝑥
4
if 𝑥 = 0
If f is continuous at 𝑥 = 0, then
sin(𝑎𝑥)
= lim 4
𝑥→0
𝑥→0
𝑥
lim
sin(𝑎𝑥)
=4
𝑥→0
𝑥
lim
Multiplying numerator and denominator by a we have
𝑎 sin(𝑎𝑥)
=4
𝑥→0
𝑎𝑥
lim
sin(𝑎𝑥)
=4
𝑥→0
𝑎𝑥
𝑎 lim
sin(𝑎𝑥)
=1
𝑥→0
𝑎𝑥
lim
𝑎=4
b.
Given 𝑓(𝑥) = sin(2𝑥)
Differentiating from first principles we have
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ→0
ℎ
𝑓 ′ (𝑥) = lim
sin 2(𝑥 + ℎ) − sin(2𝑥)
ℎ→0
ℎ
= lim
= lim
2𝑥 + 2ℎ + 2𝑥
2𝑥 + 2ℎ − 2𝑥
) sin (
)
2
2
ℎ
2 cos (
ℎ→0
= lim
4𝑥 + 2ℎ
2 ) sin(ℎ)
ℎ
2 cos (
ℎ→0
sin(ℎ)
ℎ→0
ℎ
= lim 2 cos(2𝑥 + ℎ) × lim
ℎ→0
= 2 cos 2𝑥
20
c.
Given 𝑦 =
i.
2𝑥
√1 + 𝑥 2
Using the quotient and power rule:
𝑑𝑢
𝑑𝑣
𝑑𝑦 𝑣 𝑑𝑥 − 𝑢 𝑑𝑥
𝑑𝑦
=
and
= 𝑛(𝑓(𝑥))𝑛−1 × 𝑓′(𝑥)
2
𝑑𝑥
𝑣
𝑑𝑥
𝑑𝑦
=
𝑑𝑥
1
1
√1 + 𝑥 2 (2) − 2𝑥 (2) (1 + 𝑥 2 )−2 (2𝑥)
(√1 + 𝑥 2 )
2
2𝑥 2
√1 + 𝑥 2
1 + 𝑥2
2√1 + 𝑥 2 −
=
=
2(1 + 𝑥 2 ) − 2𝑥 2
(1 + 𝑥 2 )√1 + 𝑥 2
𝑑𝑦
2
2
=
=
𝑑𝑥 (1 + 𝑥 2 )√1 + 𝑥 2 (1 + 𝑥 2 )3/2
Multiplying both sides by x we have
𝑥
𝑑𝑦
2𝑥
=
𝑑𝑥 (1 + 𝑥 2 )√1 + 𝑥 2
1
2𝑥
=(
)
2
1 + 𝑥 √1 + 𝑥 2
𝑥
ii.
𝑑𝑦
𝑦
=
𝑑𝑥 1 + 𝑥 2
From (i)
𝑑𝑦
2
=
= 2(1 + 𝑥 2 )−3/2
𝑑𝑥 (1 + 𝑥 2 )3/2
𝑑2 𝑦
3
= (2) (− ) (1 + 𝑥 2 )−5/2 (2𝑥)
2
𝑑𝑥
2
=−
=−
=−
6𝑥
(1 + 𝑥 2 )5/2
6𝑥
(1 + 𝑥 2 )2 √1 + 𝑥 2
3(2𝑥)
(1 + 𝑥 2 )2 √1 + 𝑥 2
21
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2015
Since 𝑦 =
2𝑥
√1 + 𝑥 2
𝑑2 𝑦
3𝑦
=
−
(1 + 𝑥 2 )2
𝑑𝑥 2
𝑑2 𝑦
3𝑦
+
=0
𝑑𝑥 2 (1 + 𝑥 2 )2
6.
Given 𝑦 = 3𝑥 − 7, 𝑦 + 𝑥 = 9 and 3𝑦 = 𝑥 + 3
i.
AB is the line 𝑦 = 3𝑥 − 7, AC is the line 𝑦 + 𝑥 = 9 and BC is the line
3𝑦 = 𝑥 + 3. The lines AB intersects AC at the point A therefore the coordinates
of A is found by solving these equations simultaneously.
𝑦 = 3𝑥 − 7 ……. (1)
𝑦+𝑥 =9
……. (2)
Substituting (1) into (2) we have
3𝑥 − 7 + 𝑥 = 9
4𝑥 = 16
𝑥=4
From (2) 𝑦 + 4 = 9
𝑦=5
Therefore the coordinates of A is (4, 5)
AB intersects BC at the point B therefore solving these equations gives the
coordinates of the point B.
𝑦 = 3𝑥 − 7 …… (1)
3𝑦 = 𝑥 + 3 …… (2)
Substituting (1) into (2) we have
3(3𝑥 − 7) = 𝑥 + 3
9𝑥 − 21 = 𝑥 + 3
8𝑥 = 24
𝑥=3
22
From (1) 𝑦 = 3(3) − 7 = 2
Therefore the coordinates of B is (3, 2)
AC intersects BC at the point C therefore solving these equations gives the
coordinates of the point C.
𝑦 + 𝑥 = 9 ……
(1)
3𝑦 = 𝑥 + 3 …… (2)
From (1) 𝑦 = 9 − 𝑥 …. (3)
Substituting (3) into (2) we have
3(9 − 𝑥) = 𝑥 + 3
27 − 3𝑥 = 𝑥 + 3
4𝑥 = 24
𝑥=6
𝑦 = 9−6 =3
Therefore the coordinates of C is (6, 3)
ii.
The area bounded by these three lines is given by
6
4
6
∫ 3𝑥 − 7 𝑑𝑥 + ∫ 9 − 𝑥 𝑑𝑥 − ∫
3
=[
4
3
𝑥+3
𝑑𝑥
3
3𝑥 2
𝑥2 6 1 𝑥2
4
6
− 7𝑥] + [9𝑥 − ] − [ + 3𝑥]
3
3
2
2 4 3 2
1
= [(−4) − (−7.5)] + [(36) − (28)] − [(36) − (13.5)]
3
= 4 sq. units
b.
Given that 𝑓 ′ (𝑥) = 3𝑥 2 + 8𝑥 − 3 at the point (0, −6)
i.
𝑓(𝑥) = ∫ 3𝑥 2 + 8𝑥 − 3 𝑑𝑥
= 𝑥 3 + 4𝑥 2 − 3𝑥 + 𝐶
The curve passes through the point (0, −6) therefore
−6 = 𝐶
The equation of the curve is therefore 𝑓(𝑥) = 𝑥 3 + 4𝑥 2 − 3𝑥 − 6
23
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2015
ii.
At the stationary point 𝑓 ′ (𝑥) = 0 therefore
3𝑥 2 + 8𝑥 − 3 = 0
(3𝑥 − 1)(𝑥 + 3) = 0
1
𝑥 = 3 , and − 3
1 3
1
1 2
1
When 𝑥 = 3 , 𝑦 = (3) + 4 (3) − 3 (3) − 6
𝑦=−
176
27
= −6.5
When 𝑥 = −3, 𝑦 = (−3)3 + 4(−3)2 − 3(−3) − 6
𝑦 = 12
1
Therefore the stationary points are (3 , −6.5) and (−3, 12)
𝑓 ′′ (𝑥) = 6𝑥 + 8
1
1
1
When 𝑥 = 3 , 𝑓 ′′ (3) = 6 (3) + 8 > 0
1
Therefore (3 , −6.5) is a minimum point
When 𝑥 = −3, 𝑓 ′′ (−3) = 6(−3) + 8 < 0
Therefore (−3, 12) is a maximum point.
iii.
𝑦
Max (−3, 12)
12
𝑥 3 + 4𝑥 2 − 3𝑥 − 6
−3
−2
−1
0
𝑥
1
3
1
−6
1
Min ( , −6.5)
3
24
CAPE PURE MATHEMATICS UNIT 1
SOLUTIONS FOR 2014 EXAM
Question 1
a.
p
T
T
T
T
F
F
F
F
b.
i.
q
T
T
F
F
T
T
F
F
r
T
F
T
F
T
F
T
F
𝑝→𝑞
T
T
F
F
T
T
T
T
(𝑝 → 𝑞) ∧ (𝑟 → 𝑞)
T
T
F
F
T
T
F
T
𝑟→𝑞
T
T
F
T
T
T
F
T
Given 𝑦 ⊕ 𝑥 = 𝑦 3 + 𝑥 3 + 𝑎𝑦 2 + 𝑎𝑥 2 − 5𝑦 − 5𝑥 + 16
𝑥 ⊕ 𝑦 = 𝑥 3 + 𝑦 3 + 𝑎𝑥 2 + 𝑎𝑦 2 − 5𝑥 − 5𝑦 + 16
𝑦 ⊕ 𝑥 = 𝑥 ⊕ 𝑦 therefore ⊕ is commutative in R
ii.
a.
We have 𝑓(𝑥) = 2 ⊕ 𝑥 = 23 + 𝑥 3 + 𝑎22 + 𝑎𝑥 2 − 5(2) − 5𝑥 + 16
𝑓(𝑥) = 8 + 𝑥 3 + 4𝑎 + 𝑎𝑥 2 − 10 − 5𝑥 + 16
𝑓(𝑥) = 𝑥 3 + 𝑎𝑥 2 − 5𝑥 + 4𝑎 + 14
If (𝑥 − 1) is a factor then 𝑓(1) = 0
Therefore 𝑓(1) = 13 + 𝑎(1)2 − 5(1) + 4𝑎 + 14 = 0
1 + 𝑎 − 5 + 4𝑎 + 14 = 0
5𝑎 + 10 = 0,
𝑎 = −2
b.
When 𝑎 = −2, 𝑓(𝑥) = 𝑥 3 + (−2)𝑥 2 − 5𝑥 + 4(−2) + 14
𝑓(𝑥) = 𝑥 3 − 2𝑥 2 − 5𝑥 + 6
(𝑥 − 1) is a factor of 𝑓(𝑥) therefore using long division we have.
25
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2014
𝑥2 − 𝑥 − 6
𝑥 − 1 𝑥 3 − 2𝑥 2 − 5𝑥 + 6
𝑥3 − 𝑥2
−𝑥 2 − 5𝑥
−𝑥 2 + 𝑥
−6𝑥 + 6
−6𝑥 + 6
0
3
2
2
𝑥 − 2𝑥 − 5𝑥 + 6 = (𝑥 − 1)(𝑥 − 𝑥 − 6)
= (𝑥 − 1)(𝑥 + 2)(𝑥 − 3)
Therefore factors are (𝑥 − 1), (𝑥 + 2) and (𝑥 − 3)
c.
𝑛
12 + 32 + 52 + ⋯ + (2𝑛 − 1)2 = 3 (4𝑛2 − 1)
When 𝑛 = 1,
1
L.H.S = 12 = 1, and R.H.S = 3 (4(1)2 − 1) = 1
L.H.S = R.H.S therefore result is true for 𝑛 = 1
Assume result is true for 𝑛 = 𝑘, therefore
𝑘
12 + 32 + 52 + ⋯ + (2𝑘 − 1)2 = 3 (4𝑘 2 − 1)
When 𝑛 = 𝑘 + 1
R.H.S =
𝑘+1
3
(4[𝑘 + 1]2 − 1)
L.H.S = 12 + 32 + 52 + ⋯ + (2𝑘 − 1)2 + (2(𝑘 + 1) − 1)2
𝑘
12 + 32 + 52 + ⋯ + (2𝑘 − 1)2 = 3 (4𝑘 2 − 1)
𝑘
L.H.S = 3 (4𝑘 2 − 1) + (2(𝑘 + 1) − 1)2
𝑘
= 3 (2𝑘 − 1)(2𝑘 + 1) + (2𝑘 + 1)2
𝑘
= (2𝑘 + 1) [3 (2𝑘 − 1) + (2𝑘 + 1)]
= (2𝑘 + 1) [
𝑘(2𝑘 − 1) + 3(2𝑘 + 1)
]
3
26
L. H. S =
2𝑘 + 1
(2𝑘 2 + 5𝑘 + 3)
3
=
2𝑘 + 1
(2𝑘 + 3)(𝑘 + 1)
3
=
𝑘+1
(2𝑘 + 1)(2𝑘 + 3)
3
=
𝑘+1
(4𝑘 2 + 8𝑘 + 3)
3
=
𝑘+1
(4{𝑘 2 + 2𝑘} + 3)
3
=
𝑘+1
(4[𝑘 + 1]2 − 4 + 3)
3
=
𝑘+1
(4[𝑘 + 1]2 − 1)
3
L.H.S = R.H.S therefore result is true for 𝑛 = 𝑘 + 1
Since result is true for 𝑛 = 1, 𝑘, and 𝑘 + 1 it is true for all positive integer n.
Question 2
a.
𝑥−1
Given 𝑓(𝑥) = 2𝑥 2 + 1, 𝑔(𝑥) = √
i.
a.
2
𝑓𝑓(𝑥) = 2(𝑓(𝑥))2 + 1
= 2(2𝑥 2 + 1)2 + 1
= 2(4𝑥 4 + 4𝑥 2 + 1) + 1
= 8𝑥 4 + 8𝑥 2 + 2 + 1
= 8𝑥 4 + 8𝑥 2 + 3
b.
𝑓[𝑔(𝑥)] = 2[𝑔(𝑥)]2 + 1
2
𝑥−1
= 2 (√
2
) +1
𝑥−1
= 2(
2
)+1
=𝑥−1+1
=𝑥
ii.
