# Mathematics Grade 12 KZN April 2021 Memo

> Grade 12 Mathematics memorandum from the KZN April 2021, 18 pages. Read online or get the free PDF download, question paper linked when available.

- Page: https://www.hlayiso.com/library/maths-april-qp-memo-kzn-2021-gr12/67bbc4ec-089a-46c9-9777-c257c8c03f13
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 12
- Document type: Memorandum
- Year: 2021
- Exam period: KZN April
- Curriculum: CAPS
- Pages: 18

Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 12](https://www.hlayiso.com/library/grade-12) › [Mathematics](https://www.hlayiso.com/library/grade-12/mathematics) › [Mathematics Grade 12 KZN April 2021 Memo](https://www.hlayiso.com/library/maths-april-qp-memo-kzn-2021-gr12/67bbc4ec-089a-46c9-9777-c257c8c03f13)

## About this document

Mathematics · Grade 12 · KZN April · 2021. Memorandum, 18 pages. Read online or download the PDF.

Read it free online on the page above; download the PDF from there.

## Document text

Text extracted from the PDF. Layout, diagrams and tables may not survive extraction.

### Page 1

Downloaded from hlayiso.com

ROAWACULAS =!

EDUCATION
REPUBLIC OF SOUTH AFRICA

SENIOR CERTIFICATE

MATHEMATICS

COMMON TEST

APRIL 2021

MARKS: 100

TIME: 2 hours

N.B. This question paper consists of 6 pages, an answer sheet,
1 diagram sheet and an information sheet.

Copyright Reserved

Please Turn Over

——I

### Page 2

Downloaded from hlayiso.com

Mathematics 2 April 2021 Common Test

NSC

INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.

i

This question paper consists of 7 questions.
Answer ALL questions.

Clearly show ALL calculations, diagrams, graphs, et cetera that you have used in
determining your answers.

Answers only will not necessarily be awarded full marks.

An approved scientific calculator (non-programmable and non-graphical) may be
used, unless stated otherwise.

If necessary, answers should be rounded off to TWO decimal places, unless stated
otherwise.

Diagrams are NOT necessarily drawn to scale.

Number the answers correctly according to the numbering system used in this
question paper. Write neatly and legibly.

Copyright Reserved Please Turn Over

### Page 3

Downloaded from hlayiso.com

Mathematics 3 April 2021 Common Test
NSC

QUESTION 1

Given the quadratic sequence: 44; 52; 64; 80; ...

1.1 Write down the next two terms of the sequence. (2)
1.2 Determine the n" term of the quadratic sequence. (4)
1.3 Calculate the 30" term of the sequence. (2)
1.4 Prove that the quadratic sequence will always have even terms. (3)

[11]
QUESTION 2

The 8" term of an arithmetic sequence is 31 and the sum of the first 30 terms is 1830.
Determine the first three terms of the sequence. {7]

QUESTION 3

3.1 The second term of a geometric sequence = and the ninth term is 5.
Determine the value of the common ratio. (5)

3.2. Calculate the value of m if

m
255
- ko ee
> -8)-(05) =
k=1
(4)
3.3. Given: a+ 12+ 6x + 3x7 4+ sx #0.
3.3.1 Determine the value of x for which the series converges. (3)
3.3.2 Write down the value of x for which the series is increasing. (2)
[14]

Copyright Reserved Please Turn Over

### Page 4

Downloaded from hlayiso.com

Mathematics 4 April 2021 Common Test
NSC

QUESTION 4

4.1. Given AABC and ADEF with A = D,B = Band¢ =F.

AB _ AC
Prove that ne OF (7)

4.2 Inthe figure AD, DC and BE are tangents to the circle at A, C and B respectively.
O is the centre of the circle. DE and AF intersect at H. AH produced meets BE
produced in F. AJ, BC and BJ are chords.

D

Prove that:
4.2.1 ADAH ||| AOCH. (4)
422 AO.DH
OH = — (6)
4.2.3. If BAisdrawn, then BF? = JF .AF (6)
{23]

Copyright Reserved Please Turn Over

### Page 5

Downloaded from hlayiso.com

Mathematics ns} April 2021 Common Test
NSC

QUESTION 5

In the figure AF = 2CG and FE || GB. s = :

Determine (with reasons):

AF
5.1 = 2
FG (2)
CH
5.2 = 4
HE (4)
Ee) 1 ceihcollt latches (4)
Area of AAFE
{10}
QUESTION 6
6.1 Given cos 26° = :
Without using a calculator, calculate the value of the following in terms of p.
6.1.1 cos52” (4)
6.1.2 sin71° (4)
6.2 Simplify without using into a single trigonometric ratio.
cos(-180’).tan @.cos 690 .sin (9-180") 6)

cos?(@—90°)
6.3 Show that
cos0° +cos1° + cos2” + ++»+cos178 +cos179° + cos 180° + 6sin90° = 6 (4)

