Distance from the store in km

1

2

3

4

5

7

8

10

Average number of times shopped per week

12

10

 7

 7

 6

 2

 3

 2

1.1

Strong

🗸 (1)

1.2

–0,95 (–0,9462..)

🗸 (1)

1.3

a = 11,71 (11,7132…)

b = –1,12 (–1,1176…)

yˆ = -1,12x + 11,71

🗸 value of a

🗸 value of b

🗸 equation (3)

1.4

yˆ = -1,12(6) + 11,71

= 5 times

🗸 substitition

🗸 answer

(2)

1.5

On scatter plot

🗸🗸 A line close to any 2 of the following points:

(5 ; 6) or (10 ;½ ) or

(6 ; 5) or (0 ; 11,7) (2)

[9]

2.1

Positively skewed OR skewed to the right

🗸 answer (1)

2.2

Rangeg = 2,21 – 1,39 = 0,82 m

🗸 subtract values

🗸answer (2)

2.3

🗸95 , 133,

156

 🗸160

 (2)

Intervals

Cumulative frequency

1,3 ≤ x < 1,5

24

1,5 ≤x < 1,7

95

1,7 ≤ x < 1,9

133

1,9 ≤ x < 2,1

156

2,1 ≤ x < 2,3

160

2.4

OGIVE

🗸 upper limits

🗸 cum f

🗸 shape

🗸 grounded (4)

2.5

method (using 80 to determine the height) 1,65 (accept any value between 1,6 and 1,69)

🗸 method

🗸 answer (2)

2.6.1

The mean would change by 0,1 m

🗸 answer (1)

2.6.2

No influence/change as there is no difference in variation of data.

🗸 answer (1)

[13]

3.1

M = Midpt of AC                                         [diags of rectangle bisect/]

 🗸 x-value of M

🗸 y-value of M (2)

3.2

m _BC    = 3 - 0 =     3    

               6 - p      6 - p

OR

m_BC   = 0 - 3 = - 3  

             p - 6    p - 6

🗸answer (1)

🗸answer (1)

3.3

m_AD = m_BC [AD | | BC]

m_BC = 2

   3     = 2

 6 - p

3 = 12 - 2 p

p = 4½

OR

y - y₁ = 2(x - x₁)

C(6 ;3)

y - 3 = 2(x - 6)

∴ y = 2x - 9

but y = 0

∴ x = 4½ = p

OR

🗸 m_BC = 2

🗸 equating

🗸answer (3)

🗸 m_BC = 2

 🗸 substituting (6 ; 3)

🗸answer

(3)

y = 2x + c

3 = 12 + c

- 9 = c

y = 2x - 9

0 = 2x - 9

x = 9            ∴ p = 9

      2                   2

🗸 m_BC = 2

🗸 substituting

🗸answer

(3)

3.4

DB =AC           [diag of rectangle = ]

AC = √170

\ DB =    170 or 13,04

🗸 substitution

🗸 length of AC

🗸 AC = BD

(3)

3.5

tan a = m_BC = 2

∴ a = 63,43°

🗸 tan a = m_BC

🗸 a = 63,43° (2)

3.6

In quadrilateral OFBG:

OFˆB = 63,43°                          [vert opp ∠s]

FOˆ G = GBˆ F = 90°

∴ OGˆ B = 360° -[90° + 90° + 63,43°] [sum ∠s quad = 360°]

∴ OGˆ B = 116,57°

OR

m_AB   = -½

90° +  OGˆ A = 153,43°

∴  OGˆ A  = 63,43°

OGˆ B  = 180° – 63,43°

= 116,57°

OR

FOˆG = GBˆ F = 90°

∴ GOFB is cyc quad

OGˆ B  = 180° – 63,43°  [∠s of cyc quad = 180°]

= 116,57°

OR

OFˆB = 63,43°

XOˆ G = FBˆ G = 90°

∴OGBF is a cyclic quad

∴ OGˆ B = 180° - 63,43° OGˆ B = 116,57°

🗸  size of  OFˆB

🗸 S

🗸 answer

(3)

