# Mathematics Grade 11 KZN March Test 2025 Paper & Memo

> Grade 11 Mathematics question paper and memo from the KZN March Test 2025, 12 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/kzn-maths-grade-11-march-2025-qp-and-memo-1/06d3f568-a5a7-4218-9dda-abe239fb06ef
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 11
- Document type: Question paper and memo
- Year: 2025
- Exam period: KZN March Test
- Curriculum: CAPS
- Pages: 12

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## About this document

Mathematics · Grade 11 · KZN March Test · 2025. Question paper and memorandum, 12 pages. Read online or download the PDF.

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

## Document text

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### Page 1

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Downloaded from Stanmorephysics.com

KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAI
RTIFICATE

et seeeneann

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®aesaneeses:

This question paper consists of 5 pages.

Seseuesseses”

saneeeneee®

### Page 2

Mathe: March 2025 Test

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INSTRUCTIONS AND INFORMATION
Read the following instructions carefully before answering the questions.
1. This question paper consists of 3 questions.

2. Answer ALL the questions.

3. Number the answers corre: ly. ng ( bering system used in this

question paper.

determining your answers.
Bi Answers only will NOT necess

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

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

8. Write neatly and legibly.

Copyright Reserved Please Turn Over

### Page 3

Mather Testpapers.co.za March 2025 Test
QUESTION 1
Ta Solve for x:
1. x -3x-28=0 GB)
1:42 o(1 + 25) =3x? (answer correct to TWO decimal places) (4)
113 2x? +x-3 20 (4)
2.
12 Given: Jx-3 =
vx
1.2.1 Solve for x. _ (5)
2
122 Hence, or otherwise, solve for k if Vk-20 = . 2
vet "
(6)
1.4.1 Solve for x in terms of m. (2)
1.4.2 Hence, or otherwise, determine the value(s) of m for which the above
equation will have equal roots. (3)
[29]
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### Page 4

Mathi
Testpapers.co.za
QUESTION 2
2.1 Evaluate the following expressions, without using a calculator:
S74 V121x? —¥27x°
4x
stot gh*
Dulad on
2.2 Solve for x, without

Hiaririorephysics.com

2.3 It is given that 4" =6 and 6” =8. Calculate the value of mn.

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March 2025 Test

3)

(4)

(3)
(3)

@G)
[16]

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### Page 5

(5)

(6)

(6)

Mathe: Testpapers.co.za March 2025 Test
QUESTION 3
a Given that 13cos 8 =5and a+ # =90°. By making use of a sketch and WITHOUT

the use of a calculator, calculate the value of Liat .

cosa@

32 Simplify the following to a single trigonometric term:

sin (180° +x).cos* (90°—x)+sin(—x).cos* (180°—x).
33 Evaluate WITHOUT

cos100°cos315° tan 210°

sin 675°cos150°sin170° —
jorephysies.com
3.4 Given: (1-cos* »){tans+ ——|,
sinx

320

3.4.1 Prove that (1-cos* 2)(tans+ $552) tans
sinx

3.4.2 For which value(s) of x in the interval [0°;360°) is the identity in
LION 3 .4,1)iiidefined?

d 0€ (90°; 270°). If 0<k <3, determine the value of

our answer must be a single fraction in its simplest form.

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(4)

G)

(5)
[30]

a

### Page 6

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BRINE

MARKS: 75

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These marking guidelines consist of 7 pages.

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### Page 7

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Me March 2025 Test
CAPS ~ Grade 11
Marking Guidelines
QUESTION 1
1.1.1) x7-3x=28=0
2
(x+4)(x-7)=0 OR x= ~(-3)#y(-3) -4()(28) VA correct factors OR
2(1) subst. into correct formula
aor x=7 ¥ CA answer VCA answer
GB)
2 2
L452 o(t +s) =3x
6+4x =3x°
3x7 -4x-6=0 ¥A standard form
_ -b+ Vb —4ac
2a
¥CA subst. into correct
a= quadratic formula
x=-0,90 or x=2,23 x=-0,90 or x=2,23 CAY answer VCA
answer.
NOTE: Penalty of 1 mark for incorrect rounding. (4)
1.1.3 2x°+x-320
(2x+3)(x-1)20 ¥A correct factors
¥CA critical values (can
be
on a sketch or number
line)
¥ CAY CA answer
(4)
12.1] fg-3=— aa
vx. . ”
xNx-3=2
x-3= Ea
x(x - 3) =2 x
4 P ‘
~3)=2? -3=-— ¥ A squaring both sides
x (x 3) 2 OR x * q is
*_3x-4=0 x? —3x-4=0 YA standard form
(x-4)(x+1)=0 (x-4)(x+1)=0 ¥CA factors
x=4 or x#-1 x=4 or x#-1 ¥CA YCA both answers
with rejection
(5)