{17]
Copyright Reserved Please Turn Over

### Page 6

Downloaded from hlayiso.com

Mathematics 6 April 2021 Common Test

NSC

QUESTION 7

71 Prove the following identity:

1 — sin 2x :
———_—— = sin x - cos x
sin x — cosx

7.2 Determine the general solution of:

tan 3x. -1=0

tan 24°

(3)

(5)

7.3. Determine the maximum value of 3 sin x + cos.x, without the use of a calculator. (4)

74 Given: f(x) = 2cos(x — 30°)

7.3.1 Sketch the graph of f for the domain x € [-90°; 270°] on the axes provided. (2)

7.3.2 Use the letters P and Q to indicate on the graph the solution of the equation
cos(x — 30°) = 0,5 and the x — coordinates of P and Q.

Copyright Reserved

(4)
[18]

Please Turn Over

### Page 7

Downloaded from hlayiso.com

i
i

ato
|

Please Turn Over

April 202{ Common Test

GRADE:
1x0

f
iho

|

130
|
|

a.
NSC

oo
|
|

ES i I
lo 3 7
& bn ats a

ve : a fa
an ~ 6 3
2 a g 4
= a a | z

£ g |
S z = 1° 5

dOvd 4dO-UVAL

### Page 8

Downloaded from hlayiso.com

April 2021 Common Test

8
NSC

Mathematics

GRADE:

DIAGRAM SHEET

QUESTION 4.1

QUESTION 4.2

QUESTION 5

Please Turn Over

Copyright Reserved

### Page 9

Downloaded from hlayiso.com

Mathematics

9 April 2021 Common Test
NSC

INFORMATION SHEET: MATHEMATICS

INLIGTING BLADSY
_-bt1b? —4ac
2a
A=P(l+ni) A= P(I-ni) A=P(-i)" A=P(i+i)"
T, =a+(n-l)d S, = 5 (2a+(n-1)d)
T, =ar"" s el r#l Se
or a
pol+iy 1 powl-d+i]
i i
f'(x)= lim L(x+h)- f(*)
h>0 h
d=\(x, =x)? +0,-9,)? M2572 21522)
2 2
y2 741
y=mx+e yry, =m(x-x,) m= m=tan@
x4 Xx,

(x-a) +(y—bY =r?

a b c

In AABC: — :
sinA sinB

sinC

sin(a + 8) = sina.cos f+ cos a.sin B

cos(@ + 8) = cos a.cos B—sina.sin B
cos’ asin? a

cos 2a@ = 41-2sin’? a

2cos?a-1

p=atbhx

Copyright Reserved

a? =b* +¢7 -2be.cos A area KMABC = x ab.sin C

sin(a- B) = sina.cos # —cosa.sin B

cos(a ~ £2) = cos a.cos 7 +sina.sin B

sin 2a@ = 2sina.cosa

P(A or B)= P(A) + P(B)— P(A and B)

Ae YG - xKy a)
Sa@-xy7

### Page 10

Downloaded from hlayiso.com

                                NATIONAL
                           SENIOR CERTIFICATE

                                    GRADE 12

                               MATHEMATICS

                         MARKING GUIDELINE

                               COMMON TEST

                                  APRIL 2021

 MARKS: 100

                         This memorandum consists of 9 pages.

Copyright Reserved                                              Please Turn Over

### Page 11

Downloaded from hlayiso.com
Mathematics                                          2                                 April 2021 Common Test
                                           NSC – Marking Guideline

QUESTION 1

1.1       100 ; 124                                                  AA✓✓answers                          (2)

1.2                  44          52           64            80

              1D           8          12             16

              2D                 4            4

          𝟐𝒂 = 𝟒 ∴ 𝒂 = 𝟐                                             A✓a value
          𝟑𝒂 + 𝒃 = 𝟖 ∴ 𝒃 = 𝟐                                         CA✓b value
          𝒂 + 𝒃 + 𝒄 = 𝟒𝟒 ∴ 𝒄 = 𝟒𝟎                                    CA✓c value
                                                                     CA✓nth term
          𝑻𝒏 = 𝟐𝒏𝟐 = 𝟐𝒏 + 𝟒𝟎
                                                                     OR
          OR
                                                                     A✓a value
          𝟐𝒂 = 𝟒 ∴ 𝒂 = 𝟐
                                                                     CA✓b value
          𝟑𝒂 + 𝒃 = 𝟖 ∴ 𝒃 = 𝟐
                                                                     CA✓c value
          ∴ 𝒄 = 𝑻𝟎 = 𝟒𝟎
                                                                     CA✓nth term                          (4)
          𝑻𝒏 = 𝟐𝒏𝟐 + 𝟐𝒏 + 𝟒𝟎
                                                                     OR
          OR

