NATIONAL
SENIOR CERTIFICATE

GRADE 12

MATHEMATICS P1
NOVEMBER 2010
MEMORANDUM

MARKS: 150

This memorandum consists of 27 pages.

Copyright reserved

Please turn over

Mathematics/P1

2
NSC – Memorandum

DBE/November 2010

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.
QUESTION 1
1.1.1

(3 − x)(5 − x) = 3
15 − 8 x + x 2 = 3
x 2 − 8 x + 12 = 0

Note:
If answer only : Full Marks

( x − 6)( x − 2) = 0
x = 6 or x = 2

If the candidate makes it a
linear equation, no marks

OR
(3 − x)(5 − x) = 3

For only 1 answer: 1 / 3

15 − 8 x + x 2 = 3

− (−2) ± (−2) 2 − 4(3)(−4)
x=
2(3)
2 ± 52
6
x = 1,54 or – 0,87
=

answers
(3)

completed
square form
answers

( x − 4) 2 = 4
x – 4 = 2 or x – 4 = – 2
x = 6 or x = 2
3x 2 = 2( x + 2)
3x 2 − 2 x − 4 = 0

factors

expansion

x 2 − 8 x + 12 = 0

1.1.2

expansion

(3)
Note:
No penalty for incorrect
rounding off of answers.

standard form
substitution

Substitution into incorrect
formula, no marks
answers
(4)

OR
3x 2 = 2( x + 2)

3x 2 − 2 x − 4 = 0
2
4
x2 − x =
3
3

expansion

2

1⎞
4 1
⎛
⎜x − ⎟ = +
3⎠
3 9
⎝
2

1⎞
13
⎛
⎜x − ⎟ =
3⎠
9
⎝
1
13
=±
3
3
1 ± 13
x=
3
x = 1,54 or – 0,87

completed
square

x−

Copyright reserved

13
3
answers
±

(4)
Please turn over

Mathematics/P1

1.1.3

3
NSC – Memorandum

4 + 5x > 6 x 2

DBE/November 2010

− 6 x 2 + 5x + 4 > 0

0 > 6 x 2 − 5x − 4

OR

0 > (3 x − 4)(2 x + 1)

6 x 2 − 5x − 4 < 0
(3 x − 4)(2 x + 1) < 0

5 ± 121
12
1
4
x= −
or
2
3

critical values: x =

+

0

−

1
4
<x<
2
3

+

4
3

1
–
2

−

0

factors
critical values
1
4
− and
2
3
OR

–

⎛ 1 4⎞
x ∈⎜− ; ⎟
⎝ 2 3⎠

OR

correct
inequality

1
2

OR

4
3

−

4
1
< x and x <
3
2

answer
(4)

OR

− 6 x 2 + 5x + 4 > 0
(−3 x + 4)(2 x + 1) > 0
1
4
critical values: − and
2
3
0

−

+

1
–
2

0
4
3

9 x 2 − 4 x 2 + 18 x − 6 x = 9

factors
critical values
1
4
− and
2
3

OR

1
4
⎛ 1
<x<
x ∈⎜− ;
OR
2
3
⎝ 2
3 y = 2x
2x
y=
3
2
⎛ 2x ⎞
⎛ 2x ⎞
2
x − ⎜ ⎟ + 2x − ⎜ ⎟ = 1
⎝ 3 ⎠
⎝ 3 ⎠
OR
2
4x
2x
2
x −
+ 2x −
=1
3
9
−

1.2

−

correct
inequality

–

4⎞
⎟
3⎠

OR

x=

3
or x = −3
5

4
3

−

4
1
< x and x <
3
2

answer
(4)
2x
3
substitution
y=

2

⎛ 2x ⎞
⎛ 2x ⎞
x − ⎜ ⎟ + 2x − ⎜ ⎟ = 1
⎝ 3 ⎠
⎝ 3 ⎠
2
5x
4x
+
−1 = 0
9
3
2

x=

5 x 2 + 12 x − 9 = 0
(5 x − 3)( x + 3) = 0

1
2

=

− 43 ±
− 43 ±

( 43 )2 − 4( 59 )(−1)
2( 59 )
16
9

+ 209

10
9

x = 0,6 or x = −3

simplification
standard forms
factors or
substitution
into correct
formula

x-answers
y-answers

Copyright Reserved

Please turn over

Mathematics/P1

4
NSC – Memorandum

2
or y = – 2
5
⎛3 2⎞
(x ; y) = ⎜ ; ⎟ or (−3 ; −2)
⎝5 5⎠
OR
3 y = 2x

y=

x 2 − y 2 + 2x − y = 1
4 x 2 − 4 y 2 + 8x − 4 y = 4
(2 x) 2 − 4 y 2 + 8 x − 4 y = 4

