# Mathematics Grade 12 KZN March Test 2025 Memo

> Grade 12 Mathematics memorandum from the KZN March Test 2025, 11 pages. Read online or get the free PDF download, question paper linked when available.

- Page: https://www.hlayiso.com/library/mathematics-maths-grade-12-memo-march-2025-kzn/0ea4ec9b-6e1b-4d3d-be60-b64e90798dbc
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 12
- Document type: Memorandum
- Year: 2025
- Exam period: KZN March Test
- Curriculum: CAPS
- Pages: 11

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## About this document

Mathematics · Grade 12 · KZN March Test · 2025. Memorandum, 11 pages. Read online or download the PDF.

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## Document text

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### Page 1

01234012567
6800892472
 01 63 
   0408

### Page 2

QUESTION 1

Common Test March 2025

El

S,, =3(2)° -5(21)

YA substitution

=1218 vA answer
(2)
1.12 |g. =3(22) —5(22)
= 1342 vA value of S,,
T, =S,,—-S,,
= 1342-1218
=124 ¥CA answer
_ (2)
1.13 8162 =3n? —5n YA equating
3n° -5n-8162=0 vA standard form
-(-5 ts * — 4(3)(-8162
n= ( ) ( ( M ) YCA substitution
2(3)
a=5300 or as _b4
3
N/A
YCA answer (53 only)
53 terms have to be added (4)
1.2.1 | The even-numbered terms form an AS with a=7 and d=5.
T,. of the AS
=a+(n-l)d
v —
_ 7+(39-1)5 A substitution
=197 ¥CA answer
(2)
1.2.2 | Sum of the 52 odd-numbered terms = 52 x 7 = 364 VA 52x7
Sum of the 51 even-numbered terms
n
=5[2a+(n-1)d]
2! 2(7 51-1).5
= aL (7)+(S1-1). | ¥CA substitution
= 6732 ¥CA sum of AS
Sum of first 103 terms
= 364+ 6732
= 7096 ¥CA answer
(4)
[14]

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### Page 3

Common Test March 2025

QUESTION 2
Dla 9; -27; 81 VA 9
VA -27; 81
- (2)
BL2 | Fog vA answer
Q)
2.1.3 | No, the series will not converge YCA answer (no)
r<-] OR — rdoes not lie between —1 and 1. YCA motivation
_ (2)
218 | woe r=-3 n=12 VA n=12
a(r" - 1)
|
9x((-3)" -1)
=~ 321 ¥YCA substitution
=~] 195740x ¥CA answer
(3)
~ VA
2 5 8
3 3
2"4 difference = 3 YA 2" difference = 3
_ 2nd difference 3
am 2 2 YCA value of a
3a+b=2
p=2-3(3 __5 YCA value of b
2 2
Tes =3(29)' ~5(29)+e=1166 YCA substitution in Fr.
~ 2B 2
c=-23 YCA value of ¢
Se 37 3% —23
2 2
T, =5 (1) -3(0)-23=-24 ¥CA answer
(6)
[14]

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### Page 4

Common Test March 2025

QUESTION 3
, 2x+3
3.11 | f= so
HeHS
_ 2x43
“I(e+3) VA -(x+3)
_ (2x43)
x3
_ —(2x+6-3)
7 wed
ae va 22e+3)-3)
x+
rts
aH), 3
x+3 x+3
, 3
ifs a > (2)
OR OR
2x+3
f(x) =
—x-3 3
2 -3 YA -2+ (through the
= to —x-3
3 method of long division)
ie ae vA —(x+3)
3
=-24+—
x+3 (2)
3.1.2 | For y-intercept, let x = 0:
i ae
x+3
yeaa- =-l YA y=-l
For x-intercept. let y = 0:
“__2=0 vA 3-0
webs x+3
3 -2
r+3
3 3
a YA x=-2
x 5 # 5
(3)

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### Page 5

Common Test March 2025

cm Bs)
¥CA intercepts
vA asymptotes
vA shape
(3)
3.1.4 ya-xte vA equation of straight line
Substitute (—3 ; —2): with a gradient of — 1
2=-(-3)+c
c=-5
yes-x-5 ¥CA answer
(2)
3.1.5 xeR, x#-3 YA xeER
(a) VYCA x#-3
(2)
a2 42K 2 ¥CAYCA 335-5
(b) 2 2
2)

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### Page 6

Common Test March 2025

3.2.1 [AtR: 3x+6=0
x=—2
R(-2:0) YA x-coordinate
VA y-coordinate
(2)
3.2.2. | From symmetry, the x-coordinate at S = |
y =a(x+2)(x-1) YCA y=a(x+2)(x-1)
y=ax +ax—2a
2
Substitute (-3:9} 9=4(-5] +a(-4)-20 ¥CA substitution
9= =
4
a=-4 YCA value of a
y=4xr? 4x48 YCA answer.....
OR (4)
ye=a(x+p) +q OR
1y 2
y=a{x+3) +9 vA y=af x43) 49
Subst. S(1 ; 0) OR Subst. R(-2 ; 0):
2 2:
o=a(1+5) +9 o=a{-2+3} +9 YCA substitution
Es -9=4{2)
4
a= 5 YCA value of a
ayaa) +9 ¥CA answer
(4)
[20]
QUESTION 4
ait | ft: pavox
He x= Br YA swopping x and y
x =3y ,
. Pe
? v =—_
yeti x20 ay 3
YA x20
G3)
4.1.2 | 0<x<3 OR xe[0 ;3] YAVA answer
(2)

