PROVINCIAL EXAMINATION

JUNE 2024
GRADE 11

MATHEMATICS

(PAPER 1)

TIME: 2 hours

MARKS: 100

6 pages

P.T.O.
 MATHEMATICS 2
(PAPER 1) GRADE 11

INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.

1. This question paper consists of 6 questions.

2. Answer ALL the questions.

3. Present your answers according to the instructions of each question.

4. Clearly show ALL calculations, diagrams, graphs et cetera which were used in determining the
answers.

5. Answers only will NOT necessarily be awarded full marks.

6. Use an approved scientific calculator (non-programmable and non-graphical), unless stated

otherwise.

7. If necessary, answers should be rounded-off to TWO decimal places, unless stated otherwise.

8. Diagrams are NOT necessarily drawn to scale.

9. Number the answers according to the numbering system used in this question paper.

10. Write neatly and legibly.

P.T.O.
 MATHEMATICS 3
(PAPER 1) GRADE 11

QUESTION 1

x2 + 4
1.1 Determine the value(s) of x for which the expression:
9− x

1.1.1 has real value(s). (1)

1.1.2 is not defined. (1)

1.1.3 Given, x =1, is the expression rational or irrational?

Validate your answer with an appropriate calculation. (1)

1.1.4 Determine a value for x where the expression will yield a recurring decimal
(1)
value?

1.2 Given: 2 x 2 − 3x − k = 0 . Determine x if:

1.2.1 k =5 (2)

1.2.2 k = 4 (correct to TWO decimal places) (3)

1.3 Solve for x :

1.3.1 x−4 = − x−2 (5)

1.3.2 x2 + 4 > 3x + 2 (4)

1.4 Solve simultaneously for x and y.

2 x = y + 2 and 2 x 2 = 2 − y 2 (6)

1.5 For which values of p will − 2 x 2 + 4 x − 3 = − p :

1.5.1 have two unequal roots? (4)

1.5.2 have roots that are BOTH positive? (2)

1.6 Show that the equation ( p 2 + 1) x 2 = −2 pqx − q 2 has no real roots for
p and q  Real numbers and q  0. (5)
[35]

P.T.O.
 MATHEMATICS 4
(PAPER 1) GRADE 11

QUESTION 2
2.1 Simplify WITHOUT using a calculator:

2.1.1 (2 −1 + 3−1 ) 2
(3)

2.4 x + 2 − 4 x −3
2.1.2
2 x.2 x (3)
1
−
2.1.3 3 2
[ 12 + 3 3 3 ] (4)

2.2 Solve for x WITHOUT using a calculator:

3
2.2.1 2 x = 164
(3)

2.2.2 5 x +1 + 5 x = 150 (4)

9 − 54 3 2 −2 3
2.3 WITHOUT using a calculator, show that is equal to .
6 2 4 (3)
[20]
QUESTION 3
Given the number pattern:
4 ; − 3 ; − 10 ; … ; − 227

3.1 Write down the value of the 4 th term of this pattern. (1)

3.2 Determine the general term of the number pattern in the form Tn = an + b . (1)

3.3 Calculate the number of terms of the pattern. (2)

3.4 The given number pattern above is also the FIRST differences of a quadratic number
pattern, Tn = an2 + bn + c.
Determine if the quadratic pattern will have a maximum or minimum value.
Validate your answer with an appropriate calculation. (2)
[6]
QUESTION 4

Given the quadratic number pattern:

69 ; 0 − 63 ; …

4.1 Write down the value of the next term in the pattern. (1)

4.2 Calculate an expression for the nth term of the quadratic pattern. (4)

4.3 Determine the value of the SMALLEST term in this pattern. (4)
[9]
P.T.O.
 MATHEMATICS 5
(PAPER 1) GRADE 11

QUESTION 5

Sketched below are the graphs of f ( x) = ax 2 + bx + c and g ( x) = mx + k.

