PROVINCIAL EXAMINATION

JUNE 2023
GRADE 11
MARKING GUIDELINES

MATHEMATICS (PAPER 1)

10 pages
 MATHEMATICS
MARKING GUIDELINES
(PAPER 1)

QUESTION 1

1.1 1.1.1 x {4 ; 5}  answer

 answer (2)

1.1.2 x {0 ; 3}  answer

 answer (2)

1.1.3 x {1 ; 2}  answer

 an wer (2)

1.2 1.2.1 3x2  4x  0

 x(3x  4)  0  factors
x 4
 0or  x   answers
3
NOTE: Any other valid method. (3)

1.2.2 3x – 14 = –6x2
 standard form
 6x2 + 3x – 14 = 0

 (3)  (3)2  4(6)(1 )

 x =  substitution
2(6)

 x = 1,29 or x = 1,79  answers (4)

1.2.3 ( x  1)( x  3)  12
x 2  2x  3  12
x 2  2x  15  0  standard form
( x  5)( x  3)  0 -3 5  factors

 x  5 or  x  3  answers
(4)

1.2.4 x2
2x x2
( 2  x )2  (x  2)2  squaring both sides
2  x  x2  4x  4
0  x2  3x  2  standard form
0 = (x – 2)(x – 1)  factors
 x = 2 ... or ... x = 1(NA)  answers with rejection (4)

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 MATHEMATICS
MARKING GUIDELINES
(PAPER 1)

1.2.5 x60
x6  value of x
x2y
362y  substitution
3
2y3
 3  value of y
y (3)
2

1.3 y – 1 = 2x
 y = 2x + 1 ………(1)  expr ssion r y
x2 + xy – 3x – y + 2 = 0 ………(2)
 x2 + x(2x + 1) – 3x – (2x + 1) + 2 = 0  substitution
x2 + 2x2 + x – 3x – 2x – 1 + 2 = 0
 3x2 – 4x + 1 = 0  s ndard form

(3x – 1)(x – 1) = 0
1
 x = ... or ... x = 1  -values
3
5
 y = ... or ... y = 3  y-values
3 (5)

1.4 x2  px  p2  2
 x2  px  p2  2  0
  b2  4ac
  ( p)2  4(1 ( p2  2)  substitute into 
  p2  4 p  8
 5 p2  8
 p2  0 p     5 p2  8
5 p 0 p 
 p2  0 , 5 p2  0 and
5 p  8  0
2
 5 p2  8  0

 roots are real and unequal. (3)

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 MATHEMATICS
MARKING GUIDELINES
(PAPER 1)

1.5 The 100 m fencing will make up three sides because

the existing wall will make one side.

Wall

fencing = 2x + y = 100
 y = –2x + 100
 A  xy  expression for y
A = x(–2x + 100)
A = –2x2 + 100x  expression or A
2
A = –2(x – 50x)
 A = –2(x2 – 50x – 625 – 625)  c mplete square
A = –2(x – 25)2 –625)

Maximum area: x  25  value of x

A  2(25)2  100(25)
 A  1250  value of A
1250  25 y
 y  50  value of y (7)
[39]

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 MATHEMATICS
MARKING GUIDELINES
(PAPER 1)

QUESTION 2

2.1 2.1.1
3n2.9n1
27n1
 32n2 and 33n3
3n2.32n2
33n3

33n4
33n3  33n 43n3
(3)
33n43n3  answer

37

2.1.2 x2
1x

(1  3)2  substituti n

11 3

12 33  simplifi ation

 2 3

42 3
 2 3

2(2  3)2  factorisation

 2 3

=2  answer (4)
5
2.1.3 (a2  b )  (a  b)
1
(a b) 2
5

(a  b)(a  b) 1 (a  b) 2
 (a  b) 2  factorisation
1 1 5

(a  b) 2 a  b) 2 1 (a  b) 2
 (a  b) 2  simplification
1 5
(a  b) 2  (a  b) 2
 simplification
(a  b) 3

a3  3a2b  3ab2  b3  answer (4)

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 MATHEMATICS
MARKING GUIDELINES
(PAPER 1)