          𝑻𝒏 = 𝑻𝟏 + (𝒏 − 𝟏)𝒅𝟏 + (𝒏 − 𝟏)(𝒏 − 𝟐)𝒅𝟐
                                                                     OR
          OR
                   (𝒏 − 𝟏)                                                                                (4)
          𝑻𝒏 =             [𝟐𝒂 + (𝒏 − 𝟐)𝒅] + 𝑻𝟏
                      𝟐

1.3       𝑻𝟑𝟎 = 𝟐(𝟑𝟎)𝟐 + 𝟐(𝟑𝟎) + 𝟒𝟎 = 𝟏𝟗𝟎𝟎                           CA✓substitution
                                                                     CA✓answer                            (2)

1.4       𝑻𝒏 = 𝟐𝒏𝟐 + 𝟐𝒏 + 𝟒𝟎
          𝑻𝒏 = 𝟐(𝒏𝟐 + 𝒏 + 𝟐𝟎)                                        A✓Taking out common factor of 2
          𝟐(𝒏𝟐 + 𝒏 + 𝟐𝟎) is even for all 𝒏 ∈ 𝒁                       A✓Rewriting nth term                 (3)
                                                                     A✓ is even for all 𝒏 ∈ 𝒁
                                                                     Note: Mark CA provided
                                                                     𝑻𝒏 (from1.2) is a factor of 2
                                                                                                         [11]

Copyright Reserved                                                                            Please Turn Over

### Page 12

Downloaded from hlayiso.com
Mathematics                                       3                                April 2021 Common Test
                                        NSC – Marking Guideline

QUESTION 2

2.        𝒂 + 𝟕𝒅 = 𝟑𝟏           → (𝟏)                             A✓equation (1)
          𝟏𝟓(𝟐𝒂 + 𝟐𝟗𝒅) = 𝟏𝟖𝟑𝟎
          𝟐𝒂 + 𝟐𝟗𝒅 = 𝟏𝟐𝟐        → (𝟐)                             A✓equation (2)
          𝒂 = 𝟑𝟏 − 𝟕𝒅           → (𝟑)                             CA✓making a the subject
          𝟐(𝟑𝟏 − 𝟕𝒅) + 𝟐𝟗𝒅 = 𝟏𝟐𝟐                                  CA✓correct substitution of a
          𝟔𝟐 − 𝟏𝟒𝒅 + 𝟐𝟗𝒅 = 𝟏𝟐𝟐
                        𝟏𝟓𝒅 = 𝟔𝟎
                          𝒅=𝟒                                     CA✓d value
                          𝒂=𝟑                                     CA✓a value
          𝟑 ; 𝟕 ; 𝟏𝟏 ; …                                          CA✓sequence                         [7]

QUESTION 3

3.1               𝟓                                               A✓equation (1)
          𝒂𝒓 =             → (𝟏)
                 𝟏𝟐𝟖

          𝒂𝒓𝟖 = 𝟓          → (𝟐)                                  A✓equation (2)

          𝒓𝟕 = 𝟏𝟐𝟖                                                CA✓ 𝒓𝟕 = 𝟏𝟐𝟖
          𝒓𝟕 = 𝟐 𝟕                                                CA✓exponential form
          𝒓=𝟐                                                     CA✓answer                           (5)

3.2       (−𝟖) + (−𝟖)(𝟎. 𝟓) + (−𝟖)(𝟎. 𝟓)𝟐 + ⋯                     A✓generating series
          −𝟖(𝟎. 𝟓𝒎 − 𝟏)    𝟐𝟓𝟓                                    CA✓correct substitution into
                        =−                                        correct formula
            𝟎. 𝟓 − 𝟏       𝟏𝟔

                          𝟐𝟓𝟓
          𝟎. 𝟓𝒎 − 𝟏 = −
                          𝟐𝟓𝟔
                      𝟏                                           CA✓writing in exponential
          𝟎. 𝟓𝒎 =        = 𝟎. 𝟓𝟖                                  form or using logs
                     𝟐𝟓𝟔
                                                                                                      (4)
              𝒎=𝟖                                                 CA✓answer
3.3.1     −𝟏 < 𝒓 < 𝟏                                              A✓condition for convergence
               𝒙
          −𝟏 < < 𝟏                                                A✓r value
               𝟐

          −𝟐 < 𝒙 < 𝟐                                              CA✓answer                           (3)