DBE/November 2010

(7)

y = 0,4 or y = – 2

Note:
If mathematical
breakdown
eg. if y = 2x – 3 is used,
max 3 / 7

simplification of
original quadratic
substitution
2x = 3y

(3 y )2 − 4 y 2 + 4(3 y ) − 4 y = 4

simplification

9y2 − 4y2 + 8y = 4

standard form

5y2 + 8y − 4 = 0
(5 y − 2)( y + 2) = 0
2
y=
or y = – 2
5
3
or x = −3
x=
5
⎛3 2⎞
(x ; y) = ⎜ ; ⎟ or (−3 ; −2)
⎝5 5⎠

factors or
substitution
into correct
formula
y-answers
x-answers
(7)

OR

3y = 2x
3y
x=
2
2

⎛ 3y ⎞
⎛ 3y ⎞
2
⎜ ⎟ − y + 2⎜ ⎟ − y = 1
⎝ 2 ⎠
⎝ 2 ⎠
2
9y
− y2 + 3y − y = 1
4
9y2 − 4y2 + 8y = 4
5y2 + 8y − 4 = 0
(5 y − 2)( y + 2) = 0
2
y=
or y = – 2
5
3
x=
or x = −3
5
⎛3 2⎞
(x ; y) = ⎜ ; ⎟ or (−3 ; −2)
⎝5 5⎠

Copyright Reserved

x=

3y
2

substitution
simplification
standard forms
factors or
substitution
into correct
formula
y-answers
x-answers
(7)

Please turn over

Mathematics/P1

1.3

5
NSC – Memorandum

5 2007 + 5 2010
5 2008 + 5 2009
52007 + 52007.53
= 2008
5 + 52008.5
52007 (1 + 53 )
= 2008
5 (1 + 5)
126
=
5× 6
126
=
30
21
=
5
1
=4
5
≈4

DBE/November 2010

Note: If the candidate
leaves the answer as 4,2
max 2 / 3 marks

52007 + 52007.53
52008.5 + 52008.52

Answer only of 4,2 0 / 3
marks

simplification to
1 + 53
126
or
or
2
5+5
30
21
5

answer = 4
(3)

OR

5 2007 + 5 2007.53
5 2007.5 + 52007.5 2

5 2007 + 5 2010
(divide each term by 5 2007 )
5 2008 + 5 2009
5 2007 + 5 2007.53
= 2007
5 .5 + 5 2007.52
1 + 53
=
5 + 52
126
=
30
≈4

simplification to
1 + 53
126
or
or
2
5+5
30
21
5
answer = 4

OR
let x = 2009

let x = 2007

let x = 2010

5 x − 2 + 5 x +1
5 x −1 + 5 x
5 x (5− 2 + 5)
= x −1
5 (5 + 1)
1
+5
= 25
1
+1
5
21
=
5
1
=4
5
≈4

5x + 5x +3
5 x +1 + 5 x + 2
5 x (1 + 53 )
= x
5 (5 + 52 )
1 + 125
=
5 + 25
OR
126
=
30
21
=
5
1
=4
5
≈4

5x −3 + 5x
5 x − 2 + 5 x −1
5 x (5− 3 + 1)
= x −2
5 (5 + 5−1 )

Copyright Reserved

OR

=

1
125
1
25

+1
+

126
30
21
=
5
1
=4
5
≈4
=

1
5

5 x − 2 + 5 x +1
or
5 x −1 + 5 x
5 x + 5 x +3
or
5 x +1 + 5 x + 2
5 x −3 + 5 x
5 x − 2 + 5 x −1
simplification to
1
+5
25
or
1
+1
5
1 + 125
or
5 + 25
1
125 + 1
1
1
25 + 5
answer = 4
(3)
[21]
Please turn over

Mathematics/P1

6
NSC – Memorandum

DBE/November 2010

QUESTION 2

2.1

20

∑3

n−2

n =1

Note: The mark for n = 20
can be implied in the
substitution to the formula

1
= + 1 + 3 + ... to 20 terms
3
1 20
3 −1
3
=
; r = 3; n = 20
3 −1
320 − 1
=
6

(

)

= 581130733,33

OR

581130733

1
3

OR

581130733,3

1
3
r=3
n = 20
a=

answer
(4)

Note: If leave only as
1
+ 1 + 3 + 9 + 27 + 81 + 243 + 729 + 2187 + 6561 + 19683
3
+ 59049 + 177147 + 531441 + 1594323 + 4782969
+ 14348907 + 43046721 + 129140163 + 387420489
only, then 2 / 4

Note: The 20th term is 387 420 489
Answer only: 3 / 4 marks

2.2.1

5x ; x 2 ;

x3
;…
5

x
r=
5

x
<1
5
−5< x < 5
−1 <

Note: If − 1 < x < 1

1 mark

Note: If answer is − 5 ≤ x ≤ 5
then 2 / 3

x
x2
or
5x
5
−1 < r < 1

r=

answer
(3)

Answer can be written as x ∈ (−5 ; 5)

2.2.2

2
and a = 10
5
10
S∞ =
2
1−
5
50
=
or 16,67
3

r=

Copyright Reserved

a = 10

answer
(2)

Please turn over

Mathematics/P1

2.3.1

7
NSC – Memorandum

Tn = 20 + 3(n − 1)

