# Mathematics Grade 11 EC March Test 2026 Paper & Memo

> Grade 11 Mathematics question paper and memo from the EC March Test 2026, 11 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/grade-11-mathematics-2026-ec-march-test-question-paper-and-memorandum/a17d0559-3878-486c-a9da-6d4088476a2a
- Updated: 2026-09-28
- Subject: Mathematics
- Grade: Grade 11
- Document type: Question paper and memo
- Year: 2026
- Exam period: EC March Test
- Curriculum: CAPS
- Pages: 11

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## About this document

Mathematics · Grade 11 · EC March Test · 2026. Question paper and memorandum, 11 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 Stanmorephysics.com

MATHEMATICS

MARKS : 60

TIME : 1 HOUR 10 MIN

This question paper consists of 4 pages, including formulae that might be needed

### Page 2

Grade 1Downloaded from Stanmavephysées.com 17 March 2026

INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.

1. This question paper consists of TWO questions. Answer ALL the
questions.

2: Clearly show ALL calculations that you have used in determining your
answer.

3. You may use an approved scientific calculator (non-programmable and
non-graphical), unless stated otherwise.

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

5. If necessary, round off answers to TWO decimal places, unless stated
otherwise.

6. | Number the answers correctly according umbering system used
in this question paper.

7 Write neatly and legibly.

### Page 3

Grade Downloaded from Stanmasephysies.com 17 March 2026

QUESTION 1

GB)
G)
(4)
(5)
G)
(6)
GB)
gk.6k-3,91-k
1.4 Simplify: ok-ag-k (4)
Btahimorephysics.com 31]
QUESTION 2
21 Given tana=— = and 180° < @ < 360°, determine without using a calculator and by the
use of a diagram the value(s) of:
21.1 sine (2)
2.1.2 cosa qd)
2.1.3 sin?a+cos*a (2)
2.2
(6)
2.3. Simplify witho
(5)

### Page 4

Grade 1Downloaded from Stanmavephysées.com 17 March 2026
2.4 Prove that:

2.4.1 —“(42¢0s?x) (tanx + os) = tanx (5)
sinx.
2.4.2. For which value(s) of x in the interval [0; 360] is the identity in (3)

QUESTION 3.4.1 undefined?
2.5 Determine the general solution of
2sinAcosA — cosA = 0 (5)

[29]
TOTAL MARKS 60

FORMULAE
—b+vb2 —4ac
x =
2a

cos*@ + sin*@ =1

### Page 5

Downloaded from Stanmorephysics.com

_ F Province ot the
sued EASTERN CAPE ~~
NATIONAL
SENIOR CERTIFICATE

-GRADEF 11

MARCH 2026

MATHEMATICS TERM 1 CONTROLLED TEST
MARKING GUIDELINE

MARKS: — 75

This marking guideline consists of 7 pages.

### Page 6

JG/MARCH_ 2026

Downloaded Ninane stares oregnh ysis KRSM

QUESTION 1
1.1 | Solve for x:
1.1.1 //2x@- x) =0 v x=0
x=0 or x =3 ¥Y x=3
(2)
1.1.2 | ( #1)(% -— 3) = 2x
x*=3x+x—-3= 2x
x?-—4x-3=0 Y Standard form
—btvb?2—4ac
= 2a Y Substitution
~(-4)tV (-4)?-4()(-3)
z= 2(4) v x= 4,65
x=4,65 or x =—0,65 ¥ x=-0,65
(4)
1.1.3 | —x? +24 <—-2x
x?—2x-24>0 Y Standard form
(x -—6)(x +4) 20
cv Y Critical values
x=6o0rx=-4
imorephysies.com 7 e<—4
x<-4 or x26 ¥ x26
(4)
OR OR
—x?4+24 <-2x
—x? +2x+24<0 Y Standard form
Y Critical values
¥ x<-4
¥ x26
(4)
1.1.4
Y Isolation
¥ Squaring both
7+ 3x = 4x? sides
4x? —3x-7=0
(4x —7)(x+ 1) =0 v Standard form
7
x=Q orx=—1 Y Both x values
ax=-l Y Conclusion

(5)

