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1

CAPE PURE MATHEMATICS UNIT 1

CAPE UNIT 1 2016 SOLUTIONS

Question 1

a. i. Given 𝑓(𝑥) = 2𝑥3 − 𝑥2 + 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) = 2𝑥3 − 𝑥2 − 25𝑥 − 12

𝑥 + 3 is a factor therefore using long division we have

𝑓(𝑥) = (𝑥 + 3)(2𝑥2 − 7𝑥 − 4)

= (𝑥 + 3)(𝑥 − 4)(2𝑥 + 1)

When 𝑓(𝑥) = 0, 𝑥 = −3,−1

2, 4

𝑥 + 3

2𝑥2 − 7𝑥 − 4

2𝑥3 − 𝑥2 − 25𝑥 − 12

2𝑥3 + 6𝑥2

−7𝑥2 − 25𝑥

−7𝑥2 − 21𝑥

−4𝑥 − 12

−4𝑥 − 12

0

2

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 q 𝐩 → 𝐪 𝐩˅𝐪 𝐩 ∧ 𝐪 (𝐩˅𝐪) → (𝐩 ∧ 𝐪)

T T T T T T

T F F T F F

F T T T F T

F F T F F T

ii. 𝐩 → 𝐪 and (𝐩˅𝐪) → (𝐩 ∧ 𝐪) are logically equivalent because both have the same

truth values in their output column.

Question 2

a. log2(10 − 𝑥) + log2 𝑥 = 4

log2(10 − 𝑥)𝑥 = 4

𝑥(10 − 𝑥) = 24

10𝑥 − 𝑥2 = 16

𝑥2 − 10𝑥 + 16 = 0

(𝑥 − 2)(𝑥 − 8) = 0

𝑥 = 2, 8

3

b. Given the function 𝑓(𝑥) =𝑥+3

𝑥−1, 𝑥 ≠ 1.

If the function is one-to-one then, 𝑓(𝑎) ≠ 𝑓(𝑏), 𝑎 ≠ 𝑏, (𝑎, 𝑏) ≠ 1

𝑎 + 3

𝑎 − 1≠

𝑏 + 3

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) =𝑥+3

𝑥−1

𝑓(𝑓−1(𝑥)) = 𝑓 (𝑥 + 3

𝑥 − 1) =

𝑥 + 3𝑥 − 1 + 3

𝑥 + 3𝑥 − 1 − 1

= (𝑥 + 3 + 3𝑥 − 3

𝑥 + 3 − 𝑥 + 1)

=4𝑥

4= 𝑥

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 2𝑥3 − 5𝑥2 + 4𝑥 + 6 = 0 are 𝛼, 𝛽, and 𝛾.

𝑥3 −5

2𝑥2 + 2𝑥 + 3 = 0

𝑥3 − (𝛼 + 𝛽 + 𝛾)𝑥2 + (𝛼𝛽 + 𝛽𝛾 + 𝛼𝛾)𝑥 − 𝛼𝛽𝛾 = 0

𝛼 + 𝛽 + 𝛾 =5

2, 𝛼𝛽 + 𝛽𝛾 + 𝛼𝛾 = 2, and 𝛼𝛽𝛾 = −3

4


c. ii. An equation whose roots are 1

𝛼2,

1

𝛽2 and

1

𝛾2 has

1

𝛼2+

1

𝛽2+

1

𝛾2=

𝛽2 𝛾2 + 𝛼2 𝛾2 + 𝛼2 𝛽2

𝛼2𝛽2𝛾2

=(𝛼𝛽 + 𝛽𝛾 + 𝛼𝛾)2 − 2𝛼𝛽𝛾(𝛼 + 𝛽 + 𝛾)

(𝛼𝛽𝛾)2

=(2)2 − 2(−3) (

52)

(−3)2=

19

9

1

𝛼2𝛽2+

1

𝛽2𝛾2+

1

𝛼2𝛾2=

𝛾2 + 𝛼2 + 𝛽2

𝛼2𝛽2𝛾2

=(𝛼 + 𝛽 + 𝛾)2 − 2(𝛼𝛽 + 𝛼𝛾 + 𝛽𝛾)

(𝛼𝛽𝛾)2

=(52)

2

− 2(2)

(−3)2= −

1

4

1

𝛼2𝛽2𝛾2=

1

(𝛼𝛽𝛾)2

=1

(−3)2=

1

9

𝑥3 −19

9𝑥2 −

1

4𝑥 −

1

9= 0

36𝑥3 − 76𝑥2 − 9𝑥 − 4 = 0

5

Question 3

a. i. Prove sec2 𝜃 =cosec𝜃

cosec𝜃−sin𝜃

LHS =

1sin 𝜃

1sin 𝜃 − sin 𝜃

=

1sin 𝜃

1 − sin2 𝜃sin 𝜃

= 1

1 − sin2 𝜃=

1

cos2 𝜃

= sec2 𝜃 𝐏𝐫𝐨𝐯𝐞𝐧.

ii. Given cosec𝜃

cosec𝜃−sin𝜃=

4

3

sec2 𝜃 =4

3, cos2 𝜃 =

3

4

cos 𝜃 = ±√3

2 this gives an acute angle

𝜋

6

Therefore 𝜃 =𝜋6, 5𝜋

6, 7𝜋

6, 11𝜋

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

Hence, tan 𝛼 = 1, 𝛼 =𝜋

4, and 𝑟 = √2

sin 𝜃 + cos 𝜃 = √2 sin (𝜃 +𝜋

4)

ii. The max value of 𝑓(𝜃) is √2

This occurs at (𝜃 +𝜋

4) =

𝜋

2

Therefore the smallest non-negative value of 𝜃 is 𝜋

4

6


c. Prove

tan(𝐴 + 𝐵 + 𝐶) =tan𝐴 + tan𝐵 + tan𝐶 − tan𝐴 tan𝐵 tan𝐶

1 − tan𝐴 tan𝐵 − tan𝐴 tan𝐶 − tan𝐵 tan𝐶

tan(𝐴 + 𝐵 + 𝐶) = tan(𝐴 + (𝐵 + 𝐶)) =tan𝐴 + tan(𝐵 + 𝐶)

1 − tan𝐴 tan(𝐵 + 𝐶)

=tan𝐴 +

tan𝐵 + tan𝐶1 − tan𝐵 tan𝐶

1 − tan𝐴 (tan𝐵 + tan𝐶1 − tan𝐵 tan𝐶)

=

tan𝐴 (1 − tan𝐵 tan𝐶) + tan𝐵 + tan𝐶1 − tan𝐵 tan𝐶

(1 − tan𝐵 tan𝐶) − tan𝐴 (tan𝐵 + tan𝐶)1 − tan𝐵 tan𝐶

=tan𝐴 + tan𝐵 + tan𝐶 − tan𝐴 tan𝐵 tan𝐶

1 − tan𝐴 tan𝐵 − tan𝐴 tan𝐶 − tan𝐵 tan𝐶

Proven

Question 4

a. i. Given sin 𝜃 = 𝑥, sin2 𝜃 = 𝑥2

1 − sin2 𝜃 = cos2 𝜃

cos 𝜃 = √1 − 𝑥2

tan 𝜃 =sin 𝜃

cos 𝜃

=𝑥

√1 − 𝑥2 𝐒𝐡𝐨𝐰𝐧

7

ii. Given 𝑦 = tan 2𝑡 and 𝑥 = sin 𝑡

𝑦 =2 tan 𝑡

1 − tan2 𝑡, and tan 𝑡 =

𝑥

√1 − 𝑥2

𝑦 =

2 (𝑥

√1 − 𝑥2)

1 − (𝑥

√1 − 𝑥2)2

=

2𝑥

√1 − 𝑥2

1 −𝑥2

1 − 𝑥2

=

2𝑥

√1 − 𝑥2

1 − 𝑥2 − 𝑥2

1 − 𝑥2

=2𝑥(1 − 𝑥2)

√1 − 𝑥2(1 − 2𝑥2)

𝑦 =2𝑥√1 − 𝑥2

1 − 2𝑥2

b. i. Given 𝐮 = (1

−32

) and 𝐯 = (215)

|𝐮| = √1 + 9 + 4 = √14

|𝐯| = √4 + 1 + 25 = √30

ii. cos 𝜃 =𝐮∙𝐯

|𝐮||𝐯|

=2 − 3 + 10

√14 × √30= 0.439

c. 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

8


d. 2𝑥 + 𝑦 + 3 = 0 ------ (1)

𝑥2 + 𝑦2 = 9 ------- (2)

From (1), 𝑦 = −(2𝑥 + 3) ----- (3)

Substituting (3) into (2) we have

𝑥2 + (−(2𝑥 + 3))2= 9

𝑥2 + 4𝑥2 + 12𝑥 + 9 = 9

5𝑥2 + 12𝑥 = 0

𝑥(5𝑥 + 12) = 0

𝑥 = 0,−12

5

When 𝑥 = 0, 𝑦 = −3

When 𝑥 = −12

5, 𝑦 = −2(−

12

5) − 3 =

9

5

Therefore the points of intersection are (0, −3) and (−12

5,9

5)

Question 5

a. Given ∫(𝑥 + 1)1/3 𝑑𝑥, Using the substitution 𝑢 = 𝑥 + 1 we have

𝑑𝑢

𝑑𝑥= 1, 𝑑𝑢 = 𝑑𝑥.

∫(𝑥 + 1)1/3 𝑑𝑥 = ∫(𝑢)1/3 𝑑𝑢

=(𝑢)4/3

43

+ 𝐶

=3

4𝑢4/3 + 𝐶

=3

4(𝑥 + 1)4/3 + 𝐶

9

b. V = 𝜋 ∫ 𝑥20

−1𝑑𝑦

𝑦 =3− 1,

𝑥 = (𝑦 + 1)1/3

𝑥2 = (𝑦 + 1)2/3

𝑉 = 𝜋 ∫(𝑦 + 1)2/3

0

−1

𝑑𝑦

= 𝜋 [3

5(𝑦 + 1)5/3]

0−1

=3𝜋

5 cubic units

c. Given ∫ 𝑓(𝑥)𝑎

0 𝑑𝑥 = ∫ 𝑓(𝑎 − 𝑥)

𝑎

0𝑑𝑥 𝑎 > 0

∫𝑒𝑥

𝑒𝑥 + 𝑒1−𝑥

1

0

𝑑𝑥 = ∫𝑒1−𝑥

𝑒1−𝑥 + 𝑒(1−(1−𝑥))

1

0

𝑑𝑥

= ∫𝑒1−𝑥

𝑒1−𝑥 + 𝑒𝑥

1

0

𝑑𝑥

Dividing both numerator and denominator by 𝑒𝑥 we have

= ∫𝑒1−2𝑥

𝑒1−2𝑥 + 1

1

0

𝑑𝑥

= [−1

2ln(𝑒1−2𝑥 + 1)]

10

= −1

2[(ln(𝑒−1 + 1)) − (ln(𝑒 + 1))]

= −1

2[ln |

𝑒 + 1

𝑒| − ln|𝑒 + 1|]

= −1

2[ln|𝑒 + 1| − ln 𝑒 − ln|𝑒 + 1|]

