# Mathematics Grade 12 Paper 1 NSC November 2026 Memo (English)

> Grade 12 Mathematics Paper 1 memorandum from the NSC November Exam 2026 (English), 13 pages. Read online or get the free PDF download, question paper linked when available.

- Page: https://www.hlayiso.com/library/grade-12-mathematics-2026-nsc-november-exam-unofficial-memorandum-paper-1/b0c1b243-df38-4888-b0a0-a1b69c7f75aa
- Updated: 2026-10-27
- Subject: Mathematics
- Grade: Grade 12
- Language: English
- Document type: Memorandum
- Year: 2026
- Exam period: NSC November Exam
- Paper: 1
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 13

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## About this document

Mathematics · Grade 12 · NSC November Exam · 2026. Unofficial memorandum, Paper 1, 13 pages. Unofficial marking guidelines by Hlayiso, not the DBE memorandum. Read online or download the PDF.

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## Document text

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### Page 1

UNOFFICIAL
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Unofficial memo
hlayiso mock exams
Practice Series:
Grade 12 NSC Preparation
WWW.HLAYISO.COM
NATIONAL
SENIOR CERTIFICATE
GRADE 12
MATHEMATICS P1
NOVEMBER 2026 — UNOFFICIAL MARKING GUIDELINES
MARKS: 150
These marking guidelines consist of 13 pages.

### Page 2

UNOFFICIAL
Mathematics/P1
2
NSC – Unofficial Marking Guidelines
Hlayiso/November 2026
IMPORTANT: These are unofficial marking guidelines prepared by Hlayiso for revision. They are
not the DBE’s official memorandum; the official marking guidelines may differ in mark allocation
and accepted alternatives.
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.
• Answers only will NOT necessarily be awarded full marks; where an answer-only allocation
applies it is stated in the row.
• Rounding: penalise incorrect rounding (or no rounding where TWO decimal places are required)
ONCE per question, not per item.
QUESTION 1
1.1.1
x2 +x−6 = 0
(x+3)(x−2) = 0
x = −3
or
x = 2
✓factors
✓x = −3 ✓x = 2
Answer only: full marks
(3)
1.1.2
3x2 −2x−10 = 0
x = −(−2)±
√
(−2)2 −4(3)(−10)
2(3)
= 2±
√
124
6
x = 2,19
or
x = −1,52
✓standard form
✓substitution into correct
formula
✓2,19 ✓−1,52
Note: Penalise 1 mark for
incorrect rounding. Accept
x = 1±
√
31
3
simplified first.
Answer only: max 2/4.
(4)
1.1.3
81·2x−4 = 3x
34 ·2x−4 = 3x
2x−4 = 3x−4
i.e.
(2
3
)x−4 = 1 =
(2
3
)0
x−4 = 0 ⇒x = 4
OR: 81·2x
16
= 3x ⇒
(3
2
)x = 81
16 =
(3
2
)4 ⇒x = 4
OR (logs): xlog 3
2 = log 81
16 ⇒x = 4
✓81 = 34
✓2x−4 = 3x−4 / same base
✓x = 4
OR:
✓2x−4 = 2x
16 ✓
(3
2
)x =
(3
2
)4
✓x = 4
Answer only: 1/3
(3)
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### Page 3

UNOFFICIAL
Mathematics/P1
3
NSC – Unofficial Marking Guidelines
Hlayiso/November 2026
1.1.4
√
(x+6)(10−x) = 4
√
3
(−6 ≤x ≤10)
(x+6)(10−x) = 48
−x2 +4x+60 = 48
x2 −4x−12 = 0
(x−6)(x+2) = 0
x = 6
or
x = −2
Check x = 6:
√
12·
√
4 = 4
√
3 ✓
Check x = −2:
√
4·
√
12 = 4
√
3 ✓
∴x = 6 or x = −2 (both valid)
✓squaring both sides
✓standard form
✓both values
✓both accepted after
checking / domain
Note: If a valid root is rejected,
max 3/4. Answer only: max 2/4.
(4)
1.2
y = 2x+2
(2x+2)2 +3x2 −4 = 0
4x2 +8x+4+3x2 −4 = 0
7x2 +8x = 0
x(7x+8) = 0
x = 0
or
x = −8
7
y = 2
or
y = −2
7
OR: x = y−2
2
: y2 + 3(y−2)2
4
−4 = 0
7y2 −12y−4 = 0 ⇒(7y+2)(y−2) = 0
y = −2
7 or y = 2; x = −8
7 or x = 0
✓y the subject
✓substitution
✓standard form
✓both x-values
✓both y-values (correctly
paired)
OR:
✓x subject ✓substitution
✓standard form
✓y-values ✓x-values
Note: CA for y-values from
incorrect x-values.
(5)
1.3
Let the first number be x; the second number is x+2
x(x+2) < 15
x2 +2x−15 < 0
(x+5)(x−3) < 0
−5
3
+
−
+
−5 < x < 3
✓x and x+2
✓inequality
✓standard form
✓critical values −5 and 3
✓−5 < x < 3
Note: The numbers are not
stated to be integers. A
candidate who lists only the
integers −4;−3;...;2 with
correct working: max 4/5.
Answer only: max 2/5.
(5)
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### Page 4