DBE/November 2010

Note: If
17
n=−
3
Then 1 / 2 marks

101 = 20 + (n − 1)3
84 = 3n
n = 28

Answer only:
Full marks

OR

101 = 20 + 3(n − 1)
or 101 = 3n + 17
answer
(2)

Tn = 3n + 17
101 = 3n + 17
84 = 3n

substitution
answer

n = 28

(2)
2.3.2

23 + 29 + ... to 14 terms

14
[2(23) + (14 − 1)6] OR
2
= 868
=

14
[23 + 101]
2

Note: If “to 14
terms” is left out, do
not penalise
Note: If incorrect
value for n, max 4 / 6

OR
Even numbers = 20 ; 26 ;…; 98
Tn = 6n + 14
Tn = 20 + (n − 1)6
98 = 6n + 14
84 = 6n
14 = n

S remaining

OR

OR

Note: If incorrect
formula, max 2 / 6

98 = 20 + (n − 1)6
84 = 6n

14 = n
28
14
= [2(20) + (27)(3)] − [2(20) + (13)(6)]
2
2
= 14(121) − 7(118)
= 1694 − 826
= 868
Note:
If the candidate only works
out the even numbers i.e.
826, then 3 / 6 marks

23 + 29 + ...
a = 23
n = 14
d=6
or l = 101
substitution into
correct formula
answer
(6)
OR
98 = 6n + 14 or
98 = 20 + (n − 1)
14 = n

substitution into
correct formula
1694
826
answer
(6)

If only 1694 max 1 / 6 marks

Sequence is
20; 23; 26; 29; 32; 35; 38; 41; 44; 47; 50; 53; 56; 59; 62; 65; 68; 71; 74;
77; 80; 83; 86; 89; 92; 95; 98; 101
Full marks
Sum of odd numbers
= 23 + 29 + 35 + 41 + 47 + 53 + 59 + 65 + 71 + 77 + 83 + 89 + 95 + 101
= 868

Copyright Reserved

Please turn over

(6)
[17]

Mathematics/P1

8
NSC – Memorandum

DBE/November 2010

QUESTION 3

3.1

4

9
x−9

5

37

x

x − 14
First difference : 5; x − 9; 37 − x
Second difference : x – 14; − 2x + 46
x − 14 = 46 − 2 x
3 x = 60
x = 20

37 − x
37 − x − (x − 9)

first
differences
5; x − 9;
37 − x
seconds
difference
answer

Note:
Answer only:
Full Marks

(3)

OR
x + ( x − 9) + ( x − 14) = 37

( x − 9) + ( x − 14) = 37 − x
2 x − 23 = 37 − x

3x − 23 = 37

OR

3 x = 60
x = 20

equating
manipulation
answer
(3)

3x = 60
x = 20

first
differences
5; x − 9;
37 − x
equating
answer

OR
( x − 9) − 5 = (37 − x) − ( x − 9)
x − 14 = −2 x + 46
3 x = 60
x = 20

3.2

4

(3)

9
5

11
6

2a = 6
Tn = 3n 2 + bn + c

KT1

37
17

6
Note:
If x is incorrect in 3.1 then
max 2 / 4 marks

a=3
3+b +c =4

20

a=3
Tn = 3n 2 + bn + c

b + c =1
12 + 2b + c = 9

KT2

2b + c = −3
∴9 + b = 5

b=–4
c=5
(4)

b = −4
and
c = 4 − (−1) = 5
∴Tn = 3n 2 − 4n + 5

OR
Copyright Reserved

Please turn over

Mathematics/P1

9
NSC – Memorandum

DBE/November 2010

OR

2a = 6
a=3
T0 = 5
c=5
Tn = 3n 2 + bn + 5

OR

4 = 3(1) 2 + b + 5
b = −4

2a = 6
a=3
3a + b = 5
b = −4
a+b+c = 4
3−4+c = 4
c=5

a=3
c=5
method
b=–4
(4)

Tn = 3n 2 − 4n + 5

Tn = 3n 2 − 4n + 5
OR

a + b + c = 4 Ki
4a + 2b + c = 9 Kii
16a + 4b + c = 37 Kiii
3a + b = 5
12a + 2b = 28
6a + b = 14
3a = 9
a=3
b = −4
c=5

a=3

OR

6(n − 1)(n − 2)
Tn = 4 + (n − 1)5 +
2
2
= 4 + 5n − 5 + 3n − 9n + 6
= 3n 2 − 4n + 5

c=5
method
b=–4
(4)
[7]

Tn = 3n 2 − 4n + 5

Copyright Reserved

Please turn over

Mathematics/P1

10
NSC – Memorandum

DBE/November 2010

QUESTION 4
y

f
A(2 ; 3)

f

D(0 ; 1,5)
(1 ; 0)

x

0

4.1

x=2
y=3

OR

x-asymptote = 2
y-asymptote = 3

answer
answer

(2)

