GAUTENG DEPARTMENT OF EDUCATION
PREPARATORY EXAMINATION
2020
TECHNICAL MATHEMATICS
PAPER 2
11092
TIME: 3 hours
MARKS: 150
15 pages + 2 information sheets
P.T.O.
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Technical Mathematics Grade 12 P2 Gauteng Mock 2020 QP (Eng)
Grade 12 Technical Mathematics Paper 2 question paper from the Gauteng Mock Exam 2020 (English), 17 pages. Read online or get the free PDF download, memo linked when available.
Technical Mathematics · Grade 12 · Gauteng Mock Exam · 2020 · English. Question paper, 17 pages. Read online or download the PDF.
- Subject
- Technical Mathematics
- Grade
- Grade 12
- Language
- English
- Document type
- Question paper
- Year
- 2020
- Exam period
- Gauteng Mock Exam
- Paper
- 2
- Curriculum
- CAPS
- Pages
- 17
- File size
- 504.9 KB
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TECHNICAL MATHEMATICS 2
(Paper 2) 11092/20
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, etc. that you used in determining your 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. Write neatly and legibly.
P.T.O.
TECHNICAL MATHEMATICS 3
(Paper 2) 11092/20
QUESTION 1
Points A (–1; 3), C (0; – 2) and E are the vertices of ∆ACE. D (1; – 1) is a point on line CE and B,
a point on line AC. The equation of line AE is 𝑦 = −𝑥 + 2. F is the y-intercept of AE.
𝑦
Aሺ−1; 3ሻ
F
B
O E
G 𝑥
Dሺ1; −1ሻ
Cሺ0; −2ሻ
1.1 Determine the co-ordinates of point E, the x-intercept of AE. (2)
1.2 Determine the co-ordinates of point B, if AB = BC. (2)
1.3 Show that BD || AE. (3)
1.4 ̂ C.
Determine the size of OG (3)
1 1 1
1.5 Show that OB = 2AF, if B(− 2 ; 2) is a point on AC. (3)
[13]
P.T.O.
TECHNICAL MATHEMATICS 4
(Paper 2) 11092/20
QUESTION 2
2.1 The diagram below is a representation of measurements taken from the design of a tapered
roller bearing. O (0; 0) is the centre of the circle of the rolling element. Nሺ12 ; −9ሻ and L are
points on the circle so that MN and ML are tangents to the circle. The point Kሺ0 ; 𝑎ሻ is a point
on the circle.
𝑦
Kሺ0; 𝑎ሻ
O
𝑥
L Nሺ12; −9ሻ
M
2.1.1 Determine the equation of the circle, in the form 𝑥 2 + 𝑦 2 = 𝑟 2 . (2)
2.1.2 Determine the value of 𝑎. (1)
2.1.3 Determine the equation of the tangent MN, in the form y = mx +c (3)
P.T.O.
TECHNICAL MATHEMATICS 5
(Paper 2) 11092/20
2.2 An ellipse is drawn in the diagram below. The ellipse intersects the
𝑥-axis at point Eሺ−4 ; 0ሻ and the 𝑦-axis at F.
𝑥² 𝑦²
The equation of another ellipse is given as 9 + 16 = 4.
y
x
𝑥² 𝑦²
2.2.1 Write the equation of the ellipse 9 + 16 = 4 in the form
𝑥² 𝑦²
+ 𝑏² = 1.
𝑎² (1)
x2 y2
2.2.2 Hence, draw the ellipse defined by + = 4 on the set of axes
9 16
provided. Clearly show all the intercepts with the axes. (3)
2.2.3 The drawn ellipse in QUESTION 2.2.2 intersect the x-axis at D and
the y-axis at G. Determine the co-ordinates of point F. (3)
[13]
P.T.O.
TECHNICAL MATHEMATICS 6
(Paper 2) 11092/20
QUESTION 3
In the diagram below, P(√11 ; 𝑏) is a point in the fourth quadrant of the Cartesian plane. It is given that
√11
cos α = .
6
𝑦
O 𝑥
𝛼
P ሺ√11; 𝑏ሻ
Answer the following questions WITHOUT the use of a calculator.
3.1 Show that 𝑏 = −5. (2)
3.2 Determine:
3.2.1 sec𝛼. tan𝛼 (3)
3.2.2 sinሺ180° − 𝛼ሻ (2)
3.3 Use the values in the diagram to show that cosec²𝛼 − cot 2 𝛼 = 1. (3)
[10]
QUESTION 4
4.1 Simplify each of the following WITHOUT the use of a calculator.
4.1.1 √1 − sin2 𝑥. tan𝑥. sin ሺ180° + 𝑥ሻ (5)
sin240°. sec30°
4.1.2
cos𝜋. tan315° (6)
4.2 Solve for 𝑥 if 2cos3𝑥 = cot 65,7° and 𝑥 ∈ [0° ; 360°] (5)
[16]
P.T.O.
