QUESTION 2

2.1

56 + 2 y = 64
2 y = 8            Answer only 1 mark
y = 4

✓    correct equation
✓    y-value
(2)

2.2

Time (in minutes)

Frequency 

Cumulative frequency

✓    8, 12 and 28
✓    43, 60 and 64
(2)

5 ≤ t <10

3

3

10 ≤  t < 15

5

8

15 ≤  t < 20

4

12

20 ≤  t < 25

16

28

25 ≤  t < 30

15

43

30 ≤  t < 35

17

60

35 ≤  t < 40

4

64

3.1

x + py = p
y = - x  + 1
        p
∴ A(0;1)

3.2

OA = 1

∴ OC = 4(1) = 4
∴ C (4; 0)
m_AC= - 1   = 1 - 0
            p       0 - 4
p = 4

✓     OC
✓    - 1 = 1 - 0
        p     0 - 4
✓ simplification
✓     p-value
(4)

3.3

m_EB = 4    EB ⊥ AC
y = 4x - 7½

✓      m_EB = 4
✓     equation
(2)

- x + 1 = 4x - 7½
  4
-x + 4 = 16x - 30
17x = 34
x = 2
y = ½

✓    equating 
✓    simplification
✓    x-value
✓    y-value

(4)

4x - 7½ = 0
4x = 15
         2
x = 15
       8

r = 17
     16

✓     r ²
✓    equation
(2)

4.1

x ² + 6x + 9 + y ² - 8 y + 16 = -5 + 9 + 16
(x + 3)2  + (y - 4)2   = 26
∴ M(- 3; 4)

✓completing the square
✓     x-value
✓     y-value
(3)

4.2

r = √26

✓     answer
(1)

4.3

m_AS = 5    [SA ⊥ QB]
y - 4 = 5(x + 3)
y = 5x + 19

✓      m_AS = 5
✓     subst.. m_AS &(- 3; 4)
✓     equation
(3)

x ² + 6x + 9 + (5x + 19 - 4)2   = 26
x ² + 6x + 9 + 25x ² + 150x + 225 - 26 = 0
26x ² + 156x + 208 = 0
x ² + 6x + 8 = 0
(x + 4)(x + 2) = 0
x = -4  or  x s -2
∴ y = 5(-4) + 19
= -1
Q(- 4; - 1)

✓     substitution
✓     simplification

✓    standard form
✓    factors
✓    choosing correct
✓    y-value
(6)

-1 = ^{- 1}/₅ (- 4)+ k
∴ k = ^{- 9}/₅

✓     substitute (- 4; -1)
✓     answer
(2)

tanθ = ^{- 1}/₅
∴ θ = 169⁰
OQP = 11⁰ [∠s on a str line] 
∴ OQP = ORC
∴ RCPQisa cyclic quad [converse∠s same seg]

✓     size of θ 
✓     size of OQP
✓     R

5.1.1

5.1.2

tan 513⁰.cos 27⁰ = (- tan 27⁰) cos 27⁰
= - sin 27º .cos 27⁰
     cos 27⁰
= ^{- t}/₂

✓      (- tan 27⁰ )
✓      - sin 27⁰
         cos 27⁰
✓    ^{- t}/₂

5.1.3

cos 87⁰ = cos(60⁰ + 27⁰ )
= cos 60⁰.cos 27⁰ - sin 60⁰.sin 27⁰
= ½ . √4 - t ² - √3 . ^t/₂
            2          2
=  √4 - t ² - t 3
            4

✓    60⁰ + 27⁰
✓    expansion 
✓    subst √4 - t ² & ^t/₂
                    2

✓    subst. ½ & ^√3/₂

        sin(- 2a )cos(90⁰ + a )     
sin(- a + 360⁰ ).cos(- a -180⁰ )

= (- sin 2a )(- sin a )
     (sin a )(- cos a )

= - 2 sin a cos a
           cos a
= -2 sin a

✓      (- sin 2a )
✓      (- sin a )
✓     sin a
✓      (- cos a )
✓      2 sina cos a
✓      - 2sin a
(6)

9 sin² x - 4 cos² x = 0
(3sin x - 2 cos x)(3sin x + 2 cos x) = 0
∴ 3sin x =± 2 cos x
tan x =± 2
              3
x = 33, 69⁰ +180⁰.k or x = 146, 31⁰ +180⁰.k k ∈ Z

✓     factors
✓     both equations
✓      tan x = ± 2
                       3
✓    both 33,69⁰ &146,31⁰ /- 33,69⁰
✓     180⁰.k & k ∈ Z    (5)

6.1

Amplitude= 2

✓ answer(1)

6.2

1 ≤ y ≤ 5

✓  min & max
✓ notation(2)

6.3

✓    both x- intercepts - 30⁰ &150⁰
✓    max TP & y- intercept
✓    shape (3)

6.4

- 30⁰ < x < 0⁰

✓  both c.v.^s
✓ notation(2)