= −1

2(−1) =

1

2

10


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

𝑑𝑡= 0.02𝑦

Separating variables and integrating both sides we have

∫𝑑𝑦

𝑦=∫0.02 𝑑𝑡

ln 𝑦 = 0.02𝑡 + 𝐶

𝑦 = 𝑒0.02𝑡+𝐶

𝑦 = 𝑒𝐶𝑒0.02𝑡

When 𝑡 = 0, 𝑦 = 1000, therefore

1000 = 𝑒𝐶 , 𝑦 = 1000𝑒0.02𝑡

ii. When the bacteria population is double 𝑦 = 2000,

2000 = 1000𝑒0.02𝑡

2 = 𝑒0.02𝑡

ln 2 =0.02𝑡,

𝑡 =ln 2

0.02= 34.66 hrs

Question 6

a. Given 𝑓(𝑥) = 2𝑥3 + 5𝑥2 − 𝑥 + 12

𝑓′(𝑥) = 6𝑥2 + 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

11

b. i. Given 𝑓(𝑥) = {𝑥2 + 2𝑥 + 3 𝑥 ≤ 0

𝑎𝑥 + 𝑏 𝑥 > 0

lim𝑥→0−

𝑓(𝑥) =02 + 2(0) + 3 = 3

lim𝑥→0+

𝑓(𝑥) = 𝑎(0) + 𝑏 = 𝑏

ii. For 𝑓(𝑥) to be continuous at 𝑥 = 0, lim𝑥→0−

𝑓(𝑥) = lim𝑥→0+

𝑓(𝑥)

Therefore 𝑏 = 3, and 𝑎 𝜖 𝐑.

iii. Given 𝑓′(0) = lim𝑡→0

𝑓(0+𝑡)−𝑓(0)

𝑡

For 𝑥 ≤ 0

= lim𝑡→0

𝑎(0 + 𝑡) + 3 − (𝑎(0) + 3)

𝑡

= lim𝑡→0

𝑎𝑡 + 3 − 3

𝑡

= lim𝑡→0

𝑎𝑡

𝑡= 𝑎

For 𝑥 > 0

𝑓′(0) = lim𝑡→0

(0 + 𝑡)2 + 2(0 + 𝑡) + 3 − (02 + 2(0) + 3)

𝑡

= lim𝑡→0

(𝑡)2 + 2(𝑡) + 3 − 3

𝑡

= lim𝑡→0

𝑡2 + 2𝑡

𝑡

= lim𝑡→0

𝑡 + 2 = 2

If the 𝑓(𝑥) is differentiable at 𝑥 = 0 then 𝑓′(0) = 2

Therefore 𝑎 = 2

12


c. Given 𝑓(𝑥) = √𝑥 therefore 𝑓(𝑥 + ℎ) = √𝑥 + ℎ

𝑓′(𝑥) = limℎ→0

√𝑥 + ℎ − √𝑥

ℎ×

√𝑥 + ℎ + √𝑥

√𝑥 + ℎ + √𝑥

= limℎ→0

(𝑥 + ℎ) − 𝑥

ℎ(√𝑥 + ℎ + √𝑥)

= limℎ→0

ℎ

ℎ(√𝑥 + ℎ + √𝑥)

= limℎ→0

1

(√𝑥 + ℎ + √𝑥)

=1

2√2

13


SOLUTIONS FOR 2015 EXAM

1. a. i. The inverse ~𝐩 → ~𝐪 and the contrapositive ~𝐪 → ~𝐩

ii.

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

iii. 𝐩 → 𝐪 and ~𝐪 → ~𝐩 are logically equivalent because both final columns

are the exactly same.

b. Given 𝑓(𝑥) = 𝑥3 + 𝑝𝑥2 − 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

14


ii. 𝑓(𝑥) = 𝑥3 − 6𝑥2 − 𝑥 + 30

𝑥3 − 6𝑥2 − 𝑥 + 30 = (𝑥 − 5)(𝑥2 − 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.

𝑥2 − 𝑥 − 6

𝑥3 − 6𝑥2 − 𝑥 + 30 𝑥 − 5

𝑥3 − 5𝑥2

−𝑥2 − 𝑥

−𝑥2 + 5𝑥

−6𝑥 + 30

−6𝑥 + 30

0

15

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.

b

ii. 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. 3 −4

(9)𝑥−

4

(81)𝑥= 0

3 −4

9𝑥−

4

92𝑥= 0

3(92𝑥) − 4(9𝑥) − 4 = 0 multiplying both sides by 92𝑥

(3(9𝑥) + 2)(9𝑥 − 2) = 0 factorising the equation

(3(9𝑥) + 2) = 0,

9𝑥 = −2

3, not possible

(9𝑥 − 2) = 0

9𝑥 = 2

𝑥 =log 2

log 9= 0.315

𝑎1

𝑎2

𝑎3

𝑏1

𝑏2

𝑏3

𝑐1

𝑐2

𝑐3

𝑓(𝐴) 𝑔(𝐵) 𝑔°𝑓(𝐶)

16


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

𝑥 =1

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 =log 603

log 5= 3.98 hours

3. a. i. 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

17

ii. cos 6𝑥 − cos 2𝑥 = 0

cos 6𝑥 = 4 cos3 2𝑥 − 3 cos 2𝑥

4 cos3 2𝑥 − 3 cos 2𝑥 − cos 2𝑥 = 0

4 cos3 2𝑥 − 4 cos 2𝑥 = 0

4 cos 2𝑥 (cos2 2𝑥 − 1) = 0

4 cos 2𝑥 = 0

2𝑥 =𝜋

2,3𝜋

2,5𝜋

2,7𝜋

2

𝑥 =𝜋

4,3𝜋

4,5𝜋

4,7𝜋

4

cos2 2𝑥 − 1 = 0

cos2 2𝑥 = 1

cos 2𝑥 = ±1

2𝑥 = 0, 𝜋 2𝜋, 3𝜋 4𝜋

𝑥 = 0,𝜋

2, 𝜋,

3𝜋

2, 2𝜋

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

tan 𝛼 =4

3, and 𝑟 = √32 + 42

𝛼 = tan−1 (4

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

7 − 𝑓(𝜃)=

1

7 − 5

=1

2

Minimum value of occurs when 𝑓(𝜃) = −5

1

7 − 𝑓(𝜃)=

1

7 + 5=

1

12

18


4. a. i. Given 𝐶1: 𝑥 = √10 cos 𝜃 − 3; 𝑦 = √10 sin 𝜃 + 2

𝐶2: 𝑥 = 4 cos 𝜃 + 3; 𝑦 = 4 sin 𝜃 + 2

From 𝐶1 : cos 𝜃 =𝑥+3

√10; sin 𝜃 =

𝑦−2

√10

cos2 𝜃 + sin2 𝜃 = 1

(𝑥 + 3

√10)2

+ (𝑦 − 2

√10)2

= 1

(𝑥 + 3)2 + (𝑦 − 2)2 = (√10)2

From 𝐶2 : cos 𝜃 =𝑥−3

4; sin 𝜃 =

𝑦−2

4

(𝑥 − 3

4)2

+ (𝑦 − 2

4)2

= 1

(𝑥 − 3)2 + (𝑦 − 2)2 = 42

ii. (𝑥 + 3)2 + (𝑦 − 2)2 = (√10)2 ………. (1)

From (1) (𝑦 − 2)2 = (√10)2− (𝑥 + 3)2 ……….. (2)

(𝑥 − 3)2 + (𝑦 − 2)2 = 42 …….. (3)

(𝑦 − 2)2 = 42 − (𝑥 − 3)2 ………….. (4)

Equating (2) and (4) we have

(√10)2− (𝑥 + 3)2 = 42 − (𝑥 − 3)2

10 − (𝑥2 + 6𝑥 + 9) = 16 − (𝑥2 − 6𝑥 + 9)

10 − 𝑥2 − 6𝑥 − 9 = 16 − 𝑥2 + 6𝑥 − 9

10 − 16 = 6𝑥 + 6𝑥

−6 = 12𝑥

𝑥 = −1

2

19

Substituting 𝑥 = −1

2 into (4) we have

(𝑦 − 2)2 = 42 − ((−1

2) − 3)

2

𝑦2 − 4𝑦 + 4 = 16 −49

4

𝑦2 − 4𝑦 +1

4= 0

4𝑦2 − 16𝑦 + 1 = 0

𝑦 =16 ± √162 − 16

8

𝑦 =16 ± 4√15

8

𝑦 = 3.94, 0.0635

Points of intersection are (−1

2, 3.94) and (−

1

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;

𝑥2 + (𝑦 − 3)2 = 4[(𝑥 − 5)2 + (𝑦 − 2)2]

𝑥2 + 𝑦2 − 6𝑦 + 9 = 4[𝑥2 − 10𝑥 + 25 + 𝑦2 − 4𝑦 + 4]

𝑥2 + 𝑦2 − 6𝑦 + 9 = 4𝑥2 − 40𝑥 + 100 + 4𝑦2 − 16𝑦 + 16

3𝑥2 + 3𝑦2 − 40𝑥 − 10𝑦 + 107 = 0

𝑥2 + 𝑦2 −40

3𝑥 −

10

3𝑦 +

107

3= 0

(𝑥 −20

3)2

−400

9+ (𝑦 −

5

3)2

−25

9+

107

3= 0

(𝑥 −20

3)2

+ (𝑦 −5

3)2

=104

9

This is the equation of a circle with centre (20

3,5

3) and radius

√104

3.

20


5. a. Given

𝑓(𝑥) = {sin(𝑎𝑥)

𝑥 if 𝑥 ≠ 0, 𝑎 ≠ 0

4 if 𝑥 = 0

If f is continuous at 𝑥 = 0, then

lim𝑥→0

sin(𝑎𝑥)

𝑥= lim

𝑥→04

lim𝑥→0

sin(𝑎𝑥)

𝑥= 4

Multiplying numerator and denominator by a we have

lim𝑥→0

𝑎 sin(𝑎𝑥)

𝑎𝑥= 4

𝑎 lim𝑥→0

sin(𝑎𝑥)

𝑎𝑥= 4

lim𝑥→0

sin(𝑎𝑥)

𝑎𝑥= 1

𝑎 = 4

b. Given 𝑓(𝑥) = sin(2𝑥)

Differentiating from first principles we have

𝑓′(𝑥) = limℎ→0

𝑓(𝑥 + ℎ) − 𝑓(𝑥)

ℎ

= limℎ→0

sin 2(𝑥 + ℎ) − sin(2𝑥)

ℎ

= limℎ→0

2 cos (2𝑥 + 2ℎ + 2𝑥

2 ) sin (2𝑥 + 2ℎ − 2𝑥

2 )

ℎ

= limℎ→0

2 cos (4𝑥 + 2ℎ

2 ) sin(ℎ)

ℎ

= limℎ→0

2 cos(2𝑥 + ℎ) × limℎ→0

sin(ℎ)

ℎ

= 2 cos 2𝑥

21

c. Given 𝑦 =2𝑥

√1 + 𝑥2

i. Using the quotient and power rule:

𝑑𝑦

𝑑𝑥=

𝑣𝑑𝑢𝑑𝑥

− 𝑢𝑑𝑣𝑑𝑥

𝑣2 and

𝑑𝑦

𝑑𝑥= 𝑛(𝑓(𝑥))𝑛−1 × 𝑓′(𝑥)

𝑑𝑦

𝑑𝑥=

√1 + 𝑥2(2) − 2𝑥 (12)

(1 + 𝑥2)−12(2𝑥)

(√1 + 𝑥2)2

=

2√1 + 𝑥2 −2𝑥2

√1 + 𝑥2

1 + 𝑥2

=2(1 + 𝑥2) − 2𝑥2

(1 + 𝑥2)√1 + 𝑥2

𝑑𝑦

𝑑𝑥=

2

(1 + 𝑥2)√1 + 𝑥2=

2

(1 + 𝑥2)3/2

Multiplying both sides by x we have

𝑥𝑑𝑦

𝑑𝑥=

2𝑥

(1 + 𝑥2)√1 + 𝑥2

= (1

1 + 𝑥2)

2𝑥

√1 + 𝑥2

𝑥𝑑𝑦

𝑑𝑥=

𝑦

1 + 𝑥2

ii. From (i)

𝑑𝑦

𝑑𝑥=

2

(1 + 𝑥2)3/2= 2(1 + 𝑥2)−3/2

𝑑2𝑦

𝑑𝑥2= (2) (−

3

2) (1 + 𝑥2)−5/2(2𝑥)

