SCHOOL BASED
ASSESSMENT TASK
GRADE 12
TERM 1
2
2021
MARKS: 50
TIME: 1 Hour
SASAMS WEIGHTING: 15
0
This question paper consists of 4 pages.
INSTRUCTIONS AND INFORMATION
Read the following instructions carefully before answering the questions.
1.
This task paper consists of 2 tasks (sub-sections)
2.
Answer ALL sections.
3.
Clearly show ALL calculations, diagrams, graphs, et cetera which you have 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 nongraphical),
unless stated otherwise.
6.
If necessary, answers should be rounded off to TWO decimal places, unless state
otherwise
7.
Diagrams are NOT necessarily drawn to scale.
8.
Write neatly and legibly.
QUESTION 1 : [16]
1.1.
1.2.
A quadratic pattern has a second term equal to 1, a third term equal to −6 and a
fifth term equal to −14.
1.1.1. Calculate the second difference of this quadratic pattern.
(5)
1.1.2. Hence, or otherwise, calculate the first term of the pattern.
(2)
1; 3 ; 5 are the first three terms of the first differences of a quadratic sequence. The
7th term of the quadratic sequence is35.
1.2.1. Determine the 6th and 5th terms of the quadratic sequence.
(4)
1.2.2. Determine the nth term of the quadratic sequence.
(5)
QUESTION 2 : [21]
2.1.
2.2.
Given the sequence: 2 ; 5 ; 8 ; ...
2.1.1. If the pattern continues, then write down the next TWO terms.
(1)
2.1.2. Prove that none of the terms of this sequence are perfect squares.
(5)
Consider the sequence of numbers: 2 ; 5 ; 2 ; 9 ; 2 ;13; 2 ;17 ; ...
Calculate the sum of the first 100 terms of the sequence.
(5)
2.3. A large quantity of waste material contains metal. At the successive passes through a
recovery process the mass of the metal recovered is:
•
32kg at the first pass;
•
24kg at the second pass;
•
18kg at the third pass;
•
13, 5kg at the fourth pass, and so on, to form a geometric sequence.
2.3.1. How much metal would be recovered at the eighth pass?
(Round off the answer to TWO decimal places)
(3)
2.3.2. Find the total mass of metal recovered in the first eight passes.
(Round off the answer to TWO decimal places)
(4)
2.3.3. In order to be economical at least 2, 4kg of metal must be recovered in any
pass.
Show that the eleventh pass would be uneconomical.
(3)
QUESTION 3 : [13]
3.1
Complete the statement so that it is TRUE:
The angle between the tangent to a circle and the chord drawn from the point of
contact is equal to the angle.........
(l)
3.2
In the diagram below, two unequal circles intersect at A and B. AB is
produced to C such that CD is a tangent to the circle ABD at D. F and G are
points on the smaller circle such that CGF and DBF are straight lines. AD
and AG are drawn.
D
C
Prove that:
‸
‸
‸
3.2.1
B4 = D1 + D2
(4)
3.2.2
AGCD is a cyclic quadrilateral
(4)
3.2.3
CD2 = AC.BC
(4)
TOTAL: [50]
ASSIGNMENT MEMORANDUM
QUESTION 1
1.1.1.
The 2nd , 3rd , 4th and 5th terms are:
1 ; − 6 ; T4 ; and − 14
First differences are: − 7 ; T4 + 6 ; and − 14 − T4
 d = T3 − T2 = T2 − T1
 (− 14 − T4 ) − (T4 + 6) = (T4 + 6) − (− 7)
√ −7
T4 + 6
√ − 14 − T4
√
√ setting up equation
 −20 − 2T4 = T4 + 13
 −33 = 3T4
 T4 = −11
d = −11 + 13 = 2 or − 20 − 2(− 11) = 2
√ answer
OR
√ −7
√ −7+d
√ − 7 + 2d
T5 − T2 = (T5 − T4 ) + (T4 − T3 ) + (T3 − T2 )
 −15 = (− 7 + 2d ) + (− 7 + d ) + (− 7 )
 −15 = −21 + 3d
 6 = 3d
d = 2
√ setting up equation
√ answer
(5)
1.1.2.