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### Page 7

Common Test March 2025

4.2.1

ey-{t} OR gi y=
as
gl: r=(1} gi:x=3? vA swopping x and y
y=log, x “. y=—log; x vA answer
2 (2)
4.2.2
y
¥CA shape
YA x-intercept
YCA coordinates of one more
point
(3)
1 x
4.2.3 -al1) 47
* (5)
12
Substitute (—2 ; 10): 10= a() +7 YA substitution
9a=3
1
a= 3 ¥CA answer
(2)
4.2.4 he y= +}(3) +7
- (3 3
x+l 1 x vi, ( l y 7
From y=] — +7 t =|o] ; pi=| = Ee
J (3) ~# (; 3
Translation of 1 unit to the right and 7 units downwards. YA | unit to the right
YA 7 units downwards
Answer only: (3)
full marks
[15]

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### Page 8

Common Test March 2025

QUESTION 5
: 4 4
S.1.1 sin f =— v i =_—
g : A sing ;
xtap -1° [Pythagoras]
=5? 4? ».
== 9
72.3 VA x-—value
cos 8 = 2 ¥CA answer

(3)

5.1.2  cos2f =2cos’ B-1
2
Bs}
-(3)-1
25

sil
25
OR
cos26 =cos’? B-sin? B

2

(3) -()

= 2
25 25
gol
25

OR
cos2f =1-2sin? B

wail
25

YA double Z expansion

YCA substitution

¥CA answer
(3)
OR

YA _— double Z expansion

¥CA _ substitution

¥CA answer

(3)
OR

YA _ double 2 expansion

VA _ substitution

¥CA answer
(3)

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### Page 9

Common Test March 2025

5.1.3 | sin38=sin(2B+£)
= sin2fcos P+cos2fsin B ¥A compound 2 expansion
= 2sin Bcos Bcos B+cos2fsin B YA double Z expansion
_ f4)(-3)(-3) . (-7)(4
=2 BIS 5 a5 \l5 ¥CA substitution
= 228
125 125
— ¥CA answer
125 (4)
52 sin(—180°—@) tan(180°— 0) cos(—0)
cos? (90° +0)+3sin? 0
_ sin@,—tan 0.cos@ VA sin@ VA —tan@
(-sin0)' +3sin? @ YA cos8 VA —sind
sin0
ind.—-——.cos9 i
7 sin Tr cos “A BaAd= sin@
4sin’ 0 a
__ sin’ @
4sin’ 6
-l ¥CA_ answer
4 (6)
53 1 (1-cos4x)
8
1
=—[1-cos(2x2 :
=i cos(2* 2x) ] VA gl! 20s(2x2x)]
7 sod 1 sa
= gL (t-2sin 2x) VA gl !-(1-2sin’ 2x)
= {[2sin* 2x]
Ir.
=< [sin* 2x] vA gin? 2x]
2 1), . 2
= 4 (2sin.xc0s.x) vA g(2sinxcos~)
1 .
=5(4)(sin® x.c0s? x) vA glA)(sin? x.c0s” x)
= sin’ x.cos’ x @

OR

OR

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### Page 10

g(t cosas)

Ls al! —vos{2x 2x)
- L Z . eos? 2x |
: i

= st —cos* 2x)|

Common Test March 2025

vA

vA

al! —cos(2x 2x) |

g[t=(200s° 2x-1)|

\
; = ¢[2sin* 2x]
TPs
Ja. ein?
= o[sin® 25] A [sin 2x]
1 , 2
‘ YA —
= (2sin xe0s.x) a. 4 Painzionaz)
1 . 2
= 5(4)(sin’ x.c0s? x) vA glA)(sin? x.COS: x)
= sin” x.cos” x 6)
[21]

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### Page 11

QUESTION 6

Common Test March 2025

6.1

cos(x—45°) =sin2x
cos (x - 45°) = cos(90°— 2x)
x= 45° = 90° —2x+k.360° or x-45°=360—(90°—2x) + k.360°

3x = 135° + k.360° x —45° = 270° + 2x +k.360°
x= 45°+k.120°, keZ —x =315°+ k.360°

YA co-ratio

YA both equations

YCA 45°+k.120°

x =-315°—k.360° VA keZ
x =45°+k.360°; keZ
(4)
OR OR
cos(x—45°) = sin2x
sin[ 90°—(x—45°) | = sin 2x YA co-ratio
—x+135°=2x+k.360° or —x+135°=180°-—2x+k.360° vA both equations
3x = 135°+k.360° x =45°+ k.360°
x =45°+k.120°, keZ x =45°+k.360°, keZ YCA 45°+k.120°
VA keZ
(4)
6.2.1 |360° vA answer
Q)
6.2.2 |(-135°; -1) YA x-coordinate
(a) YA y-coordinate
(2)
o. 1 YCA x-coordinate
— (= , -5) YCA y-coordinate
(2)
6.2.3 |90°<x<135° OR x € (90° ; 135°) YAY A answer
(2)
1
6.2.4 —=(cosx+sinx
wr )
= A cos + | sinx
v2 2
| cos45°.cos.x + sin 45°sin x VA cos45°.cosx+sin 45° sin x
= cos(x~ 45°) ¥A compound Z identity

. To solve for x if sin2x = cos (x = 45°):
x= 45° or x €[165°;180° | OR x = 45° or 165° < x <180°

VA x=45°

YCAYCA x [165°;180°] or

165° < x < 180°

(5)

[16]

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