= −6𝑥

(1 + 𝑥2)5/2

= −6𝑥

(1 + 𝑥2)2√1 + 𝑥2

= −3(2𝑥)

(1 + 𝑥2)2√1 + 𝑥2

22


Since 𝑦 =2𝑥

√1 + 𝑥2

𝑑2𝑦

𝑑𝑥2= −

3𝑦

(1 + 𝑥2)2

𝑑2𝑦

𝑑𝑥2+

3𝑦

(1 + 𝑥2)2= 0

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)


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)


3(3𝑥 − 7) = 𝑥 + 3

9𝑥 − 21 = 𝑥 + 3

8𝑥 = 24

𝑥 = 3

23

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)


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

∫3𝑥 − 7

4

3

𝑑𝑥 + ∫9 − 𝑥

6

4

𝑑𝑥 − ∫𝑥 + 3

3

6

3

𝑑𝑥

= [3𝑥2

2− 7𝑥]

43

+ [9𝑥 −𝑥2

2]64

−1

3[𝑥2

2+ 3𝑥]

63

= [(−4) − (−7.5)] + [(36) − (28)] −1

3[(36) − (13.5)]

= 4 sq. units

b. Given that 𝑓′(𝑥) = 3𝑥2 + 8𝑥 − 3 at the point (0, −6)

i. 𝑓(𝑥) = ∫3𝑥2 + 8𝑥 − 3𝑑𝑥

= 𝑥3 + 4𝑥2 − 3𝑥 + 𝐶

The curve passes through the point (0, −6) therefore

−6 = 𝐶

The equation of the curve is therefore 𝑓(𝑥) = 𝑥3 + 4𝑥2 − 3𝑥 − 6

24


ii. At the stationary point 𝑓′(𝑥) = 0 therefore

3𝑥2 + 8𝑥 − 3 = 0

(3𝑥 − 1)(𝑥 + 3) = 0

𝑥 =1

3, and − 3

When 𝑥 =1

3, 𝑦 = (

1

3)3

+ 4(1

3)2

− 3(1

3) − 6

𝑦 = −176

27= −6.5

When 𝑥 = −3, 𝑦 = (−3)3 + 4(−3)2 − 3(−3) − 6

𝑦 = 12

Therefore the stationary points are (1

3, −6.5) and (−3, 12)

𝑓′′(𝑥) = 6𝑥 + 8

When 𝑥 =1

3, 𝑓′′ (

1

3) = 6 (

1

3) + 8 > 0

Therefore (1

3, −6.5) is a minimum point

When 𝑥 = −3, 𝑓′′(−3) = 6(−3) + 8 < 0

Therefore (−3, 12) is a maximum point.

iii.

12

𝑦

𝑥 1 −1 −2 −3 0

−6

1

3

𝑥3 + 4𝑥2 − 3𝑥 − 6

Max (−3, 12)

Min (1

3,−6.5)

25



Question 1

a.

p q r 𝑝 → 𝑞 𝑟 → 𝑞 (𝑝 → 𝑞) ∧ (𝑟 → 𝑞)

T T T T T T

T T F T T T

T F T F F F

T F F F T F

F T T T T T

F T F T T T

F F T T F F

F F F T T T

b. i. Given 𝑦 ⊕ 𝑥 = 𝑦3 + 𝑥3 + 𝑎𝑦2 + 𝑎𝑥2 − 5𝑦 − 5𝑥 + 16

𝑥 ⊕ 𝑦 = 𝑥3 + 𝑦3 + 𝑎𝑥2 + 𝑎𝑦2 − 5𝑥 − 5𝑦 + 16

𝑦 ⊕ 𝑥 = 𝑥 ⊕ 𝑦 therefore ⊕ is commutative in R

ii. a. We have 𝑓(𝑥) = 2 ⊕ 𝑥 = 23 + 𝑥3 + 𝑎22 + 𝑎𝑥2 − 5(2) − 5𝑥 + 16

𝑓(𝑥) = 8 + 𝑥3 + 4𝑎 + 𝑎𝑥2 − 10 − 5𝑥 + 16

𝑓(𝑥) = 𝑥3 + 𝑎𝑥2 − 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) = 𝑥3 + (−2)𝑥2 − 5𝑥 + 4(−2) + 14

𝑓(𝑥) = 𝑥3 − 2𝑥2 − 5𝑥 + 6

(𝑥 − 1) is a factor of 𝑓(𝑥) therefore using long division we have.

26


𝑥3 − 2𝑥2 − 5𝑥 + 6 = (𝑥 − 1)(𝑥2 − 𝑥 − 6)

= (𝑥 − 1)(𝑥 + 2)(𝑥 − 3)

Therefore factors are (𝑥 − 1), (𝑥 + 2) and (𝑥 − 3)

c. 12 + 32 + 52 + ⋯+ (2𝑛 − 1)2 =𝑛

3(4𝑛2 − 1)

When 𝑛 = 1, L.H.S = 12 = 1, and R.H.S =1

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

12 + 32 + 52 + ⋯+ (2𝑘 − 1)2 =𝑘

3(4𝑘2 − 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

12 + 32 + 52 + ⋯+ (2𝑘 − 1)2 =𝑘

3(4𝑘2 − 1)

L.H.S =𝑘

3(4𝑘2 − 1) + (2(𝑘 + 1) − 1)2

=𝑘

3(2𝑘 − 1)(2𝑘 + 1) + (2𝑘 + 1)2

= (2𝑘 + 1) [𝑘

3(2𝑘 − 1) + (2𝑘 + 1)]

= (2𝑘 + 1) [𝑘(2𝑘 − 1) + 3(2𝑘 + 1)

3]

𝑥2 − 𝑥 − 6

𝑥 − 1 𝑥3 − 2𝑥2 − 5𝑥 + 6

𝑥3 − 𝑥2

−𝑥2 − 5𝑥

−𝑥2 + 𝑥

−6𝑥 + 6

−6𝑥 + 6

0

27

L. H. S =2𝑘 + 1

3(2𝑘2 + 5𝑘 + 3)

=2𝑘 + 1

3(2𝑘 + 3)(𝑘 + 1)

=𝑘 + 1

3(2𝑘 + 1)(2𝑘 + 3)

=𝑘 + 1

3(4𝑘2 + 8𝑘 + 3)

=𝑘 + 1

3(4{𝑘2 + 2𝑘} + 3)

=𝑘 + 1

3(4[𝑘 + 1]2 − 4 + 3)

=𝑘 + 1

3(4[𝑘 + 1]2 − 1)

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. Given 𝑓(𝑥) = 2𝑥2 + 1, 𝑔(𝑥) = √𝑥−1

2

i. a. 𝑓𝑓(𝑥) = 2(𝑓(𝑥))2 + 1

= 2(2𝑥2 + 1)2 + 1

= 2(4𝑥4 + 4𝑥2 + 1) + 1

= 8𝑥4 + 8𝑥2 + 2 + 1

= 8𝑥4 + 8𝑥2 + 3

b. 𝑓[𝑔(𝑥)] = 2[𝑔(𝑥)]2 + 1

= 2(√𝑥−1

2)

2

+ 1

= 2 (𝑥−1

2) + 1

= 𝑥 − 1 + 1

= 𝑥

ii. 𝑓−1(𝑥) = 𝑔(𝑥) Since 𝑓[𝑓−1(𝑥)] = 𝑥

28


b. Given that 𝑎3 + 𝑏3 + 3𝑎2𝑏 = 5𝑎𝑏2

We know that (𝑎 + 𝑏)3 = 𝑎3 + 3𝑎2𝑏 + 3𝑎𝑏2 + 𝑏3

Adding 3𝑎𝑏2 to both sides we have

𝑎3 + 𝑏3 + 3𝑎2𝑏 + 3𝑎𝑏2 = 3𝑎𝑏2 + 5𝑎𝑏2

(𝑎 + 𝑏)3 = 8𝑎𝑏2

(𝑎 + 𝑏)3

8= 𝑎𝑏2

(𝑎 + 𝑏

2)3

= 𝑎𝑏2

log (𝑎+𝑏

2)3= log𝑎𝑏2 [Log both sides]

3 log (𝑎 + 𝑏

2) = log 𝑎 + log 𝑏2

3 log (𝑎 + 𝑏

2) = log 𝑎 + 2 log 𝑏

c. i. 𝑒𝑥 +1

𝑒𝑥− 2 = 0

𝑒2𝑥 − 2𝑒𝑥 + 1 = 0 [Multiplying both sides by 𝑒𝑥]

Let 𝑢 = 𝑒𝑥 , 𝑢2 − 2𝑢 + 1 = (𝑢 − 1)2

𝑢 = 1, 𝑒𝑥 = 1 , 𝑥 = 0

ii. log2(𝑥 + 1) − log2(3𝑥 + 1) = 2

log2

𝑥 + 1

3𝑥 + 1= 2 log2 2

log2

𝑥 + 1

3𝑥 + 1= log2 22

𝑥 + 1

3𝑥 + 1= 4

𝑥 + 1 = 4(3𝑥 + 1)

𝑥 + 1 = 12𝑥 + 4

11𝑥 = −3

𝑥 = −3

11

29

d. √3−1

√3+1+

√3+1

√3−1+

√2−1

√2+1+

√2+1

√2−1

(√3 − 1)2+ (√3 + 1)

2

(√3 + 1)(√3 − 1)+

(√2 − 1)2+ (√2 + 1)

2

(√2 + 1)(√2 − 1)

4 − 2√3 + 4 + 2√3

3 − 1+

3 − 2√2 + 3 + 2√2

2 − 1

8

2+

6

1= 4 + 6 = 10

Question 3

a. i. cot 𝑦−cot 𝑥

cot 𝑥+cot 𝑦=

cos𝑦sin𝑦⁄ −cos𝑥

sin𝑥⁄

cos𝑥sin𝑥⁄ +

cos𝑦sin𝑦⁄

=sin 𝑥 cos 𝑦 − sin 𝑦 cos 𝑥

sin 𝑦 sin 𝑥/cos 𝑥 sin 𝑦 + sin 𝑥 cos 𝑦

sin 𝑥 sin 𝑦


sin 𝑦 sin 𝑥×

sin 𝑥 sin 𝑦

cos 𝑥 sin 𝑦 + sin 𝑥 cos 𝑦



=sin(𝑥 − 𝑦)

sin(𝑥 + 𝑦)

ii. cot𝑦−cot𝑥

cot𝑥+cot𝑦= 1, 0 ≤ 𝑦 ≤ 2𝜋,

When sin 𝑥 =1

2, sin−1 (

1

2) =

𝜋

6, cos (

𝜋

6) =

√3

2 for 0 ≤ 𝑥 ≤

𝜋

2

cot 𝑦 − cot 𝑥

cot 𝑥 + cot 𝑦=

sin 𝑥 cos 𝑦 − sin 𝑦 cos 𝑥


sin 𝑥 cos 𝑦 − sin 𝑦 cos 𝑥

cos 𝑥 sin 𝑦 + sin 𝑥 cos 𝑦= 1

(1

2) cos𝑦−sin𝑦(

√3

2)

(√3

2) sin𝑦+(

1

2) cos 𝑦

= 1 Multiplying both numerator & denominator by 2

cos 𝑦 − √3 sin 𝑦

√3 sin 𝑦 + cos 𝑦= 1

cos 𝑦 − √3 sin 𝑦 = √3 sin 𝑦 + cos 𝑦

30


0 = 2√3 sin 𝑦

sin 𝑦 = 0

𝑦 = 0, 𝜋, 2𝜋 for 0 ≤ 𝑦 ≤ 2𝜋.

