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NATIONAL
SENIOR CERTIFICATE
GRADE 12
MATHEMATICS P2
NOVEMBER 2012
MARKS: 150
TIME: 3 hours
This question paper consists of 13 pages, 1 diagram sheet and 1 information sheet.
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Mathematics Grade 12 P2 NSC November 2012 QP (Eng)
Grade 12 Mathematics Paper 2 question paper from the NSC November Exam 2012 (English), 15 pages. Read online or download the free PDF.
Mathematics · Grade 12 · NSC November Exam · 2012. Question paper, 15 pages. Read online or download the PDF.
- Subject
- Mathematics
- Grade
- Grade 12
- Document type
- Question paper
- Year
- 2012
- Exam period
- NSC November Exam
- Paper
- 2
- Curriculum
- CAPS
- Pages
- 15
- File size
- 319.0 KB
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This exam
NSC November Exam 2012
Questions in this paper
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Mathematics/P2 2 DBE/November 2012
NSC
INSTRUCTIONS AND INFORMATION
Read the following instructions carefully before answering the questions.
1. This question paper consists of 13 questions.
2. Answer ALL the questions.
3. Clearly show ALL calculations, diagrams, graphs, et cetera which you have used in
determining the answers.
4. Answers only will not necessarily be awarded full marks.
5. You may use an approved scientific calculator (non-programmable and non-
graphical), unless stated otherwise.
6. If necessary, round off answers to TWO decimal places, unless stated otherwise.
7. Diagrams are NOT necessarily drawn to scale.
8. ONE diagram sheet for QUESTION 3.2 and QUESTION 7.3 is attached at the end of
this question paper. Write your centre number and examination number on this sheet
in the spaces provided and insert the sheet inside the back cover of your ANSWER
BOOK.
9. An information sheet with formulae is included at the end of this question paper.
10. Number the answers correctly according to the numbering system used in this
question paper.
11. Write neatly and legibly.
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Mathematics/P2 3 DBE/November 2012
NSC
QUESTION 1
The scatter plot below shows the age (in years) and the average height (in centimetres) of boys
between 2 and 15 years.
Scatter Plot
180
175
170
165
160
155
150
Average height (in cm)
145
140
135
130
125
120
115
110
105
100
95
90
85
80
75
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Age of boys (in years)
[Source: www.fpnotebook.com/endo/exam/hgtmsrmnincharn.htm]
1.1 Use the scatter plot to determine the average height of a 7-year-old boy. (1)
1.2 Describe the trend in the scatter plot. (1)
1.3 What is the approximate increase in the average height per annum between the ages
of 2 and 15 years? (3)
1.4 Explain why the observed trend CANNOT continue indefinitely. (1)
[6]
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Mathematics/P2 4 DBE/November 2012
NSC
QUESTION 2
Abe plays for his school's cricket team. The number of runs scored by Abe in the eight games
that he batted in, is shown below. (Abe was given out in all of the games.)
21 8 19 7 15 32 14 12
2.1 Determine the average runs scored by Abe in the eight games. (2)
2.2 Determine the standard deviation of the data set. (2)
2.3 Abe's scores for the first three of the next eight games were 22, 35 and 2
respectively. Describe the effect of his performance on the standard deviation of this
larger set having 11 data points. (2)
2.4 Abe hopes to score an average of 20 runs in the first 16 games. What should his
average in the last five games be so that he may reach his goal? (3)
[9]
QUESTION 3
In a certain school 60 learners wrote examinations in Mathematics and Physical Sciences. The
box-and-whisker diagram below shows the marks (out of 100) that these learners scored in the
Physical Sciences examination.
Physical Sciences
3.1 Write down the range of the marks scored in the Physical Sciences examination. (1)
3.2 Use the information below to draw the box-and-whisker diagram for the Mathematics
results on DIAGRAM SHEET 1.
Minimum mark = 30
Range = 55
Upper quartile = 70
Interquartile range = 30
Median = 55 (4)
3.3 How many learners scored less than 70% in the Mathematics examination? (2)
3.4 Joe claims that the number of learners who scored between 30 and 45 in Physical
Sciences is smaller than the number of learners who scored between 30 and 55 in
Mathematics. Is Joe's claim valid? Justify your answer. (2)
[9]
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Mathematics/P2 5 DBE/November 2012
NSC
QUESTION 4
As part of an environmental awareness initiative, learners of Greenside High School were
requested to collect newspapers for recycling. The cumulative frequency graph (ogive) below
shows the total weight of the newspapers (in kilograms) collected over a period of 6 months by
30 learners.