3.3.2      𝒙 < −𝟐 𝒐𝒓 𝒙 > 𝟐                                        CACA✓✓answer                        (2)

                                                                                                     [14]

Copyright Reserved                                                                        Please Turn Over

### Page 13

Downloaded from hlayiso.com
Mathematics                                            4                            April 2021 Common Test
                                             NSC – Marking Guideline

QUESTION 4

                   A                                                   D

       P                      Q                           E                    F

 B                                C

 4.1       Draw AP = DE and AQ = DF                              ✓S Construction (or could be
           In ∆𝐀𝐁𝐂 𝐚𝐧𝐝 ∆𝐃𝐄𝐅                                      shown on diagram)
               1. AP = DE       (Construction)
               2. AQ = DF       (Construction)
                   ̂
               3. 𝐀 = 𝐃 ̂        (Given)
               ∴ ∆𝐀𝐏𝐐 ≡ ∆𝐃𝐄𝐅 (SAS)
               Now 𝐀𝐏 ̂𝐐 = 𝐃𝐄̂𝐅                                  ✓S/R
               But 𝐃𝐄̂𝐅 = 𝐁 ̂    (Given)                         ✓S
               ∴ 𝐀𝐏̂𝐐 = 𝐁 ̂                                      ✓S
               PQ || BC        (Corresponding angles =)          ✓S/R

              𝐀𝐁         𝐀𝐂
                     =            (Prop. Thm. PQ||BC)            ✓S/R
              𝐀𝐏         𝐀𝐘

              𝐀𝐁         𝐀𝐂
                     =                (Construction AP = DE      ✓R
              𝐃𝐄         𝐃𝐅
                                      and AQ = DF)                                               (7)

 4.2.1     In ∆𝐃𝐀𝐇 𝐚𝐧𝐝 ∆𝐎𝐂𝐇
              1. D𝐀̂ H = O𝐂̂H =90° (Radius ⊥Tangent)
                                                                 ✓S✓R
                 ̂ 2 is common
              2. 𝐇                                               ✓S
              3. 𝐀𝐃̂ 𝐇 = 𝐂𝐎̂H      (Remaining angles)                                            (4)
              ∴ ∆𝐃𝐀𝐇 ||| ∆𝐎𝐂𝐇      (AAA)                         ✓R(AAA)

Copyright Reserved                                                                         Please Turn Over

### Page 14

Downloaded from hlayiso.com
Mathematics                                         5                      April 2021 Common Test
                                          NSC – Marking Guideline

 4.2.2   𝐃𝐀     𝐃𝐇   𝐀𝐇
              = 𝐎𝐇 = 𝐂𝐇        (∆𝐃𝐀𝐇 ||| ∆𝐎𝐂𝐇 )                     ✓S/R
         𝐎𝐂

                 𝐃𝐇 × 𝐎𝐂
         𝐎𝐇 =                                                       ✓S
                   𝐃𝐀

         DA = DC (Tangents drawn from common point equal)           ✓S✓R
         AO = OC (Radii of a circle)                                ✓S✓R
         Therefore
                 𝐀𝐎. 𝐃𝐇
         𝐎𝐇 =
                  𝐃𝐂

                                                                                        (6)
 4.2.3   In ∆𝐀𝐁𝐅 𝐚𝐧𝐝 ∆𝐁𝐉𝐅
                  ̂ F = J𝐁
              1. 𝐁𝐀      ̂F      (Tangent – Chord Theorem)          ✓S✓R
              2. 𝐅̂ is common)                                      ✓S
                  ̂ 𝐅 = 𝐁𝐉̂𝐅
              3. 𝐀𝐁               (Remaining angles)
              ∴ ∆𝐀𝐁𝐅 ||| ∆𝐁𝐉𝐅 (AAA)                                 ✓S✓R
                𝐀𝐁   𝐁𝐅   𝐀𝐅
              ∴ 𝐁𝐉 = 𝐉𝐅 = 𝐁𝐅     (∆𝐀𝐁𝐅 ||| ∆𝐁𝐉𝐅)
                                                                    ✓S
              𝐁𝐅 𝟐 = 𝐉𝐅 . 𝐀𝐅
                                                                                        (6)
                                                                                       [23]

Copyright Reserved                                                                Please Turn Over

### Page 15

Downloaded from hlayiso.com
Mathematics                                                6                                   April 2021 Common Test
                                                 NSC – Marking Guideline
QUESTION 5

                                                                    2b
                                                                                2a
                                                        3b

                                             b                             3a

 5.1          Let AE = 2a therefore EB = 3a                                                         ✓S ✓R         (2)
              𝐀𝐅   𝟐
                 =𝟑            (Prop. Thm.; FE//GB) or (Line // one side of ∆)
              𝐅𝐆

