This document contains a mathematics assessment for NCV Level 4 students. It consists of 4 questions testing concepts in complex numbers, derivatives, functions and inequalities. The document provides instructions to markers, including adhering to marking guidelines, giving marks for reasonable answers, and providing constructive feedback. It contains worked solutions and marking allocations for each sub-question.

1. YEAR
TESTS SUB
COVER PAGE FOR ASSESSMENT TOOL MATHS AND
Name of Assessor: LANGA R
Name of Moderator LETLHAGE M.A
Learning Programme: NCV FUNDAMENTAL
Subject: MATHEMATICS
Level: LEVEL 4
Duration: 1½
Total Marks: 50
Instruction to Markers
1. Adhere to the marking guidelines.
2. Ensure consistency and accuracy when marking.
3. Acknowledge students’ own responses.
4. Provide constructive feedback to students.
T: Tests SUB: Substitute Test INT: Internal
2
1 1
2021
3
%
INTERNAL ASSIGN:
SMENT TOOL MATHS AND MATHS LIT:
LANGA R.M
LETLHAGE M.A
NCV FUNDAMENTAL
MATHEMATICS
LEVEL 4
1½HOURS
Ensure consistency and accuracy when marking.
Acknowledge students’ own responses.
Provide constructive feedback to students.
T: Tests SUB: Substitute Test INT: Internal ASSIGN: Assignment
2
1 3 P1 P2
%
INTERNAL ASSIGN:
ASSIGN: Assignment
2 3
1

2. SWGC-TVET COLLEGE SUB-TEST TOOL 2021-03-10 Page 2 of 9
INSTRUCTIONS AND INFORMATION TO MARKERS
1.
2.
3.
As an expert in this subject, kindly use your discretion in marking
this paper.
Give a mark for all reasonable answers.
Mark allocation is clearly stipulated at the bottom of each answer
in brackets and typed in italics.

3. SWGC-TVET COLLEGE SUB-TEST TOOL 2021-03-10 Page 3 of 9
QUESTION 1
√= 1 mark √ = ½ mark
1.1
1.1.1𝑖 (2 + 7𝑖) + 𝑖(3 − 6𝑖) − 16 − 𝑖
=−2 − 7𝑖 + 3𝑖 − 6𝑖 − 16 − 𝑖√
=−18 − 14𝑖√√
1- mark for simplifying
½-mark for each correct answer
(2)
1.1.2 −
= × √ − × √
= − √
= - − +
=−
3
2
−3𝑖
2
√√
½-mark each for rationalising
1-mark for simplifying
½ mark for each correct answer.
(3)
1.2 𝑥 =
±√
=
±√
√
=
−2 ± 6𝑖
4
√
+ √√ or −
½ mark for substitution in correct formula
½ mark for conversion to complex form
½ mark for each answer
(2)

4. SWGC-TVET COLLEGE SUB-TEST TOOL 2021-03-10 Page 4 of 9
1.3
( ) ( )
( °)⁵
=
5 − 2𝑖
(2𝑐𝑖𝑠120°)⁵
√√
=
, ( , °)
°×
√ √
=
, ( , )
=0,168 cis(-21,80-600 )√
= 0.168 𝑐𝑖𝑠𝑛(98,20)√
½ mark for simplifying the numerator
1 mark for converting the numerator to polar
form
1 mark for applying De Moivre’s theorem
1 mark for applying the division rule
½ mark for the modulus and ½mark for the
positive argument
(5)
1.4 𝑥 𝑖 − 𝑦 𝑖 + 6𝑥𝑖 − 𝑥𝑖 = 9𝑖 − 7𝑖
𝑥 √+𝑦 𝑖√−6𝑥√−𝑥𝑖 = −9√−7𝑖
𝑥 − 6𝑥 = −9 … . (1) √𝑦 𝑖 − 𝑥𝑖 = −7𝑖 … (2) √
𝑥 − 6𝑥 + 9 = 0𝑦 − 𝑥 = −7 ……….(2a)
(𝑥 − 3 )(𝑥 − 3 ) = 0
∴ 𝑥 = 3√
Substitute 𝑥 = 3 into equation (2a)
𝑦 − 𝑥 = −7
𝑦 − 3 = −7
𝑦 + 4 = 0
𝑦 = -4
∴ 𝑦 = ±√−4
∴ 𝑦 = ±2𝑖√√
½ mark for the simplification of each correct term
1 mark for each correct equation
1 mark for the correct x-value
½ mark for each correct y-value (5)
[15]

