# Mathematics Grade 12 EC March Test 2026 Paper & Memo

> Grade 12 Mathematics question paper and memo from the EC March Test 2026, 20 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/grade-12-mathematics-2026-ec-march-test-question-paper-and-memorandum/7a281ade-29e9-4962-a9f8-6453c684aec9
- Updated: 2026-09-28
- Subject: Mathematics
- Grade: Grade 12
- Document type: Question paper and memo
- Year: 2026
- Exam period: EC March Test
- Curriculum: CAPS
- Pages: 20

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## About this document

Mathematics · Grade 12 · EC March Test · 2026. Question paper and memorandum, 20 pages. Read online or download the PDF.

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## Document text

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### Page 1

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p. cos Province of the
eT) EASTERN CAPE
EDUCATION

OR TAMBO COASTAL DISTRICT

NATIONAL

SENIOR CERTIFICATE

GRADE 12

MATHEMATICS MARCH TEST 2026

MARKS: _ 100

TIME: 2 HOURS

This question paper consists of 11 pages

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NSC

INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.

1 This question paper consists of 8 questions three diagram sheets and formular sheet.
2 Answer ALL the questions.

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

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

5 If necessary, ALL answers should be
stated otherwise.

O decimal places, unless

6 Number the answers correctly accord bering system used in this
question paper.

7 Diagrams are NOT necessarily dra’

8 Write neatly and legibly.

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NSC
QUESTION 1
Given the quadratic pattern: 5;12;21; 32;...
1.1 Determine the general term 7, of the pattern. (4)
1.2 Calculate the value of the 25” term. (2)
1.3. Which term of the sequence has the value of 1152? (4)
1.4. Show that the difference between any two consecutive terms will always be an (3)
odd number
[13]
QUESTION 2
vam ;
The sum of the first 7 terms of an arith
2.1.1 Determine the first three terms of the sequence. (3)
2.1.2 Calculate the 20" term (7, ) using the S, formula. (3)
: —
2.2 \ (3)
For which value(s
2.3. Given a geometric series w ratio I (5)
Let S,, be the sum of the f um of the first 2n terms.
Prove that the sum of th nby S,.r”
[14]

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NSC
QUESTION 3
Given f(x) = 4 38
x1

3.1. Write down the equation:

3.1.1 Vertical asymptote qd)

3.1.2. Horizontal asymptote qd)
3.2 Calculate the x and y intercepts of f- (3)

3.3 Sketch the graph of /; showing al (4)
3.4 For which values of x is f(x) <0 ~~ (2)
[11]

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NSC
——
auesinon
The 1s below show the graphs of g(x)= ax? and h(x) =log, x . The point P

(2;8)lies on the graph of g.

tafimorephysies.com

4.1 Determine the value of a (2)
4.2 Ifh passes through (8;3), find the value of b. (2)
4.3 Write down the equation of h™' (x) in the form y=... (2)
4.4 Determine the domain of h. (2)

4.5 Determine the value(s) of x for which

45.1 g(x).h(x)<0 (2)
45.2 xh(x)20 (2)
[12]

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NSC

QUESTION 5

a(x -p)y +q by completing the square. (3)

(2)
(3)
5.3. Sketch the graph of f (3)
5.4 Write down the y-intercepts of f (2)
5.5 Write down the domain of f"! (2)
5.6 If f(x) is shifted 2 units to the rig up, write down the new (2)
equation ,
[17]
QUESTION 6
6.1 Simplify without using a calculator:
sin(x—180°).tan(x+180°).cos(90+x) (6)
sin? (180° + x)
6.2 If cos25° = V1—k? , determine the following in terms of k:
6.2.1 sin 25° (2)
6.2.2 sin50° (3)
6.2.3 cos70° (5)
Prove that we ae aes = = 2, (5)
sin0d  cos@
m+n (3)
If cos(4+B)=m and cos(4-B)=n, prove that cos 4.cos B =
[24]

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NSC

QUESTION 8

Given APQR with PQ=8units, QR=
P

8.1 Calculate the area of APOR

8.2. Calculate the length of PR

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hat

a’ =b’ +c? —2becos A

ORTCD/March 2026

(4)

[4]

(2)

(3)
5]
TOTAL 100

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NSC

3.3

a 49.