If x = p ; y = q then 1 mark

4.2

Note: If the candidate just writes down the number 2 or 3 or just
coordinates (2 ; 3), then no marks
a
+3
f ( x) =
If the asymptotes are
x−2
swopped in 4.1, then
a
a
0=
+3
f ( x) =
+2
1− 2
x −3
0 = −a + 3
a
+2
0=
a=3
1− 3
3
a=4
f ( x) =
+3
x−2
4
f ( x) =
+2
x −3
OR

a
+3
x−2
a
y−3=
x−2
( x − 2)( y − 3) = a
But (1;0) lies on the graph
∴ (−1)(−3) = a = 3
∴ ( x − 2)( y − 3) = 3
3
+3
When x = 0, y =
0−2
3
=
2
⎛ 3⎞
D⎜0 ; ⎟
⎝ 2⎠

subs in of
asymptotes
subs in (1 ; 0)

answer
(3)

y=

4.3

Copyright Reserved

equation
subs in (1 ; 0)
answer
(3)
If asymptotes swopped:
x=0
4
y=
+2
0−3
2
y=
3
⎛ 2⎞
D⎜0 ; ⎟
⎝ 3⎠

x=0
3
y=
2
(2)

Please turn over

Mathematics/P1

4.4

3 − 1,5
2−0
3
=
4
3
3
y = x+
4
2

m AD =

OR
4 y = 3x + 6
OR

y = mx +

11
NSC – Memorandum

DBE/November 2010

If asymptotes swopped:
3 − 23
m AD =
2−0
7 1
= ×
3 2
7
=
6
7
2
y = x+
6
3

3
2

3 = m(2) +

3
2

3
4
3
3
y = x+
4
2

p+0
=2
2
p=4
q+
2

3
2

3
4
answer
(3)

substitution of
point (2 ; 3) and
3
c=
2
3
4
answer
(3)

m=

4.5

substitution into
gradient

Answer only:
Full Marks

=3

q = 4 12
1⎞
⎛
Other point of intersection is ⎜ 4 ; 4 ⎟
2⎠
⎝

p+0
=2
2
q + 32
=3
2

x=4
y= 4

1
2
(4)

OR

By symmetry the rule to calculate the point of intersection is
3⎞
⎛
(x ; y) → ⎜ x + 2 ; y + ⎟
2⎠
⎝
To help with
Other point of intersection is
applying CA the
3⎞
⎛
⎜2 + 2 ; 3 + ⎟
y-coordinate will
2⎠
⎝
be 3 + (3 – y)
1⎞
⎛
= ⎜4 ; 4 ⎟
2⎠
⎝

Copyright Reserved

x-answer
y-answer

(4)

Please turn over

Mathematics/P1

12
NSC – Memorandum

OR
3
3
3
x+ =
+3
4
2 x−2
3 x( x − 2) + 6( x − 2) = 12 + 12( x − 2)
3 x 2 − 6 x + 6 x − 12 = 12 + 12 x − 24
3 x 2 − 12 x = 0
3 x ( x − 4) = 0
x = 0 and x = 4

1⎞
⎛
Other point of intersection is ⎜ 4 ; 4 ⎟
2⎠
⎝
If asymptotes swopped:

DBE/November 2010

Note:
If the candidate
does not select
the x-value
greater than 2 i.e.
a realistic
answer, max 3 / 4
marks

equating

standard form
x-values
y-value

(4)

[14]

7
2
4
x+ =
+2
6
3 x−3
7 x( x − 3) + 4( x − 3) = 4(6) + 2(6)( x − 3)
7 x 2 − 29 x = 0
x(7 x − 29) = 0
29
7
Other point of intersection is
⎛ 29 11 ⎞
⎜ ; ⎟
⎝ 7 2⎠
x = 0 or

x=

QUESTION 5

5.1

f ( x) = 4 − x − 2
y-intercept:
x = 0; y = 4 0 − 2 = −1 ; (0 ; – 1)

y-intercept

x-intercept:
4− x − 2 = 0
4

−x

−2=0
4

−x

=2

log 4 − x = log 2

log 2
−x=
OR
2 =2
log 4 OR
− 2x = 1
log 2
−x=
1
2 log 2
x=−
2
1
x=−
2
⎛ 1 ⎞
x-intercept is ⎜ − ; 0 ⎟
⎝ 2 ⎠
y=–2
−2 x

5.2

=2

4

−x

x-intercept
(4)

4− x − 2 = 0
4−x = 2
−x

=4
1
−x=
2

4

x=−

1
2

Note:
No penalty if
the answer is
not left as a
coordinate.