🗸 m       = - ½            

 🗸 S

🗸 answer

(3)

🗸 S

🗸 S

🗸 answer

(3)

🗸 S

🗸 S

🗸 answer

(3)

3.7

M is the centre

[BD is diameter]

🗸 M is centre

🗸 r = √170

            2

🗸 equation (3)

3.8

CBˆ M = BAˆ M = 45°     [diag of square bisect ∠s]

∴BC will be a tangent [converse tan chord th]

OR

🗸S

🗸 R (2)

🗸S

🗸 R

(2)

[19]

4.1

∠ in semi circle/ ∠ at centre = 2Ð on circle

🗸 R (1)

4.2

m_TS     = 7 - 2

               3 - 5

= - 5

     2

🗸 substitution

🗸 m_TS

(2)

4.3

m_TS x m_RS = -1              [TS ⊥ SR]

m_RS       = 2

                 5

y = 2 x + c

      5

2 = 2 (5) + c

      5

c = 0

y = 2 x

      5

OR

🗸 m_RS

🗸substitution m

and (5 ; 2)

🗸 equation

(3)

m_TS ´ m_RS = -1              [TS ⊥ SR]

∴ m_RS      = 2

                   5

y - y₁ = 2 (x - x₁)

            5          

y - 2 = 2 (x - 5)

     5     

y = 2 x

     5

🗸 m_RS

🗸substitution m

and (5 ; 2)

🗸 equation

(3)

🗸 r

🗸 answer (2)

🗸 substitution 

🗸 answer (2)

OR

Answer only: full marks

Answer only: only 1 coordinate

correct (1 mark)

 🗸 M

🗸 x coordinate

🗸 y coordinate(3)

M

🗸 x coordinate

🗸 y coordinate (3)

m_TM=   9 - 3   = - 1

             6 ½ - 7   12

TM :    7 = - 1 (3) + c   y = - 1 x + 29....(1)

                  12                    12        4

SR :    y = 2 x............. (2)

                 5

2 x = - 1 x + 29

5         12      4

29 x = 29

60        4

∴ x = 15

∴ y = 2 (15) = 6

         5

🗸 Equating

🗸 x coordinate

🗸 y coordinate(3)

ST =   

OR

🗸 substitution

🗸 answer

🗸 ratio (3)

🗸 area

🗸 rule

🗸 ratio (3)

 [R⊥tangent]

OR

OR

OR

🗸m_TR =-½

🗸m_KTL = 12

🗸y=12x - 29

(3)

🗸m_TR =-½

🗸m_KTL = 12

🗸substitution

(3;7)&(a;b)

(3)

🗸subst into Pyth

🗸multiplication

🗸simplification

(3)

OR

OR

🗸substitution into disctance formula

🗸substitution of b=12a-29

🗸standard form

🗸subst into formula or factorise

🗸values of a 

🗸values of b

(6) 

🗸substitution into disctance formula

🗸substitution of b=12a-29

🗸(a-3)² =1

🗸±1

🗸values of a 

🗸values of b

(6)

🗸substitution

🗸substitution of b=12a-29

🗸standard form

🗸factors

🗸values of a

🗸values of b

(6)

[23]

5.1.1

sin196° = - sin16°

= - p

🗸reduction

🗸answer

(2)

5.1.2

OR

🗸statement

🗸answer

(2)

 🗸x in terms of p 

🗸answer

(2)

5.2

sin(A + B) = cos[90° - (A + B)]

= cos[(90° - A) - B]

= cos(90° - A)cosB + sin(90° - A)sinB

= sin AcosB + cos AsinB

🗸co-ratio

🗸correct form

🗸expansion

(3)

5.3

OR

OR

🗸 √sin² 2A

🗸 cosA 🗸 - sinA

🗸 2sinAcosA

🗸 answer

(5)