2 8
2.2 RTP: – = 2
1 2 8

2 8
1 2 – 8

2 8  2 2
1 2 – 2 2

2(2 2 )  8(1  2 )
 2 2(1  2 )
 2 2(1 2)

4 288 2
 2 22 4

4 288 2
 2 24

4 28
 2 24
 simplificati n

4 22
 2 22
 factor s ion
 2
(4)

2.3

MK  LM   MK= 2
JM 2  (2  3)  22 pythag  substitution
JM  4  4 3  3  4  simplification
JM 2  3  3

JM  4 3  3  JM

1
A  (4) 4 3  3
 substitution into area formula
2
A  6,3 units2  answer (6)
[21]

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 MATHEMATICS
MARKING GUIDELINES
(PAPER 1)

QUESTION 3

3.1 T2  T1  T3  T2
4x  5  x  10x  5  (4x  5)  method
3x  5  10x  5  4x  5
3x  5  6x  10
 3x  15
x5  answer (2)

3.2 3.2.1  3n  20  106  equat ng

3n  126
 n  42 answer (2)

3.2.2  3n  20  0  T 0
20  3n
20
 n
3
n7  answ r (2)

3.2.3 Odd valued terms:

17; 11; 5; ………..  sequence
General term: Tn  6n  23  Tn
Tn  6(20)  23
Tn  97  answer (3)

3.3 3; a; 10; b; 21
 1st differences
a  3; 10  a; b 10; 21 b st
1 difference

nd
differences:

10  a  a  3  1  equating 2nd difference in

 2a  13  1 terms of a, then b.
 2a  12
a6
 value of a
and
21  b  b  10  1
 2b  31  1
 2b  30
b  15 (4)
 value of b
[13]

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 MATHEMATICS
MARKING GUIDELINES
(PAPER 1)

QUESTION 4

D(6 ; 7)

C(–1 ; 0) B
O x

A(2 ; y)

4.1 B(5; 0)  answer

NOTE: Must be in coordinate form. (2)

4.2 y  a(x  x1)(x  x2 )

7  a(6  (1))(6  5)  substitute roots and
7  7a point D(6;7)
a1  value for a
 y  1(x 1)(x 5)
y  x2  5x    y  x2  5x  x  5
 y  x2  4x  5
(3)

4.3 B(5 ; 0 C(0 ;  5)

50
mBC  0  5
5
mBC 
5  mBC
 mBC  1
 mh  1  mh
 y  y1  m(x  x1)
y  (1)  1(x  0) pt(1;0)
 y  x 1
 answer (3)

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 MATHEMATICS
MARKING GUIDELINES
(PAPER 1)

4.4 1x0  all critical values

(independent)
x5  answers

OR

x  [–1 ; 0] or [5 ; ]  all critical values

NOTE: Deduct 1 mark if brackets are incorrect in the (independent)
alternative solution.  answers (2)
[10]

QUESTION 5

5.1

 a ymptotes
2
 in ercepts
 hape

NOTE: If the candidate ca culates the intercepts and lists

the asymp otes but does not sketch the graph, award
2 mark (3)

5.2 tan 135 = –1  m  1

y – y1 = m(x – x1
y – (–1) = –1(x – 2))  subs. point (2;–1)
+ 1 = –x – 2
y = –x –  answer (3)

5.3 x  2  answer (1)

5.4 5.4.1 5 units right  answer

NOTE: Accept an answer of 5 units. (1)

5.4.2 3 units up  answer

NOTE: Accept an answer of 3 units. (1)

[9]

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 MATHEMATICS
MARKING GUIDELINES
(PAPER 1)

QUESTION 6

6.1 6.1.1 y  6  answer (1)

6.1.2 h(x)  3.2x  6

0  3.2x  6  equate to 0
6  3.2x
2  2x
x1  answer (2)

6.1.3 h(x)  3.2x  6

y = 3.2 – 6
y = –3  a swer (1)

6.1.4 x >1  answer (1)

6.2 y

 x-intercept
x
O  asymptote

–1  shape

(3)
[8]

TOTAL: 100

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