# Mathematics Grade 12 KZN March 2022 Paper & Memo

> Grade 12 Mathematics question paper and memo from the KZN March 2022, 18 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/maths-gr12-march-2022-qp-and-memo/1f286f91-1f11-4cd9-a494-de7d9156a59a
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 12
- Document type: Question paper and memo
- Year: 2022
- Exam period: KZN March
- Curriculum: CAPS
- Pages: 18

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## About this document

Mathematics · Grade 12 · KZN March · 2022. Question paper and 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

KWAZULU-NATAL PROVINCE
                        EDUCATION
                        REPUBLIC OF SOUTH AFRICA

                                        NATIONAL
                                  S ENIOR CERTIFICATE

                                        [       GRADE 12 ]
             �----------------------------------r.··········�
                                         ---                                                        ♦

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                                        MATHEMATICS �
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TIME: 2 hours

                            N .B: Th isquest ion paper _consists�f 6 pages,
                             2 d 1agram sh ee tsan d an m f orma t ,on sh ee.t

Copyright Reserved                                                                                                              Please Turn Over

### Page 2

Math i : March 2022 Ci T
“reBéwnloaded from Stanmorerbysics.com aren 2022 Common Test

INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.
1. This question paper consists of 8 questions.

2. Answer ALL questions.

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

4. Answers only will not necessarily be awarded full marks. G

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

6. If necessary, answers should be rounded off to TWO sce unless stated

otherwise. (e)
7. Diagrams are NOT necessarily drawn to scale. s
8. Number the answers correctly accordin' tothe numbering system used in this
question paper. Write neatly and laity.

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### Page 3

Math i : March 2022 Ci T
“reBéwnloaded from Stanmorerhysics.com aren 2022 Common Test

QUESTION 1

Given the quadratic sequence: 5; x; y; 29; ... and its second constant difference
is equal to 4.

1.1. Calculate the values of x and y.

= 17, determine the n' term of the quadratic sequence.

50" term of the sequence.

(4)

(4)

(2)

[10]

RS

QUESTION 2

o:

The 2™ term of an arithmetic sequence is 8 and the 7" term is eleven ieee? value of the first

term. Determine the first three terms of the sequence. XN
&
&

QUESTION 3

3.1 Given:

3.1.1 Write down the values opt three terms of the series in terms of p.

3.1.2 Determine the v efoto for which the series is converging.

3.2. The sum of the terms of a sequence is given by 4 — 4 (:) .

can three terms.

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7]

(1)

(4)

(5)
[10]

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Math i : March 2022 Ci T
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QUESTION 4

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

AB
Prove the theorem which states that — = — (7)
DE DF
A D
E F
B c
4.2 In the figure, AABC has HG || AC and ED || BC. ED and HG intersect at K.
AP =3 and BG =26C . AB=15 units,
Dc 2
A
H
E K D
B * G Cc
Determine with reasons the value of:
4.2.1 AE (3)
4.2.2 AH
(3)
4.2.3 GK
KH
(3)
4.2.4 Area of AHEK (3)
Area of AHBG
[19]

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Math i : March 2022 Ci T
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QUESTION 5

In the figure: PQ is the diameter of the circle. SP is the tangent to the circle at P.
QT intersects the circle at W and with T on the line SP. QS intersects the circle atR. P, =x.

P 1
2 x
W<3
Tr ?
Ss

5.1 Give areason why PRQ = 90°. (1)
5.2 Prove S=x. (3)
5.3. Prove that SRWT is a cyclic quadrilateral. (3)
5.4 Prove that AQWR ||| AQST. (3)
5.5 If QW =5 cm, TW =1cm, QR = 4cm and WR = 2 cn, calculate the lengths of:
5.5.1 TS (3)
5.5.2 SR (3)
[16]

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### Page 6

Math i : March 2022 Ci T
“reBéwnloaded from Stanmorerbysics.com aren 2022 Common Test

QUESTION 6
6.1 Given cos20° =p andsin14° =q

Without using a calculator, calculate the value of the following in terms of p, q or p and q.