• The x-intercepts of f are at points A and C.
• The graphs of f and g intersect at points A and B respectively.
• Point P is the turning point of f.
• Q is a point on g such that line PQ is parallel to the y-axis.

y
g

B(0;8)

A(–4;0) C(2;0) x
0

5.1 Determine the equation of g . (3)

5.2 Calculate the values of a, b and c in f ( x) = ax2 + bx + c. (5)

5.3 Show that the coordinates of P are (−1 ; 9). (2)

5.4 Determine the range of f. (1)

5.5 Determine the equation of a line p perpendicular to g passing through point C. (3)

5.6 Calculate the length of line PQ. (2)

f ( x)
5.7 For which value(s) of x is  0?
g ( x) (2)
[18]
P.T.O.
 MATHEMATICS 6
(PAPER 1) GRAAD 11

QUESTION 6

−4
Given: k ( x) = + 1 and h( x) = 2 − x − 4
x+3

6.1 Write down the equations of the asymptotes of k . (2)

6.2 Determine the x and y-intercepts of k. (2)

6.3 Write down the equation of the asymptote of h. (1)

6.4 Sketch the graph of k and h on the same axes. Clearly indicate ALL intercepts with the
axes as well as the asymptotes. (4)

6.5 If p( x) = 3h( x) , write down the equation of the asymptote of p. (1)

6.6 Write down the equation of q, the reflection of h in the y-axis. (1)

6.7 Determine the x-values for which h( x) − q( x) = 0 (1)

[12]

TOTAL: 100

END
 PROVINCIAL EXAMINATION
JUNE 2024
GRADE 11
MARKING GUIDELINES

MATHEMATICS (PAPER 1)

11 pages
 MATHEMATICS
MARKING GUIDELINES
(PAPER 1) GRADE 11

INSTRUCTIONS AND INFORMATION

A – ACCURACY
CA. – CONTINUED ACCURACY

NOTE:

• If a candidate answerd a question TWICE, mark only the first attempt.

• If a candidate crossed OUT an answer and did not redo it, mark the crossed-out answer.
• Consistent accuracy applies to ALL aspects of the marking guidelines.
• Assuming values/answers in order to solve a problem is UNACCEPTABLE.

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 MATHEMATICS
MARKING GUIDELINES
(PAPER 1) GRADE 11

QUESTION 1
1.1 1.1.1 x9 ✓ answer
(1)

1.1.2 x=9 ✓ answer

(1)

1.1.3 (1) 2 + 4
9 −1
5 ✓ answer
8
 Irrational (1)

1.1.4 x=0 ✓ answer

(1)

1.2 1.2.1 2 x 2 − 3x − k = 0
2 x 2 − 3x − 5 = 0
(2 x − 5)( x + 1) = 0
✓ factors
5
x = of x = −1 ✓ answer
2
(2)

1.2.2 2 x 2 − 3x − 4 = 0
−(−3)  (−3) 2 − 4(2)(−4) ✓ substitution
x=
2(2)
 x = 2,35 of x = −0,85
✓✓ answer
NOTE: Penalise 1 mark for incorrect
rounding in this question ONLY. (3)

1.3 1.3.1 x−4= − x−2

16 − 8 x − x 2 = x − 2 ✓ square both sides
x 2 − 9 x + 18 = 0 ✓ standard form
( x − 6)( x − 3) = 0
✓ factors
x = 3 of x  6
✓ critical values
✓ rejection (5)

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 MATHEMATICS
MARKING GUIDELINES
(PAPER 1) GRADE 11

1.3.2 x 2 + 4  3x + 2
x 2 − 3x + 2  0 ✓ standard form
( x − 2)( x − 1)  0 ✓ factors
x  1 or x  2
OR
✓✓ answers
x  (−;1)  (2; )
(4)

1.4 y + 2 = 2x
 y = 2x − 2 (1)
✓ y as subject
2x2 = 2 − y 2 (2)
 2 x 2 = 2 − (2 x − 2) 2
✓ substitution
2 x 2 = 2 − (4 x 2 − 8 x + 4)
2 x2 = 2 − 4 x2 + 8x − 4
 6x2 − 8x + 2 = 0
 3x 2 − 4 x + 1 = 0 ✓ standard form
(3 x − 1)( x − 1) = 0
1 ✓ factors
x = of x = 1
3 ✓ both x- values
4
y =− of y = 0
3 ✓ both y- values

OR
y + 2 = 2x OR
y+2
x = (1)
2 ✓ x as subject
2x2 = 2 − y 2 (2)
y+2 2
 2( ) = 2 − y2 ✓ substitution
2
y2 + 4 y + 4
2( ) = 2 − y2
4
y + 4y + 4
2
( ) = 2 − y2
2
 y + 4 y + 4 = 4 − 2 y2
2