𝑓 −1 (𝑥) = 𝑔(𝑥) Since 𝑓[𝑓 −1 (𝑥)] = 𝑥
27
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2014
b.
Given that 𝑎3 + 𝑏 3 + 3𝑎2 𝑏 = 5𝑎𝑏 2
We know that (𝑎 + 𝑏)3 = 𝑎3 + 3𝑎2 𝑏 + 3𝑎𝑏 2 + 𝑏 3
Adding 3𝑎𝑏 2 to both sides we have
𝑎3 + 𝑏 3 + 3𝑎2 𝑏 + 3𝑎𝑏 2 = 3𝑎𝑏 2 + 5𝑎𝑏 2
(𝑎 + 𝑏)3 = 8𝑎𝑏 2
(𝑎 + 𝑏)3
= 𝑎𝑏 2
8
𝑎+𝑏 3
(
) = 𝑎𝑏 2
2
𝑎+𝑏 3
)
2
log (
c.
i.
= log 𝑎𝑏 2
[Log both sides]
3 log (
𝑎+𝑏
) = log 𝑎 + log 𝑏 2
2
3 log (
𝑎+𝑏
) = log 𝑎 + 2 log 𝑏
2
𝑒𝑥 +
1
𝑒𝑥
−2=0
𝑒 2𝑥 − 2𝑒 𝑥 + 1 = 0
[Multiplying both sides by 𝑒 𝑥 ]
Let 𝑢 = 𝑒 𝑥 , 𝑢2 − 2𝑢 + 1 = (𝑢 − 1)2
𝑢 = 1, 𝑒 𝑥 = 1 , 𝑥 = 0
ii.
log 2 (𝑥 + 1) − log 2 (3𝑥 + 1) = 2
log 2
𝑥+1
= 2 log 2 2
3𝑥 + 1
log 2
𝑥+1
= log 2 22
3𝑥 + 1
𝑥+1
=4
3𝑥 + 1
𝑥 + 1 = 4(3𝑥 + 1)
𝑥 + 1 = 12𝑥 + 4
11𝑥 = −3
𝑥=−
3
11
28
d.
√3−1
√3+1
√3+1
√3−1
+
+
√2−1
√2+1
2
+
2
(√3 − 1) + (√3 + 1)
(√3 + 1)(√3 − 1)
√2+1
√2−1
2
+
2
(√2 − 1) + (√2 + 1)
(√2 + 1)(√2 − 1)
4 − 2√3 + 4 + 2√3 3 − 2√2 + 3 + 2√2
+
3−1
2−1
8 6
+ = 4 + 6 = 10
2 1
Question 3
a.
i.
ii.
cot 𝑦−cot 𝑥
cot 𝑥+cot 𝑦
cot 𝑦−cot 𝑥
cot 𝑥+cot 𝑦
=
cos 𝑦
⁄sin 𝑦−cos 𝑥⁄sin 𝑥
cos 𝑦
cos 𝑥⁄
⁄sin 𝑦
sin 𝑥 +
=
sin 𝑥 cos 𝑦 − sin 𝑦 cos 𝑥 cos 𝑥 sin 𝑦 + sin 𝑥 cos 𝑦
/
sin 𝑦 sin 𝑥
sin 𝑥 sin 𝑦
=
sin 𝑥 cos 𝑦 − sin 𝑦 cos 𝑥
sin 𝑥 sin 𝑦
×
sin 𝑦 sin 𝑥
cos 𝑥 sin 𝑦 + sin 𝑥 cos 𝑦
=
sin 𝑥 cos 𝑦 − sin 𝑦 cos 𝑥
cos 𝑥 sin 𝑦 + sin 𝑥 cos 𝑦
=
sin(𝑥 − 𝑦)
sin(𝑥 + 𝑦)
= 1, 0 ≤ 𝑦 ≤ 2𝜋,
1
1
𝜋
𝜋
When sin 𝑥 = 2 , sin−1 (2) = 6 , cos ( 6 ) =
√3
2
for 0 ≤ 𝑥 ≤
𝜋
2
cot 𝑦 − cot 𝑥 sin 𝑥 cos 𝑦 − sin 𝑦 cos 𝑥
=
cot 𝑥 + cot 𝑦 cos 𝑥 sin 𝑦 + sin 𝑥 cos 𝑦
sin 𝑥 cos 𝑦 − sin 𝑦 cos 𝑥
=1
cos 𝑥 sin 𝑦 + sin 𝑥 cos 𝑦
1
2
√3
)
2
( ) cos 𝑦−sin 𝑦(
1
√3
) sin 𝑦+( ) cos 𝑦
2
2
(
cos 𝑦 − √3 sin 𝑦
√3 sin 𝑦 + cos 𝑦
= 1 Multiplying both numerator & denominator by 2
=1
cos 𝑦 − √3 sin 𝑦 = √3 sin 𝑦 + cos 𝑦
29
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2014
0 = 2√3 sin 𝑦
sin 𝑦 = 0
𝑦 = 0, 𝜋, 2𝜋
b.
i.
for 0 ≤ 𝑦 ≤ 2𝜋.
Given 𝑓(𝜃) = 3 sin 2𝜃 + 4 cos 2𝜃 to be written in the form 𝑟 sin(2𝜃 + 𝛼)
We have 𝑟 sin(2𝜃 + 𝛼 ) = 𝑟 sin 2𝜃 cos 𝛼 + 𝑟 sin 𝛼 cos 2𝜃
𝑟 = √32 + 42 = 5
3 sin 2𝜃 + 4 cos 2𝜃 = 𝑟 sin 2𝜃 cos 𝛼 + 𝑟 sin 𝛼 cos 2𝜃
Comparing we have 𝑟 cos 𝛼 = 3, 𝑟 sin 𝛼 = 4
𝑟 sin 𝛼
4
= tan 𝛼 =
𝑟 cos 𝛼
3
4
𝛼 = tan−1 (3) = 0.927 rad
𝑓(𝜃) = 𝑟 sin(2𝜃 + 𝛼) = 5 sin(2𝜃 + 0.927)
ii.
a.
𝑓(𝜃) is at a minimum when
(2𝜃 + 0.927) =
2𝜃 =
𝜃=
3𝜋
2
3𝜋
− 0.927
2
3𝜋 0.927
−
4
2
𝜃 = 1.89 rad
b.
The maximum value of
1
7−𝑓(𝜃)
is when 𝑓(𝜃) = 5 so
1
1
1
=
=
7 − 𝑓(𝜃) 7 − 5 2
And the minimum value of
1
1
1
=
=
7 − 𝑓(𝜃) 7 − (−5) 12
30
1
7−𝑓(𝜃)
is when 𝑓(𝜃) = −5
Question 4
a.
Given the equations of 𝐿1 and 𝐿2 are 𝑥 − 𝑦 + 1 = 0 and 𝑥 + 𝑦 − 5 = 0
i.
𝐿1 and 𝐿2 intersects at the centre of the circle therefore
Solving the equations simultaneously we have
𝑥 − 𝑦 + 1 = 0 ------ (1)
𝑥 + 𝑦 − 5 = 0 ------ (2)
2𝑥 − 4 = 0
Adding (1) and (2)
𝑥=2
When 𝑥 = 2, from (2) 𝑦 = 5 − 𝑥
𝑦 = 5−2 =3
Therefore the coordinate of the centre of the circle is (2, 3)
ii.
Let A (1, 2) and B (a, b) be the coordinates of the endpoints of the diameter of the
circle and the coordinates of the it’s centre (2, 3) is the midpoint of the line AB.
Therefore in calculating the midpoint we have
𝑎+1
= 2,
2
𝑎=3
𝑏+2
=3 𝑏=4
2
Therefore B has coordinates (3, 4)
iii.
The point p moves in a circular path with centre (2, 3) and radius √2.
The equation of the path of p is given by (𝑥 − 2)2 + (𝑦 − 3)2 = (√2)
2
𝑥 2 − 4𝑥 + 4 + 𝑦 2 − 6𝑦 + 9 = 2
𝑥 2 + 𝑦 2 − 4𝑥 − 6𝑦 + 11 = 0
31
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2014
b.
1
𝑡
Given 𝑥 = 1+𝑡, and 𝑦 = 1−𝑡 2
𝑥=
1
1+𝑡
𝑥(1 + 𝑡) = 1
𝑥 + 𝑥𝑡 = 1
𝑥𝑡 = 1 − 𝑥
𝑡=
1−𝑥
𝑥
𝑦=
𝑡
𝑡
=
2
(1 + 𝑡)(1 − 𝑡)
1−𝑡
𝑦=
1
𝑡
1
×
, substituting 𝑥 =
we have
1+𝑡 1−𝑡
1+𝑡
𝑦=
𝑥𝑡
,
1−𝑡
𝑦=
c.
i.
substituting 𝑡 =
1−𝑥
we have
𝑥
1−𝑥
𝑥( 𝑥 )
1−𝑥
1−( 𝑥 )
𝑦=
1−𝑥
𝑥 − (1 − 𝑥)
𝑥
𝑦=
𝑥(1 − 𝑥)
2𝑥 − 1
Given 𝑃(3, −2, 1), 𝑄(−1, 𝜆, 5) and 𝑅(2, 1, −4)
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗
𝑃𝑄 = ⃗⃗⃗⃗⃗
𝑃𝑂 + 𝑂𝑄
3
−1
−4
⃗⃗⃗⃗⃗ = − (−2) + ( 𝜆 ) = (2 + 𝜆)
𝑃𝑄
1
5
4
⃗⃗⃗⃗⃗
𝑃𝑄 = −4𝒊 + (2 + 𝜆)𝒋 + 4𝒌
⃗⃗⃗⃗⃗ = 𝑄𝑂
⃗⃗⃗⃗⃗⃗ + 𝑂𝑅
⃗⃗⃗⃗⃗
𝑄𝑅
32
−1
2
3
⃗⃗⃗⃗⃗ = − ( 𝜆 ) + ( 1 ) = (1 − 𝜆)
𝑄𝑅
5
−4
−9
⃗⃗⃗⃗⃗ = 3𝒊 + (1 − 𝜆)𝒋 − 9𝒌
𝑄𝑅
⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗
𝑅𝑃 = 𝑅𝑂
𝑂𝑃
2
3
1
⃗⃗⃗⃗⃗
𝑅𝑃 = − ( 1 ) + (−2) = (−3)
−4
1
5
⃗⃗⃗⃗⃗ = 𝒊 − 3𝒋 + 5𝒌
𝑅𝑃
ii.
Given PQ is the hypotenuse therefore RQ and RP are perpendicular to each other.
(𝑅𝑄) ∙ (𝑅𝑃) = 0
−(3𝒊 + (1 − 𝜆)𝒋 − 9𝒌) ∙ (𝒊 − 3𝒋 + 5𝒌) = −3 − 3(𝜆 − 1) + 9(5) = 0
−3 − 3𝜆 + 3 + 45 = 0
−3𝜆 = −45,
𝜆=
Question 5
a.
Given
i.
𝑓(𝑥) = {
𝑎𝑥 + 2,
𝑎𝑥 2 ,
𝑥<3
.
𝑥≥3
lim 𝑓(𝑥) = 𝑎(32 ) = 9𝑎
𝑥→3+
lim 𝑓(𝑥) = 3𝑎 + 2
𝑥→3−
If 𝑓(𝑥) is continuous at 𝑥 = 3, then
lim 𝑓(𝑥) = lim− 𝑓(𝑥) Therefore
𝑥→3+
𝑥→3
9𝑎 = 3𝑎 + 2
6𝑎 = 2,
𝑎=
1
3
33
−45
= 15
−3
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2014
ii.
𝑔(𝑥) =
𝑥2 + 2
𝑏𝑥 2 + 𝑥 + 4
Given that lim 2𝑔(𝑥) = lim 𝑔(𝑥) we have
𝑥→1
2(
𝑥→0
12 + 2
02 + 2
=
)
𝑏(1)2 + 1 + 4
𝑏(0)2 + 0 + 4
2(
3
2
)=
𝑏+5
4
6
1
=
𝑏+5 2
12 = 𝑏 + 5
𝑏=7
b.
Let 𝑓(𝑥) =
1
√𝑥
,
𝑓(𝑥 + ℎ) =
1
√𝑥 + ℎ
Using differentiation from first principle, we have
𝑑𝑦
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
= lim
𝑑𝑥 ℎ→0
ℎ
1
1
−
𝑑𝑦
+ ℎ √𝑥
= lim √𝑥
𝑑𝑥 ℎ→0
ℎ
√𝑥 − √𝑥 + ℎ
(√𝑥 + ℎ)(√𝑥)
= lim
ℎ→0
ℎ
= lim
√𝑥 − √𝑥 + ℎ
ℎ→0 ℎ(√𝑥
+ ℎ)(√𝑥)
√𝑥 − √𝑥 + ℎ
√𝑥 + √𝑥 + ℎ
= lim [
×
]
ℎ→0 ℎ(√𝑥 + ℎ)(√𝑥)
√𝑥 + √𝑥 + ℎ
𝑥 − (𝑥 + ℎ)
= lim [
]
ℎ→0 ℎ(√𝑥 + ℎ)(√𝑥)(√𝑥 + √𝑥 + ℎ)
−ℎ
= lim [
]
ℎ→0 ℎ(√𝑥 + ℎ)(√𝑥)(√𝑥 + √𝑥 + ℎ)
34
𝑑𝑦
−1
= lim [
]
𝑑𝑥 ℎ→0 (√𝑥 + ℎ)(√𝑥)(√𝑥 + √𝑥 + ℎ)
=
−1
(√𝑥)(√𝑥)(√𝑥 + √𝑥)
=−
ii.
