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GRADE 12 MATHEMATICS PAPER 2 MEMORANDUM - AMENDED SENIOR CERTIFICATE PAST PAPERS AND MEMOS MAY/JUNE 2017

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MATHEMATICS PAPER 2
GRADE 12
SENIOR CERTIFICATE EXAMNATIONS
2017
MEMORANDUM

NOTE:

  • If a candidate answers a question TWICE, only mark the FIRST attempt.
  • If a candidate has crossed out an attempt of a question and not redone the question, mark the crossed out version.
  • Consistent accuracy applies in ALL aspects of the marking memorandum. Stop marking at the second calculation error.
  • Assuming answers/values in order to solve a problem is NOT acceptable.
  • Geometry:
    S = a mark for a correct statement (a statement mark is independent of a reason)
    R = a mark for a correct reason (a reason mark may only be awarded if the statement is correct)
    S/R = award a mark if statement and reason are both correct

QUESTION 1

TIME TAKEN (IN HOURS)  5  7 5 8 10 13 15 20 18 25 23
COST (IN THOUSANDS OF RANDS)  10 10 15 12 20 25 28 32 28 40 30

 

1.1  a = 4,806... = 4,81
b = 1,323... = 1,32
y = 4,81 + 1,32x		 ✓a = 4,81
✓b = 1,32
✓equation(3)
1.2  Cost = 25,974… = 25,97 thousand rand (calculator)
= R25 970
OR
y = 4,81 + 1,32(16)
y=25,93
Cost = R25 930		 ✓ 25,97
✓ answer (in Rands)
(2)
✓ substitution
✓ answer (in Rands)
(2)
1.3  r = 0,949…= 0,95

 ✓ answer
(1)
1.4 x = 0
y = 4,81 OR (4,80647)
∴ R4 810 OR R4806,47		 ✓ x = 0
✓ answer
(2)
[8]


QUESTION 2

2.1  modal class: 80 < x ≤ 100  ✓ correct class (1) 
2.2 
Commission earned
(in thousands of Rands) 		 Frequency  Cumulative Frequency 
20 < x ≤ 40  7 7
40 < x ≤ 60  6 13
60 < x ≤ 80  8 21
80 < x ≤ 100   10 31
100 < x ≤ 120   4 35
✓ 13 ; 21
✓ 31 ; 35 (2)
 2.3 2.3 ✓ grounded
✓ upper limits
✓ cum frequency
✓ shape
(4) 
2.4  No. of salesmen awarded bonuses: 35 – 26
= 9 salesmen 		 ✓ accept (25 – 27)
✓ accept (8 – 10)
(2) 
 2.5 Established Mean =(30x7) + (50x6) + (70x8) + (90x10) + (110x4)
                                                           35
= 68,86 thousand rand or R68 857,14
= R69 000 or 69 thousand rand		 ✓ top line using
midpts & freq
✓ 2410
✓ answer (nearest)
(3)
[12]


QUESTION 3

3

3.1 

mCD = -3 -(-5) 
             4 - 0
= -3+5 
    4 - 0
= ½ 

 ✓ substitution of C & D
✓answer
(2) 
3.2   mAD = -5 - 3
             0-(-4)
= -2
mCD x mAD = ½ x -2
=-1
∴AD⊥DC

✓ substitution of A & D
✓ mAD = -2
✓ product = –1
(3) 
3.3   3.3 ✓ correct substitution
✓ length of AB
✓ correct substitution
✓ length of BC
(4)
3 .4 

  mCD = mBF = ½     [BF ΙΙ DC]
4=½(3) + c
c =  5 
       2
y = ½x + 5 
               2
OR

y-4 = ½(x-3)
y-4=  ½x -1½
y= ½x+2 ½

 ✓ mBF =  ½

✓ substitution of B(3 ; 4)
✓equation
(3)
3.5  tan a = -2
∴a = 116.57º 
a = 90° + θ        [ext∠Δ]
∴θ = 26,57º
OR
tan a = -2 OR mAD = -2
∴ tan θ = ½
∴ θ = 26,57°
OR
Inclination of DE is β:
tan β = ½
∴ β = 26,57°
∴ODE = 63,43
∴θ = 90 ° - 63,43°
= 26,57°		 ✓tan  a = -2
✓a = 116.57º
✓θ = 26,57º
(3)

