# Mathematics Grade 12 Paper 2 Challenge Paper 2026 Memo

> Grade 12 Mathematics Paper 2 memorandum from the Challenge Paper 2026, 12 pages. Read online or get the free PDF download, question paper linked when available.

- Page: https://www.hlayiso.com/library/grade-12-mathematics-2026-challenge-paper-memorandum-paper-2/ba2ebdef-1b94-49b3-aab3-54883ffc6da6
- Updated: 2026-10-24
- Subject: Mathematics
- Grade: Grade 12
- Document type: Memorandum
- Year: 2026
- Exam period: Challenge Paper
- Paper: 2
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 12

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## About this document

Mathematics · Grade 12 · Hlayiso Challenge Paper · 2026. A deliberately harder practice Paper 2 (memorandum), 12 pages. Read online or download the PDF.

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### Page 1

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Challenge
hlayiso mock exams
Practice Series:
Grade 12 NSC Preparation
WWW.HLAYISO.COM
NATIONAL
SENIOR CERTIFICATE
GRADE 12
MATHEMATICS P2
CHALLENGE PAPER 2026
MARKING GUIDELINES
MARKS: 150
These marking guidelines consist of 12 pages.

### Page 2

Mathematics/P2
2
NSC – Marking Guidelines
Hlayiso/Challenge 2026
NOTE:
• If a candidate answers a question TWICE, only mark the FIRST attempt.
• If a candidate has crossed out an attempt of a question and not redone the question, mark the
crossed-out version.
• Consistent accuracy (CA) applies in ALL aspects of the marking guidelines. Stop marking at the
second calculation error.
• Assuming answers/values in order to solve a problem is NOT acceptable. In show that questions
the given answer earns no mark: marks are for the working that leads to it.
• Valid alternative methods that are not shown here must be marked on merit, using the same mark
allocation.
GEOMETRY
S A mark for a correct statement (a statement mark is independent of a reason).
R A mark for the correct reason (a reason mark may only be awarded if the statement is correct).
S/R Award a mark if the statement AND reason are both correct.
QUESTION 1
1.1
395+x+y
11
= 46 (sum of the 9 known values is 395)
395+x+y = 506 ⇒x+y = 111
x+(x+25) = 111 ⇒2x = 86
x = 43 and y = 68
(Check: 37 ≤43 ≤45 and 56 ≤68 ≤71, so the order
is kept.)
✓∑= 506 / equation
✓substitution y = x+25
✓x = 43 and y = 68
(3)
1.2
Min = 25; Q1 = 33; Median = 45; Q3 = 56; Max
= 71
20
30
40
50
60
70
✓min and max (whiskers)
✓Q1 and Q3 (box)
✓median
(3)
1.3
σ = 14,38 minutes
(Population standard deviation. If s = 15,08 is used:
max 1 mark.)
✓✓answer
(2)
1.4
(¯x−σ ; ¯x+σ) = (31,62 ; 60,38)
Values: 33 ; 37 ; 43 ; 45 ; 48 ; 52 ; 56
7 learners
✓interval
✓7 learners
(2)
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### Page 3

Mathematics/P2
3
NSC – Marking Guidelines
Hlayiso/Challenge 2026
1.5
New mean = 1,1(46)−5 = 45,6 minutes
New σ = 1,1(14,38) = 15,82 minutes
(Subtracting 5 does not change the spread.)
✓45,6
✓15,82 (CA from 1.3)
(2)
[12]
QUESTION 2
2.1
a = −18,42
b = 2,96
ˆy = −18,42+2,96x
✓a
✓b
✓equation
(3)
2.2
r = 0,99
Very strong positive (linear) relationship
✓r
✓description
(2)
2.3
ˆy = −18,42+2,96(27) = 61,50
About 61 cold drinks
(accept 61 or 62; calculator: 61,44)
✓substitution
✓answer (whole number)(2)
2.4
∑y = 456; ¯y = 456
8
= 57
New mean = 57+3 = 60
456+y9
9
= 60 ⇒456+y9 = 540
y9 = 84 cold drinks
✓¯y = 57
✓equation with 9 days
✓84
(3)
[10]
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### Page 4