1
1
= − 𝑥 −3/2
2
2𝑥√𝑥
Given 𝑦 =
𝑥
√1+𝑥
𝑢
Using the quotient rule when 𝑦 = 𝑣 ,
We have 𝑢 = 𝑥,
𝑢′ = 1
𝑑𝑦
=
𝑑𝑥
𝑑𝑦
𝑑𝑥
=
1
1
𝑣 ′ = (1 + 𝑥)−1/2 =
2
2√1 + 𝑥
1
)
(√1 + 𝑥)(1) − 𝑥 (
2√1 + 𝑥
2
(√1 + 𝑥)
𝑑𝑦
=
𝑑𝑥
𝑥
√1 + 𝑥
1 − 2√1 + 𝑥
1+𝑥
𝑑𝑦
=
𝑑𝑥
(√1 + 𝑥)(2√1 + 𝑥) − 𝑥
2√1 + 𝑥
1+𝑥
𝑑𝑦 2(1 + 𝑥) − 𝑥
=
𝑑𝑥
2√(1 + 𝑥)3
𝑑𝑦
𝑥+2
=
𝑑𝑥 2√(1 + 𝑥)3
Given 𝑥 = cos 𝜃,
𝑦 = sin 𝜃
𝑑𝑥
𝑑𝑦
= − sin 𝜃 ,
= cos 𝜃
𝑑θ
𝑑𝜃
𝑑𝑦 𝑑𝑦/𝑑𝜃
=
𝑑𝑥 𝑑𝑥/𝑑𝜃
=
𝑣2
𝑣 = √1 + 𝑥
𝑑𝑦 (√1 + 𝑥)(2√1 + 𝑥) − 𝑥
=
𝑑𝑥
2√1 + 𝑥(1 + 𝑥)
c.
𝑣𝑢′ −𝑢𝑣′
cos 𝜃
= − cot 𝜃
− sin 𝜃
35
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2014
Question 6
a.
i.
a.
Given
𝑑𝑦
𝑑𝑥
= 3𝑥 2 − 4𝑥 + 1
Integrating both sides we have
𝑦 = ∫ 3𝑥 2 − 4𝑥 + 1 𝑑𝑥
𝑦=3
𝑥 2+1
𝑥 1+1
−4
+𝑥+𝐶
2+1
1+1
𝑦 = 𝑥 3 − 2𝑥 2 + 𝑥 + 𝐶
When 𝑥 = −1, 𝑦 = −4
−4 = (−1)3 − 2(−1)2 + (−1) + 𝐶
−4 = −4 + 𝐶
𝐶=0
𝑦 = 𝑥 3 − 2𝑥 2 + 𝑥
b.
𝑑𝑦
At the stationary points 𝑑𝑥 = 0 therefore
3𝑥 2 − 4𝑥 + 1 = 0
(3𝑥 − 1)(𝑥 − 1) = 0
1
𝑥 = 3 , or 1
1
1 3
1 2
1
When 𝑥 = 3 , 𝑦 = (3) − 2 (3) + 3
𝑦=
1 2 1 1−6+9
4
− + =
=
27 9 3
27
27
1
4
Therefore coordinate of the stationary point is (3 , 27)
When 𝑥 = 1, 𝑦 = 13 − 2(1)2 + 1 = 0
Therefore coordinate of the stationary point is (1, 0)
36
𝑑2 𝑦
= 6𝑥 − 4
𝑑𝑥 2
1
When 𝑥 = 3 ,
𝑑2 𝑦
𝑑𝑥 2
𝑑𝑥 2
ii.
𝑑𝑥 2
1
= 6 (3) − 4 = −2
1
4
< 0 Therefore (3 , 27) is a maximum
When 𝑥 = 1,
𝑑2 𝑦
𝑑2 𝑦
𝑑2 𝑦
𝑑𝑥 2
= 6(1) − 4 = 2
> 0 Therefore (1, 0) is a minimum
y-intercept when 𝑥 = 0, 𝑦 = 0 (0, 0)
x-intercept when 𝑦 = 0, 𝑥 3 − 2𝑥 2 + 𝑥 = 0
𝑥(𝑥 2 − 2𝑥 + 1 = 0)
𝑥(𝑥 − 1)2 = 0
When 𝑦 = 0, 𝑥 = 0, 1
x-intercepts (0, 0) and (1, 0)
𝑦
1 4
max ( , )
3 27
4
27
min(1, 0)
0
𝑥
1
3
2
3
37
1
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2014
b.
i.
3
∫0 2𝑥√1 + 𝑥 2 𝑑𝑥
Using the substitution method with 𝑢 = √1 + 𝑥 2
We have
𝑑𝑢
𝑑𝑥
1
𝑥
2
√1+𝑥 2
= (1 + 𝑥 2 )−1/2 × 2𝑥 =
𝑑𝑢
𝑢 = √1 + 𝑥 2 ,
𝑑𝑥
𝑥
𝑢
𝑢
𝑥
= , 𝑑𝑥 = 𝑑𝑢
When 𝑥 = 3, 𝑢 = √1 + 32 = √10
When 𝑥 = 0, 𝑢 = √1 + 0 = 1
𝑢
𝑥
√10
2𝑥(𝑢) 𝑑𝑢
∫
1
2𝑢3 √10
2𝑢 𝑑𝑢 = [
]
3
1
√10
∫
1
2
3
2(√10)
2(1)3
=[
−
]
3
3
2
= [√103 − 1] = 20.42
3
ii.
𝑏
Volume of revolution about the x-axis is given by 𝜋 ∫𝑎 𝑦 2 𝑑𝑥 therefore from b. (i)
2
2
Volume = 𝜋 ∫0 (2𝑥√1 + 𝑥 2 ) 𝑑𝑥
2
= 𝜋 ∫ 4𝑥 2 (1 + 𝑥 2 ) 𝑑𝑥
0
2
= 𝜋 ∫ 4𝑥 2 + 4𝑥 4 𝑑𝑥
0
4𝑥 3 4𝑥 5 2
= 𝜋[
+
]
3
5 0
= 𝜋 [(
= 𝜋[
4(2)3 4(2)5
+
) − 0]
3
5
32 128
544
+
𝜋 cubic units
]=
3
5
15
38
CAPE PURE MATHEMATICS UNIT 1
SOLUTIONS FOR 2013 EXAM
Question 1
a.
i.
p
T
T
F
F
q
T
F
T
F
𝑝→𝑞
T
F
T
T
p
T
T
F
F
Q
T
F
T
F
𝑝∧𝑞
T
F
F
F
ii.
b.
~(𝑝 ∧ 𝑞)
F
T
T
T
Given 𝑦 ⊕ 𝑥 = 𝑦 2 + 𝑥 2 + 2𝑦 + 𝑥 − 5𝑥𝑦
2 ⊕ 𝑥 = 22 + 𝑥 2 + 2(2) + 𝑥 − 5𝑥(2)
2 ⊕ 𝑥 = 4 + 𝑥 2 + 4 + 𝑥 − 10𝑥
2 ⊕ 𝑥 = 𝑥 2 − 9𝑥 + 8
2⊕𝑥 = 0
𝑥 2 − 9𝑥 + 8 = 0
(𝑥 − 1)(𝑥 − 8) = 0
𝑥 = 1, 8
39
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2013
c.
When 𝑛 = 1, 51 + 3 = 8 which is divisible by 2
Therefore statement is true for 𝑛 = 1
Assume statement is true when 𝑛 = 𝑘
Therefore 5𝑘 + 3 is divisible by 2
When 𝑛 = 𝑘 + 1, we have 5𝑘+1 + 3 = 5(5𝑘 + 3) − 12
5𝑘 + 3 is assumed to be divisible by 2 and 12 is divisible by 2
Therefore 5𝑘+1 + 3 is divisible by 2
Since the statement is true for 𝑛 = 1, 𝑘, and 𝑘 + 1 it is true for all positive integer n.
d.
Given 𝑓(𝑥) = 𝑥 3 − 9𝑥 2 + 𝑝𝑥 + 16
i.
If (𝑥 + 1) is a factor then 𝑓(−1) = 0 therefore
𝑓(−1) = (−1)3 − 9(−1)2 + 𝑝(−1) + 16 = 0
−𝑝 + 6 = 0, 𝑝 = 6
ii.
𝑥 2 − 10𝑥 + 16
(𝑥 + 1)
𝑥 3 − 9𝑥 2 + 6𝑥 + 16
𝑥3 + 𝑥2
−10𝑥 2 + 6𝑥
−10𝑥 2 − 10𝑥
16𝑥 + 16
16𝑥 + 16
0
𝑥 3 − 9𝑥 2 + 6𝑥 + 16 = (𝑥 + 1)(𝑥2 − 10𝑥 + 16)
= (𝑥 + 1)(𝑥 − 2)(𝑥 − 8)
iii.
𝑥 3 − 9𝑥 2 + 6𝑥 + 16 = (𝑥 + 1)(𝑥 − 2)(𝑥 − 8) = 0
Therefore 𝑥 = −1, 2, 8
40
Question 2
a.
Given 𝑓(𝑥) = 𝑥 2 − 𝑥,
𝑥≥1
By completing the square we have
1
1
𝑥 2 − 𝑥 = (𝑥 2 − 𝑥 + 4) − 4
1 2
1
= (𝑥 − 2) − 4
1
The function has a parabolic shape with axis of symmetry 𝑥 = 2.
The domain given is 𝑥 ≥ 1, and this section of the graph is a one to one function
due to it passing the horizontal line test.
𝑦
𝑥=
1
2
𝑓(𝑥) = 𝑥 2 − 𝑥
𝑦=𝑐
1
𝑥
0
1
2
Alternatively, if we assume that 𝑓(𝑥) is not a one to one function when 𝑥 = 𝑎, or 𝑏
Then 𝑓(𝑎) = 𝑓(𝑏) where 𝑎 ≠ 𝑏.
So
𝑓(𝑎) = 𝑓(𝑏)
𝑎2 − 𝑎 = 𝑏 2 − 𝑏
𝑎2 − 𝑏 2 − 𝑎 + 𝑏 = 0
(𝑎 + 𝑏)(𝑎 − 𝑏) − (𝑎 − 𝑏) = 0
(𝑎 − 𝑏)(𝑎 + 𝑏 + 1) = 0
Therefore 𝑎 = 𝑏, or 𝑎 = −𝑏 − 1
For 𝑥 ≥ 1 which eliminates 𝑎 = −(𝑏 + 1), 𝑓(𝑎) = 𝑓(𝑏) is only true if 𝑎 = 𝑏.
Therefore the function is one to one for 𝑥 ≥ 1.
41
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2013
b.
i.
Given 𝑓(𝑥) = 3𝑥 + 2, and 𝑔(𝑥) = 𝑒 2𝑥
a.
𝑦 = 3𝑥 + 2 interchanging 𝑥 and 𝑦 we have
Let
𝑥 = 3𝑦 + 2, therefore 𝑦 =
𝑓 −1 (𝑥) =
𝑥−2
3
𝑥−2
3
Let 𝑦 = 𝑒 2𝑥 interchanging 𝑥 and 𝑦 we have
𝑥 = 𝑒 2𝑦
Natural log of both sides gives.
ln 𝑥 = ln 𝑒 2𝑦
ln 𝑥 = 2𝑦
1
1
𝑦 = 2 ln 𝑥 Therefore 𝑔−1 (𝑥) = 2 ln 𝑥
𝑓[𝑔(𝑥)] = 3[𝑔(𝑥)] + 2
b.
= 3𝑒 2𝑥 + 2
ii.
𝑦 = 3𝑒 2𝑥 + 2 interchanging 𝑥 and 𝑦 we have
𝑥 = 3𝑒 2𝑦 + 2
𝑒 2𝑦 =
𝑥−2
Natural log of both sides.
3
ln 𝑒 2𝑦 = ln (
2𝑦 = ln (
𝑥−2
)
3
𝑥−2
1
𝑥−2
) , therefore 𝑦 = ln (
)
3
2
3
[𝑓[𝑔(𝑥)]]
−1
=
𝑔−1 [𝑓 −1 (𝑥)] =
1
𝑥−2
ln (
)
2
3
1
ln(𝑓 −1 (𝑥))
2
1
𝑥−2
= ln (
)
2
3
−1
Therefore [𝑓[𝑔(𝑥)]]
= 𝑔−1 [𝑓 −1 (𝑥)]
42
c.
i.