1
2

equation
(1)

Copyright Reserved

Please turn over

Mathematics/P1

13
NSC – Memorandum

5.3

DBE/November 2010

asymptote
y-intercept or
x-intercept
shape
(decreasing)

y
1

0.5

x
-2

-1.5

-1

-0.5

0.5

1

1.5

2

(3)

2.5

-0.5

-1

-1.5

-2

-2.5

5.4

g ( x) = 4

−x

equation

−2+2

(1)

g ( x) = 4 − x

⎛1⎞
g ( x) = ⎜ ⎟
⎝4⎠
−2 x
OR g ( x) = 2

x

OR

5.5

⎛1⎞
OR g ( x) = ⎜ ⎟
⎝2⎠
−x
4 −2=3

2x

4−x = 5

4−x = 5
− x log 4 = log 5
log 5
x=−
OR x = − log 4 5 OR x = log 1 5 OR x = log 4 15
log 4
4

OR x = – 1,16 OR x =

Copyright Reserved

log 15
log 5
OR
x
=
log 14
log 4

− x log 4 = log 5

answer
(3)
[12]

Please turn over

Mathematics/P1

14
NSC – Memorandum

DBE/November 2010

QUESTION 6

6.1

f ( x) = ax 2
− 8 = a (−6) 2
− 8 = 36a
8
a=−
36
OR
2
a=−
9

6.2

answer
(2)

2
f ( x) : y = − x 2
9
2
x = − y2
9
2
9 x = −2 y
9x
−
= y2
2
− 9x
y=±
, since
2
y=− −

6.3

substitution

9x
OR
2

Note:
If candidate does not
substitute the value of a
the answer is
x
y=−
a
then 2 / 3 marks

y≤0

y = −3

−x
2

swop x and y
9x
or
2
− 9x
y=±
2
y2 = −

y=− −

9x
2

(3)

y≤0

answer
(1)

OR

y ∈ (−∞ ; 0]
6.4
shape
(third quadrant)
(concave upward)
Any point
other than
(0 ; 0) that lies
on the graph

y
8
6
4
2

-10

-8

-6

-4

O

-2
-2
-4

(-8;-6)
●

Copyright Reserved

-6

2

4

Point
corresponding
from original
graph will be
(– 8 ; – 6)
(2)

Please turn over

Mathematics/P1

6.5

15
NSC – Memorandum

y =−f
=

−1

y =−f

( x)

− 9x
2

OR

=3

OR

2
x = − y2
9
9
− x = y2
2
9x
y=− −
2

Reflection about y-axis:

y= −

−f

( x)

−x
2

9x
2

−1

( x)

answer
(3)
Note:
If candidate has
(x ; y) → (y ; – x)
then 2 / 3 marks

2
y = − x2
9
Reflection in y = x:

−1

DBE/November 2010

Note:
If candidate does not
substitute the value of a
the answer is
x
y=
a
then full marks

2
x = − y2
9

y=− −
y= −

9x
2

9x
2

(3)
[11]

Copyright Reserved

Please turn over

Mathematics/P1

16
NSC – Memorandum

DBE/November 2010

QUESTION 7

A = P(1 + i )

7.1

n

⎛ r⎞
2 P = P⎜1 + ⎟
⎝ 4⎠
⎛ r⎞
2 = ⎜1 + ⎟
⎝ 4⎠

6×4

Note:
Penalty 1 for incorrect
rounding off.

24

2P
r
and 24
4

1

r
= 2 24
4
⎛ 1
⎞
r = 4⎜⎜ 2 24 − 1⎟⎟
⎝
⎠
⎛ 1 ⎞
r = 4⎜⎜ 2 24 ⎟⎟ − 4
⎝
⎠
r = 0,1172 …
rate = 11,72% p.a. compounded quarterly

1+

OR

1

1+

r
= 2 24
4

⎛ 1 ⎞
r = 4⎜⎜ 2 24 ⎟⎟ − 4
⎝
⎠
answer
(5)

A = P(1 + i )

n

r ⎞
⎛
2 P = P ⎜1 +
⎟
⎝ 400 ⎠
r ⎞
⎛
2 = ⎜1 +
⎟
⎝ 400 ⎠

6×4

24

2P
r
and 24
400
1

1

1+

r
= 2 24
1+
400
⎛ 1
⎞
r = 400⎜⎜ 2 24 − 1⎟⎟
⎝
⎠
1
⎛
⎞
r = 400⎜⎜ 2 24 ⎟⎟ − 400
⎝
⎠
r = 11,72% p.a.

r
= 2 24
400

⎛ 241 ⎞
r = 400⎜⎜ 2 ⎟⎟ − 400
⎝
⎠
answer
(5)

7.2.1

⎛ 0,095 ⎞
A = 10000⎜1 +
⎟
12 ⎠
⎝
= R 10 402,15

5

substitution in
correct formula
answer
(2)

Copyright Reserved

Please turn over

Mathematics/P1

17
NSC – Memorandum

7.2.2

⎡ ⎛ 0,095 ⎞ − n ⎤
450 ⎢1 − ⎜1 +
⎟ ⎥
12 ⎠ ⎦⎥
⎢⎣ ⎝
10402,15 =
0,095
12
⎛ 0,095 ⎞
0,183000787 = 1 − ⎜1 +
⎟
12 ⎠
⎝
⎛ 0,095 ⎞
⎟
⎜1 +
12 ⎠
⎝

DBE/November 2010

10 402,15
substitution into
present value formula

−n

−n

= 0,816999213

Note:
Incorrect Formula
No marks

−n

⎛ 0,095 ⎞
log⎜1 +
⎟ = log 0,816999213
12 ⎠
⎝
⎛ 0,095 ⎞
− n log⎜1 +
⎟ = log 0,816999213...
12 ⎠
⎝
n = 25,63151282....