Mathematics/P2
4
NSC – Marking Guidelines
Hlayiso/Challenge 2026
QUESTION 3
3.1
mAB = 6−0
1+2 = 2 ⇒inclination of AB = 63,43◦
mAC = 6−0
1−3 = −3 ⇒inclination of AC
= 180◦−71,57◦= 108,43◦
B ˆAC = 108,43◦−63,43◦
[ext ∠of △]
B ˆAC = 45◦
✓mAB = 2 and 63,43◦
✓mAC = −3
✓108,43◦
✓45◦
(4)
3.2
mBD ×(−3) = −1 ⇒mBD = 1
3
[BD ⊥AC]
y−0 = 1
3(x+2)
y = 1
3x+ 2
3
✓m = 1
3
✓substitution of B
✓equation
(3)
3.3
AC: y−0 = −3(x−3) ⇒y = −3x+9
1
3x+ 2
3 = −3x+9 ⇒x+2 = −9x+27
x = 5
2; y = −3
(5
2
)
+9 = 3
2
D
(
21
2 ; 11
2
)
✓equation of AC
✓equating
✓D
(3)
3.4
In △ABD: A ˆDB = 90◦and ˆA = 45◦
A ˆBD = 180◦−90◦−45◦= 45◦= ˆA
∴AD = BD [sides opp equal ∠s] ∴△ABD is
isosceles
OR: AD2 = (1−2,5)2 +(6−1,5)2 = 22,5;
BD2 = (2,5+2)2 +(1,5)2 = 22,5
∴AD = BD
✓A ˆBD = 45◦
✓AD = BD (S/R)
(OR ✓✓both lengths)
(2)
3.5
BD = √22,5
DC
=
√
(3−2,5)2 +(0−1,5)2 = √2,5
Area △BDC = 1
2 BD·DC = 1
2
√22,5√2,5 = 1
2
√56,25
= 3,75 square units
OR: Area △ABC = 1
2(5)(6) = 15; DC
AC =
√2,5
√
40 = 1
4
Area △BDC = 1
4(15) = 3,75 (same height from B)
✓BD and DC
✓area formula
✓3,75
OR: ✓15 ✓1
4
✓3,75
(3)
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### Page 5

Mathematics/P2
5
NSC – Marking Guidelines
Hlayiso/Challenge 2026
3.6
BˆPC = B ˆAC, so P lies on the circle through A, B and C
[converse: ∠s in the same segment]
Centre N on the perpendicular bisector of BC: x = 1
2
B ˆNC = 2×45◦= 90◦[∠at centre = 2×∠at
circumference], so △BNC is a right-angled isosceles
△and N is 1
2BC = 5
2 above BC: N
(1
2 ; 5
2
)
r2 = NB2 =
(5
2
)2 +
(5
2
)2 = 25
2
(
x−1
2
)2 +
(
y−5
2
)2 = 25
2
(y > 0)
OR: substitute A, B, C into x2 +y2 +Dx+Ey+F = 0:
4−2D+F = 0; 9+3D+F = 0;
37+D+6E +F = 0
D = −1, F = −6, E = −5: x2 +y2 −x−5y−6 = 0
(same circle)
✓P on circle through A, B,
C (reason)
✓centre
(1
2 ; 5
2
)
✓r2 = 25
2
✓equation
OR: ✓concept
✓3 equations
✓solving
✓equation
(4)
[19]
QUESTION 4
4.1
x2 −4x+4+y2 +6y+9 = 12+4+9
(x−2)2 +(y+3)2 = 25
M(2 ; −3)
r = 5
✓completing squares
✓M
✓r = 5
(3)
4.2
mMP = 0+3
6−2 = 3
4
mTP = −4
3
[tan ⊥radius]
y−0 = −4
3(x−6)
y = −4
3x+8
✓mMP
✓mtan
✓substitution of P
✓equation
(4)
4.3
T(0 ; 8)
TP =
√
(6−0)2 +(0−8)2 =
√
100 = 10 units
OR: MT2 = 4+121 = 125; TP2 = 125−25 = 100
[tan ⊥radius]
✓T(0 ; 8)
✓substitution
✓10
(3)
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### Page 6