6.1.1 sin20° (2)
6.1.2 cos6 (6)

6.2 Simplify into a single trigonometric ratio.

1.
zsin 2x 6)
° 1 _ 2
tan(540 +\(Sa x tan x)
[14]
QUESTION 7
7.1.1 Prove the following identity:
cos 4x = 8cos*x —8 cos?x +1 (4)
7.1.2 Hence, determine, without the use of a calculator, the general solution of
16 cos* x —16cos?x+2=1 (5)
7.1.3 Write down the minimum value of the expression 16 cos* x — 16 cos? x + 2. (2)
7.2 Calculate, without the use of a calculator, the value of:
2sin? 22.5°-1
4sin 225° cos 225° (5)
sin 22.5 cos22.5
[16]
QUESTION 8
8.1 Sketch the graphs of f(x) = sin(2x) and g(x) = 2.cosx
for the domain x e[—90°; 180°]. (Use the axes provided) (6)
8.2 Use your graphs to determine the solution “ 21. (2)
[8]

Total: 100 marks
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INFORMATION SHEET: MATHEMATICS
INLIGTING BLADSY

—b+vb2 —4ac

2a
A=P(1+ni) A= P(1-ni) A=P(1-i)" A=P(1+i)"
T, =a+(n-1)d S, =F (2a+(n-2)d)
T, =ar"* S _ alr’ -1) ; r41 8, = Sj -1<r<t
. r-1 -

- xi)’ -1] p x= G+i)"]

(x) = lim f (x+h)— f(x)
hoo h
d= (X, —X,)? + (Yo - Yn)? M A*% Vat Yo
2 2
y=mx+c y—y; =m(x-x,) m= 227 m=tand
Xp —%

(x-a) +(y—b) =r?

Inaapc; —2_=-—P___¢_ a*=b2+c%-2becosA area AABC = + ab.sinC
sinA sinB sinC 2
sin(a + 8) =sin a.cos B+ cosa.sin B sin(a — f) = sin a.cos B —cosa.sin B

cos(a + 2)=cosa.cos # —sin a.sin B cos(a — £)=cosa.cos # +sina.sin 2

cos’ a-sin? a

cos2a =41-2sin? a sin 2a = 2sina.cosa

2cos* a@-1
d(«-»)
gad tn ou :
P(A) = ae) P(A or B) = P(A) + P(B) — P(A and B)
_ — YX &-xy-y)
y=a+bx yo ae

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### Page 8

% 3
EF 5
o c
E 5
5 u. 5
(6) S
8 =
s a
Fi
Ww
oO
oO

oO
= a
ro}
8
a a
<3 °
®
iG
: « ipa I.
e < “
+e]
rr} x
= ~ x l,

wn
> - - a be
= & .
S |
a N

s ng vt re) 8
= S2 2 z Fd
& zo 9 fe} &
DQ wi cr Ee Ee z
= s Of ry ry =
g§ @ £5 5 5 2
= 2 ao o o [s}

39Vd 45O0-uV3L

### Page 9

B <t3
fad a
5
—
5
oO So
nN a
x
Ss
a
$
os .
= 48 18
<x
a
fo}
8
€
So
oO
:
o
2 1g
2
®
3 8
S
c
wy
” >
= Wi a fo) + Go
fe}
x
~
3 18
Ss
S 3
eo
rr z S 8
= fe} re g
os cL a
2 wi =
4 2 a
s 2 e 8 8

BOVd 45O-UVSL

### Page 10

Downloaded from Stanmore tics. com

KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE

MARKS: 100

This memorandum consists of 9 pages.