 🗸 2 cos²A -1

🗸 cosA 🗸 - sinA

 🗸identity

🗸 answer

(5)

🗸1-2sin2A

🗸cosA 🗸- sinA

🗸 identity

🗸 answer

(5)

🗸 identity

 🗸 value of cos²B

🗸 answer

🗸 = √cos 2B + 1

                 2

🗸 value of cos²B

🗸 answer

(3)

OR

OR

                 5

🗸 answer (2)

🗸 y = 1

 🗸 answer (2)

🗸 sin²B = 1

                5

🗸 answer (2)

OR

[21]

🗸 x- intercepts/

🗸 y- intercept/

🗸 turning pts/

(3)

f (x) - 3 = 2 sin 2x - 3

∴ maximum value = 2 – 3 = –1

(2)

2 sin 2x = - cos 2x 

tan 2x = - ½

ref∠ = 26,57°

2x = 153,43° + k.180°; k ∈ Z

x = 76,72° + k.90°; k ∈ Z or x = -13.28° + k.90°, k ∈ Z

OR

2 sin 2x = - cos 2x

tan 2x = - ½

ref ∠= 26,57°

2x = 153,43° + k.360° or 333,43° + k.360°, k ∈ Z

 x = 76,72° + k.180° or 166,72° + k.180° ; k ∈ Z

🗸 tan 2x = - ½

🗸 2x = 153,43°2

or – 26,56^o

🗸 76,72° or -13,28^o

🗸 k.90°; k ∈ Z

(4)

🗸 tan 2x = - ½

🗸 2x = 153,43° &333,43°

🗸 76,72° &166,72^o

🗸 k.180°; k ∈ Z

(4) 

x ∈ (-103,28°; - 13,28°)

OR

–103,28° < x < –13,28°

🗸🗸 values

🗸notation(3)

🗸🗸 values

🗸notation

(3)

[12]

7.1

🗸 substitution into Pyth

🗸 value of DB

🗸 answer

(3)

🗸 correct ratio

🗸 OB as subject

🗸 answer

(3)

🗸 correct ratio

🗸 special angle

🗸 answer

(3)

 OR

AOB = 90º (diagonals bisect ⊥)

OB=OA

AB² = AO² + BO² [pyth]

∴AB² = 2OB²

2OB²=3²

🗸 OB = OA

🗸 Pyth

🗸 answer

(3)

7.2

OR

🗸 substitution into Pyth

🗸 length of BE

🗸correct cosine rule

🗸 cos θ as subject

🗸 simplification

(5)

🗸 substitution into Pyth

🗸 length of BE

🗸correct cosine rule

🗸 substituting

🗸 cos θ as subject

(5)

🗸 substitution into Pyth

🗸 length of BE

🗸sketch with values

🗸 3/2

🗸substitution

(5)

🗸E = 180º - 2θ

🗸sinE = sin2θ

🗸subst into sine rule

🗸diagram

🗸2sinθcosθ

(5)

🗸 substitution

🗸 x-value

🗸 substitution

🗸 answer

(4)

[12]

8.1.1

Alternate angles PQ || SR

🗸 R (1)

8.1.2(a)

Tˆ₂   = 70°                                  [∠s opp = sides]

∴Qˆ₁   = 180° - 2(70°)               [∠s/e Δ = 180°]

= 40°

🗸 S 🗸R

🗸 answer

(3)

8.1.2(b)

Pˆ₁   = 40°                                   [tangent chord th]

🗸 S 🗸R

(2)

8.2.1

AT = 20           [line from centre ⊥ to chord]

🗸S(1)

8.2.2

OR

Let OS = 7, then OT = 15 In ΔAOT:

AO² = 20² + 15²

= 625

AO = 25

In ΔAOS:

AO² = 24² + 7²

= 625

AO = 25

∴ OA = 25

OR

AO² = OS² + AS²                    [Pyth : ΔAOS]

OT ² + AT² = OS² + AS²                           [Pyth : ΔAOT]