6.5

h(x) = sin(x + 90⁰) - 2
= cos x - 2

✓✓ sin(x + 90⁰) - 2 
✓ cos x(3)

7.1

Aˆ = 90⁰ - 2θ

✓    answer (1)

7.2

sinθ  = DB
           DC

✓    answer
(1)

7.3

DC =  DB 
         sin θ

and AD = 2DB = 2DB

        DC         =  AD    in ΔADC
sin(90⁰ - 2θ )    sin θ

  DB  
  sin θ  = 2DB
cos 2θ    sin θ

       DB         = 2DB
sin θ.cos 2θ     sin θ

DB.sin θ = 2DB.sin θ.cos 2θ
2 cos 2θ = 1
2 cos 2θ - 1 = 0

✓    AD = 2DB
✓    sine rule ∆ADC
✓    subst. in sine rule

✓    cos2θ
✓    equation
(5)

[7]

8.1

Perpendicular to the chord

✓    answer
(1)

8.2

8.2.1

OM = 2x + 3 - 3
             2
OR
OM = 2x - 3
             2

8.2.2

✓    susbt. in Pyth
✓    simplification
✓    standard form linear equation
✓    x-value (4)

DM = 12
DQ = √12² + 6²
= √180
= 6√5

✓  DM
✓  subst. in Pyth
✓ answer
(3)

9.1

Aˆ  = x  [ ∠ at centre = 2∠at circumf.]

✓ S ✓R

Cˆ 2  = x  [ ∠s opp.=sides]

✓ S ✓R

Bˆ1  = x   [ tan chord theorem]

✓ S ✓R

Tˆ = x [corresp.∠s, : TF II BC]

✓ S ✓R

(8)

9.2

Tˆ  = Aˆ = x
∴ ATBE is a cyclic quad [ converse ∠s same segment]

✓    S
✓    R

(2)

[10]

Contruction:Join BN and height from N ⊥ AM and CM and height from M ⊥ AN
Area ΔAMN =½ x AM x h   [same height]
Area ΔBMN   ½ x BM x h

= AM
   BM

Area ΔAMN =½ x AN x k   [same height]
Area ΔCMN   ½ x NC x k

= AN 
   NC

Area ΔBMN Area ΔCMN  [same height, same base MN||BC]

= AM = AN
   BM    NC

ND = 2  [given]
 x      1
ND = 2x and
NE = 2  [prop theorem DE II KM or] line drawn to oneside of a Δ
 y      1

∴ NE = 2y
KM² = KN² + MN² [Pyth theorem]
= (3x)²  + (3y)²
= 9x² + 9 y²

✓  S ✓R
✓  subst in Pyth theo
✓  simplification(4)

10.2.2

DM² = DN² + M N²  [Pyth]
= (2x)²  + (3y)²
KE² = KN² + NE²   [Pyth]
= (3x)²  + (2 y)²
DM² + KE² = 4x² + 9 y² + 9x² + 4 y²
= 13(x² + y² )
DM² + KE² = 13(x² + y² )
    KM²            9(x² + y² )
= 13
    9

✓  subst in Pyth
✓  subst in Pyth
✓  value of DM² + KE²
 13(x² + y² )  
 9(x² + y² ) ✓

(4)

[13]

11.1

A₁  = B₁ [alt ∠s, AC II DB] 
A₃  = B₁ [tan chord] 
∴ A₁  = A₃
T₂  = ACB [ext ∠of a cyclic quad]
D₂  = B₂ [3rd ∠s] 

✓S/R
✓S✓R
✓S✓R
✓S

OR
A₁  = B₁ [alt 3s, AC || DB]
A₃  = B₁ [tan chord]
∴ A1  = A3
T2  = ACB [ext ∠of a cyclic quad]
∴ ΔABC|||ΔADT [∠∠∠]

OR
✓ S/R
✓S
✓S ✓R
✓R ✓R
(6)

Tˆ₄  = Tˆ₁           [vert opp∠s]
Aˆ₁  = Tˆ₄      [∠sin same seg]
A₁ = Aˆ₃       [proven] 
∴ T₁  = A₃
∴ PT is a tangent to circle ADT[converse tan chord]

✓  S/R
✓  S   ✓R
✓R
(4)

Aˆ₃  = Tˆ₁     [proven]
Pˆ = Pˆ    [common]
PTA = Dˆ₁  [3rd ∠s]
Δ APT|||Δ TPD   [∠∠∠]

✓S
✓S
✓R

OR
Aˆ₃  = Tˆ₁  [proven]
Pˆ= Pˆ    [common]
PTA = D₁  [3rd ∠s]

AP = PT       [||| Δs]
TP    PD
AP.PD = PT²
AP(AP - AD) = PT²
AP(AP - ²/₃AP) = PT²
AP. AP = PT²
       3
AP² = 3PT²

✓  S/R
✓  simplification
✓   PD i.t.o AP and AD
✓   subst in AD
(4)

[17]