application of logs

n = 25,63 months
n = 26
Accept: n = 31 (because of first 5 months)

answer
(4)

OR

⎛ 0,095 ⎞
10402,15⎜1 +
⎟
12 ⎠
⎝

n

⎛ 0,095 ⎞
10402,15⎜1 +
⎟
12 ⎠
⎝

n

⎡⎛ 0,095 ⎞ n ⎤
450 ⎢⎜1 +
⎟ − 1⎥
12 ⎠
⎢⎝
⎦⎥
⎣
=
0,095
12
⎡⎛ 0,095 ⎞ n ⎤
= 56842,10526 ⎢⎜1 +
⎟ − 1⎥
12 ⎠
⎢⎣⎝
⎥⎦

10 402,15
substitution into future
value formula

n

⎛ 0,095 ⎞
56842,10526 = 46439,95526⎜1 +
⎟
12 ⎠
⎝
⎛ 0,095 ⎞
log1,223991387 = n log⎜1 +
⎟
12 ⎠
⎝
log1,223991387
n=
⎛ 0,095 ⎞
log⎜1 +
⎟
12 ⎠
⎝
n = 25,63 months
n = 26
Accept: n = 31 (because of first 5 months)

application of logs

answer
(4)

Note: If the Present value of R 10 000 is used, then
n = 25,53 months is obtained. Max 3 / 4
marks.

Copyright Reserved

Please turn over

Mathematics/P1

7.2.3

18
NSC – Memorandum

Balance outstanding after 25 months
⎡⎛ 0,095 ⎞ 25 ⎤
450
⎟ − 1⎥
⎢⎜1 +
25
12
⎝
⎠
⎢
⎥⎦
0
,
095
⎛
⎞
⎣
= 10402,15⎜1 +
⎟ −
0,095
12 ⎠
⎝
12
= R 282,36

DBE/November 2010

correct formula
substitution into
⎡⎛ 0,095 ⎞ 25 ⎤
450 ⎢⎜1 +
⎟ − 1⎥
12 ⎠
⎣⎢⎝
⎦⎥
0,095
12
answer
(3)

OR

Balance Outstanding after 25 months
⎡⎛ 0,095 ⎞ 25 ⎤
450 ⎢⎜1 +
⎟ − 1⎥
30
12 ⎠
⎥⎦
⎢⎣⎝
⎛ 0,095 ⎞
= 10000⎜1 +
⎟ −
0,095
12 ⎠
⎝
12
= R 282,36

OR
n = 25,6315128204.... – 25
= 0,6315128204 ...
Balance Outstanding after 25 months
⎡ ⎛ 0,095 ⎞ −0,631512804 ⎤
450 ⎢1 − ⎜1 +
⎟
⎥
12 ⎠
⎥⎦
⎢⎣ ⎝
=
0,095
12
= R 282,36

OR
Present value at beginning of 25 months
⎡ ⎛ 0,095 ⎞ −25 ⎤
450 ⎢1 − ⎜1 +
⎟ ⎥
12 ⎠ ⎥⎦
⎢⎣ ⎝
= 10402,15 −
0,095
12
= R 231,84
Balance Outstanding
25
⎛ 0,095 ⎞
= 231,84⎜1 +
⎟
12 ⎠
⎝
= R 282,36

Note: Accept
If a candidate uses
– 0,63, the final
answer is R 281,68

correct formula
⎡⎛ 0,095 ⎞ 25 ⎤
450 ⎢⎜1 +
⎟ − 1⎥
12
⎝
⎠
⎣⎢
⎦⎥
0,095
12
answer
(3)

correct formula
substitution into
−0,631512804 ⎤
⎡ ⎛
0,095 ⎞
⎥
450 ⎢1 − ⎜1 +
⎟
12 ⎠
⎢⎣ ⎝
⎥⎦
0,095
12

answer
(3)
correct formula
substitution into
⎡ ⎛ 0,095 ⎞ −25 ⎤
450⎢1 − ⎜1 +
⎟ ⎥
12 ⎠ ⎥⎦
⎢⎣ ⎝
0,095
12

answer
(3)
[14]

Copyright Reserved

Please turn over

Mathematics/P1

19
NSC – Memorandum

DBE/November 2010

QUESTION 8

8.1

g ( x) = x 2 − 5
g ( x + h) − g ( x)
g ' ( x) = lim
h
h→ 0
2
( x + h) − 5 − ( x 2 − 5)
= lim
h
h→ 0
2
x + 2 xh + h 2 − 5 − x 2 + 5
= lim
h
h→ 0
2
2 xh + h
= lim
h
h→ 0
h( 2 x + h)
= lim
h
h→ 0
= lim (2 x + h)
h→ 0

Note:
If the notation is
incorrect, penalty 1
mark

formula

If candidate subtracts
and gets

expansion

substitution

x 2 + 2 xh + h 2 − 5 − x 2 − 5

in the numerator and
then candidate
corrects themselves,
max 2 / 5
Answer only: 0 / 5