3𝑥 2 + 4𝑥 + 1 ≤ 5
3𝑥 2 + 4𝑥 − 4 ≤ 0
(3𝑥 − 2)(𝑥 + 2) ≤ 0
Critical points 𝑥 = −2,
2
3
−2
𝑥 ≤ −2
3𝑥 − 2
𝑥+2
(3𝑥 − 2)(𝑥 + 2)
−2 ≤ 𝑥 ≤
−
−
+
2
3
𝑥≥
−
+
−
2
3
2
3
+
+
+
2
2
Therefore −2 ≤ 𝑥 ≤ 3 from the table is negative or from the graph −2 ≤ 𝑥 ≤ 3
is the part of the graph that is below the x-axis.
ii.
|𝑥 + 2| = 3𝑥 + 5
Squaring both sides we have
(𝑥 + 2)2 = (3𝑥 + 5)2
𝑥 2 + 4𝑥 + 4 = 9𝑥 2 + 30𝑥 + 25
8𝑥 2 + 26𝑥 + 21 = 0
8𝑥 2 + 12𝑥 + 14𝑥 + 21 = 0
4𝑥(2𝑥 + 3) + 7(2𝑥 + 3) = 0
(4𝑥 + 7)(2𝑥 + 3) = 0
7
3
𝑥 = − , not possible − only answer
4
2
Alternatively, for (𝑥 + 2) ≥ 0 we have 𝑥 + 2 = 3𝑥 + 5
2𝑥 = −3, 𝑥 = −
3
2
For (𝑥 + 2) < 0, we have −(𝑥 + 2) = 3𝑥 + 5
−𝑥 − 2 = 3𝑥 + 5
4𝑥 = −7, 𝑥 = −
43
7
not possible
4
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2013
Question 3
a.
i.
L.H.S
2 tan 𝜃
1+tan2 𝜃
substituting tan 𝜃 =
2 tan 𝜃
We have
=
1 + tan2 𝜃
sin 𝜃
cos 𝜃
sin 𝜃
2 cos 𝜃
sin 𝜃 2
1 + (cos 𝜃 )
Multiplying denominator and numerator by cos2 𝜃
2 tan 𝜃
2 sin 𝜃 cos 𝜃
=
2
1 + tan 𝜃 cos 2 𝜃 + sin2 𝜃
cos2 𝜃 + sin2 𝜃 = 1
2 tan 𝜃
= 2 sin 𝜃 cos 𝜃
1 + tan2 𝜃
2 sin 𝜃 cos 𝜃 = sin 2𝜃
ii.
Given sin 2𝜃 − tan 𝜃 = 0
2 tan 𝜃
− tan 𝜃 = 0
1 + tan2 𝜃
2 tan 𝜃 − tan 𝜃 (1 + tan2 𝜃) = 0
2 tan 𝜃 − tan 𝜃 − tan3 𝜃 = 0
tan 𝜃 − tan3 𝜃 = 0
tan 𝜃 (1 − tan2 𝜃) = 0
tan 𝜃(1 − tan 𝜃)(1 + tan 𝜃) = 0
tan 𝜃 = 0, 1, −1
𝜃 = tan−1 (0) = 0, 𝜋, 2𝜋
Acute angle for 𝜃 = tan−1 (1) =
𝜋 3𝜋 5𝜋 7𝜋
𝜃 = 0, ,
, ,
4 4 4 4
44
𝜋
4
b.
i.
Given 𝑓(𝜃) = 3 cos 𝜃 − 4 sin 𝜃
𝑟 cos(𝜃 + 𝛼) = 𝑟 cos 𝜃 cos 𝛼 − 𝑟 sin 𝜃 sin 𝛼
𝑟 cos 𝛼 = 3,
𝑟 sin 𝛼 = 4
𝑟 sin 𝛼
4
= tan 𝛼 =
𝑟 cos 𝛼
3
4
𝛼 = tan−1 ( ) = 0.927
3
𝑟 = √32 + 42 = 5
3 cos 𝜃 − 4 sin 𝜃 = 5 cos(𝜃 + 0.927)
ii.
a.
𝑓(𝜃) = 5 cos(𝜃 + 0.927)
Therefore maximum value of 𝑓(𝜃) is 5 (−1 ≤ cos 𝜃 ≤ 1)
b.
Minimum value of
Therefore
iii.
a.
1
8+𝑓(𝜃)
is when 𝑓(𝜃) is maximum
1
1
1
=
=
8 + 𝑓(𝜃) 8 + 5 13
Given that A, B and C are the angles of a triangle where their sum is π.
𝐴+𝐵+𝐶 =𝜋
𝐴 = 𝜋 − (𝐵 + 𝐶) Taking sine of the angles
sin 𝐴 = sin[𝜋 − (𝐵 + 𝐶)]
sin[𝜋 − (𝐵 + 𝐶)] = sin 𝜋 cos(𝐵 + 𝐶) − sin(𝐵 + 𝐶) cos 𝜋
sin 𝜋 = 0,
cos 𝜋 = −1 Therefore
sin[𝜋 − (𝐵 + 𝐶)] = 0 − sin(𝐵 + 𝐶) (−1)
sin[𝜋 − (𝐵 + 𝐶)] = sin(𝐵 + 𝐶)
sin 𝐴 = sin(𝐵 + 𝐶)
b.
from (iii).a. sin 𝐴 = sin(𝐵 + 𝐶) therefore
sin 𝐵 = sin(𝐴 + 𝐶) and sin 𝐶 = sin(𝐴 + 𝐵) so
sin 𝐴 + sin 𝐵 + sin 𝐶 = sin(𝐵 + 𝐶) + sin(𝐴 + 𝐶) + sin(𝐴 + 𝐵)
45
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2013
Question 4
a.
i.
Given 𝑥 2 + 𝑦 2 − 6𝑥 − 4𝑦 + 4 = 0
By completing the square we have
−6 2
−6 2
−4 2
−4 2
(𝑥 2 − 6𝑥 + ( ) ) − ( ) + (𝑦 2 − 4𝑦 + ( ) ) − ( ) + 4 = 0
2
2
2
2
(𝑥 2 − 6𝑥 + 9) − 9 + (𝑦 2 − 4𝑦 + 4) − 4 + 4 = 0
(𝑥 − 3)2 + (𝑦 − 2)2 = 9 = 32
The equation of a circle is given by (𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 𝑟 2
Where (𝑎, 𝑏) is the centre and r its radius.
Therefore the circle has centre (3, 2) and radius √9 = 3
ii.
The gradient of the line between the centre (3, 2) and a point on the
a.
2−2
circumference (6, 2) is given by 6−3 = 0,
Therefore equation of the normal to the circle at (6, 2) is given by 𝑦 = 2
The tangent is perpendicular to the normal therefore the line 𝑦 = 2 at
b.
(6, 2) is perpendicular to 𝑥 = 6, which is a vertical line parallel to the
y-axis.
b.
Given 𝑥 = 𝑡 2 + 𝑡, 𝑦 = 2𝑡 − 4
From 𝑦 = 2𝑡 − 4, 𝑡 =
Substituting 𝑡 =
𝑦+4
2
𝑦+4
2
into 𝑥 = 𝑡 2 + 𝑡
𝑦+4 2 𝑦+4
We have 𝑥 = (
) +
2
2
𝑥=
𝑦 2 + 8𝑦 + 16 𝑦 + 4
+
4
2
𝑥=
𝑦 2 + 8𝑦 + 16 + 2(𝑦 + 4)
4
𝑥=
𝑦 2 + 10𝑦 + 24
4
4𝑥 = 𝑦 2 + 10𝑦 + 24
46
c.
i.
Given 𝐴(3, −1, 2), 𝐵(1, 2, −4) and 𝐶(−1, 1, −2)
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗
𝐴𝐵 = ⃗⃗⃗⃗⃗
𝐴𝑂 + 𝑂𝐵
3
1
= − (−1) + ( 2 )
2
−4
−2
=( 3 )
−6
⃗⃗⃗⃗⃗
𝐴𝐵 = −2𝒊 + 3𝒋 − 6𝒌
⃗⃗⃗⃗⃗ = 𝐵𝑂
⃗⃗⃗⃗⃗ + 𝑂𝐶
⃗⃗⃗⃗⃗
𝐵𝐶
1
−1
= −( 2 )+( 1 )
−4
−2
−2
= (−1)
2
⃗⃗⃗⃗⃗ = −2𝒊 − 𝒋 + 2𝒌
𝐵𝐶
ii.
Given 𝒓 = −16𝒋 − 8𝒌, if 𝒓 is perpendicular to the plane through A, B, and C
Then 𝒓 ∙ 𝑨𝑩 = 0 and 𝒓 ∙ 𝑩𝑪 = 0
0
−2
𝒓 ∙ 𝑨𝑩 = (−16) ∙ ( 3 )
−8
−6
= (0 × −2) + (−16 × 3) + (−8 × −6)
= 0 − 48 + 48 = 0
Therefore r is perpendicular to ⃗⃗⃗⃗⃗
𝐴𝐵
0
−2
𝒓 ∙ 𝑩𝑪 = (−16) ∙ (−1)
−8
2
= (0 × −2) + (−16 × −1) + (−8 × 2)
= 0 + 16 − 16 = 0
⃗⃗⃗⃗⃗
Therefore r is perpendicular to 𝐵𝐶
47
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2013
iii.
The vector equation of a plane is given by 𝒓 ∙ 𝒏 = 𝒂 ∙ 𝒏
where r is any vector (𝑥𝒊 + 𝑦𝒋 + 𝑧𝒌) on the plane, n is a vector normal to
the plane and a is the position vector for a point on the plane.
Using 𝒏 = −16𝒋 − 8𝒌 and 𝒂 = 3𝒊 − 𝒋 + 2𝒌 we have
𝒓∙𝒏=𝒂∙𝒏
(𝑥𝒊 + 𝑦𝒋 + 𝑧𝒌) ∙ (−16𝒋 − 8𝒌) = (3𝒊 − 𝒋 + 2𝒌) ∙ (−16𝒋 − 8𝒌)
−16𝑦 − 8𝑧 = −16 + 16
−16𝑦 − 8𝑧 = 0
2𝑦 + 𝑧 = 0
Question 5
a.
Given 𝑓(𝑥) = {
i.
𝑥 + 2,
𝑥2,
𝑥<2
.
𝑥>2
lim 𝑓(𝑥) = lim+ 𝑥 2
𝑥→2+
𝑥→2
= 22 = 4
lim 𝑓(𝑥) = lim− 𝑥 + 2
𝑥→2−
𝑥→2
=2+2=4
lim 𝑓(𝑥) = lim− 𝑓(𝑥)
𝑥→2+
𝑥→2
lim 𝑓(𝑥) = 4
𝑥→2
ii.
𝑓(𝑥) is not continuous at 𝑥 = 2 because 𝑓(2) is undefined.
48
b.
𝑥 2 + 2𝑥 + 3
Let 𝑦 =
(𝑥 2 + 2)3
Using the quotient rule 𝑦 =
Let 𝑢 = 𝑥 2 + 2𝑥 + 3
𝑣 = (𝑥 2 + 2)3
𝑢
𝑣
,
𝑑𝑦
𝑑𝑥
=
𝑣𝑢′ −𝑢𝑣′
𝑣2
𝑢′ = 2𝑥 + 2
𝑣 ′ = 3(𝑥 2 + 2)2 × 2𝑥
𝑣 ′ = 6𝑥(𝑥 2 + 2)2
𝑑𝑦 (𝑥 2 + 2)3 × (2𝑥 + 2) − (𝑥 2 + 2𝑥 + 3) × (6𝑥(𝑥 2 + 2)2 )
=
((𝑥 2 + 2)3 )2
𝑑𝑥
𝑑𝑦 (𝑥 2 + 2)2 [(𝑥 2 + 2)(2𝑥 + 2) − 6𝑥(𝑥 2 + 2𝑥 + 3)]
=
(𝑥 2 + 2)6
𝑑𝑥
𝑑𝑦 2𝑥 3 + 2𝑥 2 + 4𝑥 + 4 − 6𝑥 3 − 12𝑥 2 − 18𝑥
=
(𝑥 2 + 2)4
𝑑𝑥
𝑑𝑦 −4𝑥 3 − 10𝑥 2 − 14𝑥 + 4
=
(𝑥 2 + 2)4
𝑑𝑥
c.
Given 𝑥 = 1 − 3 cos 𝜃 ,
𝑦 = 2 sin 𝜃
𝑑𝑥
= 0 − 3(− sin 𝜃) = 3 sin 𝜃
𝑑𝜃
𝑑𝑦
= 2 cos 𝜃
𝑑𝜃
𝑑𝑦 𝑑𝑦/𝑑𝜃
=
𝑑𝑥 𝑑𝑥/𝑑𝜃
𝑑𝑦 2 cos 𝜃
=
𝑑𝑥 3 sin 𝜃
𝑑𝑦 2
= cot 𝜃
𝑑𝑥 3
d.
i.
𝑦 = 𝑥 2 + 3 ------ (1)
𝑦 = 4𝑥
------- (2)
4𝑥 = 𝑥 2 + 3
Substitute 𝑦 = 4𝑥 into (1)
𝑥 2 − 4𝑥 + 3 = 0
49
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2013
(𝑥 − 1)(𝑥 − 3) = 0
𝑥 = 1, or 3
When 𝑥 = 1, 𝑦 = 4(1) = 4
𝑥 = 3, 𝑦 = 4(3) = 12
𝑃(1, 4) and 𝑄(3, 12)
ii.
Area of the shaded region is given by A
3
3
𝐴 = ∫ 4𝑥 𝑑𝑥 − ∫ 𝑥 2 + 3 𝑑𝑥
1
1
𝑥3
3
3
𝐴 = [2𝑥 2 ] − [ + 3𝑥]
1
1
3
𝐴 = [2(32 ) − 2] − [(
33
1
+ 3(3)) − ( + 3)]
3
3
1
𝐴 = 16 − 18 + 3 3
4
= 3 sq. units
Question 6
a.
i.