### Page 7

Downloaded NismmESkaresoreyah ysies eRsN

JG/MARCH_ 2026

m=3

12
Y Dividing by 3
Y Adding the half
square on both
sides
Y Introducing root
on both sides
¥ x=30rx=1
(4)
13 za
(a)|x+y=50 ¥ x+y=50
x-5=7(y7-5) ¥ x-S=7(y—5)
(2)
b))x+ty=50 eevee (1)
K=—SE TVS) cccvewss
From (1)
YSH=50—K —s-swaxaesed
Sub (3) into (2) v Equation...3
x—5=7((50—x) —-5) Y Substitution
x—-5=7(50-—x—-5) {
x—-5=7(45 —x) E@fimorephysics.com ¥ Simplification
x—-5 = 315 —7x
7%+% =315+5
8x = 320
x = 40 Y Value of x
y =50-—40
y=10 Y Value of y
(5)
1.4.1 i m+tvVm2-9
2
A= (5)? -9
A= 16 ¥v A= 16
-. The roots are real, rational and unequal. v Real
Q)
1.4.2} A<0
m-9<0 ¥ m’?-9<0or
(m — 3)(m A<0
C.v v Factors

VV -3<x<3

(4)

[32]

### Page 8

Downloaded Ninanes OrERN 'SKOG ‘KES

JG/MARCH_ 2026

QUESTION 2
2.1.1 |aoyRer
| g0n+2
WUT }faxsy"*3 srt Prime bases
Yay (2x52)nt2
fo gnt3ygnt3 5n-3 Addition and
= Gnt2 ants subtractions of
= 2nt3—n-2 y gn+3+n-1-2n-4 exponents for
= 21% 5-2 same bases
2i
~ 52
_ 2 Answer
~ 35 G3)
2-1,2:| 3 *8= 5.g*F1
3xt1
__ 3%.33-5.3%.31 Rewriting with
~ 3%,31 prime bases
__ 3%(33-5.31) A
= 3x31 fahimorephysics.com Factorising
_ 33-5.31
="
_ 27-15 Simplification
~ 3
_2
Answer
i (4)
2.1.3 | V¥45-V20
3V5
_ 9x5—-V4x5
= jeu Ve Breaking surds
Simplification
Answer

(3)

### Page 9

Downloaded NinamExscaresoreynh ysios ‘RSM

JG/MARCH_ 2026

2.2__| Solve for x:
2.2.1 | A357
a v Applying laws of
exponents
v¥ Equating the
exponents
I v Answer
GB)
2 =
2.22 | 3 —5x3+6=0
1 1 v Factors
oe — = 1.
(x 2) (x. 3) 9 ¥Y x3=2 and
x3=2 or x3=3 erin
| 3 13
xP2=2"9 or x72 = 31%
£=2° or x= 3" v¥ x=8
x=8 or x=27 Vi 4 = 27
; (4)
1
x3—5x3+6=0 Ok
1 1 pe
(x3)? —5x3+6=0 Eafimorephysies.com
3 Y Fact
Let x3=k actors
¥ k=2ork=3
¥Y x=8
¥ x=27
(4)

### Page 10

Sownloaded stares SMOG ‘KRSM JG/MARCH 2026
QUESTION 3
3.1 | Ssing+3=0
5sinB = —3
sing-= -2 , cosB <0 4
Vv sing =—=
x Cp
r y
v Diagram
y=-3
r=5
re=x2+y?
(5)? =x? + (—3)2
x= 16
va? = VI6 # apa
x=i4 v 3
x=-4
3) (-4
tanB.cosB __ ee , a
2 ~~ 2 Zahmorephysies.com 5
3 v Answer
* to (6)
3.2
3.2.1 | cos42° = p
v y=y1-p?
a On. 1-p?
3sin42° = 3 (: 7 7 meee
=3/1—-p2 (3)
3.2.2 | cos132° = cos(180° — 48°)
= —cos48°
—_ (=) Y —cos48°
7 v —J1-p?
aii ha (2)

### Page 11

Downloaded

MAGIOR

JG/MARCH_ 2026

3.2.3 | tan318° = tan(360° — 42°)
= —tan 42° ¥ —tan42°
P y _ vie?
= vip? D
P 2)
33
3.3.1 | 2cos* x + 2sinx.cosx.tanx
= 2cos?x +2sinx.cosx. ——~ vu
cosx cosx
= 2cos*x + 2sin? x
= 2(cos* x + sin? x) Y Factorising
= 2(1)
=2 v Answer
G3)
3.3.2 | cos(360°—x).tan(—x)
sin? (x-180°)
__ cos x.-tanx Y cosx
(-sin x)? v —tanx
cos x.tanx A
= -——_ v —sinx
sin’ x sinx
cosx sin Vio oa
__ PcoaaE cosx
sin?x
=-— v Answer
sinx (5)
3.4 cos225°.sin(—135°) — sin330° Ti
tan225° ~
cos225°.sin(—135°) — sin330°
LHS =
tan225°
__ €0S225°.—sin 135° — sin330° Y —cos45°
v¥ —sin 135°
or
—sin 45°
¥ —sin30°
v tan45°
Y Special angles
(5)
[26]

TOTAL MARKS: 75

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