3y2 + 4 y = 0 ✓ standard form

y (3 y + 4) = 0
✓ factors
4
y =0 of y=−
3 ✓ both y-values
1
x =1 of x= ✓ both x-values (6)
3
1.5 1.5.1 For unequal roots:   0 ✓ condition of 
4
 MATHEMATICS
MARKING GUIDELINES
(PAPER 1) GRADE 11

 2 x 2 − 4 x + (3 − p) = 0
 = b 2 − 4ac
  = (−4) 2 − 4(2)(3 − p) ✓ substitution into 
 = 16 − 24 + 8 p (4)
  = −8 + 8 p ✓ expression for 
 −8 + 8 p  0
8p  8
 p 1
✓ answer

1.5.2 From 1.5.1

3 − p  0 (a  0 and  b  0) ✓ c0
 − p  −3
✓ answer
1  p  3
NOTE: Any other valid method (2)

1.6 ( p 2 + 1) x 2 = −2 pqx − q 2
✓ standard form
 ( p 2 + 1) x 2 + 2 pqx + q 2 = 0
 = (2 pq) 2 − 4( p 2 + 1)q 2 ✓ substitute into 
 = 4 p q − 4 p q − 4q
2 2 2 2 2
✓ simplification
  = −4q 2 ✓ expression for 
 q 2  0, q  ; q  0
 −4q 2  0 ✓ explanation
 Roots non-real (5)
[35]

5
 MATHEMATICS
MARKING GUIDELINES
(PAPER 1) GRADE 11

QUESTION 2
2.1 2.1.1 (2 −1 + 3−1 ) 2
1 1 ✓ converting exponents
= ( + )2
2 3
3+ 2
=( )
6
5 ✓ simplification
= ( )2
6
25
= ✓ answer
36 (3)

2.1.2 2 .4 x + 2 − 4 x − 3
2 x .2 x
2 .4 x .4 2 − 4 x .4 − 3
=
4x
1
4 x (32 − )
64 ✓✓ factors
= x
4
1
= 32 −
64
2 048 − 1
=
64
2 047
= ✓ answer
64
63
NOTE: Accept 31 (3)
64

1
2.1.3 −
3 2 [ 12 + 3 (3 3)]
1 1
−
= 3 2 [2 3 + (3 3) 3 ] ✓ simplify brackets
1 1 1 1
−
= 3 [2.3 + 3 .3 ]
2 2 3 6 ✓ surds as rational
1 1 1
− exponents
= 3 [2.3 + 3 ]
2 2 2

1 1
−
= 3 .3 (2 + 1)
2 2
✓ factors
= 30 (3)
=3
✓ answer (4)

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 MATHEMATICS
MARKING GUIDELINES
(PAPER 1) GRADE 11

2.2.1 3
2 x = 16
4

3
x4 = 8
3 4 4

x4 3
= 83 ✓ method

 x = (3 8 ) 4 ✓ x as subject
 x = 16 ✓ answer
(3)
NOTE: Any other valid method.

2.2.2 5 x +1 + 5 x = 150
5 x.51 + 5 x = 150
 5 x (51 + 1) = 150 ✓ factors
 5 (6) = 150
x

5 x = 25 ✓ simplification
5 = 5
x 2
✓ 25 as base of 5
x = 2
✓ answer (4)
NOTE: Any other valid method.

2.3 9 − 54
6 2
(9 − 9.2.3 ) 2
=  ✓ 
2
6 2 2 2
9 2 − 3 2 . 3. 2
=
6.2
9 2 − 3.2 3 ✓ simplification
=
12
3(3 2 − 2 3 ) ✓ factorisation
=
12
3 2 −2 3
=
4 (3)
[20]

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 MATHEMATICS
MARKING GUIDELINES
(PAPER 1) GRADE 11

QUESTION 3

3.1 T4 = −17 ✓ answer (1)

3.2 Tn = −7n + 11 ✓ answer (1)

3.3 Tn = −7n + 11
 −227 = −7n + 11 ✓ substitute Tn
− 238 = −7n
✓ answer
 n = 34 (2)

3.4 2a = −7
7 ✓ value for a
a=−
2
✓ conclusion
 maximum
(2)
NOTE: Any other valid method.
[6]