∫ 𝑥(1 − 𝑥)2 𝑑𝑥
Let 𝑢 = 1 − 𝑥, 𝑥 = 1 − 𝑢,
𝑑𝑢
𝑑𝑥
= −1, 𝑑𝑥 = −𝑑𝑢
∫(1 − 𝑢)𝑢2 (−1)𝑑𝑢
∫ 𝑢3 − 𝑢2 𝑑𝑢 =
𝑢4 𝑢3
− +𝐶
4
3
∫ 𝑥(1 − 𝑥)2 𝑑𝑥 =
(1 − 𝑥)4 (1 − 𝑥)3
−
+𝐶
4
3
=−
1
[(1 − 𝑥)3 (1 + 3𝑥)] + 𝐶
12
50
ii.
Given 𝑓(𝑡) = 2 cos 𝑡 , 𝑔(𝑡) = 4 sin 5𝑡 + 3 cos 𝑡
∫[𝑓(𝑡) + 𝑔(𝑡)] 𝑑𝑡 = ∫ 2 cos 𝑡 + 4 sin 5𝑡 + 3 cos 𝑡 𝑑𝑡
= ∫ 5 cos 𝑡 + 4 sin 5𝑡 𝑑𝑡
4
∫[𝑓(𝑡) + 𝑔(𝑡)] 𝑑𝑡 = 5 sin 𝑡 − cos 5𝑡 + 𝐶
5
∫ 𝑓(𝑡) 𝑑𝑡 + ∫ 𝑔(𝑡) 𝑑𝑡 = ∫ 2 cos 𝑡 𝑑𝑡 + ∫ 4 sin 5𝑡 + 3 cos 𝑡 𝑑𝑡
4
= 2 sin 𝑡 + 𝐴 + (− cos 5𝑡) + 3 sin 𝑡 + 𝐵
5
4
∫ 𝑓(𝑡) 𝑑𝑡 + ∫ 𝑔(𝑡) 𝑑𝑡 = 5 sin 𝑡 − cos 5𝑡 + 𝐶
5
A, B, and C are merely constants of the integrals therefore 𝐶 = 𝐴 + 𝐵
b.
i.
Length of rectangle is x, width of rectangle is 2r and
length of semi-circle is
2𝜋𝑟
2
= 𝜋𝑟
Perimeter of track is given by 2𝑥 + 2𝑟 + 𝜋𝑟 = 600
𝑟(2 + 𝜋) = 600 − 2𝑥
𝑟=
ii.
600 − 2𝑥
2+𝜋
Area of track is given by
1
𝐴 = 𝑥(2𝑟) + 2 (𝜋𝑟 2 )
𝜋
𝐴 = 2𝑥𝑟 + 2 𝑟 2
600 − 2𝑥
𝜋 600 − 2𝑥 2
𝐴 = 2𝑥 (
)+ (
)
2+𝜋
2 2+𝜋
1200𝑥 − 4𝑥 2
𝜋 600 − 2𝑥 2
𝐴=(
)
)+ (
2+𝜋
2 2+𝜋
𝑑𝐴
1
𝜋
=
(2)(600 − 2𝑥)(−2)]
[1200 − 8𝑥 +
𝑑𝑥 2 + 𝜋
2(2 + 𝜋)
𝑑𝐴
1
2𝜋
(600 − 2𝑥)]
=
[1200 − 8𝑥 −
(2 + 𝜋)
𝑑𝑥 2 + 𝜋
51
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2013
When
𝑑𝐴
1
2𝜋
(600 − 2𝑥)] = 0
= 0,
[1200 − 8𝑥 −
(2
𝑑𝑥
2+𝜋
+ 𝜋)
1200 − 8𝑥 −
2𝜋
(600 − 2𝑥) = 0
(2 + 𝜋)
(2 + 𝜋)(1200 − 8𝑥) − 2𝜋(600 − 2𝑥) = 0
2400 − 16𝑥 + 1200𝜋 − 8𝜋𝑥 − 1200𝜋 + 4𝜋𝑥 = 0
2400 − 16𝑥 − 4𝜋𝑥 = 0
16𝑥 + 4𝜋𝑥 = 2400
4𝑥(4 + 𝜋) = 2400
4𝑥 =
2400
4+𝜋
600
𝑥 = 4+𝜋 ≈ 84 metres
𝑑2 𝐴
1
2𝜋
1
4𝜋
=
) (−2)] =
− 8]
[−8 − (
[
2
(2 + 𝜋)
𝑑𝑥
2+𝜋
2+𝜋 2+𝜋
8>
4𝜋
2+𝜋
Therefore
𝑑2 𝐴
<0
𝑑𝑥 2
600
Therefore 𝑥 = 4+𝜋 give the maximum area.
c.
i.
Let 𝑦 = −𝑥 sin 𝑥 − 2 cos 𝑥 + 𝐴𝑥 + 𝐵
𝑦 ′ = −[𝑥 cos 𝑥 + sin 𝑥] − 2(− sin 𝑥) + 𝐴
= −𝑥 cos 𝑥 − sin 𝑥 + 2 sin 𝑥 + 𝐴
= sin 𝑥 − 𝑥 cos 𝑥 + 𝐴
𝑦 ′′ = cos 𝑥 − [cos 𝑥 + 𝑥(− sin 𝑥)]
= cos 𝑥 − cos 𝑥 + 𝑥 sin 𝑥
𝑦′′ = 𝑥 sin 𝑥
52
ii.
𝑦 = −𝑥 sin 𝑥 − 2 cos 𝑥 + 𝐴𝑥 + 𝐵
When 𝑥 = 0, 𝑦 = 1, We have
1 = −2 + 𝐵
𝐵=3
When 𝑥 = 𝜋, 𝑦 = 6, 𝐵 = 3 we have
6 = −𝜋 sin 𝜋 − 2 cos 𝜋 + 𝜋𝐴 + 3
6 = 2 + 𝜋𝐴 + 3
𝜋𝐴 = 1
1
𝐴=𝜋
1
The specific solution is 𝑦 = −𝑥 sin 𝑥 − 2 cos 𝑥 + 𝜋 𝑥 + 3
53
CAPE PURE MATHEMATICS UNIT 1
SOLUTIONS TO 2012 EXAM
Question 1
a.
Given 𝑓(𝑥) = 2𝑥 3 − 𝑝𝑥 2 + 𝑞𝑥 − 10
i.
𝑥 − 1 is a factor of 𝑓(𝑥) therefore 𝑓(1) = 0
𝑓(1) = 2(1)3 − 𝑝(1)2 + 𝑞(1) − 10 = 0
2 − 𝑝 + 𝑞 − 10 = 0
𝑝 − 𝑞 = −8 ------- (1)
When 𝑓(𝑥) is divided by 𝑥 + 1 it gives a remainder of −6 therefore
𝑓(−1) = 2(−1)3 − 𝑝(−1)2 + 𝑞(−1) − 10 = −6
−2 − 𝑝 − 𝑞 − 10 = −6
𝑝 + 𝑞 = −6 -------- (2)
Adding equations (1) and (2) we have
2𝑝 = −14,
therefore 𝑝 = −7
Substituting 𝑝 = −7 into (2) give
−7 + 𝑞 = −6, therefore 𝑞 = 1
𝑓(𝑥) = 2𝑥 3 + 7𝑥 2 + 𝑥 − 10
ii.
2𝑥 2 + 9𝑥 + 10
𝑥 − 1 2𝑥 3 + 7𝑥 2 + 𝑥 − 10
2𝑥 3 − 2𝑥 2
9𝑥 2 + 𝑥
9𝑥 2 − 9𝑥
10𝑥 − 10
10𝑥 − 10
0
54
2𝑥 3 + 7𝑥 2 + 𝑥 − 10 = (𝑥 − 1)(2𝑥2 + 9𝑥 + 10)
= (𝑥 − 1)(𝑥 + 2)(2𝑥 + 5)
Therefore the factors of 𝑓(𝑥) are (𝑥 − 1), (𝑥 + 2), and (2𝑥 + 5)
b.
2
Given (√𝑥 + √𝑦) = 16 + √240
2
(√𝑥 + √𝑦) = 𝑥 + 2√𝑥𝑦 + 𝑦
𝑥 + 𝑦 + 2√𝑥𝑦 = 16 + √240
Therefore 𝑥 + 𝑦 = 16 --------
(1)
𝑦 = 16 − 𝑥 ------- (2)
2√𝑥𝑦 = √240 --------
(3)
√240 = √4 × 60 = 2√60
2√𝑥𝑦 = 2√60 -----Therefore
𝑥𝑦 = 60
(4)
--------
(5)
Substituting (2) into (5) we have
𝑥(16 − 𝑥) = 60
16𝑥 − 𝑥 2 = 60
𝑥 2 − 16𝑥 + 60 = 0
(𝑥 − 6)(𝑥 − 10) = 0
𝑥 = 6, 10
When 𝑥 = 6, 𝑦 = 16 − 6 = 10
When 𝑥 = 10, 𝑦 = 16 − 10 = 6
𝑥 = 6, 𝑦 = 10 and 𝑥 = 10, 𝑦 = 6
55
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2012
c.
Given |3𝑥 − 7| ≤ 5
Squaring both sides we have
(3𝑥 − 7)2 ≤ 25
|3𝑥 − 7| ≤ 5
9𝑥 2 − 42𝑥 + 49 ≤ 25
9𝑥 2 − 42𝑥 + 24 ≤ 0
2
3
3(3𝑥 2 − 14𝑥 + 8) ≤ 0
4
3(3𝑥 − 2)(𝑥 − 4) ≤ 0
2
Critical points 𝑥 = 3 , 4
𝑥≤
3𝑥 − 2
𝑥−4
(3𝑥 − 2)(𝑥 − 4)
2
3
−
−
+
2
≤𝑥≤4
3
−
+
−
𝑥≥4
+
+
+
2
Therefore 3 ≤ 𝑥 ≤ 4
Alternatively, when (3𝑥 − 7) ≥ 0 we have 3𝑥 − 7 ≤ 5
3𝑥 ≤ 12, 𝑥 ≤ 4
When (3𝑥 − 7) ≤ 0 we have −(3𝑥 − 7) ≤ 5
−3𝑥 + 7 ≤ 5
−3𝑥 ≤ −2
2
𝑥≥3
2
Therefore 3 ≤ 𝑥 ≤ 4
ii.
|3𝑥 − 7| + 5 ≤ 0
|3𝑥 − 7| ≥ 0 Modulus always give the value of the function as positive and
5 is also greater than zero, therefore if we add the two together a number greater
than zero will be the result.
Therefore |3𝑥 − 7| + 5 cannot be less than zero for any real value of x and this
function will not intersect the x-axis resulting in us having no real solution.
56
Question 2
a.
Given 𝑓(𝑥) → 𝑥 2 − 3
i.
𝑓(𝑓(𝑥)) = [𝑓(𝑥)]2 − 3
= (𝑥 2 − 3)2 − 3
= 𝑥 4 − 6𝑥 2 + 9 − 3
= 𝑥 4 − 6𝑥 2 + 6
ii.
𝑓(𝑓(𝑥)) = 𝑓(𝑥 + 3)
𝑥 4 − 6𝑥 2 + 6 = (𝑥 + 3)2 − 3
𝑥 4 − 6𝑥 2 + 6 = 𝑥 2 + 6𝑥 + 9 − 3
𝑥 4 − 6𝑥 2 + 6 = 𝑥 2 + 6𝑥 + 6
𝑥 4 − 7𝑥 2 − 6𝑥 = 𝑥(𝑥 3 − 7𝑥 − 6)
𝑥 + 1 is a factor of 𝑥 3 − 7𝑥 − 6
(−1)3 − 7(−1) − 6 = −1 + 7 − 6 = 0
𝑥 + 2 is a factor of 𝑥 3 − 7𝑥 − 6
(−2)3 − 7(−2) − 6 = −8 + 14 − 6 = 0
𝑥 − 3 is a factor of 𝑥 3 − 7𝑥 − 6
(3)3 − 7(3) − 6 = 27 − 21 − 6 = 0
Therefore 𝑥 4 − 7𝑥 2 − 6𝑥 = 𝑥(𝑥 + 1)(𝑥 + 2)(𝑥 − 3)
𝑥 = 0, −1, −2, 3
b.
i.
Given 𝛼 𝑎𝑛𝑑 𝛽 are the roots the equation 4𝑥 2 − 3𝑥 + 1 = 0
(𝑥 − 𝛼)(𝑥 − 𝛽) = 𝑥 2 − (𝛼 + 𝛽)𝑥 + 𝛼𝛽
4𝑥 2 − 3𝑥 + 1 = 0 Dividing both sides by 4 we have
3
1
3
1
𝑥 2 − 4 𝑥 + 4 = 0 Therefore 𝛼 + 𝛽 = 4 𝑎𝑛𝑑 𝛼𝛽 = 4
ii.
(𝛼 + 𝛽)2 = 𝛼 2 + 2𝛼𝛽 + 𝛽 2
𝛼 2 + 𝛽 2 = (𝛼 + 𝛽)2 − 2𝛼𝛽
3 2
1
= (4) − 2 (4)
1
= 16
57
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2012
iii.
Given
2
𝛼2
and
2
𝛽2
are the roots of a quadratic equation we have
2
2
2𝛽 2 + 2𝛼 2
+
=
𝛼 2 𝛽2
𝛼 2 𝛽2
Sum of roots
=
2(𝛽 2 + 𝛼 2 )
(𝛼𝛽)2
1
1
)
16
=
= 8
1
1 2
( )
( )
16
4
2(
=2
Product of roots
2
2
4
× 2=
2
(𝛼𝛽)2
𝛼
𝛽
=
4
1 2
( )
4
= 4 × 16 = 64
Therefore the quadratic equation is 𝑥 2 − 2𝑥 + 64 = 0
c.
i.