= 2x

2x + h
answer
(5)

OR
g ( x) = x 2 − 5
g ( x + h) − g ( x)
g ' ( x) = lim
h
h→ 0
2
( x + h) − 5 − ( x 2 − 5)
= lim
h
h→ 0
2
( x + h) − x 2
= lim
h
h→ 0
( x + h + x)( x + h − x)
= lim
h
h→ 0
h( 2 x + h)
= lim
h
h→ 0
= lim (2 x + h)
h→ 0

= 2x

formula
substitution
expansion

2x + h
answer
(5)

8.2

6

x
+4 x
2
1
1
y = x6 + 4x 2
2
1
−
dy
5
= 3x + 2 x 2
dx
y=

Copyright Reserved

Note:
dy
If
or y ′ is left out, penalty 1 mark
dx

If a candidate shows evidence of how to
differentiate from an incorrect function
which involves breakdown, then max 1 / 3

+ 4x

1
2

3x 5
2x

−

1
2

(3)

Please turn over

Mathematics/P1

8.3

20
NSC – Memorandum

DBE/November 2010

b
x
2
g ( x) = ax + bx −1
g ( x) = ax 2 +

g ′( x) = 2ax − bx −2
b
0 = 2a (4) − 2
(4)
b
8a =
16
b = 128a
b
96 = a (4) 2 +
4
1
96 = 16a + (128a )
4
96 = 48a
a=2
b = 256

Note:
In the equation
g ′( x) = 0 ; = 0 must be
shown in the equation.

g ′( x) = 2ax − bx −2
0 = g ′( x)
b
2a (4) − 2
(4)

subs (4 ; 96)

a=2
b = 256

(6)

OR
b
x2
b
g ′(4) = 8a −
=0
16
b
g (4) = 16a + = 96
4
b
32a − = 0
4
48a = 96
g ′( x) = 2ax −

a=2
b = 256

Copyright Reserved

b
x2
b
g ′(4) = 8a −
16
g ′( x) = 0
b
g (4) = 16a + = 96
4
a=2
b = 256
(6)
[14]
g ′( x) = 2ax −

Please turn over

Mathematics/P1

21
NSC – Memorandum

DBE/November 2010

QUESTION 9
y

h

C(– 2 ; 0)

D(6 ; 0)

A

x

O

y=

g′(x)

E
B

9.1

The y-intercept of g is E(0 ; – 4)

9.2

OR
x = 0 and y = – 4
y = a( x + 2)( x − 6)

answer
(1)

− 4 = a(0 + 2)(0 − 6)
− 4 = −12a
1
a=
3
1
y = ( x + 2)( x − 6)
3
1
4
y = x2 − x − 4
3
3
OR

setting up of equation
subs (0 ; – 4)

a=

1
3

y=

1 2 4
x − x−4
3
3
(4)

g ′(0) = −4 = c

g ′( x) = ax 2 + bx − 4
g ′(−2) = 0
4a − 2b − 4 = 0
b = 2a − 2
g ′′(2) = 0
2a (2) + b = 0
b = −4a
2a − 2 = −4a
1
a=
3
4
b=−
3
1 2 4
y = x − x−4
3
3
Copyright Reserved

substitution x = – 2
and g ′(x) = 0
g ′′(2) = 0

a=

1
3

y=

1 2 4
x − x−4
3
3
(4)

Please turn over

Mathematics/P1

22
NSC – Memorandum

OR
c = −4
4a − 2b − 4 = 0
36a + 6b − 4 = 0
48a − 16 = 0
a=

setting up of equation
simultaneous equation

1
3

b=−

DBE/November 2010

4
3

a=

1
3

y=

1 2 4
x − x−4
3
3
(4)

1
4
y = x2 − x − 4
3
3

OR

y = a( x + 2)( x − 6)

= a( x 2 − 4 x − 12)

setting up of equation
ax 2 − 4ax − 12a

= ax 2 − 4ax − 12a
− 12a = −4
1
a=
3
1
4
y = x2 − x − 4
3
3

1
3
1
4
y = x2 − x − 4
3
3

a=

(4)
OR
dy
= 2ax + b
dx
0 = 2 a ( 2) + b
b = −4a
EITHER
subs (6 ; 0)
0 = 36a + 6b − 4
4 = 36a + 6b
2 = 18a + 3b
2 = 18a + 3(−4a)
2 = 6a
a=

1
3

b=−
y=

4
3

OR
0 = 4a − 2b − 4
0 = 4a − 2(−4a) − 4
12a = 4
1
a=
3
4
b=−
3
1
4
y = x2 − x − 4
3
3

b = −4 a

simultaneous equation
1
a=
3

y=

1 2 4
x − x−4
3
3
(4)

1 2 4
x − x−4
3
3

Copyright Reserved

Please turn over

Mathematics/P1

9.3

At turning point g ′( x) = 0
x = – 2 and x = 6

23
NSC – Memorandum

DBE/November 2010

Answer only:
Full marks

g ′( x) = 0
x = 6 and x = – 2
(2)

