MATHEMATICS PAPER 2

GRADE 12

MEMORANDUM

NATIONAL SENIOR CERTIFICATE

NOVEMBER 2016

 

NOTE:

  • If a candidate answered a question TWICE, mark only the FIRST
  • If a candidate has crossed out an attempt to answer a question and did not redo it, mark the crossed-out version.
  • Consistent accuracy applies in ALL aspects of the marking memorandum. Stop marking at the second calculation
  • Assuming answers/values in order to solve a problem is NOT

QUESTION 1

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

 scatter

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]


QUESTION 2

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

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]

             

 

QUESTION 3

3

3.1

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

3.1

 ? x-value of M

? y-value of M (2)

3.2

BC    3 - 0 =     3    

               6 - p      6 - p

OR

mBC   0 - 3- 3  

             p - 6    p - 6

?answer (1)

?answer (1)

3.3

mAD = mBC [AD | | BC]

mBC = 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

? mBC = 2

? equating

?answer (3)

? mBC = 2

 ? substituting (6 ; 3)

?answer

(3)

 

 

y = 2x + c

3 = 12 + c

- 9 = c

y = 2x - 9

0 = 2x - 9

x =            ∴ p = 9

      2                   2

? mBC = 2

 

? substituting

?answer

(3)

3.4

DB =AC           [diag of rectangle = ]

3.4

AC = √170

\ DB =    170 or 13,04

? substitution

? length of AC

? AC = BD

(3)

3.5

tan a = mBC = 2

∴ a = 63,43°

? tan a = mBC

? 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

mAB   = -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

3.7M is the centre

[BD is diameter]

3.7b

? 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

3.8

        

?S

? R (2)

?S

? R

(2)

[19]

 

QUESTION 4

4

4.1

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

? R (1)

4.2

mTS     = 7 - 2

               3 - 5

= - 5

     2

? substitution

? mTS

(2)

4.3

mTSmRS = -1              [TS ⊥ SR]

mRS       = 2

                 5

y = 2 x + c

      5

2 = 2 (5) + c

      5

c = 0

y = 2 x

      5

OR

? mRS

 

?substitution m

and (5 ; 2)

 

? equation

(3)

 

 

mTS ´ mRS = -1              [TS ⊥ SR]

∴ mRS      = 2

                   5

y - y1 = 2 (x - x1)

            5          

y - 2 = 2 (x - 5)

     5     

y = 2 x

     5

 

? mRS

 

?substitution m

and (5 ; 2)

? equation

(3)

 
4.4.1  4.41

? r

? answer (2)

? substitution 

? answer (2)

 
4.4.2 

4.42

OR

4.42b

Answer only: full marks

Answer only: only 1 coordinate

correct (1 mark)

 

 ? M

? x coordinate

? y coordinate(3)

M

? x coordinate

? y coordinate (3)

 

mTM =   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 = (15) = 6

         5

? Equating

? x coordinate

? y coordinate(3)

4.4.3

ST =   4.43

OR

4.43b

? substitution

? answer

? ratio (3)

? area

rule

? ratio (3)

4.4.4

 4.44[R⊥tangent]

OR

4.44b

OR

4.44c

OR

4.44d

?mTR =-½

?mKTL = 12

?y=12x - 29

(3)

?mTR =-½

?mKTL = 12

?substitution

(3;7)&(a;b)

(3)

?subst into Pyth

?multiplication

?simplification

(3)

4.4.5

 4.45

OR

4.45b

OR

4.45c

?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]

  

QUESTION 5

5.1.1

sin196° = - sin16°

= - p

?reduction

?answer

(2)

5.1.2

 5.12

OR

5.12b

?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

 5.3

OR

5.3b

OR

5.3c

 

 

? √sin2 2A

? cosA ? - sinA

? 2sinAcosA

? answer

(5)

 ? 2 cos2 A -1

? cosA ? - sinA

 ?identity

? answer

(5)

?1-2sin2A

?cosA ?- sinA

? identity

? answer

(5)

5.4.1

5.41

? identity

 ? value of cos2B

? answer

? = √cos 2B + 1

                 2

? value of cos2B

? answer

(3)

5.4.2

 5.42

OR

5.42b

OR

5.42c

 ? sin2B = 1

                 5

? answer (2)

? y = 1

 ? answer (2)

? sin2B = 1

                5

? answer (2)

5.43

 5.43

OR

5.43b

5.43ans

[21]


QUESTION 6

6.1   6.1

? x- intercepts/

? y- intercept/

? turning pts/

(3)

 
6.2 

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

∴ maximum value = 2 – 3 = –1

? ? answer

(2)

6.3 

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 – 26,56o

? 76,72° or -13,28o

? k.90°; k  Z

(4)

? tan 2x = - ½

? 2x = 153,43° &333,43°

? 76,72° &166,72o

? k.180°; k  Z

(4) 

6.4

x  (-103,28°; - 13,28°)

OR

–103,28° < x < –13,28°

 

?? values

?notation(3)

?? values

?notation

(3)

[12]

 

 

QUESTION 7

7

7.1

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

7.1b

? OB = OA

? Pyth

? answer

(3)

7.2

 7.2

OR

7.2c

 

? 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)

7.3 7.3

? substitution

? x-value

? substitution

? answer

(4)

[12]

 

QUESTION 8 

8

8.1.1

Alternate angles PQ || SR

? R (1)

8.1.2(a)

2   = 70°                                  [∠s opp = sides]

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

= 40°

? S ?R

? answer

(3)

8.1.2(b)

1   = 40°                                   [tangent chord th]

? S ?R

(2)

 

8.2

 8.2

8.2.1

AT = 20           [line from centre ⊥ to chord]

?S(1)

8.2.2

8.22

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]

 

QUESTION 9 

9

9.1.1

tangent chord theorem

? R (1)

9.1.2

corresponding ∠s/e; FB || DC

? R (1)

9.2

1 = BCˆ D

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

? S

? R (2)

9.3

2  = Eˆ                   [∠s in the same segment]

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

? S

? S

(2)

9.4

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

2   = x - y  OR   3 + Bˆ = x     [from 9.3 and 9.4]

1  = x - y                      [from 9.2 and 9.3]

∴Bˆ 2   = Cˆ1

OR

In Δ BFE and D BEC

1  = Eˆ2                            [= x]

Fˆ = Bˆ3 + Bˆ4                  [tan - chord theorem]

∴ΔDBFE///DCBE    [∠,∠,∠]

∴Bˆ 2  = Cˆ1

? S

? S

? S

(3)

? identifying Δ ’s

? S

? S

(3)

[9]

 

QUESTION 10

10.1      10  

 

10.1

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

? constr

10.1b

(6)

Proof :

 10.1 

 

10.2

10.2

 

10.2.1 Corresponding ∠s/e; GF || LK ? R (1)
10.2.2(a)

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

LM   KM               y        x

2GH = 2                             [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]

2   = 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

10.23

 

? S

?substitution

? simplificatio n

(3)

[20]

 

TOTAL MARKS

150

 

Last modified on Tuesday, 15 June 2021 07:13