✓tan a = -2
✓tan θ = ½
✓θ = 26,57º
(3)

✓β = 26,57º
✓ODE = 63,43
✓θ = 26,57º
(3)
3.6   x2  + y2 = r2 
(4) 2 +(-3) 2 = 25
x2 + y2 = 25		 ✓ r2 = 25
✓ equation (2)
[17] 


QUESTION 4

4

4.1   4.1 ✓ substitution M & P
✓ x-value of N
✓ y-value of N
(3)
4.2   4.2 ✓ substitution N & P
✓ r = √13
✓ LHS of eq
✓ RHS of eq
(4)
✓ substitution N & M
✓ r = √13
✓ LHS of eq
✓ RHS of eq
(4)
4.3 4.3 ✓ correct substitution
✓ NMm
✓ MRm
✓ substitution of
MRm&(–3 ;–2)
✓ equation
(5)
4.4  Symmetry of a kite: S(–3 ; 4)
OR
PSM = 90º P [∠ in semi circle]
PS ⊥ SM
∴ S(–3 ; 4)
OR
(NS)2 = (radius)2 
(-3+1)2 + (y-1)2 = 13
(y-1)2 = 9
y-1 = ± 3
y=4 OR y≠-2
∴ S(–3 ; 4) 		 ✓ x-value of S
✓ y-value of S
(2)
✓ x-value of S
✓ y-value of S
(2)
✓ x-value of S
✓ y-value of S
(2)
4.5 

4.5

4.5b

 ✓ equating lengths
✓ simplification
✓ y-value of R
✓ x-value of R
(4)

✓ yR = 1
✓ horizontal line OR R lies on y = 1
✓ equating
✓ x-value of R (x < −4,6)
(4)

✓ y =  2  x + 6
          3
✓ equating
✓ x-value of R (x < −4,6)
✓ y-value of R
(4)
4.6 

4.6

4.6b

 4.6ans

4.6ans2
    [22]


QUESTION 5

5.1.1 

 tanA = sinA
            cosA
=  2p
    p
=2
OR
tanA =  2p
             p
=2

5.11

 ✓identity
✓value of tan A
(2)
✓   y 
     x
✓ value of tan A
(2) 
5.1.2  sin2 A + cos2 A = 1
(2p)2 + p2 = 1
4p2 + p2 = 1
5p2 = 1 
p2 =  1 
         5
∴p = -√ 1 
            5		 ✓ (2p)2 + p2 = 1
✓ simplification of LHS
✓ answer
(3)
5.2  2sin2 x -5sin + 2 = 0
(2sinx -1)(sinx-2) = 0
sinx = ½ or sinx =2(no solution)
ref ∠ = 30°
∴x= 30° + k360° or x=150° + k360°; k∈Z		 ✓ factors or formula
✓ both equations
✓ no solution/geen opl
✓30°+k.360.30°
✓ 150°+k360°;
✓ k∈Z
(6)
5.3.1  sin(x+300°) = sinxcos300° + cosxsin300° ✓ expansion (1) 
5.3.2

sin(x+300°) - cos(x-150°)
=sinx+300°+ cos x sin300° - (cosxcos-150°+sinxsin150°)
=sinx+60° - cos x sin60° - (cosxcos-30°+sinxsin30°)
=sinx+60° - cos x sin60° - cosxcos-30°- sinxsin30°
=½sinx-√3 cosx + √3  cosx-½sinx
              2               2
=0
OR
sin(x+300°) - cos(x-150°)
=sinx+300°+ cos x sin300° - (cosxcos-150°+sinxsin150°)
=sinx+60° - cos x sin60° - (cosxcos-30°+sinxsin30°)
=sinx+60° - cos x sin60° - cosxcos-30°- sinxsin30°
=sinx+30° - cos x sin60° - cosxcos-60°- sinxsin30°
=0