 5.2          Let AF = 2b and FG = 3b
              Then CG = b      (Given AF = 2CG)                                                     ✓S

              𝐂𝐇       𝐂𝐆
                 = 𝐆𝐅 = 𝟑𝒃
                            𝒃
                                     (Prop. Thm.; GH//FE) or (Line // one side of ∆)                ✓S ✓R
              𝐇𝐄

                  𝐂𝐇 𝟏
              ∴     =                                                                               ✓S            (4)
                  𝐇𝐄 𝟑
 5.3          𝐀𝐄       𝐀𝐅   𝐅𝐄   𝟐
                   = 𝐀𝐆 = 𝐆𝐁 = 𝟓 …. (Prop. Thm; FE || GB) or (Line // one side of ∆)
              𝐀𝐁

               ̂𝐁 = 𝐆𝐅̂𝐄 ….. (Corresp Angles ; FE || GB)
              𝐂𝐆                                                                                    ✓S/R
              Let FE = 2x and GB = 5x
              Then
                                𝟏              ̂𝐁
              𝐀𝐫𝐞𝐚 𝐨𝐟 ∆ 𝐁𝐂𝐆     𝟐
                                  (𝒃)(𝟓𝒙) 𝐬𝐢𝐧 𝐂𝐆                                     2b             ✓S
                              =
              𝐀𝐫𝐞𝐚 𝐨𝐟 ∆ 𝐀𝐅𝐄 𝟏 (𝟐𝒃)(𝟐𝒙) 𝐬𝐢𝐧 𝐀𝐅̂𝐄
                                𝟐                                                         2x        ✓S
                      𝟏               ̂𝐁
                        (𝒃)(𝟓𝒙) 𝐬𝐢𝐧 𝐂𝐆
              =       𝟐
                𝟏                        ̂𝐁)
                  (𝟐𝒃)(𝟐𝒙) 𝐬𝐢𝐧 (𝟏𝟖𝟎° − 𝐂𝐆
                𝟐                                                                                   ✓S
                                                         b                           5x
                   𝟓
              =                                                                                                   (4)
                   𝟒
              OR                                                                                    OR

                                      𝟏                                                             ✓S/R
              𝐀𝐫𝐞𝐚 𝐨𝐟 ∆ 𝐁𝐂𝐆 = 𝟔 𝐀𝐫𝐞𝐚 𝐨𝐟 ∆𝐀𝐁𝐂 …(Equal Heights)
                                     𝟏
              𝐀𝐫𝐞𝐚 𝐨𝐟 ∆ 𝐀𝐅𝐄 = 𝟑 𝐀𝐫𝐞𝐚 𝐨𝐟 ∆𝐀𝐄𝐂 …(Equal Heights)                                       ✓S/R
                                     𝟐
              𝐀𝐫𝐞𝐚 𝐨𝐟 ∆ 𝐀𝐄𝐂 = 𝟓 𝐀𝐫𝐞𝐚 𝐨𝐟 ∆𝐀𝐁𝐂 …(Equal Heights)
                                                                                                    ✓S
                               𝟐
              𝐀𝐫𝐞𝐚 𝐨𝐟 ∆ 𝐀𝐅𝐄 =    𝐀𝐫𝐞𝐚 𝐨𝐟 ∆𝐀𝐁𝐂
                              𝟏𝟓                                                                    ✓S
              𝐀𝐫𝐞𝐚 𝐨𝐟 ∆ 𝐁𝐂𝐆 𝟏𝟓 𝟓
                            =    =                                                                                (4)
              𝐀𝐫𝐞𝐚 𝐨𝐟 ∆ 𝐀𝐅𝐄 𝟏𝟐 𝟒
                                                                                                                 [10]

Copyright Reserved                                                                                    Please Turn Over

### Page 16

Downloaded from hlayiso.com
 Mathematics                                                   7                           April 2021 Common Test
                                                     NSC – Marking Guideline
 QUESTION 6

6.1.1
                         128°
                32°
                      64°
                                                                               A✓diagram
                                             p
         √𝑝2 − 1

                                                                52°
                                                         26°
                                         1                      13°