b. i. 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

𝑟 cos 𝛼= tan 𝛼 =

4

3

𝛼 = tan−1 (4

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) =3𝜋

2

2𝜃 =3𝜋

2− 0.927

𝜃 =3𝜋

4−

0.927

2

𝜃 = 1.89 rad

b. The maximum value of 1

7−𝑓(𝜃) is when 𝑓(𝜃) = 5 so

1

7 − 𝑓(𝜃)=

1

7 − 5=

1

2

And the minimum value of 1

7−𝑓(𝜃) is when 𝑓(𝜃) = −5

1

7 − 𝑓(𝜃)=

1

7 − (−5)=

1

12

31

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

2= 3 𝑏 = 4

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

𝑥2 − 4𝑥 + 4 + 𝑦2 − 6𝑦 + 9 = 2

𝑥2 + 𝑦2 − 4𝑥 − 6𝑦 + 11 = 0

32


b. Given 𝑥 =1

1+𝑡, and 𝑦 =

𝑡

1−𝑡2

𝑥 =1

1 + 𝑡

𝑥(1 + 𝑡) = 1

𝑥 + 𝑥𝑡 = 1

𝑥𝑡 = 1 − 𝑥

𝑡 =1 − 𝑥

𝑥

𝑦 =𝑡

1 − 𝑡2=

𝑡

(1 + 𝑡)(1 − 𝑡)

𝑦 =1

1 + 𝑡×

𝑡

1 − 𝑡, substituting 𝑥 =

1

1 + 𝑡 we have

𝑦 =𝑥𝑡

1 − 𝑡, substituting 𝑡 =

1 − 𝑥

𝑥 we have

𝑦 =𝑥 (

1 − 𝑥𝑥 )

1 − (1 − 𝑥

𝑥 )

𝑦 =1 − 𝑥

𝑥 − (1 − 𝑥)𝑥

𝑦 =𝑥(1 − 𝑥)

2𝑥 − 1

c. i. Given 𝑃(3,−2, 1), 𝑄(−1, 𝜆, 5) and 𝑅(2, 1, −4)

𝑃𝑄⃗⃗⃗⃗ ⃗ = 𝑃𝑂⃗⃗⃗⃗ ⃗ + 𝑂𝑄⃗⃗⃗⃗⃗⃗

𝑃𝑄⃗⃗⃗⃗ ⃗ = −(3

−21

) + (−1𝜆5

) = (−4

2 + 𝜆4

)

𝑃𝑄⃗⃗⃗⃗ ⃗ = −4𝒊 + (2 + 𝜆)𝒋 + 4𝒌

𝑄𝑅⃗⃗⃗⃗ ⃗ = 𝑄𝑂⃗⃗⃗⃗⃗⃗ + 𝑂𝑅⃗⃗⃗⃗ ⃗

33

𝑄𝑅⃗⃗⃗⃗ ⃗ = −(−1𝜆5

) + (21

−4) = (

31 − 𝜆−9

)

𝑄𝑅⃗⃗⃗⃗ ⃗ = 3𝒊 + (1 − 𝜆)𝒋 − 9𝒌

𝑅𝑃⃗⃗⃗⃗ ⃗ = 𝑅𝑂⃗⃗⃗⃗ ⃗ + 𝑂𝑃⃗⃗⃗⃗ ⃗

𝑅𝑃⃗⃗⃗⃗ ⃗ = −(21

−4) + (

3−21

) = (1

−35

)

𝑅𝑃⃗⃗⃗⃗ ⃗ = 𝒊 − 3𝒋 + 5𝒌

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 =−45

−3= 15

Question 5

a. Given 𝑓(𝑥) = {𝑎𝑥 + 2, 𝑥 < 3

𝑎𝑥2, 𝑥 ≥ 3.

i. lim𝑥→3+

𝑓(𝑥) = 𝑎(32) = 9𝑎

lim𝑥→3−

𝑓(𝑥) = 3𝑎 + 2

If 𝑓(𝑥) is continuous at 𝑥 = 3, then

lim𝑥→3+

𝑓(𝑥) = lim𝑥→3−

𝑓(𝑥) Therefore

9𝑎 = 3𝑎 + 2

6𝑎 = 2,

𝑎 =1

3

34


ii. 𝑔(𝑥) =𝑥2 + 2

𝑏𝑥2 + 𝑥 + 4

Given that lim𝑥→1

2𝑔(𝑥) = lim𝑥→0

𝑔(𝑥) we have

2 (12 + 2

𝑏(1)2 + 1 + 4) =

02 + 2

𝑏(0)2 + 0 + 4

2 (3

𝑏 + 5) =

2

4

6

𝑏 + 5=

1

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

𝑑𝑥= lim

ℎ→0

𝑓(𝑥 + ℎ) − 𝑓(𝑥)

ℎ

𝑑𝑦

𝑑𝑥= lim

ℎ→0

1

√𝑥 + ℎ−

1

√𝑥ℎ

= limℎ→0

√𝑥 − √𝑥 + ℎ

(√𝑥 + ℎ)(√𝑥)

ℎ

= limℎ→0

√𝑥 − √𝑥 + ℎ

ℎ(√𝑥 + ℎ)(√𝑥)

= limℎ→0

[√𝑥 − √𝑥 + ℎ

ℎ(√𝑥 + ℎ)(√𝑥)×

√𝑥 + √𝑥 + ℎ

√𝑥 + √𝑥 + ℎ]

= lim ℎ→0

[𝑥 − (𝑥 + ℎ)

ℎ(√𝑥 + ℎ)(√𝑥)(√𝑥 + √𝑥 + ℎ)]

= lim ℎ→0

[−ℎ

ℎ(√𝑥 + ℎ)(√𝑥)(√𝑥 + √𝑥 + ℎ)]

35

𝑑𝑦

𝑑𝑥= lim

ℎ→0[

−1

(√𝑥 + ℎ)(√𝑥)(√𝑥 + √𝑥 + ℎ)]

=−1

(√𝑥)(√𝑥)(√𝑥 + √𝑥)

= −1

2𝑥√𝑥= −

1

2𝑥−3/2

ii. Given 𝑦 =𝑥

√1+𝑥

Using the quotient rule when 𝑦 =𝑢

𝑣,

𝑑𝑦

𝑑𝑥=

𝑣𝑢′−𝑢𝑣′

𝑣2

We have 𝑢 = 𝑥, 𝑣 = √1 + 𝑥

𝑢′ = 1 𝑣′ =1

2(1 + 𝑥)−1/2 =

1

2√1 + 𝑥

𝑑𝑦

𝑑𝑥=

(√1 + 𝑥)(1) − 𝑥 (1

2√1 + 𝑥)

(√1 + 𝑥)2

𝑑𝑦

𝑑𝑥=

√1 + 𝑥1 −

𝑥

2√1 + 𝑥1 + 𝑥

𝑑𝑦

𝑑𝑥=

(√1 + 𝑥)(2√1 + 𝑥) − 𝑥

2√1 + 𝑥1 + 𝑥

𝑑𝑦

𝑑𝑥=

(√1 + 𝑥)(2√1 + 𝑥) − 𝑥

2√1 + 𝑥(1 + 𝑥)

𝑑𝑦

𝑑𝑥=

2(1 + 𝑥) − 𝑥

2√(1 + 𝑥)3

𝑑𝑦

𝑑𝑥=

𝑥 + 2

2√(1 + 𝑥)3

c. Given 𝑥 = cos 𝜃, 𝑦 = sin 𝜃

𝑑𝑥

𝑑θ= − sin 𝜃 ,

𝑑𝑦

𝑑𝜃= cos 𝜃

𝑑𝑦

𝑑𝑥=

𝑑𝑦/𝑑𝜃

𝑑𝑥/𝑑𝜃

=cos𝜃

− sin𝜃= −cot 𝜃

36


Question 6

a. i. a. Given 𝑑𝑦

𝑑𝑥= 3𝑥2 − 4𝑥 + 1

Integrating both sides we have

𝑦 = ∫3𝑥2 − 4𝑥 + 1𝑑𝑥

𝑦 = 3𝑥2+1

2 + 1− 4

𝑥1+1

1 + 1+ 𝑥 + 𝐶

𝑦 = 𝑥3 − 2𝑥2 + 𝑥 + 𝐶

When 𝑥 = −1, 𝑦 = −4

−4 = (−1)3 − 2(−1)2 + (−1) + 𝐶

−4 = −4 + 𝐶

𝐶 = 0

𝑦 = 𝑥3 − 2𝑥2 + 𝑥

b. At the stationary points 𝑑𝑦

𝑑𝑥= 0 therefore

3𝑥2 − 4𝑥 + 1 = 0

(3𝑥 − 1)(𝑥 − 1) = 0

𝑥 =1

3, or 1

When 𝑥 =1

3, 𝑦 = (

1

3)3

− 2(1

3)2

+1

3

𝑦 =1

27−

2

9+

1

3=

1 − 6 + 9

27=

4

27

Therefore coordinate of the stationary point is (1

3,

4

27)

When 𝑥 = 1, 𝑦 = 13 − 2(1)2 + 1 = 0

Therefore coordinate of the stationary point is (1, 0)

37

𝑑2𝑦

𝑑𝑥2= 6𝑥 − 4

When 𝑥 =1

3,

𝑑2𝑦

𝑑𝑥2 = 6(1

3) − 4 = −2

𝑑2𝑦

𝑑𝑥2< 0 Therefore (

1

3,

4

27) is a maximum

When 𝑥 = 1, 𝑑2𝑦

𝑑𝑥2= 6(1) − 4 = 2

𝑑2𝑦

𝑑𝑥2> 0 Therefore (1, 0) is a minimum

ii. y-intercept when 𝑥 = 0, 𝑦 = 0 (0, 0)

x-intercept when 𝑦 = 0, 𝑥3 − 2𝑥2 + 𝑥 = 0

𝑥(𝑥2 − 2𝑥 + 1 = 0)

𝑥(𝑥 − 1)2 = 0

When 𝑦 = 0, 𝑥 = 0, 1

x-intercepts (0, 0) and (1, 0)

1 0

1

3

2

3

𝑥

4

27

𝑦

max (1

3,4

27)

min(1, 0)

38


b. i. ∫ 2𝑥√1 + 𝑥23

0 𝑑𝑥

Using the substitution method with 𝑢 = √1 + 𝑥2

We have 𝑑𝑢

𝑑𝑥=

1

2(1 + 𝑥2)−1/2 × 2𝑥 =

𝑥

√1+𝑥2

𝑢 = √1 + 𝑥2, 𝑑𝑢

𝑑𝑥=

𝑥

𝑢, 𝑑𝑥 =

𝑢

𝑥𝑑𝑢

When 𝑥 = 3, 𝑢 = √1 + 32 = √10

When 𝑥 = 0, 𝑢 = √1 + 0 = 1

∫ 2𝑥(𝑢)√10

1

𝑢𝑥𝑑𝑢

∫ 2𝑢2√10

1

𝑑𝑢 = [2𝑢3

3]√10

1

= [2(√10)

3

3−

2(1)3

3]

=2

3[√103 − 1] = 20.42

ii. Volume of revolution about the x-axis is given by 𝜋 ∫ 𝑦2𝑏

𝑎𝑑𝑥 therefore from b. (i)

Volume = 𝜋 ∫ (2𝑥√1 + 𝑥2)2

0

2𝑑𝑥

= 𝜋 ∫ 4𝑥2(1 + 𝑥2)2

0

𝑑𝑥

= 𝜋 ∫ 4𝑥2 + 4𝑥42

0

𝑑𝑥

= 𝜋 [4𝑥3

3+

4𝑥5

5]20

= 𝜋 [(4(2)3

3+

4(2)5

5) − 0]

= 𝜋 [32

3+

128

5] =

544

15𝜋 cubic units

39



Question 1

a. i.

p q 𝑝 → 𝑞

T T T

T F F

F T T

F F T

ii.

p Q 𝑝 ∧ 𝑞 ~(𝑝 ∧ 𝑞)

T T T F

T F F T

F T F T

F F F T

b. Given 𝑦 ⊕ 𝑥 = 𝑦2 + 𝑥2 + 2𝑦 + 𝑥 − 5𝑥𝑦

2 ⊕ 𝑥 = 22 + 𝑥2 + 2(2) + 𝑥 − 5𝑥(2)

2 ⊕ 𝑥 = 4 + 𝑥2 + 4 + 𝑥 − 10𝑥

2 ⊕ 𝑥 = 𝑥2 − 9𝑥 + 8

2 ⊕ 𝑥 = 0

𝑥2 − 9𝑥 + 8 = 0

(𝑥 − 1)(𝑥 − 8) = 0

𝑥 = 1, 8

40


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) = 𝑥3 − 9𝑥2 + 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.