Mathematics/P2
6
NSC – Marking Guidelines
Hlayiso/Challenge 2026
4.4
TQ = TP = 10
[tans from same pt]
x2 +(y−8)2 = 100 ⇒x2 +y2 −16y−36 = 0
Circle: x2 +y2 −4x+6y−12 = 0
Subtract: 4x−22y−24 = 0 ⇒x = 11y+12
2
Substitute:
(11y+12
2
)2
+y2 −2(11y+12)+6y−12 = 0
125y2 +200y = 0 ⇒25y(5y+8) = 0
y = 0 (point P) or y = −8
5
Q
(
−14
5 ; −8
5
)
= (−2,8 ; −1,6)
OR: Q is the reflection of P in MT. Foot of ⊥from P
on MT: MT: y = −11
2 x+8; line through P ⊥MT:
y = 2
11(x−6)
intersection (1,6 ; −0,8) is the midpoint of PQ ⇒
Q(−2,8 ; −1,6)
✓TQ = 10 (S/R)
✓equation of circle centre T
✓x in terms of y
✓standard form
✓Q
OR:
✓concept ✓MT
✓⊥line
✓midpoint ✓Q
(5)
4.5
MˆPT = M ˆQT = 90◦
[tan ⊥radius]
∴MT is a diameter
[converse: ∠in semi-circle]
Centre =
(2+0
2
; −3+8
2
)
=
(
1 ; 5
2
)
;
r2 = MT2
4
= 125
4
(x−1)2 +
(
y−5
2
)2
= 125
4
✓MT diameter (reason)
✓centre
✓equation
(3)
4.6
The parallel tangent touches the circle at P′, the other
end of diameter PM:
P′ = (2(2)−6 ; 2(−3)−0) = (−2 ; −6)
−6 = −4
3(−2)+k ⇒k = −26
3
−26
3 < k < 8
OR: substitute y = −4
3x+k into the circle:
25x2 −(24k +108)x+9k2 +54k −108 = 0
∆> 0 ⇒3k2 +2k −208 < 0 ⇒(3k +26)(k −8) < 0
✓P′(−2 ; −6)
✓k = −26
3
✓answer (CA from 4.2)
OR: ✓standard form
✓factors ✓answer
(3)
[21]
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### Page 7

Mathematics/P2
7
NSC – Marking Guidelines
Hlayiso/Challenge 2026
QUESTION 5
5.1
sinAcosB+cosAsinB = 4
5
…(1)
sinAcosB−cosAsinB = 1
5
…(2)
(1)+(2): sinAcosB = 1
2; (1)−(2): cosAsinB = 3
10
tanA
tanB = sinAcosB
cosAsinB =
1
2
3
10
= 5
3
✓both expansions
✓1
2 and 3
10
✓sinAcosB
cosAsinB
✓5
3
(4)
5.2
cos(540◦−2x) = cos(180◦−2x) = −cos2x
sin(x−450◦) = sin(x−90◦) = −cosx;
cos(−x) = cosx
tan(x−180◦) = tanx; tan(540◦+x) = tanx
Expression = −cos2x
−cos2 x +tan2 x
= cos2 x−sin2 x
cos2 x
+ sin2 x
cos2 x
= cos2 x
cos2 x = 1
✓−cos2x
✓−cosx and cosx
✓tanx.tanx
✓cos2x
cos2 x
✓cos2 x−sin2 x
✓1
(6)
5.3
(cos15◦−sin15◦)2 =
cos2 15◦+sin2 15◦−2sin15◦cos15◦= 1−sin30◦= 1
2
cos15◦> sin15◦, so cos15◦−sin15◦=
√
1
2 =
√
2
2
OR: =
√
2(cos45◦cos15◦−sin45◦sin15◦) =
√
2cos60◦=
√
2
2
OR: cos(45◦−30◦)−sin(45◦−30◦) =
√
6+
√
2
4
−
√
6−
√
2
4
=
√
2
2
✓1−sin30◦
✓1
2
✓positive root
√
2
2
OR: ✓
√
2cos45◦form
✓cos60◦✓answer
(3)
[13]
QUESTION 6
6.1
LHS = 1−(1−2sin2 x)+2sinxcosx
1+(2cos2 x−1)+2sinxcosx
= 2sin2 x+2sinxcosx
2cos2 x+2sinxcosx
= 2sinx(sinx+cosx)
2cosx(cosx+sinx)
= sinx
cosx = tanx = RHS
✓1−2sin2 x and
2cos2 x−1
✓2sinxcosx
✓factorising numerator
✓factorising denominator(4)
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### Page 8