✓ 2nd expansion
✓✓ reduction
✓ special angle values
✓ answer
(5)

2nd expansion
✓✓  reduction
✓ co-ratios
✓ answer
(5)
5.4   5.4 ✓ identity of tan x
✓sinx + cosx
        cosx
✓sin 2x + cos 2x
        cosx
✓ sin 2x + cos 2x = 1
✓ simplify
(5)
✓ identity of tan x
✓sin 2x + cos 2x
        cosx
✓ sin 2x + cos 2x = 1
✓ simplify
✓ multiplication
(5) 
5.5.1   5.51 ✓ square both sides
✓ sin2x+cos2x=1
✓ sin 2x
(3)
5.5.2  From 5.5.1
sin x + cosx = √1+sin2x
∴ max value: sinx + cosx = √1+1
=√2
OR
Maximum value of 1 + sin 2x = 1 + 1
= 2
∴ maximum value of sinx + cosx = √2
OR
(sin x cos x)2 = sin2 x+2sinxcosx + cos2 x
=1+sin2x
∴ max value: (sinx + cosx)2 = 1+1 =2
∴ max value of sinx + cosx = √2

 ✓ max of sin 2x =1

✓ answer (2)
✓ max of sin 2x =1
✓ answer (2)
✓ max of sin 2x =1
✓ answer
(2)

 
    [27]


QUESTION 6

6

Related Items

6.1  Period = 180°  ✓ answer (1) 
6.2 °−75  ✓ answer (1) 
6.3   6.3 ✓cos45°.cosx+sin45°.sinx
✓ cos(x-45°)
✓✓ answer
(4)
    [6] 


QUESTION 7

7

7.1   KC = 6cm ✓ answer (1) 
7.2

Let P be the point of intersection of KL and CB
KP= sin60°
KC
KP=6sin60°
KP= 3√3or 5,20
∴ KL=8+3√3 or 13,20 cm

7.2

 ✓ trig ratio
✓ length of KP
✓ answer
(3)
7.3  DK2=62 + 122
DK = √180 or 6√5 or 13,42cm
sinKDL = sinDLK 
   KL            DK
sinKDL = KL
sinDLK    DK
= 8+3√3  or 13,20 or 0,98
     6√5        13,42		 ✓ DK=6√5
✓ use of sine rule
✓sinKDL = KL
   sinDLK    DK
✓ answer
(4) 
    [8]


QUESTION 8

8

8.1  L-100º [ext∠cyclic quad = int opp∠]
OR
N1=80° [∠s on straight line]
L = 100° [opp ∠s of cyclic quad] 		 ✓S ✓R
(2) 
✓S ✓R
(2)
8.2  N1=80° [∠s on straight line]
O1=160º [∠ at centre=2 × ∠ at circumference]
OR
refelx KOM=200º [∠ at centre=2 × ∠ at circumference]
O1=160º [∠s around a pt] 		 ✓S
✓S ✓R
(3)
✓S ✓R
✓S
(3)
8.3   M1 = 360º-(100º+55º+160º)
∴M1 = 45º		 ✓S
✓S
(2) [7]


QUESTION 9

9

9.1.1  ∠ in semi-circle  ✓answer (1)
9.1.2  Opp ∠s of quad = 180°  ✓answer (1)
9.2.1  OF ⊥ AC [line from centre bisects chord]
∴ AC || GO [co-interior∠s = 180°
OR/ corresp ∠s =] 		 ✓ S ✓ R
✓ R
(3)
9.2.2  G1 = A2 [corresp ∠s AC ΙΙ GO]
A2 = B1 [∠s in same segment]
∴G1 = B1 
OR
G1 = B2 [ext ∠cyclic quad]
but ΔABF = ΔCBF [s,∠,s]
∴B2 = B1
∴G1 = B1