𝑥3 − 9𝑥2 + 6𝑥 + 16 = (𝑥 + 1)(𝑥2 − 10𝑥 + 16)

= (𝑥 + 1)(𝑥 − 2)(𝑥 − 8)

iii. 𝑥3 − 9𝑥2 + 6𝑥 + 16 = (𝑥 + 1)(𝑥 − 2)(𝑥 − 8) = 0

Therefore 𝑥 = −1, 2, 8

𝑥2 − 10𝑥 + 16

(𝑥 + 1) 𝑥3 − 9𝑥2 + 6𝑥 + 16

𝑥3 + 𝑥2

−10𝑥2 + 6𝑥

−10𝑥2 − 10𝑥

16𝑥 + 16

16𝑥 + 16

0

41

Question 2

a. Given 𝑓(𝑥) = 𝑥2 − 𝑥, 𝑥 ≥ 1

By completing the square we have

𝑥2 − 𝑥 = (𝑥2 − 𝑥 +1

4) −

1

4

= (𝑥 −1

2)2

−1

4

The function has a parabolic shape with axis of symmetry 𝑥 =1

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.

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 = 𝑏2 − 𝑏

𝑎2 − 𝑏2 − 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.

𝑦 = 𝑐

0

𝑥 =1

2

1

𝑓(𝑥) = 𝑥2 − 𝑥

𝑥

𝑦

1

2

42


b. i. Given 𝑓(𝑥) = 3𝑥 + 2, and 𝑔(𝑥) = 𝑒2𝑥

a. Let 𝑦 = 3𝑥 + 2 interchanging 𝑥 and 𝑦 we have

𝑥 = 3𝑦 + 2, therefore 𝑦 =𝑥 − 2

3

𝑓−1(𝑥) =𝑥 − 2

3

Let 𝑦 = 𝑒2𝑥 interchanging 𝑥 and 𝑦 we have

𝑥 = 𝑒2𝑦 Natural log of both sides gives.

ln 𝑥 = ln 𝑒2𝑦

ln 𝑥 = 2𝑦

𝑦 =1

2ln 𝑥 Therefore 𝑔−1(𝑥) =

1

2ln 𝑥

b. 𝑓[𝑔(𝑥)] = 3[𝑔(𝑥)] + 2

= 3𝑒2𝑥 + 2

ii. 𝑦 = 3𝑒2𝑥 + 2 interchanging 𝑥 and 𝑦 we have

𝑥 = 3𝑒2𝑦 + 2

𝑒2𝑦 =𝑥−2

3 Natural log of both sides.

ln 𝑒2𝑦 = ln (𝑥 − 2

3)

2𝑦 = ln (𝑥 − 2

3) , therefore 𝑦 =

1

2ln (

𝑥 − 2

3)

[𝑓[𝑔(𝑥)]]−1

=1

2 ln (

𝑥 − 2

3)

𝑔−1[𝑓−1(𝑥)] =1

2ln(𝑓−1(𝑥))

=1

2ln (

𝑥 − 2

3)

Therefore [𝑓[𝑔(𝑥)]]−1

= 𝑔−1[𝑓−1(𝑥)]

43

c. i. 3𝑥2 + 4𝑥 + 1 ≤ 5

3𝑥2 + 4𝑥 − 4 ≤ 0

(3𝑥 − 2)(𝑥 + 2) ≤ 0

Critical points 𝑥 = −2, 2

3

𝑥 ≤ −2 −2 ≤ 𝑥 ≤

2

3 𝑥 ≥

2

3

3𝑥 − 2 − − +

𝑥 + 2 − + +

(3𝑥 − 2)(𝑥 + 2) + − +

Therefore −2 ≤ 𝑥 ≤2

3 from the table is negative or from the graph −2 ≤ 𝑥 ≤

2

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

𝑥2 + 4𝑥 + 4 = 9𝑥2 + 30𝑥 + 25

8𝑥2 + 26𝑥 + 21 = 0

8𝑥2 + 12𝑥 + 14𝑥 + 21 = 0

4𝑥(2𝑥 + 3) + 7(2𝑥 + 3) = 0

(4𝑥 + 7)(2𝑥 + 3) = 0

𝑥 = −7

4, not possible −

3

2 only answer

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 = −7

4 not possible

−2 2

3

44


Question 3

a. i. L.H.S 2 tan 𝜃

1+tan2 𝜃 substituting tan 𝜃 =

sin𝜃

cos𝜃

We have 2 tan𝜃

1 + tan2 𝜃=

2sin𝜃cos 𝜃

1 + (sin 𝜃cos 𝜃

)2

Multiplying denominator and numerator by cos2 𝜃

2 tan𝜃

1 + tan2 𝜃=

2 sin𝜃 cos 𝜃

cos2 𝜃 + sin2 𝜃 cos2 𝜃 + sin2 𝜃 = 1

2 tan𝜃

1 + tan2 𝜃= 2 sin 𝜃 cos 𝜃

2 sin 𝜃 cos 𝜃 = sin 2𝜃

ii. Given sin 2𝜃 − tan𝜃 = 0

2 tan𝜃

1 + tan2 𝜃− tan𝜃 = 0

2 tan 𝜃 − tan𝜃 (1 + tan2 𝜃) = 0

2 tan 𝜃 − tan𝜃 − tan3 𝜃 = 0

tan 𝜃 − tan3 𝜃 = 0

tan 𝜃 (1 − tan2 𝜃) = 0

tan 𝜃(1 − tan𝜃)(1 + tan𝜃) = 0

tan 𝜃 = 0, 1, −1

𝜃 = tan−1(0) = 0, 𝜋, 2𝜋

Acute angle for 𝜃 = tan−1(1) =𝜋

4

𝜃 = 0,𝜋

4,3𝜋

4,5𝜋

4,7𝜋

4

45

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 sin𝛼

𝑟 cos𝛼= tan𝛼 =

4

3

𝛼 = tan−1 (4

3) = 0.927

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 1

8+𝑓(𝜃) is when 𝑓(𝜃) is maximum

Therefore 1

8 + 𝑓(𝜃)=

1

8 + 5=

1

13

iii. a. 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

sin𝐵 = sin(𝐴 + 𝐶) and sin 𝐶 = sin(𝐴 + 𝐵) so

sin 𝐴 + sin𝐵 + sin 𝐶 = sin(𝐵 + 𝐶) + sin(𝐴 + 𝐶) + sin(𝐴 + 𝐵)

46


Question 4

a. i. Given 𝑥2 + 𝑦2 − 6𝑥 − 4𝑦 + 4 = 0

By completing the square we have

(𝑥2 − 6𝑥 + (−6

2)2

) − (−6

2)2

+ (𝑦2 − 4𝑦 + (−4

2)2

) − (−4

2)2

+ 4 = 0

(𝑥2 − 6𝑥 + 9) − 9 + (𝑦2 − 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 = 𝑟2

Where (𝑎, 𝑏) is the centre and r its radius.

Therefore the circle has centre (3, 2) and radius √9 = 3

ii. a. The gradient of the line between the centre (3, 2) and a point on the

circumference (6, 2) is given by 2−2

6−3= 0,

Therefore equation of the normal to the circle at (6, 2) is given by 𝑦 = 2

b. The tangent is perpendicular to the normal therefore the line 𝑦 = 2 at

(6, 2) is perpendicular to 𝑥 = 6, which is a vertical line parallel to the

y-axis.

b. Given 𝑥 = 𝑡2 + 𝑡, 𝑦 = 2𝑡 − 4

From 𝑦 = 2𝑡 − 4, 𝑡 =𝑦 + 4

2

Substituting 𝑡 =𝑦+4

2 into 𝑥 = 𝑡2 + 𝑡

We have 𝑥 = (𝑦 + 4

2)2

+𝑦 + 4

2

𝑥 =𝑦2 + 8𝑦 + 16

4+

𝑦 + 4

2

𝑥 =𝑦2 + 8𝑦 + 16 + 2(𝑦 + 4)

4

𝑥 =𝑦2 + 10𝑦 + 24

4

4𝑥 = 𝑦2 + 10𝑦 + 24

47

c. i. Given 𝐴(3, −1, 2), 𝐵(1, 2, −4) and 𝐶(−1, 1, −2)

𝐴𝐵⃗⃗⃗⃗ ⃗ = 𝐴𝑂⃗⃗⃗⃗ ⃗ + 𝑂𝐵⃗⃗ ⃗⃗ ⃗

= −(3

−12

) + (12

−4)

= (−23

−6)

𝐴𝐵⃗⃗⃗⃗ ⃗ = −2𝒊 + 3𝒋 − 6𝒌

𝐵𝐶⃗⃗⃗⃗ ⃗ = 𝐵𝑂⃗⃗ ⃗⃗ ⃗ + 𝑂𝐶⃗⃗⃗⃗ ⃗

= −(12

−4) + (

−11

−2)

= (−2−12

)

𝐵𝐶⃗⃗⃗⃗ ⃗ = −2𝒊 − 𝒋 + 2𝒌

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

𝒓 ∙ 𝑨𝑩 = (0

−16−8

) ∙ (−23

−6)

= (0 × −2) + (−16 × 3) + (−8 × −6)

= 0 − 48 + 48 = 0

Therefore r is perpendicular to 𝐴𝐵⃗⃗⃗⃗ ⃗

𝒓 ∙ 𝑩𝑪 = (0

−16−8

) ∙ (−2−12

)

= (0 × −2) + (−16 × −1) + (−8 × 2)

= 0 + 16 − 16 = 0

Therefore r is perpendicular to 𝐵𝐶⃗⃗⃗⃗ ⃗

48


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) = {𝑥 + 2, 𝑥 < 2

𝑥2, 𝑥 > 2.

i. lim𝑥→2+


𝑥2

= 22 = 4

lim𝑥→2−


𝑥 + 2

= 2 + 2 = 4

lim𝑥→2+


𝑓(𝑥)

lim𝑥→2

𝑓(𝑥) =4

ii. 𝑓(𝑥) is not continuous at 𝑥 = 2 because 𝑓(2) is undefined.