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### Page 8

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Me March 2025 Test
CAPS — Grade 11
Marking Guidelines
1.22| Jezaoe
k-17
k-171=4 YCA equating k-17 to value of x
¥CA answer
k=21 (2)
13 | gys2x-3 VA y=2x-3
x?—15=-5x(2x-3)-(2x-3)° VCA substitution
x? =15=—10x? +15x—(4x? -12x+9)
x? =15=-10x? +15x=4: YCA simplification
x? +10x7 +4x° -15x
15x? -27x-6=0
5x? -9x-2=0
(5x+1)(x-2)=0 CA factors
1
on Of oF YCA both x—values
y=2(-2)}-3 or y=2(2)-3
17 7
as y= YCA both y—values
OR (6)
28 . OR
2 vA xa
2
(=) -15= -5(23)-¥ ¥ CA substitution
2 61,2
y +6y49 15 Sy Isy__ 2
4 2
2 —5y-
(0) gas) af SPP) ag
y? +6y+9-60=-10)" -30y—4y" ¥CA simplification
15y? +36y—51=0
sy? +12y-17=0
(Sy+17)(y-1)=0 ¥CA factors
¥CA both y—values
YCA both x—values
(6)

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### Page 9

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Me March 2025 Test
CAPS — Grade 11
Marking Guidelines
LA.l 5x eme+18=0 x" +2mx+36=0
2 1
=m+,|(m) - 43) 18 2
LP ey 2 (8) OR x= ~2m* y(2m) -4(1)(36) | vA correct substitution
{3} 2(1) into quadratic formula
2
~2m+ 4m’ —144
x=—m£m —36 a YCA simplified answer
(2)
1.4.2 | A=m? -36 A=4m’? -144
b’ -4ac =0
sm? —36=0 Y¥CA A equated to 0
(m+6)(m-6)=0 YCA factors
m=-6 or m=6 ' ¥CA both answers
NOTE: Penalise a mark i: i n factors m= +36 (3)
[29]

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### Page 10

Me Testpapers.co.za
CAPS — Grade I1

March 2025 Test

Marking Guidelines
QUESTION 2
211) Vi2he 3278
-4°x
_ Ily=3x v A numerator: square & cube
~ Ig v A denominator
=-8 ¥CA answer
d GB)
2.1,2:| igt.6r.o*
161.3%
ke k3
_ (2°) Aa} (3° A splitting to prime bases
ey
_ Beare ah ae ¥CA applying the exponential laws
aka 37k correctly

= DUES SP ~34+2-2k+k

=2134 ill

¥CA simplification

¥CA answer

(4)

Peer

vA isolating the variable

vA raising both sides of equation to
an exponent of —3

¥CA answer
G3)

2.2.2

vA both equations after transposing

YA 5‘ =5° and no solution
v A answer

(3)

2.3

vA raising both sides to exponent n
YA substituting 8 for 6”

v A answer

(3)

[16]

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### Page 11

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Me March 2025 Test
CAPS ~ Grade 11
Marking Guidelines
QUESTION 3
3.1 | 13cosB=5
r 13 12 .
y=r—x° [Pythagoras] VA diagram
= 3? —5? >x
y= 12 v¥ Ay—value
12
tanB 5 ¥CA correct substitution for
cosa 12 tan B
13 YCA correct substitution for
~ 12,13 cosa
5 12
= fe or 2 .
5 5
¥CA answer
(5)
3.2 | sin(180°+.x).cos* (90°-x)+sin(—x).cos* (180° —x)
F x Ae F 2 VA -sinx VA sin? x
=(-sinx)(sinx) +(—sin x)(—cos x) . i
VA -sinx VA (—cosx)
=~sin x(sin’ x-+cos* x) YCA factors
=—sin x(1)
=-sinx ¥CA answer
(6)
3.3 cos 100° cos 315° tan 210°
sin 675° cos150°sin 170°
__ (~cos80°) (cos 45°)( tan 30°) vA numerator
_ (=8in452)( cos VA denominator
Y A cos80° =sinl10° OR
sin 10° = cos80°
Y CA special angle values
¥CA simplification
2
== ¥CA answer
2 (6)

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### Page 12

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March 2025 Test
CAPS ~ Grade 11
Marking Guidelines
cos x
3.4.1 | LHS= (1 =cos* x){ tan x+— }
sin x
: sinx cosx P sinx
2 sir xf += } YA sin’ x YA
cosx sinx cos x
.» _( sin? x+cos* x ,
=sin* x] —————_ VA LCD (cosx.sinx )
cos x.sin x

+2 1
=Sin* x] ————
cos x.sin x

YA sin? x+cos* x=1

k?4+9

_ sinx
cosx : is out of 4 marksin || vA simplification to sins
=tanx aper. It should be OES
=RHS as in the memo. (5)
3.4.2 | Undefined when sinx =0a
“x E {0° ; 90°; AV AY AY
(Any 3 values given)
Q)
3.5. | x =r?-y? [Pythagoras
=(k +9) -(-6k)°
=k* +18k? +81-36k? ¥ A diagram in quadrant 3
=k*-18k? +81
-(e-9)
x= +(e -9)
vx=-(k-9) VAx-— value
ie. 1 _ 6k ¥CA correct substitution
cosO -(k°-9)  —(k?-9)

¥CA simplification to a
single fraction

Copyright Reserved

__(k-3)(k=3)
~ (k=3)(k+3)
= (t-3) ¥CA answer
(k +3) (5)
[30]
TOTAL: 75

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