5. SWGC-TVET COLLEGE SUB-TEST TOOL 2021-03-10 Page 5 of 9
QUESTION 2
2.1𝑓(𝑥) = 5 − 2𝑥
𝑓(𝑥 + ℎ) = 5 − 2(𝑥 + ℎ)²
=5 − 2(𝑥 + 2𝑥ℎ + ℎ )
𝑓 (𝑥) = lim
→
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ
=lim →
( ( )
√
=
lim → -√
=lim →
²
√
=lim →
( )
= lim → −4𝑥 − 2ℎ √
=−4𝑥√
½ mark for substitution
½ mark for expanding
½ mark for simplifying the numerator
½ mark for simplification for diving by h
1 mark for the correct answer
(3)
2.2
2.2.1
𝑓(𝑥) = 4𝑥 −
5
√𝑥
–
𝑎
𝑥
–
ln 𝑥
4
− 3
= 4𝑥 − 5𝑥 √- a𝑥 √− 𝑙𝑛𝑥 − 3
𝑓 (𝑥) = 12𝑥 √+ 𝑥 √+𝑎𝑥 √− √ -0√
=12𝑥 + √+ √−
√
= 12𝑥 +
10
3√𝑥
+
𝑎
𝑥
−
1
4𝑥
½ mark for term 2 and term3
½ mark for each derivative
½ mark for each positive exponent
1 mark for changing to surd form
(5)
2.2.2
𝑦 =
𝑒
tan 4𝑥
½ - for applying the quotient rule

6. SWGC-TVET COLLEGE SUB-TEST TOOL 2021-03-10 Page 6 of 9
𝑑𝑦
𝑑𝑥
=
𝑔(𝑥). 𝑓 (𝑥)√−𝑓(𝑥). 𝑔′(𝑥)
[𝑔(𝑥)]²
√
√ √
𝑑𝑦
𝑑𝑥
(𝑡𝑎𝑛4𝑥)(4𝑒 ) − √(𝑒 )(4𝑠𝑒𝑐²4𝑥)
(𝑡𝑎𝑛4𝑥)²√
=
( ² )
( )²
1 mark each for 𝑓′(𝑥)
1 mark each for 𝑔′(𝑥)
½ mark for [𝑔(𝑥)]²
(3)
2.2.3
𝑦 = (3𝑥 − 5)⁵
= 5(3𝑥 − 5) (6𝑥) √√
= 30𝑥 ( 3𝑥 − 5)⁴ √
ALTERNATIVE
𝑦 = (3𝑥 − 5)⁵
𝐿𝑒𝑡 𝑢 = 3𝑥 − 5
𝑑𝑢
𝑑𝑥
= 6𝑥√
∴ 𝑦 = 𝑢
𝑑𝑦
𝑑𝑢
= 5𝑢⁴
√
𝑑𝑦
𝑑𝑥
=
𝑑𝑢
𝑑𝑥
×
𝑑𝑦
𝑑𝑢
= (6𝑥) 5(𝑢 )
= 30𝑥(3𝑥 − 5)⁴√
1 mark for differentiating exponent
1 mark for differentiating the base
1 mark for simplifying
½ mark for
½ mark for
1 mark for the chain rule
1 mark for simplifying
(3)
[14]