1
3
5
7
9
log10 (3) + log10 (5) + log10 (7) + log10 (9) + log10 (10)
1
3
5
7
9
log10 [(3) × (5) × (7) × (9) × (10)]
1
log10 (10) = log10 10−1
= −1
99
ii.
𝑟
1
2
98
99
∑ log10 (
) = log10 ( ) + log10 ( ) + ⋯ + log10 ( ) + log10 (
)
𝑟+1
2
3
99
100
𝑟=1
1
2
98 99
= log10 [( ) × ( ) × … × ( ) (
)]
2
3
99 100
= log10 (
1
)
100
= log10 10−2
= −2
58
Question 3
a.
i.
Given cos(𝐴 + 𝐵) = cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵 and cos 2𝜃 = 2 cos2 𝜃 − 1
cos 3𝜃 = cos(2𝜃 + 𝜃)
= cos 2𝜃 cos 𝜃 − sin 2𝜃 sin 𝜃
= (2 cos2 𝜃 − 1) cos 𝜃 − (2 sin 𝜃 cos 𝜃) sin 𝜃
[sin 2𝜃 = 2 sin 𝜃 cos 𝜃]
= (2 cos2 𝜃 − 1) cos 𝜃 − 2 sin2 𝜃 cos 𝜃
= cos 𝜃 [(2 cos2 𝜃 − 1) − 2 sin2 𝜃]
1
= 2 cos 𝜃 [cos2 𝜃 − sin2 𝜃 − 2]
ii.
1
L.H.S = 2 [sin 6𝜃 − sin 2𝜃]
Using the factor formulae sin 𝐴 − sin 𝐵 = 2 cos (
1
2
1
[sin 6𝜃 − sin 2𝜃] = [2 cos (
2
6𝜃+2𝜃
2
6𝜃−2𝜃
) sin (
2
= (2 cos 2 2𝜃 − 1) sin 2𝜃
sin 6𝜃 − sin 2𝜃 = 0
(2 cos 2 2𝜃 − 1) sin 2𝜃 = 0
2 cos 2 2𝜃 − 1 = 0
cos2 2𝜃 =
1
2
cos 2𝜃 = ±
2𝜃 =
𝜃=
1
√2
𝜋
,
4
0≤𝜃≤
𝜋
2
3𝜋
4
𝜋 3𝜋
,
8 8
sin 2𝜃 = 0
0≤𝜃≤
2𝜃 = 0, 𝜋
𝜃 = 0,
𝜋
2
𝜃 = 0,
𝜋
,
2
𝜋
,
8
3𝜋
8
59
2
) sin (
𝐴−𝐵
2
)
)]
[cos 4𝜃 = 2 cos2 2𝜃 − 1]
= cos 4𝜃 sin 2𝜃
iii.
𝐴+𝐵
𝜋
2
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2012
b.
Given cot 2 𝜃 + cos 𝜃 = 0
cos 2 𝜃
2
+ cos 𝜃 = 0
sin2 𝜃
2 cos 2 𝜃 + sin2 𝜃 cos 𝜃 = 0
2 cos 2 𝜃 + (1 − cos 2 𝜃) cos 𝜃 = 0
2 cos 2 𝜃 + cos 𝜃 − cos3 𝜃 = 0
cos3 𝜃 − 2 cos2 𝜃 − cos 𝜃 = 0
cos 𝜃 (cos 2 𝜃 − 2 cos 𝜃 − 1) = 0
cos 𝜃 = 0
cos2 𝜃 − 2 cos 𝜃 − 1 = 0
Using the quadratic formula 𝑥 =
−𝑏±√𝑏2 −4𝑎𝑐
cos 𝜃 =
−(−2) ± √(−2)2 − 4(1)(−1)
2
cos 𝜃 =
2 ± √8
,
2
cos 𝜃 =
2 ± 2√2
= 1 ± √2
2
cos 𝜃 = 1 − √2,
√8 = √2 × 4 = 2√2
cos 𝜃 ≠ 1 + √2
cos 𝜃 = 0, 1 − √2
Question 4
a.
i.
2𝑎
Given 𝑦 = 3 sec 𝜃 , and 𝑥 = 3 tan 𝜃
sec 2 𝜃 = 1 + tan2 𝜃
𝑦
𝑥
sec 𝜃 = , tan 𝜃 =
3
3
𝑦 2
𝑥 2
( ) =1+( )
3
3
𝑦2
𝑥2
=1+
9
9
𝑦2 = 9 + 𝑥2
60
ii.
𝑦 2 = 9 + 𝑥 2 ------ (1)
𝑦 = √10𝑥
------ (2)
𝑦 2 = 10𝑥
--------- (3)
10𝑥 = 9 + 𝑥 2
squaring (2)
Substitute (3) into (1)
𝑥 2 − 10𝑥 + 9 = 0
(𝑥 − 1)(𝑥 − 9) = 0
𝑥 = 1, 9
When 𝑥 = 1, 𝑦 = √10 point (1, √10)
When 𝑥 = 9, 𝑦 = √90 = 3√10 point (9, 3√10)
b.
i.
𝒑 = −3𝒊 + 4𝒋 and 𝒒 = −𝒊 + 6𝒋
ii.
𝒑 − 𝒒 = (−3𝒊 + 4𝒋) − (−𝒊 + 6𝒋)
= −2𝒊 − 2𝒋
iii.
𝒑 ∙ 𝒒 = (−3𝒊 + 4𝒋) ∙ (−𝒊 + 6𝒋)
= 3 + 24 = 27
iv.
𝒑 ∙ 𝒒 = |𝒑| × |𝒒| cos 𝜃
cos 𝜃 =
𝒑∙𝒒
|𝒑| × |𝒒|
|𝒑| = √(−3)2 + (4)2
= √25 = 5
|𝒒| = √(−1)2 + (6)2
= √37
cos 𝜃 =
27
5 × √37
cos 𝜃 = 0.888
𝜃 = cos −1 0.888 = 27.4𝑜
61
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2012
Question 5
a.
i.
Let 𝑓(𝑥) =
𝑥 3 +8
𝑥 2 −4
𝑓(𝑥) is discontinuous when 𝑥 2 − 4 = 0
Therefore 𝑥 = 2, −2
ii.
lim
𝑥 3 +8
𝑥→−2 𝑥 2 −4
𝑥 3 + 8 = (𝑥 + 2)(𝑥 2 − 2𝑥 + 4)
(𝑥 + 2)(𝑥 2 − 2𝑥 + 4)
lim
𝑥→−2
(𝑥 + 2)(𝑥 − 2)
(𝑥 2 − 2𝑥 + 4) (−2)2 − 2(−2) + 4
=
𝑥→−2
(𝑥 − 2)
−2 − 2
lim
=
ii.
lim
12
= −3
−4
2𝑥 3 +4𝑥
𝑥→0 sin 2𝑥
2𝑥 3 + 4𝑥
2𝑥
lim
𝑥→0 sin 2𝑥
2𝑥
𝑥+2
𝑥→0 sin 2𝑥
2𝑥
lim
lim 𝑥 + 2
𝑥→0
sin 2𝑥
lim 2𝑥
=
0+2
1
𝑥→0
2𝑥 3 + 4𝑥
=2
𝑥→0 sin 2𝑥
lim
62
b.
Given 𝑓(𝑥) = {
i.
a.
b.
𝑥 2 + 1, 𝑥 > 1,
4 + 𝑝𝑥, 𝑥 < 1.
lim 𝑓(𝑥) = 12 + 1 = 2
𝑥→1+
lim 𝑓(𝑥) = 4 + 𝑝(1)
𝑥→1−
When lim 𝑓(𝑥) exist lim+ 𝑓(𝑥) = lim− 𝑓(𝑥)
𝑥→1
𝑥→1
𝑥→1
Therefore 4 + 𝑝 = 2
𝑝 = −2
ii.
c.
𝑓(1) = 2 For 𝑓 to be continuous at 𝑥 = 1.
𝑣
Given 𝑀 = 𝑢𝑡 2 + 𝑡 2
𝑑𝑀
= 2𝑢𝑡 + (−2𝑣𝑡 −3 )
𝑑𝑡
𝑑𝑀
2𝑣
= 2𝑢𝑡 − 3
𝑑𝑡
𝑡
When 𝑀 = −1, 𝑡 = 1 therefore
−1 = 𝑢 + 𝑣
When
𝑑𝑀
𝑑𝑡
=
------ (1)
35
4
, 𝑡 = 2 therefore
35
2𝑣
= 2𝑢(2) − 3
4
2
35
𝑣
= 4𝑢 −
4
4
35 = 16𝑢 − 𝑣 ----- (2)
Add eq. (1) and (2) we have
34 = 17𝑢, 𝑢 = 2
From (1) when 𝑢 = 2 we have
−1 = 2 + 𝑣, 𝑣 = −3
Therefore 𝑢 = 2, 𝑣 = −3
63
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2013
Question 6
a.
i.
𝑦 = √4𝑥 2 − 7
Given
𝑑𝑦 1
= (4𝑥 2 − 7)−1/2 × 8𝑥
𝑑𝑥 2
𝑑𝑦
4𝑥
=
𝑑𝑥 √4𝑥 2 − 7
√4𝑥 2 − 7
𝑑𝑦
= 4𝑥
𝑑𝑥
𝑦 = √4𝑥 2 − 7
𝑦
ii.
𝑑𝑦
𝑑𝑥
=
𝑑𝑦
= 4𝑥
𝑑𝑥
4𝑥
√4𝑥 2 −7
𝑢
𝑑𝑦
𝑣
𝑑𝑥
using the quotient rule 𝑦 = ,
𝑑𝑢
𝑑𝑣
−𝑢
𝑑𝑥
𝑑𝑥
𝑣2
𝑑𝑢
=4
𝑑𝑥
𝑢 = 4𝑥,
𝑣 = √4𝑥 2 − 7
𝑑2 𝑦
=
𝑑𝑥 2
=
𝑣
𝑑𝑣
4𝑥
=
𝑑𝑥 √4𝑥 2 − 7
√4𝑥 2 − 7(4) − 4𝑥 (
4𝑥
)
√4𝑥 2 − 7
2
(√4𝑥 2 − 7)
4(√4𝑥 2 − 7)(√4𝑥 2 − 7) − (4𝑥)(4𝑥)
√4𝑥 2 − 7
=
4𝑥 2 − 7
=
=
=
4(4𝑥 2 − 7) − 16𝑥 2
(4𝑥 2 − 7)√4𝑥 2 − 7
16𝑥 2 − 28 − 16𝑥 2
(4𝑥 2 − 7)√4𝑥 2 − 7
−28
(4𝑥 2 − 7)√4𝑥 2 − 7
64
(√4𝑥 2 − 7)
𝑦
𝑑2 𝑦
28
=− 2
2
𝑑𝑥
4𝑥 − 7
𝑑2 𝑦
28
=− 2
2
𝑑𝑥
4𝑥 − 7
2
𝑑𝑦 2
4𝑥
( ) =(
)
𝑑𝑥
√4𝑥 2 − 7
=
16𝑥 2
4𝑥 2 − 7
𝑑2 𝑦
𝑑𝑦 2
28
16𝑥 2
𝑦 2+( ) =− 2
+ 2
𝑑𝑥
𝑑𝑥
4𝑥 − 7 4𝑥 − 7
Therefore 𝑦
b.
i.
𝑑𝑦
𝑑𝑥
=
16𝑥 2 − 28
4𝑥 2 − 7
=
4(4𝑥 2 − 7)
=4
4𝑥 2 − 7
𝑑2𝑦
𝑑𝑦 2
+( ) =4
𝑑𝑥 2
𝑑𝑥
= 3𝑥 2 − 6𝑥
Integrating both sides we have
𝑦 = ∫ 3𝑥 2 − 6𝑥 𝑑𝑥
𝑦 = 𝑥 3 − 3𝑥 2 + 𝐶
When 𝑥 = −1, 𝑦 = 0
0 = (−1)3 − 3(−1)2 + 𝐶
0 = −1 − 3 + 𝐶
𝐶=4
𝑦 = 𝑥 3 − 3𝑥 2 + 4
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SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2012
ii.
𝑑𝑦
𝑑𝑥
= 3𝑥 2 − 6𝑥
At the stationary points
𝑑𝑦
𝑑𝑥
= 0 therefore
3𝑥 2 − 6𝑥 = 0
3𝑥(𝑥 − 2) = 0
𝑥 = 0, 2
When 𝑥 = 0, 𝑦 = 4 (0, 4)
When 𝑥 = 2, 𝑦 = 23 − 3(22 ) + 4
𝑦 = 8 − 12 + 4 = 0
(2, 0)
Therefore the stationary points are (0, 4) and (2, 0)
iii.
𝑑2𝑦
𝑑𝑥 2
= 6𝑥 − 6
When 𝑥 = 0,
𝑑2 𝑦
𝑑𝑥 2
= −6
Therefore (0, 4) is a maximum
When 𝑥 = 2,
𝑑2𝑦
𝑑𝑥 2
= 6(2) − 6 = 6
Therefore (2, 0) is a minimum
66
iv.