49

b. Let 𝑦 =𝑥2 + 2𝑥 + 3

(𝑥2 + 2)3

Using the quotient rule 𝑦 =𝑢

𝑣,

𝑑𝑦

𝑑𝑥=

𝑣𝑢′−𝑢𝑣′

𝑣2

Let 𝑢 = 𝑥2 + 2𝑥 + 3 𝑢′ = 2𝑥 + 2

𝑣 = (𝑥2 + 2)3 𝑣′ = 3(𝑥2 + 2)2 × 2𝑥

𝑣′ = 6𝑥(𝑥2 + 2)2

𝑑𝑦

𝑑𝑥=

(𝑥2 + 2)3 × (2𝑥 + 2) − (𝑥2 + 2𝑥 + 3) × (6𝑥(𝑥2 + 2)2)

((𝑥2 + 2)3)2

𝑑𝑦

𝑑𝑥=

(𝑥2 + 2)2[(𝑥2 + 2)(2𝑥 + 2) − 6𝑥(𝑥2 + 2𝑥 + 3)]

(𝑥2 + 2)6

𝑑𝑦

𝑑𝑥=

2𝑥3 + 2𝑥2 + 4𝑥 + 4 − 6𝑥3 − 12𝑥2 − 18𝑥

(𝑥2 + 2)4

𝑑𝑦

𝑑𝑥=

−4𝑥3 − 10𝑥2 − 14𝑥 + 4

(𝑥2 + 2)4

c. Given 𝑥 = 1 − 3 cos 𝜃 , 𝑦 = 2 sin 𝜃

𝑑𝑥

𝑑𝜃= 0 − 3(− sin 𝜃) = 3 sin 𝜃

𝑑𝑦

𝑑𝜃= 2 cos 𝜃

𝑑𝑦

𝑑𝑥=

𝑑𝑦/𝑑𝜃

𝑑𝑥/𝑑𝜃

𝑑𝑦

𝑑𝑥=

2 cos 𝜃

3 sin𝜃

𝑑𝑦

𝑑𝑥=

2

3cot 𝜃

d. i. 𝑦 = 𝑥2 + 3 ------ (1)

𝑦 = 4𝑥 ------- (2)

4𝑥 = 𝑥2 + 3 Substitute 𝑦 = 4𝑥 into (1)

𝑥2 − 4𝑥 + 3 = 0

50


(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

𝐴 = ∫4𝑥

3

1

𝑑𝑥 − ∫𝑥2 + 3

3

1

𝑑𝑥

𝐴 = [2𝑥2]31

− [𝑥3

3+ 3𝑥]

31

𝐴 = [2(32) − 2] − [(33

3+ 3(3)) − (

1

3+ 3)]

𝐴 = 16 − 18 + 31

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

4−

𝑢3

3+ 𝐶

∫𝑥(1 − 𝑥)2 𝑑𝑥 =(1 − 𝑥)4

4−

(1 − 𝑥)3

3+ 𝐶

= −1

12[(1 − 𝑥)3(1 + 3𝑥)] + 𝐶

51

ii. Given 𝑓(𝑡) = 2 cos 𝑡 , 𝑔(𝑡) = 4 sin 5𝑡 + 3 cos 𝑡

∫[𝑓(𝑡) + 𝑔(𝑡)] 𝑑𝑡 = ∫2 cos 𝑡 + 4 sin5𝑡 + 3 cos 𝑡 𝑑𝑡

= ∫5 cos 𝑡 + 4 sin5𝑡 𝑑𝑡

∫[𝑓(𝑡) + 𝑔(𝑡)] 𝑑𝑡 = 5 sin 𝑡 −4

5cos 5𝑡 + 𝐶

∫𝑓(𝑡) 𝑑𝑡 + ∫𝑔(𝑡) 𝑑𝑡 = ∫2 cos 𝑡 𝑑𝑡 + ∫4 sin 5𝑡 + 3 cos 𝑡 𝑑𝑡

= 2 sin 𝑡 + 𝐴 +4

5(− cos 5𝑡) + 3 sin 𝑡 + 𝐵

∫𝑓(𝑡) 𝑑𝑡 + ∫𝑔(𝑡) 𝑑𝑡 = 5 sin 𝑡 −4

5cos 5𝑡 + 𝐶

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 =600 − 2𝑥

2 + 𝜋

ii. Area of track is given by

𝐴 = 𝑥(2𝑟) +1

2(𝜋𝑟2)

𝐴 = 2𝑥𝑟 +𝜋

2𝑟2

𝐴 = 2𝑥 (600 − 2𝑥

2 + 𝜋) +

𝜋

2(600 − 2𝑥

2 + 𝜋)2

𝐴 = (1200𝑥 − 4𝑥2

2 + 𝜋) +

𝜋

2(600 − 2𝑥

2 + 𝜋)2

𝑑𝐴

𝑑𝑥=

1

2 + 𝜋[1200 − 8𝑥 +

𝜋

2(2 + 𝜋)(2)(600 − 2𝑥)(−2)]

𝑑𝐴

𝑑𝑥=

1

2 + 𝜋[1200 − 8𝑥 −

2𝜋

(2 + 𝜋)(600 − 2𝑥)]

52


When 𝑑𝐴

𝑑𝑥= 0,

1

2 + 𝜋[1200 − 8𝑥 −

2𝜋

(2 + 𝜋)(600 − 2𝑥)] = 0

1200 − 8𝑥 −2𝜋

(2 + 𝜋)(600 − 2𝑥) = 0

(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

𝑥 =600

4+𝜋≈ 84 metres

𝑑2𝐴

𝑑𝑥2=

1

2 + 𝜋[−8 − (

2𝜋

(2 + 𝜋)) (−2)] =

1

2 + 𝜋[

4𝜋

2 + 𝜋− 8]

8 >4𝜋

2 + 𝜋 Therefore

𝑑2𝐴

𝑑𝑥2< 0

Therefore 𝑥 =600

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

53

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

𝐴 =1

𝜋

The specific solution is 𝑦 = −𝑥 sin 𝑥 − 2 cos 𝑥 +1

𝜋𝑥 + 3

54


SOLUTIONS TO 2012 EXAM

Question 1

a. Given 𝑓(𝑥) = 2𝑥3 − 𝑝𝑥2 + 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) = 2𝑥3 + 7𝑥2 + 𝑥 − 10

ii.

2𝑥2 + 9𝑥 + 10

𝑥 − 1 2𝑥3 + 7𝑥2 + 𝑥 − 10

2𝑥3 − 2𝑥2

9𝑥2 + 𝑥

9𝑥2 − 9𝑥

10𝑥 − 10

10𝑥 − 10

0

55

2𝑥3 + 7𝑥2 + 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. Given (√𝑥 + √𝑦)2= 16 + √240

(√𝑥 + √𝑦)2= 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 ------ (4)

Therefore 𝑥𝑦 = 60 -------- (5)


𝑥(16 − 𝑥) = 60

16𝑥 − 𝑥2 = 60

𝑥2 − 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

56


c. Given |3𝑥 − 7| ≤ 5

Squaring both sides we have

(3𝑥 − 7)2 ≤ 25

9𝑥2 − 42𝑥 + 49 ≤ 25

9𝑥2 − 42𝑥 + 24 ≤ 0

3(3𝑥2 − 14𝑥 + 8) ≤ 0

3(3𝑥 − 2)(𝑥 − 4) ≤ 0

Critical points 𝑥 =2

3, 4

𝑥 ≤

2

3

2

3≤ 𝑥 ≤ 4

𝑥 ≥ 4

3𝑥 − 2 − − +

𝑥 − 4 − + +

(3𝑥 − 2)(𝑥 − 4) + − +

Therefore 2

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

𝑥 ≥2

3

Therefore 2

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.

2

3 4

|3𝑥 − 7| ≤ 5

57

Question 2

a. Given 𝑓(𝑥) → 𝑥2 − 3

i. 𝑓(𝑓(𝑥)) = [𝑓(𝑥)]2 − 3

= (𝑥2 − 3)2 − 3

= 𝑥4 − 6𝑥2 + 9 − 3

= 𝑥4 − 6𝑥2 + 6

ii. 𝑓(𝑓(𝑥)) = 𝑓(𝑥 + 3)

𝑥4 − 6𝑥2 + 6 = (𝑥 + 3)2 − 3

𝑥4 − 6𝑥2 + 6 = 𝑥2 + 6𝑥 + 9 − 3

𝑥4 − 6𝑥2 + 6 = 𝑥2 + 6𝑥 + 6

𝑥4 − 7𝑥2 − 6𝑥 = 𝑥(𝑥3 − 7𝑥 − 6)

𝑥 + 1 is a factor of 𝑥3 − 7𝑥 − 6

(−1)3 − 7(−1) − 6 = −1 + 7 − 6 = 0

𝑥 + 2 is a factor of 𝑥3 − 7𝑥 − 6

(−2)3 − 7(−2) − 6 = −8 + 14 − 6 = 0

𝑥 − 3 is a factor of 𝑥3 − 7𝑥 − 6

(3)3 − 7(3) − 6 = 27 − 21 − 6 = 0

Therefore 𝑥4 − 7𝑥2 − 6𝑥 = 𝑥(𝑥 + 1)(𝑥 + 2)(𝑥 − 3)

𝑥 = 0,−1,−2, 3

b. i. Given 𝛼 𝑎𝑛𝑑 𝛽 are the roots the equation 4𝑥2 − 3𝑥 + 1 = 0

(𝑥 − 𝛼)(𝑥 − 𝛽) = 𝑥2 − (𝛼 + 𝛽)𝑥 + 𝛼𝛽

4𝑥2 − 3𝑥 + 1 = 0 Dividing both sides by 4 we have

𝑥2 −3

4𝑥 +

1

4= 0 Therefore 𝛼 + 𝛽 =

3

4 𝑎𝑛𝑑 𝛼𝛽 =

1

4

ii. (𝛼 + 𝛽)2 = 𝛼2 + 2𝛼𝛽 + 𝛽2

𝛼2 + 𝛽2 = (𝛼 + 𝛽)2 − 2𝛼𝛽

= (3

4)2

− 2(1

4)

=1

16

58


iii. Given 2

𝛼2 and 2

𝛽2 are the roots of a quadratic equation we have

Sum of roots 2

𝛼2+

2

𝛽2=

2𝛽2 + 2𝛼2

𝛼2𝛽2

=2(𝛽2 + 𝛼2)

(𝛼𝛽)2

=2 (

116

)

(14)2 =

18

(116

)

= 2

Product of roots 2

𝛼2×

2

𝛽2=

4

(𝛼𝛽)2

=4

(14)2

= 4 × 16 = 64

Therefore the quadratic equation is 𝑥2 − 2𝑥 + 64 = 0

c. i. log10 (1

3) + log10 (

3

5) + log10 (

5

7) + log10 (

7

9) + log10 (

9

10)

log10 [(1

3) × (

3

5) × (

5

7) × (

7

9) × (

9

10)]

log10 (1

10) = log10 10−1

= −1

ii. ∑ log10 (𝑟

𝑟 + 1)

99

𝑟=1

= log10 (1

2) + log10 (

2

3) + ⋯+ log10 (

98

99) + log10 (

99

100)

= log10 [(1

2) × (

2

3) × …× (

98

99) (

99

100)]

= log10 (1

100)

= log10 10−2

= −2

59

Question 3

a. i. Given cos(𝐴 + 𝐵) = cos𝐴 cos 𝐵 − sin𝐴 sin𝐵 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]

= 2 cos 𝜃 [cos2 𝜃 − sin2 𝜃 −1

2]

ii. L.H.S =1

2[sin 6𝜃 − sin 2𝜃]

Using the factor formulae sin 𝐴 − sin𝐵 = 2 cos (𝐴+𝐵

2) sin (

𝐴−𝐵

2)

1

2[sin 6𝜃 − sin 2𝜃] =

1

2[2 cos (

6𝜃+2𝜃

2) sin (

6𝜃−2𝜃

2)]

= cos 4𝜃 sin 2𝜃 [cos 4𝜃 = 2 cos2 2𝜃 − 1]