Mathematics/P2
8
NSC – Marking Guidelines
Hlayiso/Challenge 2026
6.2
Undefined if cosx = 0 or sinx+cosx = 0, i.e.
tanx = −1
cosx = 0: x = −90◦or 90◦
tanx = −1: x = −45◦or 135◦
✓cosx = 0 and
sinx+cosx = 0
✓±90◦
✓−45◦and 135◦
(3)
6.3
[ f(x)]2 −2 f(x)−3 = 0
( f(x)−3)(f(x)+1) = 0
f(x) = 3 or f(x) = −1
f(x) = −1: tanx = −1 ⇒x = 135◦+k.180◦; but then
sinx+cosx = 0 and f(x) is undefined (6.2). No
solution.
f(x) = 3: tanx = 3
x = 71,57◦+k.180◦, k ∈Z
(If tanx = −1 is not rejected: max 4/6. Penalise 1 mark
if k ∈Z is omitted.)
✓standard form
✓factors
✓tanx = −1 considered
✓rejected with reason
✓tanx = 3
✓71,57◦+k.180◦, k ∈Z (6)
[13]
QUESTION 7
7.1
a = 1
b = 2 (period 180◦)
p = −30◦
✓a
✓b
✓p
(3)
7.2
sin2x = cos(x−30◦) = sin(90◦−(x−30◦)) =
sin(120◦−x)
2x = 120◦−x+k.360◦⇒x = 40◦+k.120◦
or
2x = 180◦−(120◦−x)+k.360◦⇒x = 60◦+k.360◦,
k ∈Z
x ∈{−80◦; 40◦; 60◦; 160◦}
OR: cos(90◦−2x) = cos(x−30◦):
90◦−2x = ±(x−30◦)+k.360◦
✓co-function
✓40◦+k.120◦
✓60◦+k.360◦
✓all four values
(4)
7.3
40◦≤x ≤60◦or 160◦≤x ≤180◦
(accept strict inequalities at 40◦, 60◦, 160◦if ≥is not
used: max 2/3)
✓40◦≤x ≤60◦
✓160◦≤x
✓x ≤180◦
(3)
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### Page 9

Mathematics/P2
9
NSC – Marking Guidelines
Hlayiso/Challenge 2026
7.4
−sinx = cos(x+90◦) = cos((x+120◦)−30◦) =
g(x+120◦)
Translate g 120◦to the LEFT
(OR 240◦to the right)
✓cos(x+90◦)
✓120◦left / 240◦right
(2)
[12]
QUESTION 8
8.1
In △PQR: tanα = h
QR ⇒QR =
h
tanα ; similarly QS
=
h
tanα
In △QRS: d2 = QR2 +QS2 −2QR.QScos2θ
[cosine rule]
d2 = 2QR2(1−cos2θ) = 2QR2(2sin2 θ) =
4QR2 sin2 θ
d = 2QRsinθ
(θ acute)
h
tanα =
d
2sinθ ⇒h = d tanα
2sinθ
OR: QR = QS, so QM ⊥RS and R ˆQM = θ;
sinθ = d/2
QR , QR =
d
2sinθ ; h = QRtanα
✓QR in terms of h
✓QR = QS
✓cosine rule
✓1−cos2θ = 2sin2 θ
✓QR =
d
2sinθ
OR: ✓QM ⊥RS ✓θ
✓sinθ ✓QR ✓h
(5)
8.2
h = 80tan32◦
2sin50◦= 32,63 m
✓substitution
✓answer
(2)
8.3
QR =
80
2sin50◦= 52,22 m
QM = QRcos50◦= 33,56 m
(QM ⊥RS, isosceles
△)
tanϕ =
h
QM = 32,63
33,56
⇒ϕ = 44,19◦
(OR tanϕ = tanα
cosθ = tan32◦
cos50◦)
✓QM
✓tanϕ
✓44,19◦(accept
44,18◦–44,20◦)
(3)
[10]
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### Page 10