Weight of newspaper collected (in kilograms)
4.1 Determine the modal class of the weight of the newspapers collected. (1)
4.2 Determine the median weight of the newspapers collected by this group of learners. (1)
4.3 How many learners collected more than 60 kilograms of newspaper? (2)
[4]
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Mathematics/P2 6 DBE/November 2012
NSC
QUESTION 5
ABCD is a rhombus with A(– 3 ; 8) and C(5 ; – 4). The diagonals of ABCD bisect each other
at M. The point E(6 ; 1) lies on BC.
y
A(– 3 ; 8)
θ
B
M
E(6 ; 1)
P Q R S T x
D
C(5 ; – 4)
5.1 Calculate the coordinates of M. (2)
5.2 Calculate the gradient of BC. (2)
5.3 Determine the equation of the line AD in the form y = mx + c. (3)
5.4 Determine the size of θ, that is BÂC . Show ALL calculations. (6)
[13]
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Mathematics/P2 7 DBE/November 2012
NSC
QUESTION 6
A circle centred at N(3 ; 2) touches the x-axis at point L. The line PQ, defined by the equation
4 4
y = x + , is a tangent to the same circle at point A.
3 3
y Q
A
N(3 ; 2)
B
K x
O L
P
6.1 Why is NL perpendicular to OL? (1)
6.2 Determine the coordinates of L. (1)
6.3 Determine the equation of the circle with centre N in the form
(x – a)2 + (y – b)2 = r2
(3)
6.4 Calculate the length of KL. (3)
6.5 Determine the equation of the diameter AB in the form y = mx + c. (4)
Show that the coordinates of A are ; .
7 16
6.6 (3)
5 5
6.7 Calculate the length of KA. (3)
6.8 Why is KLNA a kite? (2)
6.9 Show that AB̂K = 45°. (3)
6.10 If the given circle is reflected about the x-axis, give the coordinates of the centre of
the new circle. (1)
[24]
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Mathematics/P2 8 DBE/November 2012
NSC
QUESTION 7
Consider the diagram below where A(– 5 ; 2), B(– 4 ; 1) and C(– 3 ; 3) are the vertices of
∆ABC.
y
10
8
6
A/
4
B/
C
A C/
2
B x
-12 -10 -8 -6 -4 -2 0 2 4 6 8 10 12
-2
-4
7.1 Describe the single transformation of ∆ABC to ∆A/B/C/. (2)
7.2 Write down the general rule of the transformation in QUESTION 7.1. (2)
7.3 ∆ A/B/C/ is enlarged by a scale factor of 2 to form ∆ A//B//C//. Draw the enlargement
on DIAGRAM SHEET 1. (2)
7.4 Write down the general rule of the transformation in QUESTION 7.3. (1)
7.5 ∆ABC is reflected about the x-axis and then it is reflected about the y-axis to form
∆DEF.
7.5.1 Write down the coordinates of D, where D is the image of A after the
transformation described above. (2)
7.5.2 Write down the general rule of this transformation in the form:
( x ; y ) → ( ; ) → ( ; ) . (2)
7.5.3 Describe a single transformation that ∆ABC undergoes to form ∆DEF. (2)
[13]
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Mathematics/P2 9 DBE/November 2012
NSC
QUESTION 8
Answer this question WITHOUT using a calculator.
8.1 The point P(k ; 8) lies in the first quadrant y
such that OP = 17 units and TÔP = α P(k ; 8)
as shown in the diagram alongside. 17
α x
O T
8.1.1 Determine the value of k. (2)
8.1.2 Write down the value of cos α . (1)
8.1.3 If it is further given that α + β = 180°, determine cos β . (2)
8.1.4 Hence, determine the value of sin( β − α ). (4)
1 − cos 2 x − sin x
8.2 Consider the expression:
sin 2 x − cos x
1 − cos 2 x − sin x
8.2.1 Prove that: = tan x (4)
sin 2 x − cos x
8.2.2 The above expression is undefined if sin 2 x − cos x = 0 . Solve this
equation in the interval 0° ≤ x ≤ 360° . (4)
[17]
QUESTION 9
sin 2 θ
9.1 Simplify as far as possible: (5)
sin(180° − θ ). cos(90° + θ ) + tan 45°
sin 104°(2 cos 2 15° − 1)
9.2 Simplify without the use of a calculator:
tan 38°. sin 2 412° (8)
[13]
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Mathematics/P2 10 DBE/November 2012
NSC
QUESTION 10
The graphs of f ( x) = sin( x + 30°) and g ( x) = −2 cos x for − 90° ≤ x ≤180° are given below.