= (2 cos2 2𝜃 − 1) sin 2𝜃

iii. sin 6𝜃 − sin 2𝜃 = 0

(2 cos2 2𝜃 − 1) sin 2𝜃 = 0

2 cos2 2𝜃 − 1 = 0

cos2 2𝜃 =1

2

cos 2𝜃 = ±1

√2 0 ≤ 𝜃 ≤

𝜋

2

2𝜃 =𝜋

4,

3𝜋

4

𝜃 =𝜋

8,3𝜋

8

sin 2𝜃 = 0 0 ≤ 𝜃 ≤𝜋

2

2𝜃 = 0, 𝜋

𝜃 = 0,𝜋

2

𝜃 = 0,𝜋

2,

𝜋

8,

3𝜋

8

60


b. Given cot2 𝜃 + cos 𝜃 = 0

2cos2 𝜃

sin2 𝜃+ cos 𝜃 = 0

2 cos2 𝜃 + sin2 𝜃 cos 𝜃 = 0

2 cos2 𝜃 + (1 − cos2 𝜃) cos 𝜃 = 0

2 cos2 𝜃 + cos 𝜃 − cos3 𝜃 = 0

cos3 𝜃 − 2 cos2 𝜃 − cos 𝜃 = 0

cos 𝜃 (cos2 𝜃 − 2 cos 𝜃 − 1) = 0

cos 𝜃 = 0

cos2 𝜃 − 2 cos 𝜃 − 1 = 0

Using the quadratic formula 𝑥 =−𝑏±√𝑏2−4𝑎𝑐

2𝑎

cos 𝜃 =−(−2) ± √(−2)2 − 4(1)(−1)

2

cos 𝜃 =2 ± √8

2, √8 = √2 × 4 = 2√2

cos 𝜃 =2 ± 2√2

2= 1 ± √2

cos 𝜃 = 1 − √2, cos 𝜃 ≠ 1 + √2

cos 𝜃 = 0, 1 − √2

Question 4

a. i. Given 𝑦 = 3 sec 𝜃 , and 𝑥 = 3 tan 𝜃

sec2 𝜃 = 1 + tan2 𝜃

sec 𝜃 =𝑦

3, tan 𝜃 =

𝑥

3

(𝑦

3)2

= 1 + (𝑥

3)2

𝑦2

9= 1 +

𝑥2

9

𝑦2 = 9 + 𝑥2

61

ii. 𝑦2 = 9 + 𝑥2 ------ (1)

𝑦 = √10𝑥 ------ (2)

𝑦2 = 10𝑥 --------- (3) squaring (2)

10𝑥 = 9 + 𝑥2 Substitute (3) into (1)

𝑥2 − 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

62


Question 5

a. i. Let 𝑓(𝑥) =𝑥3+8

𝑥2−4

𝑓(𝑥) is discontinuous when 𝑥2 − 4 = 0

Therefore 𝑥 = 2,−2

ii. lim𝑥→−2

𝑥3+8

𝑥2−4

𝑥3 + 8 = (𝑥 + 2)(𝑥2 − 2𝑥 + 4)

lim𝑥→−2

(𝑥 + 2)(𝑥2 − 2𝑥 + 4)

(𝑥 + 2)(𝑥 − 2)

lim𝑥→−2

(𝑥2 − 2𝑥 + 4)

(𝑥 − 2)=

(−2)2 − 2(−2) + 4

−2 − 2

=12

−4= −3

ii. lim𝑥→0

2𝑥3+4𝑥

sin 2𝑥

lim𝑥→0

2𝑥3 + 4𝑥2𝑥

sin 2𝑥2𝑥

lim𝑥→0

𝑥 + 2

sin2𝑥2𝑥

lim𝑥→0

𝑥 + 2

lim𝑥→0

sin2𝑥2𝑥

=0 + 2

1

lim𝑥→0

2𝑥3 + 4𝑥

sin 2𝑥= 2

63

b. Given 𝑓(𝑥) = {𝑥2 + 1, 𝑥 > 1,4 + 𝑝𝑥, 𝑥 < 1.

i. a. lim𝑥→1+

𝑓(𝑥) = 12 + 1 = 2

b. lim𝑥→1−

𝑓(𝑥) = 4 + 𝑝(1)

When lim𝑥→1

𝑓(𝑥) exist lim𝑥→1+


𝑓(𝑥)

Therefore 4 + 𝑝 = 2

𝑝 = −2

ii. 𝑓(1) = 2 For 𝑓 to be continuous at 𝑥 = 1.

c. Given 𝑀 = 𝑢𝑡2 +𝑣

𝑡2

𝑑𝑀

𝑑𝑡= 2𝑢𝑡 + (−2𝑣𝑡−3)

𝑑𝑀

𝑑𝑡= 2𝑢𝑡 −

2𝑣

𝑡3

When 𝑀 = −1, 𝑡 = 1 therefore

−1 = 𝑢 + 𝑣 ------ (1)

When 𝑑𝑀

𝑑𝑡=

35

4, 𝑡 = 2 therefore

35

4= 2𝑢(2) −

2𝑣

23

35

4= 4𝑢 −

𝑣

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

64


Question 6

a. i. Given 𝑦 = √4𝑥2 − 7

𝑑𝑦

𝑑𝑥=

1

2(4𝑥2 − 7)−1/2 × 8𝑥

𝑑𝑦

𝑑𝑥=

4𝑥

√4𝑥2 − 7

√4𝑥2 − 7𝑑𝑦

𝑑𝑥= 4𝑥

𝑦 = √4𝑥2 − 7

𝑦𝑑𝑦

𝑑𝑥= 4𝑥

ii. 𝑑𝑦

𝑑𝑥=

4𝑥

√4𝑥2−7

using the quotient rule 𝑦 =𝑢

𝑣,

𝑑𝑦

𝑑𝑥=

𝑣𝑑𝑢

𝑑𝑥−𝑢

𝑑𝑣

𝑑𝑥

𝑣2

𝑢 = 4𝑥, 𝑑𝑢

𝑑𝑥= 4

𝑣 = √4𝑥2 − 7 𝑑𝑣

𝑑𝑥=

4𝑥

√4𝑥2 − 7

𝑑2𝑦

𝑑𝑥2=

√4𝑥2 − 7(4) − 4𝑥 (4𝑥

√4𝑥2 − 7)

(√4𝑥2 − 7)2

=

4(√4𝑥2 − 7)(√4𝑥2 − 7) − (4𝑥)(4𝑥)

√4𝑥2 − 74𝑥2 − 7

=4(4𝑥2 − 7) − 16𝑥2

(4𝑥2 − 7)√4𝑥2 − 7

=16𝑥2 − 28 − 16𝑥2

(4𝑥2 − 7)√4𝑥2 − 7

=−28

(4𝑥2 − 7)√4𝑥2 − 7

65

(√4𝑥2 − 7)𝑑2𝑦

𝑑𝑥2= −

28

4𝑥2 − 7

𝑦𝑑2𝑦

𝑑𝑥2= −

28

4𝑥2 − 7

(𝑑𝑦

𝑑𝑥)2

= (4𝑥

√4𝑥2 − 7)2

=16𝑥2

4𝑥2 − 7

𝑦𝑑2𝑦

𝑑𝑥2+ (

𝑑𝑦

𝑑𝑥)2

= −28

4𝑥2 − 7+

16𝑥2

4𝑥2 − 7

=16𝑥2 − 28

4𝑥2 − 7

=4(4𝑥2 − 7)

4𝑥2 − 7= 4

Therefore 𝑦𝑑2𝑦

𝑑𝑥2+ (

𝑑𝑦

𝑑𝑥)2= 4

b. i. 𝑑𝑦

𝑑𝑥= 3𝑥2 − 6𝑥

Integrating both sides we have

𝑦 = ∫3𝑥2 − 6𝑥 𝑑𝑥

𝑦 = 𝑥3 − 3𝑥2 + 𝐶

When 𝑥 = −1, 𝑦 = 0

0 = (−1)3 − 3(−1)2 + 𝐶

0 = −1 − 3 + 𝐶

𝐶 = 4

𝑦 = 𝑥3 − 3𝑥2 + 4

66


ii. 𝑑𝑦

𝑑𝑥= 3𝑥2 − 6𝑥

At the stationary points 𝑑𝑦

𝑑𝑥= 0 therefore

3𝑥2 − 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

𝑑𝑥2= 6𝑥 − 6


𝑑𝑥2 = −6

Therefore (0, 4) is a maximum


𝑑𝑥2= 6(2) − 6 = 6

Therefore (2, 0) is a minimum

67

iv. The curve meets the x-axis when 𝑦 = 0 therefore

𝑥3 − 3𝑥2 + 4 = 0

The minimum point has 𝑦 = 0, 𝑥 = 2

Therefore (𝑥 − 2)2 is a factor

(𝑥 − 2)2(𝑥 − 𝑎) = 𝑥3 − 3𝑥2 + 4

Equating the constants we have

−4𝑎 = 4, 𝑎 = −1

𝑥3 − 3𝑥2 + 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.

0 2 𝑥

𝑦

4

−2

2

𝑃(−1, 0)

𝑄(2, 0)

𝑚𝑎𝑥 (0,4)

𝑚𝑖𝑛 (2,0)

−1

68


SOLUTIONS TO 2011 EXAM

Question 1

a. i. (√75 + √12)2− (√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

(33)14 × (32)

38 × (34)

18

334 × 3

34 × 3

12

3(34+34+12)

32 = 9

b. 𝑓(𝑥) = 𝑥3 + 𝑚𝑥2 + 𝑛𝑥 + 𝑝

Q 0 1 2

(0, 4)

f(x)

x

69

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) = 𝑥3 − 𝑥2 − 4𝑥 + 4

iii. 𝑥3 − 𝑥2 − 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.

70


c. i. Given √log2 𝑥 = log2 √𝑥

√log2 𝑥 = log2(𝑥)12

√log2 𝑥 =1

2log2 𝑥

Let 𝑦 = log2 𝑥 therefore

√𝑦 =1

2𝑦 Squaring both sides we have

𝑦 =1

4𝑦2

4𝑦 = 𝑦2

𝑦2 − 4𝑦 = 0

𝑦(𝑦 − 4) = 0

𝑦 = 0, 4

When 𝑦 = 0, log2 𝑥 = 0

𝑥 = 20 = 1

When 𝑦 = 4, log2 𝑥 = 4

𝑥 = 24 = 16

Therefore 𝑥 = 1, 16

ii. Given 𝑥2 − |𝑥| − 12 < 0

Because of |𝑥| = {𝑥, 𝑥 ≥ 0−𝑥, 𝑥 < 0

We have 𝑥2 − 𝑥 − 12 < 0, for 𝑥 ≥ 0

(𝑥 + 3)(𝑥 − 4) < 0

Critical points 𝑥 = −3, 4

71

𝑥2 − 𝑥 − 12 < 0, has inequality for values of 𝑥 between −3 < 𝑥 < 4

We have 𝑥2 + 𝑥 − 12 < 0, for 𝑥 < 0

(𝑥 − 3)(𝑥 + 4) < 0

Critical points 𝑥 = 3,−4, therefore for 𝑥2 + 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 𝑥2 − 𝑝𝑥 + 24 = 0

(𝑥 − 𝛼)(𝑥 − 𝛽) = 0, 𝑥2 − (𝛼 + 𝛽)𝑥 + 𝛼𝛽

a. 𝛼 + 𝛽 = 𝑝

b. 𝛼𝛽 = 24,

𝛼2 + 𝛽2 = (𝛼 + 𝛽)2 − 2𝛼𝛽

= 𝑝2 − 2(24)

= 𝑝2 − 48

ii. Given 𝛼2 + 𝛽2 = 33,

𝑝2 − 48 = 33

𝑝2 − 81 = 0

(𝑝 − 9)(𝑝 + 9) = 0

𝑝 = 9, or − 9

𝑥2 − 𝑥 − 12

−3 4

𝑥2 + 𝑥 − 12

−4 3

72


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.