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### Page 11

March 2022 Common Test

“atteSwnloaded from Stanworephysigsacom

QUESTION 1
1.1 5 x y 29
Nv IN N £
1“ Diff x5 yrx 29-y
\ VA
\ LZ
NE we
2™ Diff y— 2x45 29-2y+x
y-2x+5=4 V sett; . sta:
y-2x=-1.. (1) AY setting up equation 1” line
29-—2y+x=4 . . sty
-2y+x=-25.. (2) AY setting up equation 1“ line
(1) x2: 2y—4x =-2... (3)
3) + (2):
Oe ‘ 7 . x=9 CAY x — value
syad7 CAY y-—value
(4)
1.2
2a=4 .a=2 AY a-—value
3a+b=4 -«b=-2 CAWb — value
a+b+c=5 «c=5 CAV c — value
Ty = 2n? -2n+5 CAYn™ term (4)
OR OR
2a=4 .a=2 Ava -— value
3a+b=5 «b=-2 CAYb — value
“2cC=Ty=5 CAYc — value
Ty, = 2n*—2n+5 CA“ n™ term (4)
OR
OR
n-1)(n-2
OR OR
n-1
7, =! 2 20+ (n—2)d) +7,
1.3 Ts59 = 2(50)? — 2(50) + 5 = 4905 CAY substitution (from 1.2)
CAY answer (2)
[10]

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### Page 12

Mathenpy

ics

ownloaded from Stanworephysias.com

March 2022 Common Test

QUESTION 2

2 a+d=8 > (1) AY equation (1)
a+6d=11a > (2) AVY equation (2)
d=8-a > (3) CAY making d/a the subject

a+6(8—a)=11a
a+48—-6a=11la

CAY correct substitution of d/a

48 = 16a

a=3 CAV a/d value

d=5 CAV d/a value

3;8; 13;.. CAY sequence
7]

QUESTION 3
3.1.1 1.2.4 AV All three terms qd)
"p’ p?

3.1.2 -1<r<i1 AY condition for convergence
-1<-<1 CAV r value in terms of p
p<-lorp>1 CACAY V answers (4)

3.2 1)"

S,=4 4(;5)
1)" AY first t 1
-4-4(_) = irst term value

T, =5S,=4 4(;5) 2

12
T,+Tz=S2,=4- 4(;5) =3 AY Sum of first two terms

1y> 1
Ty +1, +73 =5S;=4-4(5) =35 Av Sum of first 3 terms
“T,=1 CA“ second term value
1
613 = 2 CA” third term value (5)
[10]

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QUESTION 4

March 2022 Common Test

4.1

On AB mark off a point P such that AP = DE and
on AC mark off a point Q such that AQ = DF.
Join PQ.
In AABC and ADEF

1. AP=DE (Construction)

2. AQ=DF (Construction)

3. A=D (Given)

-. AAPQ = ADEF (SAS)

Now APQ = DEF

But DEF = B (Given)

: APQ=B

PQ || BC (Corresponding angles =)
AB _ AC P. Thm. PQ||BC

AP = AQ (Prop. Thm. PQ||BC)
AB AC Cc ion AP = DE
DE DF (Construction =

and AQ = DF)

YS Construction

vS/IR
vS

vs
vS/IR

vS/R

YR

(7]

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March 2022 Common Test

4.2
15
6
3c 2a
>
B 26 G c
4.2.1 AE _ AD _ 3
15 ac 57 Prop. Thm. ED || BC) VSYR
AE = 9 units Y Answer GB)
422 | AWG
i @ = 30" (Prop. Thm. HG || AC) VSYR
AH = 5 units Y Answer (G3)
4.2.3 HE = 4 units and BH = 6 units CAVHE = 4 units
or = = = $5... (Prop. Thm. EK || BG) CAY SAYR (3)
4.2.4 | AHEK _ 2 : ; YS/R
ante = 57 (As with equal altitudes) G)
auBe 1075 .. (As with equal altitudes) YSIR
Area of AHEK — AHEK | AHEG _ 2 2 4
Area of AHBG  AHEG * HBG” 5* 5 25 v Answer
OR OR
1 . og Y Area of numerator
AHEK _ 2 .(4)(2c) sin EHK V Area of
SHBG 5 (10)(5c) sin EAK Denominator
AHEK 4 v Answer (3)
AHBG 25 OR
OR
2 . . ¥S/IR
AHEK = 5 (AHEG) ... (As with equal altitudes)
= = (2 AHBG) ... (As with equal altitudes) vSIR
AreaofAHEK 4 V
———— ——— =— Answer
Areaof AHBG 25
(3)
[19}