7. SWGC-TVET COLLEGE SUB-TEST TOOL 2021-03-10 Page 7 of 9
QUESTION 3
3.1
𝑔(𝑥) = 2𝑥 − 𝑥 − 𝑝𝑥 + 20
√
𝑔(−4) = 2(−4)³ − (−4) − 𝑝(−4) + 20
= −128 − 16 + 4𝑝 + 20
−124 + 4𝑝
Since R = 4
∴ −124 + 4𝑝 = 4√
∴ 4𝑝 = 128
∴ 𝑝 = 32√
½ mark for substituting
½ mark for equating to 4
1 mark for the answer (2)
3.2
𝑓(𝑥) = −2𝑥 − 3𝑥 + 13𝑏𝑥 + 20
𝑙𝑒𝑡 𝑥 + 𝑎 = 0
𝑥 = −𝑎
√
𝑓(−𝑎) = −2(−𝑎) − 3(−𝑎) + 13𝑏(−𝑎) + 20
= 2𝑎 − 3𝑎 − 13𝑎𝑏 + 20
Since R = -56
√
2𝑎 − 3𝑎 − 13𝑎𝑏 + 20 = −56 … … (1)
𝑓(𝑥) = −2𝑥 − 3𝑥 + 13𝑏𝑥 + 20
𝑙𝑒𝑡 𝑥 − 𝑎 = 0
𝑥 = 𝑎
𝑓(−𝑎) = −2(𝑎) − 3(𝑎) + 13𝑏(𝑎) + 20√
= −2𝑎 − 3𝑎 + 13𝑎𝑏 + 20
Since x-a is a factor of f(x) R =0
√
−2𝑎 − 3𝑎 + 13𝑎𝑏 + 20 = 0 … … (2)
Add equation 1+2
2𝑎 − 3𝑎 − 13𝑎𝑏 + 20 = −56
½ mark for substitution
1 mark for the equation (1)
½ mark for substitution
1 mark for the equation (2)

8. -2𝑎 − 3𝑎 + 13𝑎𝑏 + 20 = 0
−6𝑎 + 40 =
−6𝑎 = −96
𝑎 = 16
𝑎 = ±4√
Substitute 𝑥 = 4 𝑜𝑟 𝑥 = −4 into equation (1
2𝑎 − 3𝑎 − 13𝑎𝑏 + 20
2(4) − 3(4) − 13(4)𝑏
128 – 48 - 52b +20=-56
−52𝑏 = −156
= 𝑏 = 3√
3.3.1 𝒇(𝒙) =
𝟏
𝟑
ˣ
𝒚 =
𝟏
𝟑
ˣ
𝒇 𝟏
: x=
𝟏
𝟑
ʸ√
3.3.3 Decreasing function. √
As x-increases, y decreases√
3.3.4 Domain {𝑥: 𝑥 > 0, 𝑥𝜖𝑅} 𝑂𝑅 (0
3.3.5 Range{𝑦: 𝑦 > 0; 𝑦𝜖𝑅} 𝑂𝑅 (0, ∞
−56
96
into equation (1)
20 = −56
( ) + 20 = −56
156
1 mark for two values of a
1 mark for the value of b
1-mark for the correct answer
Graph of 𝑓(𝑥) = ˣ
½- mark for label
½-mark for shape
1 –mark for point (0,1)
Graph of𝑓 : x= ʸ
½- mark for label
½-mark for shape
1 –mark for point (1,0
Graph of 𝑦 = 𝑥
½- mark for label
½-mark for shape
½- mark for decreasing function
½- mark for the reason
0, ∞) √ 1-mark for correct answer
∞) √ mark for correct answer
(5)
(1)
mark for shape
(5)
mark for decreasing function
(1)
mark for correct answer
(1)
mark for correct answer
(1)
[15]

9. SWGC-TVET COLLEGE SUB-TEST TOOL 2021-03-10 Page 9 of 9
QUESTION 4
4.1
4.1.1 40 pears 1 mark for the correct answer (1)
4.1.2 Yes the point (20;60) lies within the
feasible region 2 mark for the correct answer (2)
4.2𝑥 + 𝑦 ≤ 100
𝑦 ≤ 40
𝑥 ≥ 10
𝑦 ≤
3
2
𝑥
½ mark for the correct constrain
1 mark for the correct constrain
½ mark for the correct constrain
1 mark for the correct constrain (3)
[6]