Notesale: Turn your study into money

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NATIONAL
SENIOR CERTIFICATE

GRADE 12

MATHEMATICS P1
FEBRUARY/MARCH 2013

MARKS: 150
TIME: 3 hours

This question paper consists of 10 pages, 1 diagram sheet and 1 information sheet
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–Mar
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1
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2
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3
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4
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5
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6
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7
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1
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8
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9
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10
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11
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QUESTION 1
1
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2

Solve for x:
1
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1

(x

1
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2

x 2 + x − 13 = 0 (Leave your answer correct to TWO decimal places
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1
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3 x = 81 − 3 x

(4)

1
...
4

( x + 1)(4 − x) > 0

(3)

Given:

2

)

− 9 (2 x + 1) = 0

(3)

2 x + 2 x + 2 = −5 y + 20

1
...
1

Express 2 x in terms of y
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2
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2
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(3)
[21]

QUESTION 2
2
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2
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1

Determine the value of p
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1
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(3)

2
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3

Why does the sum to infinity for this series exist?

(1)

2
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4

Calculate S ∞

(3)

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2
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–Mar
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2
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1

Write down the next term of the sequence
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2
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(3)

2
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3

Calculate the smallest value of n for which the sum of the first n terms
of the sequence will be greater than 10 140
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3

Calculate

∑ (3k + 5)

(3)
[22]

k =1

QUESTION 3
Consider the sequence: 3 ; 9 ; 27 ;
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Vusi disagrees and says that the fourth term of the sequence is 57
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1

Explain why Jacob and Vusi could both be correct
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2

Jacob and Vusi continue with their number patterns
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2
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2
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3

y

x

0

4
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(1)

4
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(1)

4
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4

Sketch the graph of f
other point
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Indicate the x-intercept and ONE
(3)

4
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6

Prove that:

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(2)

in the form y =
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(2)

= f (2 x) − f (−2 x) for all values of x
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–Mar
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x− p
C(2 ; 6) is the point of intersection of the asymptotes of g
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2 

Sketched below is the graph of g ( x) =

y

g

C(2 ; 6)

g

B(5/2 ; 0)

x

0

a
+q
x− p

5
...
2

F is the reflection of B across C
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(4)
(2)
[6]

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QUESTION 6
S(1 ; 18) is the turning point of the graph of f ( x) = ax 2 + bx + c
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The graph of g ( x) = −2 x + 8 has an x-intercept at T
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y
S(1 ; 18)

f
R

g

P

T

x

0

6
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(2)

6
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Show ALL your
working
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3

If f ( x) = −2 x 2 + 4 x + 16 , calculate the coordinates of R
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4

Use your graphs to solve for x where:
6
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1

f ( x) ≥ g ( x)

(2)

6
...
2

− 2x 2 + 4x − 2 < 0

(4)
[16]

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2013

QUESTION 7
7
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2

Raeesa invests R4 million into an account earning interest of 6% per annum,
compounded annually
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7
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1

7
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2

She withdraws an allowance of R30 000 per month
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How many
such withdrawals will Joanne be able to make?
If Joanne withdraws R20 000 per month, how many withdrawals will she
be able to make?

(6)

(3)
[12]

QUESTION 8
Jeffrey invests R700 per month into an account earning interest at a rate of 8% per annum,
compounded monthly
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Jeffrey and his friend's investments
are worth the same at the end of 12 months
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[3]

QUESTION 9
9
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2

Determine

9
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–Mar
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1

Write down the coordinates of the y-intercept of f
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2

Show that (2 ; 0) is the only x-intercept of f
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3

Calculate the coordinates of the turning points of f
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4

Sketch the graph of f in your ANSWER BOOK
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(3)
[14]

QUESTION 11
A rectangular box is constructed in such a way that the length (l) of the base is three times as
long as its width
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The material used to construct the sides of the box costs R50 per square
metre
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Let the width of the box be x metres
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1

Determine an expression for the height (h) of the box in terms of x
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2

Show that the cost to construct the box can be expressed as C =

11
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(3)

1 200
+ 600 x 2
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–Mar
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Their boundary lines are represented graphically in the
sketch below
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The constraints are:
0 ≤ y ≤ 50
0 ≤ x ≤ 40
2 x + y ≤ 60
35 ≤ x + y ≤ 60
y
60

55

50

A

B

D

C

45

40

35

J

30

25

E

20

15

I

10

5

H
0

5

10

15

20

25

F

G
30

35

40

x
45

50

55

60

12
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(2)

12
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(3)

12
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4

If P = 4 x + y for (x ; y) in the feasible region, determine the maximum value of P
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5

If the objective function C = kx + y is minimised at J only, determine ALL possible
values of k
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1
y
60

55

50

A

B

D

C

45

40

35

J

30

25

E

20

15

I

10

5

H
0

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5

10

15

20

25

F

G
30

35

40

x
45

50

55

60

Mathematics/P1

DBE/Feb
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2013
NSC

INFORMATION SHEET: MATHEMATICS

x=

− b ± b 2 − 4ac
2a

A = P (1 + ni )

A = P (1 − ni )
n

n

∑1 = n

∑i =
i =1

i =1

n(n + 1)
2

Tn = ar n −1

Sn =

[

]

(

x (1 + i ) − 1
F=
i
n

)

a r n −1
r −1
P=

A = P(1 − i ) n

A = P(1 + i ) n

Tn = a + (n − 1)d

Sn =

;

r ≠1

S∞ =

n
(2a + (n − 1)d )
2

a
; −1 < r < 1
1− r

x[1 − (1 + i )− n ]
i

f ( x + h) − f ( x )
h
h→ 0

f ' ( x) = lim

d = ( x 2 − x1 ) 2 + ( y 2 − y1 ) 2

 x + x2 y1 + y 2 

;
M  1
2 
 2

y − y1 = m( x − x1 )

y = mx + c

m=

y 2 − y1
x 2 − x1

m = tan θ

( x − a )2 + ( y − b )2 = r 2
a
b
c
=
=
sin A sin B sin C

In ∆ABC:

area ∆ABC =

a 2 = b 2 + c 2 − 2bc
...
sin C
2

sin (α + β ) = sin α
...
sin β

sin (α − β ) = sin α
...
sin β

cos(α + β ) = cos α
...
sin β

cos(α − β ) = cos α
...
sin β

cos 2 α − sin 2 α

cos 2α = 1 − 2 sin 2 α
2 cos 2 α − 1


sin 2α = 2 sin α

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