TECHNICAL MATHEMATICS 7
(Paper 2) 11092/20
QUESTION 5
The diagram below shows the graph of 𝑓ሺ𝑥ሻ = 𝑎𝑠𝑖𝑛𝑥 for 𝑥 ∈ [0°; 360°].
5.1 Write down the value of a. (1)
5.2 Sketch the graph of 𝑔ሺ𝑥ሻ = tan ሺ𝑥 − 45°ሻ for 𝑥 ∈ [0° ; 360°] on the
same system of axes as 𝑓ሺ𝑥ሻ in the ANSWER BOOK. Clearly show the
co-ordinates of all points of intersection with the axes, asymptotes and all turning points
and end points of the graph. (4)
5.3 Use the graphs of 𝑓 and 𝑔 to answer the following questions.
5.3.1 Write down the period of 𝑔. (1)
5.3.2 Give the equation(s) for the asymptote(s) of 𝑔. (2)
5.3.3 Determine 𝑓ሺ270°ሻ − 𝑔ሺ270°ሻ. (1)
5.3.4 For which values of x will 𝑔ሺ𝑥ሻ > 0? (3)
5.3.5 Determine the value of x where 𝑔ሺ𝑥ሻ = 𝑓ሺ𝑥ሻ + 3. (1)
[13]
P.T.O.
TECHNICAL MATHEMATICS 8
(Paper 2) 11092/20
QUESTION 6
The main cable configuration (i.e. an arrangement of parts or elements in a particular form) of a mobile
crane is shown in the diagram below.
AE = 7,5 m; AD = 9,6 m and BC = 2,3 m. AD ̂ B = 61° and BĈD = 78°.
A
7,5 m 9,6 m
B
2,3 m
61°
78°
E D C
Determine the following, correct to TWO decimal places.
6.1 ̂E
The size of AD (2)
6.2 ̂C
The size of BD (1)
6.3 The length of BD (3)
6.4 The length of AB (3)
[9]
P.T.O.
TECHNICAL MATHEMATICS 9
(Paper 2) 11092/20
QUESTION 7
In the diagram below, O is the centre of the circle and AB is a chord such that OE ⊥ AB.
AB = 16 cm, ED = x and OD = 6 cm.
A
1 2
1 2
E C
𝑥 D 6 O
B
7.1 Calculate the length of the radius of the circle. (4)
7.2 Calculate the value of 𝑥. (1)
7.3 Calculate the length of AC. (3)
[8]
P.T.O.
TECHNICAL MATHEMATICS 10
(Paper 2) 11092/20
QUESTION 8
8.1 Complete the following: The exterior angle of a cyclic quadrilateral is … (1)
8.2 In the diagram below, A, B, C and P are points on a circle. AB = AC and
AB̂ C = 80°. AP and BC produced meets at D.
A
2 P
1
1
2 3 D
2
3
1
C
80°
1 2
B
8.2.1 Determine, with reasons, another THREE angles equal to 80°. (5)
̂ C = 20°. Calculate the size of ̂
8.2.2 It is also given that PA B1 . (3)
8.3 Prove, using appropriate calculations that AB is a tangent to a circle through points B, D
and P. (3)
[12]
P.T.O.
TECHNICAL MATHEMATICS 11
(Paper 2) 11092/20
QUESTION 9
In the diagram below, A, G and B are points on DE of ∆DEF. BC || EF and
AC || DF. DA = 4 cm, BG = 3 cm, BE = 6 cm and CF = 8 cm.
D
4
A
G
3
B C
6 8
E F
9.1 Calculate, with reasons, the lengths of:
9.1.1 GC (3)
9.1.2 AG (2)
AB
9.2 Write down the value of DE , in its simplest form. (1)
1
9.3 The area of a triangle can be calculated by Area ∆ = 2 ሺbase) (⊥ height).
If the height of ∆DEF = 3 × the height of ∆ABC, calculate the ratio between the areas
of ∆ABC and ∆DEF. (3)
[9]
P.T.O.
TECHNICAL MATHEMATICS 12
(Paper 2) 11092/20
QUESTION 10
In the diagram below, AC is the diameter of the circle and SA is a tangent to the circle at point A.
AC || BT with B on the circumference of the circle and T a point on SA.
AT intersects BC at S.
C
B
1
3 2
1
S 1 2
T A
10.1 ̂ C = 90°.
Give a reason why AB (1)
10.2 ̂ T = 90°.
Give a reason why CA (1)
10.3 Prove that ∆ACS ||| ∆BCA. (4)
10.4 Show that AC² = BC.CS. (2)
10.5 Calculate the length of BC if AC = 75 mm and CS = 125 mm. (2)
10.6 Determine the length of TS if AT = 36 mm. (3)
[13]
P.T.O.