           𝐜𝐨𝐬 𝟓𝟐° = 𝐜𝐨𝐬[𝟐(𝟐𝟔°)]
                   = 𝟐𝐜𝐨𝐬 𝟐 𝟐𝟔° − 𝟏                                            A✓writing as double angle
                       𝟏                                                       A✓expansion
                   = 𝟐(𝑷)𝟐 −𝟏
                                                                               CA✓answer                            (4)
6.1.2      𝐬𝐢𝐧 𝟕𝟏° = 𝐬𝐢𝐧(𝟒𝟓° + 𝟐𝟔° )                                           A✓𝐬𝐢𝐧(𝟒𝟓° + 𝟐𝟔° )
                  = 𝐬𝐢𝐧 𝟒𝟓° 𝐜𝐨𝐬 𝟐𝟔° + 𝐜𝐨𝐬 𝟒𝟓° 𝐬𝐢𝐧 𝟐𝟔°                          A✓compound angle
                    √𝟐 𝟏      √𝟐 √𝒑𝟐 − 𝟏                                       expansion
                  =     . +       .                                            CA CA ✓✓each term
                     𝟐 𝒑       𝟐      𝒑
                                                                                                                    (4)
                         ° ).                    °                °)              𝒔𝒊𝒏 𝜽
6.2        𝐜𝐨𝐬(−𝟏𝟖𝟎             𝐭𝐚𝐧 𝜽. 𝐜𝐨𝐬 𝟔𝟗𝟎 . 𝐬𝐢 𝐧(𝜽 − 𝟏𝟖𝟎                  A✓𝒄𝒐𝒔 𝜽
                                  𝐜𝐨𝐬 𝟐 (𝜽 − 𝟗𝟎° )                             A✓𝐜𝐨𝐬 𝟑𝟎° .
                            𝒔𝒊𝒏 𝜽                                              A✓− 𝐬𝐢𝐧 𝜽
               𝐜𝐨𝐬(𝟏𝟖𝟎° ) × 𝒄𝒐𝒔 𝜽 . 𝐜𝐨𝐬 𝟑𝟎° . (− 𝐬𝐢𝐧 𝜽)
           =
                                    𝐬𝐢𝐧𝟐 𝜽
                                                                                    √𝟑
                         𝐬𝐢𝐧 𝜽 √𝟑                                              CA✓ 𝟐      or 0,866
               (−𝟏) ×
                         𝐜𝐨𝐬 𝜽 . ( 𝟐 )(− 𝐬𝐢𝐧 𝜽)
           =
                                𝐬𝐢𝐧𝟐 𝜽
            √𝟑 𝐬𝐢𝐧𝟐 𝜽
           = 𝟐 𝐜𝐨𝐬 𝜽
             𝐬𝐢𝐧𝟐 𝜽

                 √𝟑                                                            CA✓answer                            (5)
           =
               𝟐 𝐜𝐨𝐬 𝜽

6.3     𝐋𝐇𝐒 = 𝐜𝐨𝐬 𝟎° + 𝐜𝐨𝐬 𝟏° + 𝐜𝐨𝐬 𝟐° + ⋯ + 𝐜𝐨𝐬 𝟏𝟕𝟖° +                        A✓−𝒄𝒐𝒔 𝟐°
        𝐜𝐨𝐬 𝟏𝟕𝟗° + 𝒄𝒐𝒔 𝟏𝟖𝟎° + 𝟔𝐬𝐢𝐧 𝟗𝟎°                                         A✓−𝒄𝒐𝒔 𝟏°
        = 𝐜𝐨𝐬 𝟎° + 𝐜𝐨𝐬 𝟏° + 𝐜𝐨𝐬 𝟐° + … − 𝐜𝐨𝐬 𝟐° − 𝒄𝒐𝒔 𝟏°                       A✓− 𝒄𝒐𝒔 𝟎°
                     − 𝒄𝒐𝒔 𝟎° + 𝟔𝐬𝐢𝐧 𝟗𝟎°                                       A✓All terms cancel except 6          (4)
        = 𝟔 = 𝐑𝐇𝐒

        OR
        𝐋𝐇𝐒 = ( 𝐜𝐨𝐬 𝟎° + 𝐜𝐨𝐬 𝟏𝟖𝟎° ) + (𝐜𝐨𝐬 𝟏° + 𝐜𝐨𝐬 𝟏𝟕𝟗° ) +                   A✓( 𝐜𝐨𝐬 𝟎° + 𝐜𝐨𝐬 𝟏𝟖𝟎°)
        (𝐜𝐨𝐬 𝟐° + 𝐜𝐨𝐬 𝟏𝟕𝟖° ) … … … + 𝟔𝐬𝐢𝐧 𝟗𝟎°                                  A✓(𝐜𝐨𝐬 𝟏° + 𝐜𝐨𝐬 𝟏𝟕𝟗° )
        𝐋𝐇𝐒 = (𝟎) + (𝟎) + (𝟎) … … … + 𝟔𝐬𝐢𝐧 𝟗𝟎°                                 A✓(𝐜𝐨𝐬 𝟐° + 𝐜𝐨𝐬 𝟏𝟕𝟖°
        LHS = 6                                                                A✓All terms cancel except 6
                                                                                                                   [17]