√ method
 T1 = 1 − (− 9 )
 T1 = 10
OR
2a = 2
a = 1
√
T1 = 10
5a + b = T3 − T2
 5(1) + b = −7
 b = −7 − 5 = −12
√ method
4a + 2b + c = T2 or 9a + 3b + c = T3
 4(1) + 2(− 12 ) + c = 1
 c = 21
 Tn = n 2 − 12n + 21
 T1 = (1) − 12(1) + 21 = 10
2
√
T1 = 10
(2)
1.2.1.
1.2.2.
Tn = 2n − 1 for the first difference sequence
T6 = 2(6) − 1 = 11
 T6 of original seq. = 35 − 11 = 24
√ 6 term of 1 diff seq.
T5 = 2(5) − 1 = 9
√ 5 term of 1 diff seq.
 T5 of original seq. = 24 − 9 = 15
√ 5 term of quadr. seq.
(4)
2a = 2
a = 1
th
th
√ 6 term of quadr. seq.
th
th
√ value of a
c=0
√ value of c
Tn = an + bn + c
2
35 = 1(7 ) + b(7 ) + 0
− 14 = 7b
 b = −2
2
 Tn = n − 2n
2
√ T7 = 35
√ value of b
√ answer
OR
36a + 6b + c = 24 → (1)
25a + 5b + c = 15 → (2)
49a + 7b + c = 35 → (3)
(1) − (2) : 11a + b = 9 → (4)
(3) − (1) : 13a + b = 911 → (5)
(5) − (4) : 2a = 2
st
√ 11a + b
√ 13a + b
√ value of a
√ value of b
√ answer
st
a = 1
 b = −2
(5)
 Tn = n 2 − 2n
[16]
QUESTION 2
2.1.1.
11 ; 14
2.1.2.
Tn = 3n − 1
√√ for n term
Now if 3n − 1 = n 2
√ equating n term to n
√ answer (both terms )
(1)
th
th
Then n 2 − 3n + 1 = 0
3 5
   a perfect square
2
 n is irrational
But n must be natural
 No term is a perfect square
.............(since   a perfect square)
√ value of n
√ for concluding n is
(2 + 2 + ... + 2)+ (5 + 9 + 13 + ...)
√ (2 + 2 + ... + 2)
n =
2.2.
for 50 terms
for 50 terms
50
50
i =1
i =1
=  (2 ) +  (4i + 1)
50
= 2(50 ) + 2(5) + 49(4)
2
= 100 + 25(10 + 196 )
= 100 + 5 150
= 5 250
irrational and for
deducing n is natural
(5)
for 50 terms
√ (5 + 9 + 13 + ...)
for 50 terms
√
setting up  expression
√ substitution
√ answer
(5)
2.3.1.
2.3.2.
3
a = 32 ; r = ; n = 8
4
7
3
T8 = 32 
4
2 187
=
= 4, 27
512
Sn =
(
a 1− r
1− r
n
)
√ value of r
√ substitution
√ answer
(3)
√ formula
2
  3 8 
32 1 −   
  4  
S8 = 
3
1−
4
  3 8 
= 128 1 −   
  4  
= 115,19
2.3.3.
10
3
T11 = 32 
4
= 1, 80
 2, 4
Hence, the 11th pass is uneconomical.
√ substitution
√ answer
√ correct rounding − off
(4)
√ substitution
√
value of T11
√ conclusion
(3)
[21]
QUESTION 3
3.1 . . . in the alternate segment.
3.2.1
(1)
B4 = A1 + D2
Exterior angle of triangle
✓✓S/R
D1 = A1
Tan-chord theorem
✓✓S/R
∴B4 = D1 + D2
(4)
3.2.2
G2 = B2
Angles on the same segment
✓S/R
B2 = B4
Vertically opposite angles
✓S/R
G2 = B4
Proven
✓S
Exterior angle of the quad. = interior
✓R
= D1 + D2
AGCD is a cyclic quad.
opposite/
Converse exterior angle of a cyclic quad.
(4)
3.2.3 C1 = C1
A1 = D1
∴ 𝐷1 + 𝐷2 = 𝐵4
∆𝐶𝐴𝐷///∆𝐶𝐷𝐵
∴
Common
✓
Proven/Tan-chord theorem
✓
Proven
✓
A.A.A.
✓
𝐴𝐷 𝐶𝐷 𝐴𝐶
=
=
𝐵𝐷 𝐵𝐶 𝐶𝐷
CD2 = AC.BC
(4)
[11]
TOTAL: [50]