Mathematics/P2
10
NSC – Marking Guidelines
Hlayiso/Challenge 2026
QUESTION 9
9.1
Construction: Draw diameter AOD and join DB.
D ˆAS = 90◦
[tan ⊥radius]
∴B ˆAS = 90◦−D ˆAB
A ˆBD = 90◦
[∠in semi-circle]
∴A ˆDB = 90◦−D ˆAB
[∠s of △]
∴B ˆAS = A ˆDB
A ˆDB = A ˆCB
[∠s in the same segment]
∴B ˆAS = A ˆCB
OR: Construction: join OA and OB. Let B ˆAS = x.
O ˆAS = 90◦[tan ⊥radius] ⇒O ˆAB = 90◦−x = O ˆBA
[∠s opp equal radii]
A ˆOB = 180◦−2(90◦−x) = 2x [∠s of △];
A ˆCB = 1
2A ˆOB = x [∠at centre = 2×∠at
circumference]
✓construction
✓S/R
✓S/R
✓S
✓S/R
✓conclusion
OR: ✓construction
✓S/R ✓S/R
✓2x (S)
✓S/R ✓conclusion
(6)
9.2.1
A ˆBD = P ˆAD = 35◦
[tan-chord theorem]
✓S ✓R
(2)
9.2.2
D ˆBC = A ˆBD = 35◦
[equal chords subtend equal ∠s]
A ˆBC = 35◦+35◦= 70◦
✓S/R
✓answer
(2)
9.2.3
A ˆOC = 2A ˆBC = 140◦
[∠at centre = 2×∠at
circumference]
O ˆAC = O ˆCA
[∠s opp equal radii]
O ˆAC = 180◦−140◦
2
= 20◦
OR: A ˆDC = 180◦−70◦= 110◦[opp ∠s of cyclic
quad]; D ˆAC = 35◦[∠s opp equal sides]; P ˆAC = 70◦
and O ˆAP = 90◦[tan ⊥radius] ⇒O ˆAC = 20◦
✓140◦(S/R)
✓R
✓20◦
(3)
[13]
QUESTION 10
10.1
Construction: Draw DG ∥BF with G on AC.
In △ADG: AE = ED and EF ∥DG
∴AF = FG
[midpt theorem / line ∥one side of △]
In △CBF: DG ∥BF
FG
GC = BD
DC = 3
2
[line ∥one side of △]
Let FG = 3k, GC = 2k; AF = 3k
AF : FC = 3k : (3k +2k) = 3 : 5
✓construction
✓AF = FG (S/R)
✓S ✓R
✓3 : 5
(5)
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Mathematics/P2
11
NSC – Marking Guidelines
Hlayiso/Challenge 2026
10.2
In △ADG: EF = 1
2DG
[midpt theorem]
In △CBF: △CDG ||| △CBF
[∠∠∠; DG ∥BF]
DG
BF = CD
CB = 2
5 ⇒BF = 5
2DG
BE = BF −EF = 5
2DG −1
2DG = 2DG
BE : EF = 2 : 1
2 = 4 : 1
✓EF = 1
2DG (S/R)
✓2
5 (S/R)
✓BE = 2DG
✓4 : 1
(4)
10.3
area △AEF
area △ADC =
1
2AE.AFsin ˆA
1
2AD.ACsin ˆA = 1
2 × 3
8 = 3
16
area △ADC
area △ABC = DC
BC = 2
5
[same height]
area △AEF
area △ABC = 3
16 × 2
5 = 3
40
(CA from 10.1)
✓area rule with sin ˆA
✓3
16
✓2
5
✓3
40
(4)
[13]
QUESTION 11
11.1
C ˆAB = x
[tan-chord theorem]
O ˆCA = C ˆAO = x
[∠s opp equal radii]
(accept C ˆAD, C ˆAO for C ˆAB; accept A ˆCO)
✓S ✓R
✓S ✓R
(4)
11.2
O ˆCD = 90◦
[tan ⊥radius] and ˆE = 90◦(given)
∴OC ∥AE
[corresp ∠s =]
E ˆAC = A ˆCO = x
[alt ∠s; OC ∥AE]
∴E ˆAC = C ˆAB, i.e. AC bisects E ˆAB
OR: A ˆCB = 90◦[∠in semi-circle];
E ˆCA = 180◦−90◦−x = 90◦−x [∠s on str line]; in
△AEC: E ˆAC = 180◦−90◦−(90◦−x) = x [∠s of △]
✓OC ∥AE (S/R)
✓S ✓R
OR:
✓90◦(S/R)
✓90◦−x
✓x
(3)
11.3
In △DCB and △DAC:
ˆD = ˆD
[common]
D ˆCB = D ˆAC = x
[tan-chord theorem / 11.1]
∴△DCB ||| △DAC
[∠∠∠]
✓S
✓S
✓R
(3)
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Mathematics/P2
12
NSC – Marking Guidelines
Hlayiso/Challenge 2026
11.4
DC
DA = DB
DC
[||| △s] ⇒DC2 = DB.DA
64 = 4DA ⇒DA = 16; AB = 12, OC = 6, DO = 10
In △DOC and △DAE: OC ∥AE, so △DOC ||| △DAE
[∠∠∠]
AE
OC = DA
DO ⇒AE = 6×16
10
= 9,6 units
OR: sin ˆD = OC
OD = 6
10 (in △OCD); AE =
DAsin ˆD = 16×0,6 = 9,6
✓DC2 = DB.DA (S/R)
✓DA = 16 and OC = 6
✓ratio from ||| / sin ˆD
✓9,6
(4)
[14]
TOTAL: 150
Copyright reserved

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