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### Page 15

March 2022 Common Test

“atteSwnloaded from Stanworephysigsacom

QUESTION 5
5.1 Angle subtended by the diameter on a circle/ VR (1)
Angle in a semi-circle
5.2 PQR = 90° — x ... (Sum of angles of A) vs (3)
SPQ=90° _... (Diameter perpendicular to tangent) YSYR
S=x ... (Sum of angles of A)
5.3 QWR=x ... (Angles in the same segment are equal) | ¥S/R
QWR=S=x YQWR=S=x
SRWT is acyclic quad. ... (Converse of exterior angle of | VR
cyclic quad equal to interior (3)
opposite angle)
54 In A’s QWR and QST
1) QWR=S=x ........ (Proven) vs
2) Q is common YS
3) Ty = WRQ........0...... (Remaining angles of A’s) (G3)
AQWR |\i AQST .... (AAA) yR
5.5.1 QW _ WR_ QR vS/R
OS Tor = gr AQWR || AQST)
24 ¥ substitution
ST 6 CAY Answer (3)
ST=3 cm
5.5.2 = =2 ___ (AQWR || AQST) CAY SYR
QS = 7,5 cm CAY Answer (3)
[16]

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### Page 16

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ics

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March 2022 Common Test

QUESTION 6
6.1.1
35° 140
70
1
1-p?
20 40
P 10°
38° 152
76
1
q
X 28°
14
v1-q@ 7
sin20° = 1-p? MY AY Answer (2)
6.1.2 cos 6 = cos(20° — 14°) AY Writing as difference

= cos 20 cos 14 + sin 20 sin 14

=p.j1-q*+/1-p?.q

AY Expansion
A Y (cos 20°) value

CAV J1— GCA J1 — p?

AY (sin 14’) value (6)

6.2

dsin 2x

tan(540° + x) ( — tan? x)

cos? x

5(2 sin x cos x)

1— sin?x
tan x.(——,—
cos?x

5 (2 sin x cos x)

sinx cos?x

Av expansion of sin 2x
Av reduction to tan x

Av1—sin?x = cos?x

sinx

cos X ‘cos?x Avtanx =

- Cosx cos x
= _|sinxcosx x —

sinx CAY simplifying Vcos? x
= ycos?x
=cosx CAY answer (6)
[14]
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QUESTION 7

March 2022 Common Test

711

cos 4x

= cos(2.2x)

= 2cos*2x-1

= 2(2cos?x—1)?-1

= 2(4cos* x —4cos?x+1)-1

=8 cos*x—8cos?x+1

AY cos 4x = cos (2.2x)
Av expansion of cos 4x
AY expansion of cos 2x

AY squaring bracket

(4)
7.1.2 | 2cos4x=1
cos 4x = ; AY setting up equation
4x = 60 + 360k or 4x = 300° /-60' + 360k | CA”both solutions
° ° ° Av 90k
x=15 +90k or x=75 /-15 +90k;keZ AVkEZ (5)
CAY x =15°,75 /-15"
7.1.3 | 16 cost x—16cos*x+2
=2cos 4x AY
:. Minimum Value = —2 AY (2)
7.2 2sin? 22.5°-1
4sin 22.5’ cos 22.5°
—(1— 2sin? 22.5") Av re-writing numerator and
“2.2 sin 22.5 cos 22.5" denominator
—cos45° AY Numerator
= >sin45" AY Denominator
1 AV
-1 Wh Special angle values/ co - ratios
=, v2
21
v2
1 1 AY Answer (5)
=--=x1=-—
2 2
[16]

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QUESTION 8

| y

March 2022 Common Test

30 60 9
+1 t
at
8.1 f: x-int., Maximum and Minimum and Shape AAAVVVY
gi y-int., Maximum and Minimum and Shape AAAY VY (6)
8.2 | g(*) >4
0) > FG)
g(x) = f(x); f(x) >0
x €(0°;180") ;x #90 AAV Answer (2)
[8]
Total Marks: 100

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