(Fs
|

aH

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NSC

5.3

o

cy

&

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NSC

ORTCD/March 2026

dfimorephysles.com

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NSC
INFORMATION SHEET
A= Pil—ni) A= Pii-i)" A= Pil+i)"
. =F ba tn—pa]
T.=ar"' s, -4=!) jrel S. pont -l<ral
Pes =

3 t+ iy po tt-t+)"]
i

— im LEt-I@)
es se

d= (x, =P +, —¥F

y=mrrc Ia

mew m-=tan@
(x~ of +b) =r?

a? =b? +c7 ~2becos A

area MABC = jebsinc
sin(a + 8)=sin a.cos 8 +cosa.sing sinla—$)=sina.cos B—cosa.sinB

cos(a+ f)=cosa.cos f-sina.sin# cos(a—£)=cosa.cos B+sina.sinB
cos’ a-sin? a

cos2a = <1~2sin* a sin 2a = 2sina.cosa
2cos’ a=1

Le, ~xy

a n
P(A) = P\A or 8) ~ PCA) + P{B) ~ PUA and B)
; _Le-70-P)
prath 6 Sa-3

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ee Province of the
heed EASTERN CAPE
EDUCATION

—

OR TAMBO COASTAL DISTRICT

NATIONAL

SENIOR CERTIFICATE

GRADE 12

TERM 1 CONTROLLED TEST 2026

MATHEMATICS

MARKING GUIDELINES
This marking guidelines consist of 8 pages.

NOTE
e  Ifacandidate answers a question TWICE, only mark the FIRST attempt.
e  Ifacandidate has crossed out an attempt of a question and not redone the question, mark the crossed out
version.
¢ Consistent accuracy applies in ALL aspects of the Marking Guidelines. Stop marking at the calculation
error.
e Assuming answers/values in order to solve a problem is NOT acceptable.

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QUESTION 1
1.1 V value of a (4)
¥ value of b
¥ value of c
Y equation
T, =1.n' +4.n+0
T, =n +4n
ma (Ry P4425) Y answer (2)
T,, = 725
13 T, =n +4n=1152 ¥ substitution (4)
Vv ;
n? +4n-1152=0 ‘awe
(n—32)(n+36)=0 Y answer
n#-36
n=32
Ty =1152
14 T,.,-T, = (n+l) +4(n41)-(n? +4n) ¥ substitution (3)
¥ simplification
T,,,-T, =n’ +2n+1+4n+4-n? —4n v answer
TT, =2n45
2n is always an even number
adding 5 to any even number gives odd number
T,,, -T,, =2n+5 is always an odd number
[13]

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QUESTION 2
211 \Anay 311 —5(1 Y first term @)
RFS, = 30) -5() =-l ¥ second term
UTI) 5 4 v third term
WAT? 3(2) —5(2) 30) -S5(U1
ipa, -5, 222-52) 300-500,
— 3(3) -5(3) 3(2) -5(2
Fass, 50) 30¥ 50)
17255355
ae 3(20)° —5(20) 3(19)° -5(19) Y Sip GB)
Dy = Sry — Sto = 2 ~ 2 =56 Y Sy
v¥ answer
2.2 -l<k-1<1 . V¥-1<k-1<1 (3)
-l+l<k<14l \ 7 VY —-141<k<14+1
-l<k<2 i. V-1<k<2
Y Soy (5)
1 bs vS,,-S,
ahimorephysics.com
¥ factor“
—_
Y factor r”
¥ simplification
[14]

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wl. Yx=l (1)
3.1.2 : Vy=3 (1)
3.2 3 4 2 ¥ substitution for | (3)
lf (3) at 3 x-intercepts
for x—int let y=o vsimplly
0O= a +3
x-1
-3(x-1)=2
v¥asymptotes (4)
vy-intercepts
¥ point fixing
graph
v¥ shape
v ¥ answer (2)
[11]