The curve meets the x-axis when 𝑦 = 0 therefore
𝑥 3 − 3𝑥 2 + 4 = 0
The minimum point has 𝑦 = 0, 𝑥 = 2
Therefore (𝑥 − 2)2 is a factor
(𝑥 − 2)2 (𝑥 − 𝑎) = 𝑥 3 − 3𝑥 2 + 4
Equating the constants we have
−4𝑎 = 4, 𝑎 = −1
𝑥 3 − 3𝑥 2 + 4 = (𝑥 − 2)2 (𝑥 + 1)
Therefore the curve meets the x-axis at 𝑥 = 2, and − 1
𝑃(−1, 0) and 𝑄(2, 0)
v.
𝑦
4
𝑚𝑎𝑥 (0,4)
2
𝑃(−1, 0)
𝑚𝑖𝑛 (2,0)
−1
−2
0
2
𝑄(2, 0)
67
𝑥
CAPE PURE MATHEMATICS UNIT 1
SOLUTIONS TO 2011 EXAM
Question 1
a.
i.
2
(√75 + √12) − (√75 − √12)
2
[(√75 + √12) + (√75 − √12)][(√75 + √12) − (√75 − √12)]
(2√75)(2√12)
(2√25 × 3)(2√4 × 3)
2 × 5√3 × 2 × 2√3
40 × 3 = 120
ii.
271/4 × 93/8 × 811/8
1
3
1
(33 )4 × (32 )8 × (34 )8
3
3
1
34 × 34 × 32
3 3 1
3(4+4+2)
32 = 9
b.
𝑓(𝑥) = 𝑥 3 + 𝑚𝑥 2 + 𝑛𝑥 + 𝑝
f(x)
(0, 4)
Q
0
1
68
2
x
i.
When 𝑥 = 0, 𝑓(0) = 𝑝, therefore 𝑝 is the y-intercept
From the graph the y-intercept is where the curve cuts the y-axis.
Therefore 𝑝 = 4
ii.
From the graph when 𝑦 = 0, 𝑥 = 1, 2 therefore
𝑓(1) = 1 + 𝑚 + 𝑛 + 4 = 0
𝑚 + 𝑛 = −5 ------ (1)
𝑓(2) = 23 + 𝑚(22 ) + 𝑛(2) + 4 = 0
4𝑚 + 2𝑛 = −12 ----- (2)
2𝑚 + 𝑛 = −6 ---- (3)
dividing (2) by 2
−𝑚 = 1 (1) Subtract (2)
𝑚 = −1
From (1) when 𝑚 = −1 we have
−1 + 𝑛 = −5
𝑛 = −4
So 𝑓(𝑥) = 𝑥 3 − 𝑥 2 − 4𝑥 + 4
iii.
𝑥 3 − 𝑥 2 − 4𝑥 + 4 = (𝑥 − 1)(𝑥 − 2)(𝑥 + 𝑎)
Equating the constants we have
4 = (−1)(−2) × 𝑎
4 = 2𝑎
𝑎=2
Therefore the third factor is 𝑥 + 2
𝑥+2=0
𝑥 = −2
The x coordinate of the point Q is −2.
69
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2011
c.
i.
Given √log 2 𝑥 = log 2 √𝑥
1
√log 2 𝑥 = log 2 (𝑥)2
1
√log 2 𝑥 = log 2 𝑥
2
Let 𝑦 = log 2 𝑥 therefore
1
√𝑦 = 2 𝑦 Squaring both sides we have
𝑦=
1 2
𝑦
4
4𝑦 = 𝑦 2
𝑦 2 − 4𝑦 = 0
𝑦(𝑦 − 4) = 0
𝑦 = 0, 4
When 𝑦 = 0, log 2 𝑥 = 0
𝑥 = 20 = 1
When 𝑦 = 4, log 2 𝑥 = 4
𝑥 = 24 = 16
Therefore 𝑥 = 1, 16
ii.
Given 𝑥 2 − |𝑥| − 12 < 0
Because of |𝑥| = {
𝑥, 𝑥 ≥ 0
−𝑥, 𝑥 < 0
We have 𝑥 2 − 𝑥 − 12 < 0, for 𝑥 ≥ 0
(𝑥 + 3)(𝑥 − 4) < 0
Critical points 𝑥 = −3, 4
70
𝑥 2 − 𝑥 − 12
−3
𝑥 2 + 𝑥 − 12
4
−4
3
𝑥 2 − 𝑥 − 12 < 0, has inequality for values of 𝑥 between −3 < 𝑥 < 4
We have 𝑥 2 + 𝑥 − 12 < 0, for 𝑥 < 0
(𝑥 − 3)(𝑥 + 4) < 0
Critical points 𝑥 = 3, −4, therefore for 𝑥 2 + 𝑥 − 12 < 0, has inequality
for values of 𝑥, −4 < 𝑥 < 3
Taking the union of both sets we have inequality for values of 𝑥, −4 < 𝑥 < 4
Question 2
a.
i.
Given 𝛼 and 𝛽 are the roots of 𝑥 2 − 𝑝𝑥 + 24 = 0
(𝑥 − 𝛼)(𝑥 − 𝛽) = 0, 𝑥 2 − (𝛼 + 𝛽)𝑥 + 𝛼𝛽
a.
𝛼+𝛽 =𝑝
b.
𝛼𝛽 = 24,
𝛼 2 + 𝛽 2 = (𝛼 + 𝛽)2 − 2𝛼𝛽
= 𝑝2 − 2(24)
= 𝑝2 − 48
ii.
Given 𝛼 2 + 𝛽 2 = 33,
𝑝2 − 48 = 33
𝑝2 − 81 = 0
(𝑝 − 9)(𝑝 + 9) = 0
𝑝 = 9, or − 9
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SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2011
b.
Given 𝑓(2𝑥 + 3) = 2𝑓(𝑥) + 3 and 𝑓(0) = 6
i.
When 𝑥 = 0, we have
𝑓(3) = 2𝑓(0) + 3
= 2(6) + 3
= 12 + 3 = 15
ii.
When 𝑥 = 2, we have
𝑓(2(3) + 3) = 2𝑓(3) + 3
𝑓(6 + 3) = 2(15) + 3
𝑓(9) = 30 + 3 = 33
iii.
When 𝑥 = −3 we have
𝑓(2(−3) + 3) = 2𝑓(−3) + 3
𝑓(−6 + 3) = 2𝑓(−3) + 3
𝑓(−3) = 2𝑓(−3) + 3
−3 = 2𝑓(−3) − 𝑓(−3)
𝑓(−3) = −3
c.
An even number can be express as 2𝑛 where 𝑛 is an integer.
A odd number can be express as 2𝑛 − 1 where 𝑛 is an integer.
For two consecutive numbers one must be even and the other odd, therefore
The product of two consecutive integers can be 𝑘(𝑘 + 1) = 2𝑛(2𝑛 − 1).
Where 2𝑛(2𝑛 − 1) = 2[𝑛(𝑛 − 1)].
Two times any number makes it even. Therefore the product of two consecutive integers
is an even integer.
72
d.
Given to prove that 𝑛(𝑛2 + 5) is divisible by 6
When 𝑛 = 1 we have
1(12 + 5) = 6 which is divisible by 6
Therefore the statement is true for 𝑛 = 1
Assume statement is true when 𝑛 = 𝑘, therefore
𝑘(𝑘 2 + 5) is divisible by 6
When 𝑛 = 𝑘 + 1 we have
(𝑘 + 1)[(𝑘 + 1)2 + 5]
(𝑘 + 1)(𝑘 2 + 2𝑘 + 1 + 5)
(𝑘 + 1)(𝑘 2 + 2𝑘 + 6)
𝑘 3 + 2𝑘 2 + 6𝑘 + 𝑘 2 + 2𝑘 + 6
𝑘 3 + 3𝑘 2 + 8𝑘 + 6
𝑘 3 + 5𝑘 + 3𝑘 2 + 3𝑘 + 6
𝑘(𝑘 2 + 5) + 3𝑘(𝑘 + 1) + 6
We assumed 𝑘(𝑘 2 + 5) is divisible by 6, 𝑘(𝑘 + 1) is an even integer which when
multiplied by 3 is divisible by 6 and 6 is divisible by 6.
Therefore when 𝑛 = 𝑘 + 1 the statement is true.
Since the statement is true for 𝑛 = 1, 𝑘 and 𝑘 + 1, it is true for all positive integer n.
Question 3
a.
𝒂 = 𝑎1 𝒊 + 𝑎2 𝒋 and 𝒃 = 𝑏1 𝒊 + 𝑏2 𝒋 with |𝒂| = 13 and |𝒃| = 10
i.
|𝒂| = √(𝑎1 )𝟐 + (𝑎2 )𝟐 = 13 and |𝒃| = √(𝑏1 )𝟐 + (𝑏2 )𝟐 = 10
(𝒂 + 𝒃) ∙ (𝒂 − 𝒃) = [(𝑎1 + 𝑏1 )𝒊 + (𝑎2 + 𝑏2 )𝒋] ∙ [(𝑎1 − 𝑏1 )𝒊 + (𝑎2 − 𝑏2 )𝒋]
= ((𝑎1 )2 − (𝑏1 )2) + ((𝑎2 )2 − (𝑏2 )2 )
= ((𝑎1 )2 + (𝑎2 )2 ) − ((𝑏1 )2 + (𝑏2 )2 )
= 132 − 102 = 169 − 100
= 69
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SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2011
ii.
2𝒃 − 𝒂 = 11𝒊 and so 𝒂 = 2𝒃 − 11𝒊
(𝒂 + 𝒃) = 2𝒃 − 11𝒊 + 𝒃 and (𝒂 − 𝒃) = 2𝒃 − 11𝒊 − 𝒃
(𝒂 + 𝒃) = 3𝒃 − 11𝒊 𝑎𝑛𝑑 (𝒂 − 𝒃) = 𝒃 − 11𝒊
(𝒂 + 𝒃) = 3( 𝑏1 𝒊 + 𝑏2 𝒋) − 11𝒊 and (𝒂 − 𝒃) = ( 𝑏1 𝒊 + 𝑏2 𝒋) − 11𝒊
(𝒂 + 𝒃) = (3𝑏1 − 11)𝒊 + 3𝑏2 𝒋 and (𝒂 − 𝒃) = (𝑏1 − 11)𝒊 + 𝑏2 𝒋
(𝒂 + 𝒃) ∙ (𝒂 − 𝒃) = [(3𝑏1 − 11)(𝑏1 − 11)] + [3𝑏2 × 𝑏2 ] = 69
= 3𝑏1 2 − 44𝑏1 + 121 + 3𝑏2 2 = 69
= 3(𝑏1 2 + 𝑏2 2 ) − 44𝑏1 + 52 = 0
|𝑏| = √(𝑏1 2 + 𝑏2 2 ) = 10 𝑠𝑜 𝑏1 2 + 𝑏2 2 = 100
= 3(100) − 44𝑏1 + 52 = 0
44𝑏1 = 352
𝑏1 = 8
𝑏1 2 + 𝑏2 2 = 100
𝑏2 = √100 − 82 = ±6
Therefore 𝑏 = 8𝒊 + 6𝒋 or 8𝒊 − 6𝒋
𝒂 = 2𝒃 − 11𝒊
𝒂 = 2(8𝒊 + 6𝒋) − 11𝒊 = 5𝒊 + 12𝒋
𝒂 = 2(8𝒊 − 6𝒋) − 11𝒊 = 5𝒊 − 12𝒋
b.
i.
Given the line L has equation 𝑥 − 𝑦 + 1 = 0 and the circle C has equation
𝑥 2 + 𝑦 2 − 2𝑦 − 15 = 0
The general equation of a circle is given by 𝑥 2 + 𝑦 2 − 2𝑓𝑥 − 2𝑔𝑦 + 𝑐 = 0
where f and g are the coordinates of the centre of the circle.
From the equation of the circle the coordinate of the centre is (0, 1)
Therefore from the line equation 𝑥 − 𝑦 + 1 = 0 when 𝑥 = 0 and 𝑦 = 1
74
We have 0 − 1 + 1 = 0 therefore it is shown that the line L passes through the
centre of the circle.
ii.
L intersects C at P and Q therefore we solve simultaneously the equations of
L and C.
𝑥 2 + 𝑦 2 − 2𝑦 − 15 = 0 -------- (1)
𝑥 − 𝑦 + 1 = 0 -------- (2)
From (2) 𝑦 = 𝑥 + 1 ------- (3)
Substitute (3) into (1) we have
𝑥 2 + (𝑥 + 1)2 − 2(𝑥 + 1) − 15 = 0
𝑥 2 + 𝑥 2 + 2𝑥 + 1 − 2𝑥 − 2 − 15 = 0
2𝑥 2 − 16 = 0
𝑥2 = 8
𝑥 = ±√8 = ±2√2
𝑦 = 1 ± 2√2
The coordinates of P and Q are (2√2, 1 + 2√2) and (−2√2, 1 − 2√2)
iii.
Given the parametric equations 𝑥 = 𝑏 + 𝑎 cos 𝜃 and 𝑦 = 𝑐 + 𝑎 sin 𝜃.
We have cos 𝜃 =
𝑥−𝑏
𝑎
and sin 𝜃 =
𝑥−𝑏 2
cos2 𝜃 + sin2 𝜃 = 1 so (
𝑎
𝑦−𝑐
𝑎
𝑦−𝑐 2
) +(
𝑎
) =1
(𝑥 − 𝑏)2 + (𝑦 − 𝑐)2 = 𝑎2
The equation of C written in this form is given by
𝑥 2 + (𝑦 − 1)2 − 1 − 15 = 0
𝑥 2 + (𝑦 − 1)2 = 16 = 42
Therefore 𝑏 = 0, 𝑐 = 1, and 𝑎 = 4
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SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2011
iv.