 Copyright Reserved                                                                                  Please Turn Over

### Page 17

Downloaded from hlayiso.com
Mathematics                                           8                                April 2021 Common Test
                                            NSC – Marking Guideline
QUESTION 7

7.1      𝟏 − 𝒔𝒊𝒏 𝟐 𝒙
        𝒔𝒊𝒏 𝒙 − 𝒄𝒐𝒔 𝒙
              𝒔𝒊𝒏𝟐 𝒙−𝟐 𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔 𝒙+𝒄𝒐𝒔𝟐 𝒙
        =                                                        A✓ sin2 x + cos2x = 1
                     𝒔𝒊𝒏 𝒙−𝒄𝒐𝒔 𝒙
                                                                 A✓ 2 sin x cos x
              (𝒔𝒊𝒏 𝒙−𝒄𝒐𝒔 𝒙)𝟐
        =                                                        A✓ (sin x – cos x)2
               𝒔𝒊𝒏 𝒙−𝒄𝒐𝒔 𝒙
        = 𝒔𝒊𝒏 𝒙 − 𝒄𝒐𝒔 𝒙
        = RHS

                                                                                                           (3)
7.2                 𝟏
        𝐭𝐚𝐧 𝟑𝒙 .         −𝟏=𝟎
                 𝐭𝐚𝐧 𝟐𝟒°

                                                                 A✓transposing and forming
        𝐭𝐚𝐧 𝟑𝒙 = 𝐭𝐚𝐧 𝟐𝟒°
                                                                 equation
        𝟑𝒙 = 𝟐𝟒° + 𝟏𝟖𝟎𝒌 ; 𝒌 ∈ 𝒁                                  A✓3𝑥 = 24° A✓180k                         (5)
                                                                 A✓ 𝑘 ∈ 𝑍
        𝒙 = 𝟖° + 𝟔𝟎𝒌 ; 𝒌 ∈ 𝒁
                                                                 CA✓ 𝑥 = 8° + 60𝑘
                                                                 Note: If calculator is used
                                                                 maximum marks may be
                                                                 allocated
7.3        √𝟑      𝟏                                                    3         1       
        𝟐 [ 𝐬𝐢𝐧 𝒙 + 𝐜𝐨𝐬 𝒙]                                       A✓ 2      sin x + cos x 
           𝟐       𝟐                                                    2          2      
        = 𝟐[𝐬𝐢𝐧 𝟔𝟎° 𝐬𝐢𝐧 𝒙 + 𝐜𝐨𝐬 𝟔𝟎° 𝐜𝐨𝐬 𝒙]                       A✓ [sin 60o sin x + cos 60o cos x]
        = 𝟐[𝐜𝐨𝐬(𝒙 − 𝟔𝟎°)]
                                                                 A✓ cos (x − 60o)
         Maximum value is 2, since maximum value of             CA✓ answer
        cos (x − 60o) = 1
                                                                                                           (4)
        OR                                                       OR
           √𝟑      𝟏                                                  3               
        𝟐 [ 𝐬𝐢𝐧 𝒙 + 𝐜𝐨𝐬 𝒙]                                                      1
           𝟐       𝟐                                             A✓ 2   sin x + cos x 
                                                                      2        2      
        = 𝟐[𝐜𝐨𝐬 𝟑𝟎° 𝐬𝐢𝐧 𝒙 + 𝐬𝐢𝐧 𝟑𝟎° 𝐜𝐨𝐬 𝒙]
                                                                 A✓ 𝟐[𝐜𝐨𝐬 𝟑𝟎° 𝐬𝐢𝐧 𝒙 +
        = 𝟐[𝐬𝐢𝐧(𝒙 + 𝟑𝟎°)]                                        𝐬𝐢𝐧 𝟑𝟎° 𝐜𝐨𝐬 𝒙]
         Maximum value is 2, since maximum value of             A✓ sin (x + 30o)                          (4)
        sin (x + 30o) = 1                                        CA✓ answer

Copyright Reserved                                                                            Please Turn Over

### Page 18

Downloaded from hlayiso.com
Mathematics                                                 9                                   April 2021 Common Test
                                                  NSC – Marking Guideline

7.4.1
                                         y

                                    2

                                  1.5

                            P       1                         Q

                                  0.5

                                                                                                            x
              -90    -60    -30
                                  O          30       60     90    120       150   180    210      240    270