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QUESTION 4
41 [g(x)e ax’ r substitution (2)
; r answer
8=a(2) =4a
@=2
g(x)=2x°
4.2 | h(x) =log, x ¥ substitution (2)
¥ answer
3=log,8
b=8=2°
b=3
h(x) =log, x
43 | y=log,x vinterchange (2)
x=lo variable.
ey v¥ answer
yah
yas
4.4 Domain =x/x > 0 diimorephYsles.com v V answer (2)
4.5.1 | 0<x<1 v VY answer (2)
4.5.2 | x21 ¥ ¥ answer (2)
[12]
QUESTION 5
5.1 v factor 2 (3)
¥ complete
square
¥ answer
5.2.1 vx-value (2)
vy-value
5.2.2 vy=0 (3)
Y factorisation
v Values of x

(x-3)(x+1)=0

x=30rx=-1

(3:0), (-10)

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v intercepts with (3)
axis
v indication of y-
intercepts
v shape
=, | 7 (03) 2)
anmorepn com 4
(0;3), (0;-1) 4 ¥ (0;-1)
4.5 | Domain=x/x>-8 ¥ value (2)
¥ notation
5.6 | f'(x)=f(x-2)+5 ¥ substitution Q)
2 v answer
f'(x)=2(x-2) -6+5
f'(x) = 2x? -8x+8-1
f'(x)=2x? -8x+7
[17]

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QUESTION 6
6.1 | sin(x—180°).tan(x+180°).cos(90+x) ¥-sinx (6)
sin’ (180° +x) vtanx
(=sin x).(tan x).(—sin x) v-sinx
= 7 _\2 ¥ simplification
(-sinx) Y answer
_ (-sin.x)' (tan.x)
(-sinx)"
= tanx
6.2
cos25°= MITA _ ah
1 Iyp
opp =?
(np)’ =(opp)" +(adj)°
(1)? =opp'+ (VIF)
opp? =K*
opp=k
6.2.1} . 1. opp _k ¥ substitution (2)
a hyp 1 Y answer
sin 25° =k
¥sin2(25) (3)
v double angle
v Answer
¥cos(45°+ 25°) | (5)
¥ compound
angle formular
¥ substitute first
term
¥ substitution in
second term
¥ Simplification

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6.3 | LHS v¥ sinxcosx (5)
'sin30. cos30_ sin3@cos@—cos36sin0 ¥ numerator |
sind) cosO sin @cosO v’simplification
a v express
Ne (30-9) denominator as
ta double angle
—(2sin0cos0
2. panrang) ¥ simplification
=2. gn 20 =2
sin 20
LHS = RHS
6.4 | cos(4+B)=cos Acos B-sin Asin B = m.....(1) 7 eauetion 4 (3)
cos(A—B)=cos Acos B +sin Asin B = n......(2) PR saliieton
(1) +(2)
2cos AcosB=m+n
cos Acos B= mean
[24]
afimoreph 4 ics.com
QUESTION 7
AB=c and BC=a | construction —_| (4)

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RTP: =b° +c’ —2becos A v Pythagoras
Construction: draw perpendicular height h from vertex Hiseieae for both
@ dicular to B. ranges
aati lad teular'to ¥ substitute by
BD=c-AD
~ B2— B2 2
AACD: b° =h’ + AD YcosA
ABCD : a? = h? + DB? =h? +(c- AD)
@=h’ +c —2c.4D+ AD*
@ =h? + AD? +c? —2cAD
a@ =b’ +c —2cAD
cos A= adj. = ae
hyp b
—
AD =becosA
a =b> +c’ —2becos A
QESTION 8 2
8.1 1 . of ¥ substitution (2)
Area of APQR=- (8)(12)sin 110% Japiswer
Area of APQR = 45.11 square unit
8.2 PR? =(8 2 AL12 2 ~2(8)(12)cos! ¥ substitution (3)
; hay ey 2] ¥ simplification
PQ* = 273.67 v answer
POQ=16.54
[5]
TOTAL 100

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