✓ S ✓ R
✓ S ✓ R
(4)

✓ S ✓ R
✓ R
✓ S
(4)
9.3 OF:FB = 3:2  DB=2r
∴DO = 5k and DF =8k   OR  DF= 2r -  2 r =  8 r
                                                            5        5
∴DG = DO =  r       [line ΙΙ side of Δ]
  DA     DF    8 r 
                     5
∴DG =  5 
  DA     8		 ✓ S ✓ R
✓ S
(3)
    [12]


QUESTION 10

10

10.1  Tangent-chord theorem  ✓ R
(1) 
10.2.1

A2 + A3 =B1 + B2  [∠s opp = sides]
S3 = B1 + B2  [ext ∠ cyclic quad]
∴S3 = A2 + A3 
∴ AB ΙΙ ST  [corresp ∠2=]
OR
RTS = BAS [ext ∠ cyclic quad]
BAS = ABT [∠s opp = sides]
AB ΙΙ ST  [corresp ∠s]

 S ✓ R
✓ S ✓ R
✓ R
(5)
✓ S ✓ R
✓ S ✓ R
✓ R
(5)
10.2.2  B2 = x [tan chord theorem]
x + T4 = B1 + B2 [corresp ∠s AB ΙΙ ST]
T4 = B1 
B1 = A1 [tan chord theorem]
T4 = A1 		 ✓ S ✓ R
✓ S ✓ R
✓ R
(5)
10.2.3  T4 = A1 [proven in 10.2.2]
∴RTAP is a cyclic quadrilateral [line subtends =∠s ]
 		 ✓ S
✓ R
(2) 
    [13]


QUESTION 11

11

11.1  Constr: On sides AB and AC of ΔABC, mark points G and H respectively such that AG = DE and AH = DF. Draw GH
Proof:
ΔAGH = ΔDEF [s,∠,s]
∴ AGH = E
=B        [B=E given]
∴ GH ΙΙ BC [corresp ∠s =]
∴ AG = AH  [line ΙΙ side of Δ]
   AB    AC
∴ DE = DF  [constr]
   AB     AC		 ✓ construction
✓ S/R
✓ S
✓ S /R
✓ S ✓ R
(6) 
11.2.1(a) 11.2
AP = PC [diag ΙΙm bisect each other]
But TP = PS [given]
AP - TP = PC - PS
∴ AT = SC(		 ✓S
✓S OR
S
(2)
11.2.1(b)

In PSR and PBA:
P1 = P3 [vertically opp ∠s ]
B1 = S1 [∠s in same segment]
PSR ΙΙΙ PBA  [sum ∠s Δ]
∴ΔPSR ΙΙΙ PBA  [∠∠∠]

 ✓S ✓R
✓S ✓R
✓R
(5)
✓S ✓R
✓S ✓R
✓S
(5)
11.2.2(a) PR  = PS [ΙΙΙ Δs]
PA      PB
∴PR  = TR  = PS [Given PR  = TR  ]
  PA      AD     PB            PA      AD
∴PR  = TR  = TP [PS = TP;PB=PD]
  PA      AD     PD
∴ΔRPT ΙΙΙ ΔAPD [sides of Δ in prop]		 ✓ S (all 3
ratios)
✓ S
✓ R
(3) 
11.2.2(b) T1 = D2  [ΙΙΙ Δs]
∴ATRD is a cyclic quad [converse:ext ∠ of cyclic quad]		 ✓ S
✓ R
(2) 
    [18]


TOTAL: 150

Last modified on Tuesday, 29 June 2021 09:54
Published in Grade 12 Amended Senior Certificate Exam Past Papers and Memos 2017

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