                                  -0.5

                                   -1

                                  -1.5

                                   -2

                                                                            A✓for both x – intercepts
                                                                            A✓for both turning points
                                                                                                                (2)
 7.4.2   𝐜𝐨𝐬(𝒙 − 𝟑𝟎° ) = 𝟎, 𝟓                                               A✓𝒙 − 𝟑𝟎° = 𝟔𝟎°
         𝟐𝐜𝐨𝐬(𝒙 − 𝟑𝟎° ) = 𝟏                                                 and 𝒙 − 𝟑𝟎° = −𝟔𝟎°
         𝒙 − 𝟑𝟎° = 𝟔𝟎° 𝒐𝒓 𝒙 − 𝟑𝟎° = −𝟔𝟎°                                    CA✓ 𝟗𝟎° 𝐚𝐧𝐝 −𝟑𝟎°
         𝒙 = 𝟗𝟎° 𝒂𝒕 𝑸 𝒐𝒓 𝒙 = −𝟑𝟎° 𝒂𝒕 𝑷                                      CACA✓✓for P and Q
                                                                                                                 (4)
                                                                                                                [18]

Copyright Reserved                                                                                       Please Turn Over

## Related documents

- [MATHS P1 GR12 MEMO SEPT 2021 English IsiXhosa](https://www.hlayiso.com/library/maths-p1-gr12-memo-sept-2021-english-isixhosa/6e14ca2d-44f7-417b-b314-e40962f91676) — Mathematics · Memo · 2021 · Eng
- [Maths-April-QP Memo-EC-2021-Gr12](https://www.hlayiso.com/library/maths-april-qp-memo-ec-2021-gr12/e1469b3f-b9fe-40c2-9c23-509439908f72) — Mathematics · Memo · 2021
- [Maths-April-QP Memo-Gauteng-2021-Gr12](https://www.hlayiso.com/library/maths-april-qp-memo-gauteng-2021-gr12/e8fb9e00-d9bb-4e33-a21d-a354cdc802ad) — Mathematics · Memo · 2021
- [MATHS P1 QP GR12 SEPT 2021 Afrikaans](https://www.hlayiso.com/library/maths-p1-qp-gr12-sept-2021-afrikaans/ecc6ca73-048f-4257-8f9a-db12f35059e1) — Mathematics · QP · 2021 · Afr
- [MATHS P1 QP GR12 SEPT 2021 English](https://www.hlayiso.com/library/maths-p1-qp-gr12-sept-2021-english/12f7db49-5bd7-414f-af58-65ed09d4c3e4) — Mathematics · QP · 2021 · Eng
- [Mathematics P2 Nov 2020 ANSBK Afr Eng (2) final](https://www.hlayiso.com/library/mathematics-p2-nov-2020-ansbk-afr-eng-2-final/199754f9-9f9b-4bd9-8890-beebee25863f) — Mathematics · QP · 2021 · Eng

## More from Grade 12 Mathematics

- [MATHS P1 ANSWER BOOK GR12 SEPT 2025 English Afrikaans watermark](https://www.hlayiso.com/library/maths-p1-answer-book-gr12-sept-2025-english-afrikaans-watermark/d78c3fd3-dcd5-4f3f-911b-e5635b4686cc) — Addendum · 2025 · Afr
- [Maths Grade 12 NSC P2 QP September 2025 Gauteng](https://www.hlayiso.com/library/maths-grade-12-nsc-p2-qp-september-2025-gauteng/1d6bf7fd-d288-4a01-99f7-2c87e247760c) — QP · 2025
- [Maths Grade 12 NSC P2 QP September 2025 Afr Limpopo](https://www.hlayiso.com/library/maths-grade-12-nsc-p2-qp-september-2025-afr-limpopo/f2f8d481-5d87-4e89-bb21-6077bbcf403d) — QP · 2025 · Afr
- [Maths Grade 12 NSC P1 QP September 2025 Afr Northern Cape](https://www.hlayiso.com/library/maths-grade-12-nsc-p1-qp-september-2025-afr-northern-cape/f5b3a4ce-fa91-4475-96e0-0654a87e1405) — QP · 2025 · Afr
- [Maths Grade 12 NSC P2 QP September 2025 Eng Northern Cape](https://www.hlayiso.com/library/maths-grade-12-nsc-p2-qp-september-2025-eng-northern-cape/4ede8207-ca82-4419-bae1-fae734498cc7) — QP · 2025 · Eng
- [MATHS P2 ANSWER BOOK GR12 JUNE 2024 Afr+English](https://www.hlayiso.com/library/maths-p2-answer-book-gr12-june-2024-afr-english/4a71e975-4ac4-4f55-8e23-0b8155a5b0c6) — Addendum · 2024 · Eng

## Collection

- [Grade 12 past papers](https://www.hlayiso.com/grade-12/past-papers)

---

Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. When citing, link the page URL above, not a PDF or /api/ URL.