The graphs intersect at point P and point Q.
y
2
g
1
f Q
x
-90 -60 -30 0 30 60 90 120 150 180
P
-1
-2
10.1 Calculate f (0) – g(0). (1)
10.2 Calculate the x-coordinates of point P and point Q. (7)
10.3 For which values of x will f ( x) ≥ g ( x) ? (2)
10.4 Graph h is obtained by the following transformation of f: h( x) = 2 f ( x + 60°) .
Describe the relationship between g and h. (2)
[12]
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Mathematics/P2 11 DBE/November 2012
NSC
QUESTION 11
ABCD is a parallelogram with AB = 3 units, BC = 2 units and AB̂C = θ for 0° < θ ≤ 90° .
3
A B
θ
2
3
D C
11.1 Prove that the area of parallelogram ABCD is 6 sin θ . (3)
11.2 Calculate the value of θ for which the area of the parallelogram is 3 3 square units. (3)
11.3 Determine the value of θ for which the parallelogram has the maximum area. (2)
[8]
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Mathematics/P2 12 DBE/November 2012
NSC
QUESTION 12
A hot-air balloon H is directly above point B on the ground. Two ropes are used to keep the
hot-air balloon in position. The ropes are held by two people on the ground at point C and
point D. B, C and D are in the same horizontal plane. The angle of elevation from C to H is x.
CD̂B = 2 x and CB̂D = 90° − x . The distance between C and D is k metres.
H
θ
B
90° – x
2x
x D
C k
12.1 Show that CB = 2k sin x. (5)
12.2 Hence, show that the length of rope HC is 2k tan x. (3)
12.3 If k = 40 m, x = 23° and HD = 31,8 m, calculate θ, the angle between the two ropes. (4)
[12]
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Mathematics/P2 13 DBE/November 2012
NSC
QUESTION 13
The face of a standard clock is positioned such that the centre is at the origin. At a certain time,
the end of the minute hand is at the point P(2 ; 4). 37 minutes later, the end of the minute hand
is at the point P/ (a ; b).
y
P
D/
D
x
O
P/
13.1 Determine the value of a and b. (6)
13.2 OD is the position of the hour hand when the minute hand is at P and OD/ is the
position of the hour hand when the minute hand is at P/. Calculate the angle between
OD and OD/. (4)
[10]
TOTAL: 150
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Mathematics/P2 DBE/November 2012
NSC
CENTRE NUMBER:
EXAMINATION NUMBER:
DIAGRAM SHEET 1
QUESTION 3.2
Physical Science
Mathematics
QUESTION 7.3
y
10
8
6
A/
4
B/
C
A C/
2
B x
-12 -10 -8 -6 -4 -2 0 2 4 6 8 10 12
-2
-4
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Mathematics/P2 DBE/November 2012
NSC
INFORMATION SHEET
− b ± b 2 − 4ac
x=
2a
A = P (1 + ni ) A = P (1 − ni ) A = P(1 − i ) n A = P(1 + i ) n
n(n + 1) n
(2a + (n − 1)d )
n n
∑1 = n
i =1
∑i =
i =1 2
Tn = a + (n − 1)d Sn =
2
Tn = ar n −1 Sn =
(
a r n −1 ) ; r ≠1 S∞ =
a
; −1 < r < 1
r −1 1− r
F=
[
x (1 + i ) − 1
n
] P=
x[1 − (1 + i )− n ]
i i
f ( x + h) − f ( x )
f ' ( x) = lim
h→ 0 h
x + x y + y2
d = ( x 2 − x1 ) 2 + ( y 2 − y1 ) 2 M 1 2 ; 1
2 2
y 2 − y1
y = mx + c y − y1 = m( x − x1 ) m= m = tan θ
x 2 − x1
( x − a )2 + ( y − b )2 = r 2
a b c
In ∆ABC: = = a 2 = b 2 + c 2 − 2bc. cos A
sin A sin B sin C
1
area ∆ABC = ab. sin C
2
sin (α + β ) = sin α . cos β + cos α . sin β sin (α − β ) = sin α . cos β − cos α . sin β
cos(α + β ) = cos α . cos β − sin α . sin β cos(α − β ) = cos α . cos β + sin α . sin β
cos 2 α − sin 2 α
cos 2α = 1 − 2 sin 2 α sin 2α = 2 sin α . cos α
2 cos 2 α − 1
( x ; y ) → ( x cos θ − y sin θ ; y cos θ + x sin θ )
n 2
∑ fx ∑ (x − x )
i
x= σ = i =1
2
n n
n( A)
P ( A) = P(A or B) = P(A) + P(B) – P(A and B)
n(S )
yˆ = a + bx b=
∑ (x − x )( y − y )
∑ (x − x) 2
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