73

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(𝑘2 + 5) is divisible by 6

When 𝑛 = 𝑘 + 1 we have

(𝑘 + 1)[(𝑘 + 1)2 + 5]

(𝑘 + 1)(𝑘2 + 2𝑘 + 1 + 5)

(𝑘 + 1)(𝑘2 + 2𝑘 + 6)

𝑘3 + 2𝑘2 + 6𝑘 + 𝑘2 + 2𝑘 + 6

𝑘3 + 3𝑘2 + 8𝑘 + 6

𝑘3 + 5𝑘 + 3𝑘2 + 3𝑘 + 6

𝑘(𝑘2 + 5) + 3𝑘(𝑘 + 1) + 6

We assumed 𝑘(𝑘2 + 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 = 𝑎1𝒊 + 𝑎2𝒋 and 𝒃 = 𝑏1𝒊 + 𝑏2𝒋 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

74


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( 𝑏1𝒊 + 𝑏2𝒋) − 11𝒊 and (𝒂 − 𝒃) = ( 𝑏1𝒊 + 𝑏2𝒋) − 11𝒊

(𝒂 + 𝒃) = (3𝑏1 − 11)𝒊 + 3𝑏2𝒋 and (𝒂 − 𝒃) = (𝑏1 − 11)𝒊 + 𝑏2𝒋

(𝒂 + 𝒃) ∙ (𝒂 − 𝒃) = [(3𝑏1 − 11)(𝑏1 − 11)] + [3𝑏2 × 𝑏2] = 69

= 3𝑏12 − 44𝑏1 + 121 + 3𝑏2

2 = 69

= 3(𝑏12 + 𝑏2

2) − 44𝑏1 + 52 = 0

|𝑏| = √(𝑏12 + 𝑏2

2) = 10 𝑠𝑜 𝑏12 + 𝑏2

2 = 100

= 3(100) − 44𝑏1 + 52 = 0

44𝑏1 = 352

𝑏1 = 8

𝑏12 + 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

𝑥2 + 𝑦2 − 2𝑦 − 15 = 0

The general equation of a circle is given by 𝑥2 + 𝑦2 − 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

75

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.

𝑥2 + 𝑦2 − 2𝑦 − 15 = 0 -------- (1)

𝑥 − 𝑦 + 1 = 0 -------- (2)

From (2) 𝑦 = 𝑥 + 1 ------- (3)

Substitute (3) into (1) we have

𝑥2 + (𝑥 + 1)2 − 2(𝑥 + 1) − 15 = 0

𝑥2 + 𝑥2 + 2𝑥 + 1 − 2𝑥 − 2 − 15 = 0

2𝑥2 − 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 =

𝑦−𝑐

𝑎

cos2 𝜃 + sin2 𝜃 = 1 so (𝑥−𝑏

𝑎)2+ (

𝑦−𝑐

𝑎)2= 1

(𝑥 − 𝑏)2 + (𝑦 − 𝑐)2 = 𝑎2

The equation of C written in this form is given by

𝑥2 + (𝑦 − 1)2 − 1 − 15 = 0

𝑥2 + (𝑦 − 1)2 = 16 = 42

Therefore 𝑏 = 0, 𝑐 = 1, and 𝑎 = 4

76


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

𝑓2 + (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

𝑓 − 0= −1

𝑔 − 1 = −𝑓

𝑓 = 1 − 𝑔

Substituting 𝑓 = 1 − 𝑔 into 𝑓2 + 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 )

The possible equations are (𝑥 − 2√2)2+ (𝑦 − (1 − 2√2))

2

= 16

(𝑥 − 2√2)2+ (𝑦 − 1 + 2√2)

2= 16

and (𝑥 − (−2√2))2

+ (𝑦 − (1 + 2√2))2

= 16

(𝑥 + 2√2)2+ (𝑦 − 1 − 2√2)

2= 16

77

Question 4

a. i. Given 8 cos4 𝜃 − 10 cos2 𝜃 + 3 = 0

Let 𝑥 = cos2 𝜃 then

8𝑥2 − 10𝑥 + 3 = 0

8𝑥2 − 6𝑥 − 4𝑥 + 3 = 0

2𝑥(4𝑥 − 3) − (4𝑥 − 3) = 0

(2𝑥 − 1)(4𝑥 − 3) = 0

𝑥 =1

2 or

3

4

cos2 𝜃 =1

2, so cos 𝜃 = ±

1

√2

The acute angle is 𝜃 = cos−1 (1

√2) =

𝜋

4

0 ≤ 𝜃 ≤ 𝜋 therefore 𝜃 is in the first and second quadrants .

The angles are 𝜃 =𝜋

4 or 𝜃 = 𝜋 −

𝜋

4=

3𝜋

4 for the second quadrant.

cos2 𝜃 =3

4, so cos 𝜃 = ±

√3

2

The acute angle is 𝜃 = cos−1 (√3

2) =

𝜋

3

The angles are 𝜃 =𝜋

6 or 𝜃 = 𝜋 −

𝜋

6=

5𝜋

6 for the second quadrant.

𝜃 =𝜋

4,3𝜋

4,𝜋

6 or

5𝜋

6

b. i. 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

6= cos 𝜃 , so 𝐵𝑅 = 6 cos 𝜃

𝑅𝐶

8= sin 𝜃, so 𝑅𝐶 = 8 sin 𝜃

𝐵𝐶 = 6 cos 𝜃 + 8 sin 𝜃

78


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 sin𝛼

𝑅 cos𝛼= tan𝛼 =

6

8

tan𝛼 =6

8=

3

4

𝛼 = tan−1 (3

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

sin(𝜃 + 0.644) =7

10

𝜃 + 0.644 = sin−1 (7

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−cos2𝜃

sin 2𝜃=

2 sin2 𝜃

2 sin𝜃 cos𝜃

=sin𝜃

cos 𝜃= tan 𝜃

79

ii. a. 1−cos4𝜃

sin 4𝜃=

2 sin2 2𝜃

2 sin 2𝜃 cos2𝜃

=sin2𝜃

cos 2𝜃= tan2𝜃

b. 1−cos6𝜃

sin 6𝜃=

2 sin2 3𝜃

2 sin 3𝜃 cos3𝜃

=sin3𝜃

cos3𝜃= tan3𝜃

iii. From the above identities it can be seen that

1 − cos 2𝑟𝜃

sin 2𝑟𝜃= tan 𝑟𝜃

Therefore 1 − cos 2𝑟𝜃 = tan 𝑟𝜃 sin 2𝑟𝜃

1 = tan 𝑟𝜃 sin 2𝑟𝜃 + cos 2𝑟𝜃

∑tan𝑟𝜃 sin 2𝑟𝜃 + cos 2𝑟𝜃

𝑛

𝑟=1

= ∑1

𝑛

𝑟=1

= 𝑛

Question 5

a. lim𝑥→−2

𝑥2+5𝑥+6

𝑥2−𝑥−6

lim𝑥→−2

(𝑥 + 2)(𝑥 + 3)

(𝑥 + 2)(𝑥 − 3)

lim𝑥→−2

(𝑥 + 3)

(𝑥 − 3)=

−2 + 3

−2 − 3= −

1

5

b. Given 𝑓(𝑥) = {𝑥2 + 1 if 𝑥 ≥ 2

𝑏𝑥 + 1 if 𝑥 < 2

i. 𝑓(2) = 22 + 1 = 5

ii. lim𝑥→2+


(𝑥2 + 1) = 22 + 1 = 5

iii. lim𝑥→2−


(𝑏𝑥 + 1 ) = 2𝑏 + 1

80


iv. if f is continuous at 𝑥 = 2 then

lim𝑥→2+


𝑓(𝑥)

5 = 2𝑏 + 1

4 = 2𝑏 => 𝑏 = 2

c. Given 𝑦 = 𝑝𝑥3 + 𝑞𝑥2 + 3𝑥 + 2 and at 𝑇(1, 2) 𝑑𝑦

𝑑𝑥= 7

We have 𝑑𝑦

𝑑𝑥= 3𝑝𝑥2 + 2𝑞𝑥 + 3

When 𝑥 = 1,𝑑𝑦

𝑑𝑥= 7, 7 = 3𝑝 + 2𝑞 + 3 ------ (1)

4 = 3𝑝 + 2𝑞 --------- (2)

When 𝑥 = 1, 𝑦 = 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

−3 = 𝑝 + 𝑞

−3 = 10 + 𝑞 => 𝑞 = −13

Therefore the equation is 𝑦 = 10𝑥3 − 13𝑥2 + 3𝑥 + 2

ii. The gradient of the tangent at T is 𝑑𝑦

𝑑𝑥= 7

Therefore the gradient of the normal is −7

The equation of the normal is given by 𝑦 − 2 = −1

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

81

Question 6

a. i. Given 𝑦 = 𝑥(𝑥2 − 12), 𝑦 = 𝑥3 − 12𝑥

We have 𝑑𝑦

𝑑𝑥= 3𝑥2 − 12

At the stationary points 𝑑𝑦

𝑑𝑥= 0, 3𝑥2 − 12 = 0

3(𝑥2 − 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. At the origin 𝑥 = 0, 𝑑𝑦

𝑑𝑥= −12 this is the gradient of the tangent.

The gradient of the normal is therefore −1

12, and the equation of the normal at the

origin is given by 𝑦 − 0 = −1

12(𝑥 − 0), 𝑦 = −

1

12𝑥.

iii. The curve 𝑦 = 𝑥(𝑥2 − 12) cuts the x-axis when 𝑦 = 0 therefore

𝑥(𝑥2 − 12) = 0, and 𝑥 = 0, ±√12 = ±2√3

The area between the curve and the positive x-axis is given by

𝐴 = − ∫ 𝑥(𝑥2 − 12)𝑑𝑥

√12

0

𝐴 = ∫ (12𝑥 − 𝑥3)

√12

0

𝑑𝑥

𝐴 = [6𝑥2 −𝑥4

4]√12

0

𝐴 = 6(√12)2−

(√12)4

4

= 6(12) −144

4= 36 sq. units

82


b. i. Using the result ∫ 𝑓(𝑥)

𝑎

0

𝑑𝑥 = ∫𝑓(𝑎 − 𝑥)

𝑎

0

𝑑𝑥

We have ∫ 𝑥 sin𝑥

𝜋

0

𝑑𝑥 = ∫(𝜋 − 𝑥) sin(𝜋 − 𝑥)

𝜋

0

𝑑𝑥

sin(𝜋 − 𝑥) = sin 𝑥 therefore ∫ 𝑥 sin 𝑥

𝜋

0

𝑑𝑥 = ∫(𝜋 − 𝑥) sin 𝑥

𝜋

0

𝑑𝑥

ii. a. ∫ 𝑥 sin𝑥

𝜋

0

𝑑𝑥 = ∫(𝜋 − 𝑥) sin 𝑥

𝜋

0

𝑑𝑥

= ∫(𝜋 sin 𝑥 − 𝑥 sin 𝑥)

𝜋

0

𝑑𝑥

= ∫ πsin 𝑥

𝜋

0

𝑑𝑥 − ∫𝑥 sin𝑥

𝜋

0

𝑑𝑥

= 𝜋 ∫ sin𝑥

𝜋

0

𝑑𝑥 − ∫ 𝑥 sin 𝑥

𝜋

0

𝑑𝑥

𝑏. ∫ 𝑥 sin𝑥

𝜋

0

𝑑𝑥 = 𝜋 ∫ sin 𝑥

𝜋

0

𝑑𝑥 − ∫𝑥 sin 𝑥

𝜋

0

𝑑𝑥

2∫ 𝑥 sin𝑥

𝜋

0

𝑑𝑥 = 𝜋 ∫ sin 𝑥

𝜋

0

𝑑𝑥

= 𝜋[− cos 𝑥]𝜋0

= 𝜋[(− cos 𝜋) − (− cos(0))]

= 𝜋[(−(−1)) − (−1)]

2∫ 𝑥 sin𝑥

𝜋

0

𝑑𝑥 = 2𝜋

∫ 𝑥 sin𝑥

𝜋

0

𝑑𝑥 = 𝜋

cape pure mathematics unit 1 cape unit 1 2016 … · 1 cape pure mathematics unit 1 cape unit 1...

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