Let OT = 15x. Then OS = 7x

But AS = 24    [line from centre ^ to chord]

(15x)² + 400 = (7x)² + 576 225x² + 400 = 49x² + 576

176x² = 176

x = 1

∴AO =√225 + 400

= 25

🗸 equating

🗸 AS = 24

🗸 substitution

OS = 7/15OT 

🗸 OT

🗸 radius

(5)

🗸🗸 testing in

ΔAOT

🗸🗸 testing in

ΔAOS

🗸conclusion

(5)

🗸equating

🗸AS=24

🗸substitution

🗸x=1

🗸radius

(5)

🗸AS=24

🗸substitution

OS=7/15OT

🗸equating

🗸subst pyth

🗸radius

(5)

[12]

9.1.1

tangent chord theorem

🗸 R (1)

9.1.2

corresponding ∠s/e; FB || DC

🗸 R (1)

9.2

Eˆ₁ = BCˆ D

∴ BCDE = cyclic quad [converse ext ∠ cyc quad]

🗸 S

🗸 R (2)

9.3

Dˆ₂  = Eˆ₂                    [∠s in the same segment]

Dˆ₂  = FBˆ D                 [alt ∠s, BF ||CD||CD]

🗸 S

🗸 S

(2)

9.4

Bˆ ₃   = y  OR       Bˆ₃ = Cˆ₂  [∠s in the same segment]

Bˆ₂   = x - y  OR   Bˆ₃ + Bˆ₂  = x     [from 9.3 and 9.4]

Cˆ₁  = x - y                      [from 9.2 and 9.3]

∴Bˆ ₂   = Cˆ₁

OR

In Δ BFE and D BEC

Eˆ₁  = Eˆ₂                            [= x]

Fˆ = Bˆ₃ + Bˆ₄                  [tan - chord theorem]

∴ΔDBFE///DCBE    [∠,∠,∠]

∴Bˆ 2  = Cˆ1

🗸 S

🗸 S

🗸 S

(3)

🗸 identifying Δ ’s

🗸 S

🗸 S

(3)

[9]

10.1

Constr : Join S to R and T to Q and draw h₁ from S ⊥ PT and h₂ from T ⊥ PS

🗸 constr

(6)

Proof :

GL = FK     OR   GL = 2x     [prop theorem; GF | | LK]

LM   KM               y        x

2GH = 2x                              [LH = HG]

  y         x

∴GH = y

🗸S 🗸 R

🗸 GL = 2GH

(3)

10.2.2(b)

K₁ = GFM                 [corresponding∠s; GF || LK]

LKM or  → K₁ = MHˆ F                 [ext ∠cyclic quad]

MHˆF  =  GFM

In ΔMFH and ΔMGF:

Mˆ   =  Mˆ                                  [common]

MHˆF  =  GFˆM                         [proven]

∴ΔMFH | | | ΔMGF                     [∠∠∠]

OR

K₁ = GFˆM                [corresponding∠s; GF || LK]

LKˆM or K₁ = MHˆF                 [ext ∠cyclic quad]

MHˆF  =  GFˆM

In ΔMFH and ΔMGF:

Mˆ   =  Mˆ                                  [common]

MHˆF  =  GFˆM                           [proven]

Fˆ₂   = Gˆ                                    [∠s of Δ = 180°]

∴ΔMFH | | | ΔMGF

🗸S 🗸 R

🗸 S

🗸 S

🗸 R

(5)

🗸S 🗸 R

🗸 S

🗸 S

🗸 S

(5)

10.2.2(c)

∴GF = MF                            [ ||| Δs]

  FH     MH

= 3x

   2y

🗸S 🗸R

(2)

10.2.3

MF = MG                            [ ||| Δs]

MH     MF

3x = 3y                                     [from 10.2.2(c)]

2y    3x

🗸 S

🗸substitution

🗸 simplificatio n

(3)

[20]

TOTAL MARKS

150