TECHNICAL MATHEMATICS 13
(Paper 2) 11092/20
QUESTION 11
The diagram below, shows a cross section of a water pipe partially filled with water. The radius of the
pipe is 1,5 m and the length is 5 m. The segment ABCD represents the water level. M is the centre of the
̂ C = 125° and MB = 0,75 m.
circle, AM
M
1,5 m
125°
0,75m
A C
B
D
11.1 Calculate the length of AB, correct to ONE decimal place. (2)
11.2 Calculate the area of ∆AMC. (3)
11.3 Convert 125° to radians. (1)
11.4 Calculate the area of sector ADCM. (3)
11.5 Hence calculate the area of segment ABCD. (2)
11.6 Calculate the volume of water in the pipe if it is given that the volume of a
prism = area of the base x height. (2)
[13]
P.T.O.
TECHNICAL MATHEMATICS 14
(Paper 2) 11092/20
QUESTION 12
12.1 A wheel has a circumferential velocity of 10,3 metres per second.
12.1.1 Calculate the diameter of the wheel if the wheel turns at 13 revolutions per
minute. (3)
12.1.2 Calculate the angular velocity of the wheel in radians per second. (3)
12.2 The diagram below, shows a lampshade with the following dimensions.
CE = 23 cm and EF = 40 cm. AB = 1 cm and AC = 2 cm. The lampshade is open at the
top and the bottom. The following formulae may be used:
1
Volume of a cone = 3 𝜋𝑟²ℎ
Surface area of a cone = 𝜋𝑟 2 + 𝜋𝑟𝑠
(Use 𝜋 = 3,14ሻ
12.2.1 Give the radius, DE, of the bigger circle at the bottom of the lampshade. (1)
12.2.2 Calculate the radius, BC of the smaller circle at the top of the lampshade. (2)
12.2.3 Calculate the surface area of the lampshade. (3)
12.2.4 Calculate the volume of the lampshade. (4)
[16]
P.T.O.
TECHNICAL MATHEMATICS 15
(Paper 2) 11092/20
QUESTION 13
Polystyrene is used to manufacture cornices. A cross section of a cornice is shown in the diagram below.
The width, BC, of the cornice is 25 cm and is divided into 5 sections of equal widths.
The length of the ordinates are:
10 cm ; 10 cm ; 7,3 cm ; 4,1 cm ; 6 cm ; 1 cm
Determine the area of the cross section, ABCD, by using the mid-ordinate rule. (5)
[5]
TOTAL: 150
END
TECHNICAL MATHEMATICS 16
(Paper 2) 11092/20
INFORMATION SHEET: TECHNICAL MATHEMATICS
− b b 2 − 4ac b 4ac − b2
x= x=− y=
2a 2a 4a
𝑎 𝑥 = 𝑏 ⇔ 𝑥 = 𝑙𝑜𝑔𝑎 𝑏 , 𝑎 > 0, 𝑎 ≠ 1 and 𝑏 > 0
A = P(1 + ni) A = P(1 − ni) A = P(1 − i) n A = P(1 + i) n
m
im
ieff = 1 + − 1
m
f ( x + h) − f ( x )
f / ( x ) = lim
h→0 h
xn+1
x dx = n +1 + C , n −1
n
1 𝑎𝑥
∫ 𝑥 𝑑𝑥 = 𝑙𝑛ሺ 𝑥ሻ + 𝐶, 𝑥 > 0 ∫ 𝑎 𝑥 𝑑𝑥 = 𝑙𝑛 𝑎 + 𝐶, 𝑎 > 0
x + x2 y1 + y2
d = ( x2 − x1 ) 2 + ( y2 − y1 ) 2 M 1 ;
2 2
y 2 − y1
y = mx + c 𝑦 − 𝑦1 = 𝑚ሺ𝑥 − 𝑥1 ሻ m= m = tan
x 2 − x1
x2 y2
+ =1
a2 b2
a b c
In ABC: = = a 2 = b2 + c 2 − 2bc. cosA
sin A sin B sin C
1
area of ΔABC = 2 𝑎𝑏 ⋅ sin C
sin2 θ + cos2 θ = 1 1 + tan2 θ = sec 2 θ cot 2 θ + 1 = cosec 2 𝜃
TECHNICAL MATHEMATICS 17
(Paper 2) 11092/20
rad = 180
Angular velocity = 𝜔 = 2𝜋𝑛 = 3600 𝑛 where n = rotation frequency
Circumferential velocity = 𝜈 = 𝜋𝐷𝑛 where D = diameter and n = rotation frequency
s = r where r = radius and 𝜃 = central angle in radians
𝑟𝑠 𝑟2𝜃
Area of 𝑎 sector = 2 = 2 where r = radius, s = arc length and
𝜃 = central angle in radians
4ℎ2 − 4𝑑ℎ + 𝑥 2 = 0 where h = height of segment, d = diameter of circle and
x = length of chord
o +o
AT = a 1 n + o2 + o3 + o4 + . . . + on−1 where a = equal parts, oi = ith ordinate and
2
n = number of ordinates
OR
o +o
AT = a ( m1 + m2 + m3 + . . . + mn ) where a = equal parts, m1 = 1 2 and
2
n = number of ordinates
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