QUESTION 4

4.1 69 0 − 63 T4
− 69 − 63 − 57
6 6
next term (T4 ) is: − 120 ✓ answer (1)

4.2 Tn = an2 + bn + c
T2 = a(2) 2 + b(2) + c = 4a + 2b + c
T1 = a(1) 2 + b(1) + c = a + b + c
T2 − T1 = 3a + b
2a = 6
✓ 2 nd difference
a = 3
✓ value for a
3a + b = −69
3(3) + b = −69
9 + b = −69
 b = −78 ✓ value for b
a + b + c = 69
3 + (−78) + c = 69 ✓ value for c
 c = 144
Tn = 3n 2 − 78n + 144 (4)

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 MATHEMATICS
MARKING GUIDELINES
(PAPER 1) GRADE 11

4.3 Smallest term (minimum) is at the turning point of the

quadratic equation:
− b − (−78) 78 ✓ substitute into A.O.S
n = = = = 13
2a 2(3) 6 ✓ value for n
Tn = 3n − 78n + 144
2

T13 = 3n 2 − 78n + 144

T13 = 3(13) 2 − 78(13) + 144
✓ substitution
T13 = −363
✓ answer (4)
[9]

QUESTION 5
5.1 The points A( − 4 ; 0) and B(0 ; 8) lie on g.

8−0 ✓ substitute for mAB

m AB =
0 − (−4)
✓ value of mAB
m AB =2
y = mx + c
 8 = 2(0) + c  A(0;8) ✓ value of c
c = 8
 g ( x) = 2 x + 8
NOTE: Any other valid method. (3)

5.2 f ( x) = a( x − x1 )( x − x2 )
f ( x) = a( x + 4)( x − 2) ✓ substitute points A
subst (0;8) and C
8 = a (0 + 4)(0 − 2)
8 = a (−8)
a = −1 ✓ value of a
f ( x) = −( x + 4)( x − 2)
f ( x) = −( x 2 + 2 x − 8)
f ( x) = − x 2 − 2 x + 8
✓ standard form (A)
 a = −1
b = −2 ✓ value of b
c =8 ✓ value of c (5)

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 MATHEMATICS
MARKING GUIDELINES
(PAPER 1) GRADE 11

5.3 f ( x) = − x 2 − 2 x + 8
− b − (−2) ✓ correct substitution
x = =
2a 2(−1)
 x = −1
 f (−1) = −(−1) 2 − 2(−1) + 8 ✓ correct substitution
 f (−1) = 9
NOTE: Any other valid method. (2)

5.4 y9 ✓ answer (1)

5.5 mg = 2
1 ✓
 mp = − mp
2
y − y1 = m( x − x1 )
1 ✓ substitute point C
 y − 0 = − ( x − 2)
2
1
 y = − x +1
2 ✓ answer (3)

5.6 f (−1) = 9 ✓ f (−1) = 9 and

g (−1) = 6 g (−1) = 6
 f (−1) − g (−1)
 PQ = 9 − 6
 PQ = 3 ✓ answer
OR
OR

DP(−1;9) ✓ using TP and g (−1)

g (−1) = 6
 PQ = 9 − 6 ✓ answer
 PQ = 3 (2)

5.7 x  2 ; x  −4 ✓✓ answer (A) (2)

[18]

QUESTION 6
6.1 x = −3 ✓ answer
y =1 ✓ answer
(2)

10
 MATHEMATICS
MARKING GUIDELINES
(PAPER 1) GRADE 11

6.2 −4
f ( x) = +1
x+3
−4
0= +1
x+3
−4
−1 =
x+3
− x − 3 = −4
− x = −1
x =1
✓ x-intercept
−4
y= +1
0+3
−4
y= +1
3
1 ✓ y- intercept
y =−
3 (2)

6.3 y = −4 ✓ answer (1)

6.4

✓ shape of k

✓ k and h intercepts

✓ asymptotes k and h

✓ shape of h

(4)

6.5 p ( x) = 3h( x)
 p( x) = 3(2 − x − 4)
 p( x) = 3.2 − x − 12
asymptote: y = −12 ✓ answer
NOTE: Answer only, full marks. (1)

6.6 q( x) = 2 x − 4 ✓ answer (1)

6.7 x=0 ✓ answer (1)

[12]

TOTAL: 100
11