If only 1 value given,
max 1 / 2

9.4

−2+6
2
x=2
x=

−2+6
2
answer
x=

Note:
Answer only
Full marks

OR

(2)

x-value of point of inflection of g is at A.
g ′′( x) = 0
2x 4
− =0
3 3
2x − 4 = 0
2x = 4
x=2

2x − 4 = 0
answer

(2)
x=

2( )
answer

OR
b
x=−
2a
x=

4
3
1
3

2(

1
16
2
g ′( x) = ( x − 2) −
3
3
x=2

OR

)

x=2

9.5

4
3
1
3

g / ( x) > 0 for x < −2 , so g is increasing for x < −2 .
g / ( x) < 0 for x > −2 , so g is decreasing for x > −2 .
∴ g has a local maximum at x = − 2 because the graph is
increasing followed by decreasing

(2)
1
16
2
g ′( x) = ( x − 2) −
3
3
answer
(2)
g / ( x) > 0
g is incr for x < −2
g is decr for x > −2
(3)

OR

g

g'
−2

6

−2

6

g / ( x) > 0 for x < −2
g / ( x) < 0 for x > −2
max at x = – 2
(3)

∴ g has a local maximum at x = − 2
OR

Copyright Reserved

Please turn over

Mathematics/P1

24
NSC – Memorandum

g'(x)=0
g'(x) > 0
g

g'(x) < 0
x = −2

OR
g ′(−2) = 0
g ′′(−2) < 0 so graph is concave down at x = – 2, so g has a local
maximum

Copyright Reserved

DBE/November 2010

g / ( x) > 0 for x < −2
g / ( x) < 0 for x > −2
max at x = – 2
(3)

g ′(−2) = 0
g ′′(−2) < 0
max at x = – 2
(3)
[12]

Please turn over

Mathematics/P1

25
NSC – Memorandum

DBE/November 2010

QUESTION 10

10.1

V = π r 2h + 2 ×

1 4 3
× πr
2 3

4
V = π r h + π r3
3
4
π
= π r 2h + π r 3
6
3
π 4
π r 2h = − π r 3
6 3
π
4πr 3
h=
−
6πr 2 3πr 2
1
4r
h= 2 −
6r
3
S = 2 × 2π r 2 + 2π rh
2

10.2

S = 4π r 2 + 2π rh

10.3

4r ⎞
⎛ 1
S = 4π r 2 + 2π r ⎜ 2 − ⎟
3⎠
⎝ 6r
2
π 8π r
S = 4π r 2 + −
3r
3
4
π
= πr 2 +
3
3r
π
4
S = πr 2 + r −1
3
3
dS 8πr π
=
− 2 =0
dr
3
3r
1
8r = 2
r
3
8r = 1

r=

1
2
2

Then

4 ⎛1⎞ π
π ⎜ ⎟ + (2)
3 ⎝2⎠
3
S = π square metres
= 3,14 square metres
S=

volume equation
substitution of

π
6

4πr 3
π
−
h=
6πr 2 πr 2
(3)

surface area
equation
substitution of h
simplification
(3)

π

r −1

3
1 ⎞
dS π ⎛
= ⎜ 8r − 2 ⎟
dr 3 ⎝
r ⎠
or
dS π
= 8r − r − 2
dr 3
dS
=0
dr
1
8r = 2
r
1
r=
2

(

)

S=π
(6)
[12]

Copyright Reserved

Please turn over

Mathematics/P1

26
NSC – Memorandum

DBE/November 2010

QUESTION 11

11.1

x, y ∈ N 0
x + 2 y ≤ 28

y≤−

or

3x + y ≤ 24

x
+ 14
2

y ≤ −3 x + 24

or

Note:
If inequality signs incorrect
or equal signs used: max 3 / 4
marks

First
inequality
Second
inequality
(4)
graph of

11.2

x + 2 y ≤ 28

24
23
22
21
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1

y

1

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28

graph
of

3 x + y ≤ 24

feasible
region
(quadrilate
ral)
(3)

3x + y = 24
11.5 y = x

Feasible
Region

P(6 ; 6)

x + 2y = 28

11.5 FR

x

11.3.1

8

11.3.2

14

11.4

Maximise x + y
Use search line with gradient – 1

answer
(1)
answer
(1)
4 Type
A
12 Type

4 Type A
12 Type B

Copyright reserved

B
(2)

Please turn over

Mathematics/P1

11.5

27
NSC – Memorandum

x≥ y
y≤x
New Feasible region (triangle) in diagram
Maximise x + y.
Maximum at (6 ; 6)
Answer: 6 + 6 = 12 braai stands
Machine Time = x + 2y
= 6 + 2×6
= 6 + 12
= 18 hours

DBE/November 2010

y≤x
Note:
Answer only of machine
time 18 hours and braai
stands 12
Full marks

(6 ; 6)
12
18
hours
(5)
[16]
TOTAL:

Copyright reserved

150