Let the circle 𝐶2 has centre (𝑓, 𝑔) and radius 4 therefore 𝐶2 has equation
(𝑥 − 𝑓)2 + (𝑦 − 𝑔)2 = 16
The circle 𝐶2 touches the line L at the centre of C (0, 1) and has the same radius
Therefore (0 − 𝑓)2 + (1 − 𝑔)2 = 16
𝑓 2 + (1 − 𝑔)2 = 16
Also the line through the centres of the circle is perpendicular to the line L
therefore it has gradient −1.
𝑔−1
= −1
𝑓−0
𝑔 − 1 = −𝑓
𝑓 =1−𝑔
Substituting 𝑓 = 1 − 𝑔 into
𝑓 2 + 𝑔2 − 2𝑔 − 15 = 0 we have
(1 − 𝑔)2 + (1 − 𝑔)2 = 16
2(1 − 𝑔)2 = 16
(1 − 𝑔)2 = 8
1 − 𝑔 = ±√8 = ±2√2
𝑔 = 1 − 2√2, 𝑜𝑟 1 + 2√2
𝑓 =1−𝑔
𝑓 = 1 − (1 − 2√2) = 2√2
𝑓 = 1 − (1 + 2√2) = −2√2
Therefore the centres are (2√2, −1 − 2√2) and (−2√2, 1 + 2√2 )
2
2
The possible equations are (𝑥 − 2√2) + (𝑦 − (1 − 2√2)) = 16
2
2
(𝑥 − 2√2) + (𝑦 − 1 + 2√2) = 16
2
and
2
(𝑥 − (−2√2)) + (𝑦 − (1 + 2√2)) = 16
2
2
(𝑥 + 2√2) + (𝑦 − 1 − 2√2) = 16
76
Question 4
a.
i.
Given 8 cos 4 𝜃 − 10 cos2 𝜃 + 3 = 0
Let 𝑥 = cos 2 𝜃 then
8𝑥 2 − 10𝑥 + 3 = 0
8𝑥 2 − 6𝑥 − 4𝑥 + 3 = 0
2𝑥(4𝑥 − 3) − (4𝑥 − 3) = 0
(2𝑥 − 1)(4𝑥 − 3) = 0
1
𝑥 = 2 or
3
4
1
cos2 𝜃 = 2 , so cos 𝜃 = ±
1
√2
1
𝜋
√2
4
The acute angle is 𝜃 = cos −1 ( ) =
0 ≤ 𝜃 ≤ 𝜋 therefore 𝜃 is in the first and second quadrants .
The angles are 𝜃 =
𝜋
4
𝜋
or 𝜃 = 𝜋 − 4 =
3
cos2 𝜃 = 4 , so cos 𝜃 = ±
√3
𝜃=
b.
i.
𝜋
6
4
for the second quadrant.
√3
2
The acute angle is 𝜃 = cos −1 ( 2 ) =
The angles are 𝜃 =
3𝜋
𝜋
𝜋
3
or 𝜃 = 𝜋 − 6 =
5𝜋
6
for the second quadrant.
𝜋 3𝜋 𝜋
5𝜋
, , or
4 4 6
6
From the diagram angle 𝑄𝑅𝑆 𝑖𝑠 90𝑜 therefore triangle 𝑆𝑅𝐶 is similar to
triangle 𝑅𝑄𝐵 and as a result angle 𝑅𝑆𝐶 is equal to angle 𝑄𝑅𝐵.
𝐵𝐶 = 𝐵𝑅 + 𝑅𝐶
𝐵𝑅
= cos 𝜃 ,
6
𝑅𝐶
= sin 𝜃,
8
so 𝐵𝑅 = 6 cos 𝜃
so 𝑅𝐶 = 8 sin 𝜃
𝐵𝐶 = 6 cos 𝜃 + 8 sin 𝜃
77
SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2011
ii.
Given |𝐵𝐶| = 7 then
6 cos 𝜃 + 8 sin 𝜃 = 7
Using 𝑅 sin(𝜃 + 𝛼) = 6 cos 𝜃 + 8 sin 𝜃
We have 𝑅 sin(𝜃 + 𝛼) = 𝑅 sin 𝜃 cos 𝛼 + 𝑅 sin 𝛼 cos 𝜃
Therefore 8 = 𝑅 cos 𝛼 𝑎𝑛𝑑 6 = 𝑅 sin 𝛼
𝑅 sin 𝛼
6
= tan 𝛼 =
𝑅 cos 𝛼
8
tan 𝛼 =
6 3
=
8 4
3
𝛼 = tan−1 (4) = 36.87𝑜 or 0.644 rad
𝑅 = √62 + 82 = 10
6 cos 𝜃 + 8 sin 𝜃 = 10 sin(𝜃 + 0.644)
10 sin(𝜃 + 0.644) = 7
7
sin(𝜃 + 0.644) = 10
7
𝜃 + 0.644 = sin−1 (10)
𝜃 + 0.644 = 44.42𝑜 or 0.775 rad
𝜃 = 0.775 − 0.644
𝜃 = 0.131 rad or 7.55𝑜
iii.
𝐵𝐶 = 6 cos 𝜃 + 8 sin 𝜃 = 10 sin(𝜃 + 0.644)
Therefore the maximum value of BC is 10 because sin(𝜃 + 0.644) has a
maximum value of 1. So |𝐵𝐶| = 15 is NOT possible
c.
i.
1−cos 2𝜃
sin 2𝜃
=
=
2 sin2 𝜃
2 sin 𝜃 cos 𝜃
sin 𝜃
= tan 𝜃
cos 𝜃
78
ii.
a.
1−cos 4𝜃
=
sin 4𝜃
=
b.
1−cos 6𝜃
=
iii.
2 sin 2𝜃 cos2 𝜃
sin 2𝜃
= tan 2𝜃
cos 2𝜃
=
sin 6𝜃
2 sin2 2𝜃
2 sin2 3𝜃
2 sin 3𝜃 cos 3𝜃
sin 3𝜃
= tan 3𝜃
cos 3𝜃
From the above identities it can be seen that
1 − cos 2𝑟𝜃
= tan 𝑟𝜃
sin 2𝑟𝜃
Therefore 1 − cos 2𝑟𝜃 = tan 𝑟𝜃 sin 2𝑟𝜃
1 = tan 𝑟𝜃 sin 2𝑟𝜃 + cos 2𝑟𝜃
𝑛
𝑛
∑ tan 𝑟𝜃 sin 2𝑟𝜃 + cos 2𝑟𝜃 = ∑ 1 = 𝑛
𝑟=1
𝑟=1
Question 5
a.
lim
𝑥 2 +5𝑥+6
𝑥→−2 𝑥 2 −𝑥−6
(𝑥 + 2)(𝑥 + 3)
𝑥→−2 (𝑥 + 2)(𝑥 − 3)
lim
(𝑥 + 3) −2 + 3
1
=
=−
𝑥→−2 (𝑥 − 3)
−2 − 3
5
lim
b.
2
Given 𝑓(𝑥) = {𝑥 + 1 if 𝑥 ≥ 2
𝑏𝑥 + 1 if 𝑥 < 2
i.
ii.
iii.
𝑓(2) = 22 + 1 = 5
lim 𝑓(𝑥) = lim+(𝑥 2 + 1) = 22 + 1 = 5
𝑥→2+
𝑥→2
lim 𝑓(𝑥) = lim−(𝑏𝑥 + 1 ) = 2𝑏 + 1
𝑥→2−
𝑥→2
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SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2011
iv.
if f is continuous at 𝑥 = 2 then
lim 𝑓(𝑥) = lim− 𝑓(𝑥)
𝑥→2+
𝑥→2
5 = 2𝑏 + 1
4 = 2𝑏 => 𝑏 = 2
c.
Given 𝑦 = 𝑝𝑥 3 + 𝑞𝑥 2 + 3𝑥 + 2 and at 𝑇(1, 2)
𝑑𝑦
𝑑𝑥
=7
𝑑𝑦
We have 𝑑𝑥 = 3𝑝𝑥 2 + 2𝑞𝑥 + 3
When 𝑥 = 1,
𝑑𝑦
7 = 3𝑝 + 2𝑞 + 3 ------
(1)
4 = 3𝑝 + 2𝑞 ---------
(2)
2 = 𝑝 + 𝑞 + 3 + 2 -----
(3)
−3 = 𝑝 + 𝑞
--------
(4)
Multiple (4) by 2
−6 = 2𝑝 + 2𝑞 ------
(5)
Subtract (5) from (2)
10 = 𝑝
𝑑𝑥
= 7,
When 𝑥 = 1, 𝑦 = 2
−3 = 𝑝 + 𝑞
−3 = 10 + 𝑞 =>
𝑞 = −13
Therefore the equation is 𝑦 = 10𝑥 3 − 13𝑥 2 + 3𝑥 + 2
ii.
The gradient of the tangent at T is
𝑑𝑦
𝑑𝑥
=7
Therefore the gradient of the normal is −7
1
The equation of the normal is given by 𝑦 − 2 = − 7 (𝑥 − 1)
7𝑦 − 14 = 1 − 𝑥
7𝑦 + 𝑥 = 15
iii.
The line 𝑥 = 1 cuts the x-axis at 𝑥 = 1, therefore coordinates of M is (1, 0)
the normal has equation 7𝑦 + 𝑥 = 15 therefore coordinates for N
when 𝑦 = 0, 𝑥 = 15,
M and N is on the x-axis therefore length of 𝑀𝑁 = 15 − 1 = 14
80
Question 6
a.
i.
Given 𝑦 = 𝑥(𝑥 2 − 12), 𝑦 = 𝑥 3 − 12𝑥
We have
𝑑𝑦
𝑑𝑥
= 3𝑥 2 − 12
𝑑𝑦
At the stationary points 𝑑𝑥 = 0, 3𝑥 2 − 12 = 0
3(𝑥 2 − 4) = 0
𝑥 = 2, −2
When 𝑥 = 2, 𝑦 = 2(22 − 12) = −16
When 𝑥 = −2, 𝑦 = −2((−2)2 − 12) = 16
Therefore the stationary points have coordinates (2, −16) and (−2, 16)
ii.
𝑑𝑦
At the origin 𝑥 = 0,
𝑑𝑥
= −12 this is the gradient of the tangent.
1
The gradient of the normal is therefore − 12, and the equation of the normal at the
origin is given by 𝑦 − 0 = −
iii.
1
12
(𝑥 − 0), 𝑦 = −
1
12
𝑥.
The curve 𝑦 = 𝑥(𝑥 2 − 12) cuts the x-axis when 𝑦 = 0 therefore
𝑥(𝑥 2 − 12) = 0, and 𝑥 = 0, ±√12 = ±2√3
The area between the curve and the positive x-axis is given by
√12
𝐴 = − ∫ 𝑥(𝑥 2 − 12)𝑑𝑥
0
√12
𝐴 = ∫ (12𝑥 − 𝑥 3 ) 𝑑𝑥
0
𝐴 = [6𝑥 2 −
𝑥 4 √12
]
4 0
4
(√12)
𝐴 = 6(√12) −
4
2
= 6(12) −
144
= 36 sq. units
4
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SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2011
𝑎
b.
i.
𝑎
Using the result ∫ 𝑓(𝑥) 𝑑𝑥 = ∫ 𝑓(𝑎 − 𝑥) 𝑑𝑥
0
0
𝜋
𝜋
We have ∫ 𝑥 sin 𝑥 𝑑𝑥 = ∫(𝜋 − 𝑥) sin(𝜋 − 𝑥) 𝑑𝑥
0
0
𝜋
𝜋
sin(𝜋 − 𝑥) = sin 𝑥 therefore ∫ 𝑥 sin 𝑥 𝑑𝑥 = ∫(𝜋 − 𝑥) sin 𝑥 𝑑𝑥
0
𝜋
ii.
a.
0
𝜋
∫ 𝑥 sin 𝑥 𝑑𝑥 = ∫(𝜋 − 𝑥) sin 𝑥 𝑑𝑥
0
0
𝜋
= ∫(𝜋 sin 𝑥 − 𝑥 sin 𝑥) 𝑑𝑥
0
𝜋
𝜋
= ∫ πsin 𝑥 𝑑𝑥 − ∫ 𝑥 sin 𝑥 𝑑𝑥
0
0
𝜋
𝜋
= 𝜋 ∫ sin 𝑥 𝑑𝑥 − ∫ 𝑥 sin 𝑥 𝑑𝑥
0
𝜋
𝑏.
0
𝜋
𝜋
∫ 𝑥 sin 𝑥 𝑑𝑥 = 𝜋 ∫ sin 𝑥 𝑑𝑥 − ∫ 𝑥 sin 𝑥 𝑑𝑥
0
0
𝜋
0
𝜋
2 ∫ 𝑥 sin 𝑥 𝑑𝑥 = 𝜋 ∫ sin 𝑥 𝑑𝑥
0
0
= 𝜋[− cos 𝑥]
𝜋
0
= 𝜋[(− cos 𝜋) − (− cos(0))]
= 𝜋[(−(−1)) − (−1)]
𝜋
2 ∫ 𝑥 sin 𝑥 𝑑𝑥 = 2𝜋
0
𝜋
∫ 𝑥 sin 𝑥 𝑑𝑥 = 𝜋
0
82