UNOFFICIAL
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4
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QUESTION 2
2.1.1
d = −2
3 −(−1) = 1
3
✓d = 1
3
(1)
2.1.2
Tn = −1+(n−1)
(1
3
)
= 27
n−1 = 84 ⇒n = 85
✓substitution into Tn with
Tn = 27
✓n = 85
(2)
2.1.3
Tn = −1+(n−1)1
3 = n
3 −4
3
Last 65 terms: T21 to T85
85
∑
n=21
(n
3 −4
3
)
or
85
∑
n=21
[
−1+(n−1)1
3
]
OR: T21 = 17
3 :
65
∑
k=1
(17
3 +(k −1)1
3
)
=
65
∑
k=1
k +16
3
OR (from 27 down):
65
∑
k=1
(
27−(k −1)1
3
)
✓general term
✓lower limit 21
✓upper limit 85
OR:
✓general term ✓1 ✓65
Note: CA from 2.1.2. Value
(not required): 3185
3 .
(3)
2.2.1
T1 = 44 = 256
T2 = 43 = 64
✓256
✓64
(2)
2.2.2
4−n+5 =
1
1024
4−n+5 = 4−5
−n+5 = −5 ⇒n = 10
i.e. the 10th term
OR: 256
(1
4
)n−1 =
1
1024 ⇒
(1
4
)n−1 =
(1
4
)9 ⇒n = 10
✓equating
✓4−5 / same base (or logs)
✓n = 10
Answer only: 1/3
(3)
2.2.3
r = 64
256 = 1
4
(−1 < r < 1)
S∞= 256
1−1
4
= 1024
3
= 341,33
✓r = 1
4
✓substitution into S∞
✓1024
3
or 341,33
Note: CA from 2.2.1.
(3)
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QUESTION 3
3.1
2a = −2 ⇒a = −1
Maximum at n = 8: −b
2a = 8 ⇒−b
−2 = 8 ⇒b = 16
T8 = −(8)2 +16(8)+c = 44
c = −20
Tn = −n2 +16n−20
OR: Tn = a(n−8)2 +44 with a = −1
Tn = −(n2 −16n+64)+44 = −n2 +16n−20
✓a = −1
✓b = 16
✓substitution of (8; 44)
✓c = −20
OR:
✓a = −1 ✓turning-point
form
✓expansion ✓answer
(4)
3.2
Let the amount in the 1st minute be a litres; d = 2
Tank half full after 36 min and empty after 54 min:
S36 = 1
2S54
36
2 [2a+35(2)] = 1
2 · 54
2 [2a+53(2)]
36a+1260 = 27a+1431
9a = 171 ⇒a = 19 litres
OR: S36 = S54 −S36 (second half drains in the last 18
min):
18(2a+70) = 9[2(a+72)+17(2)] ⇒36a+1260 =
18a+1602 ⇒a = 19
Check: S36 = 1944 ℓ, S54 = 3888 ℓ
✓d = 2
✓S36 = 1
2S54
✓S36 ✓S54 substituted
✓a = 19 litres
Note: Assumes the tank was
full at the start. Answer only:
0/5.
(5)
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Hlayiso/November 2026
QUESTION 4
4.1
f(x) = 5−x =
(1
5
)x and the base 0 < 1
5 < 1
✓
(1
5
)x with 0 < b < 1
Note: Accept: as x increases,
f(x) decreases; or f ′(x) < 0 for
all x.
(1)
4.2
f −1: x = 5−y
−y = log5 x
y = −log5 x
(or y = log 1
5 x)
✓swop x and y
✓y = −log5 x
(2)
4.3
x
y
O
y = x
f
f −1
(0;1)
(1;0)
(−1;5)
(5;−1)
Asymptote of f: y = 0 (the x-axis)
Asymptote of f −1: x = 0 (the y-axis)
f:
✓decreasing shape through
(0;1)
✓asymptote y = 0
shown/stated (graph does not
cut the x-axis)
f −1:
✓decreasing shape through
(1;0)
✓asymptote x = 0
shown/stated (graph does not
cut the y-axis)
Note: Graphs must be
symmetrical about y = x (not
required to draw it). CA from
4.2.
(4)
4.4
−log5 x ≥−1
log5 x ≤1
0 < x ≤5
or
x ∈(0; 5]
✓0 < x
✓x ≤5
Note: x ≤5 only: 1/2.
(2)
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UNOFFICIAL
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QUESTION 5
5.1
T(−2; 0)
(g(x) = 0: 2x+4 = 0)
✓T(−2;0)
(1)
5.2
TP = 6 ⇒P(4; 0)
y = a(x+2)(x−4)
Substitute S(−1;−10):
−10 = a(−1+2)(−1−4) = −5a
a = 2
f(x) = 2(x2 −2x−8) = 2x2 −4x−16
∴a = 2; b = −4; c = −16
OR: 4a−2b+c = 0; 16a+4b+c = 0;
a−b+c = −10
solving simultaneously gives a = 2, b = −4, c = −16
✓P(4;0)
✓y = a(x+2)(x−4)
✓substitution of S
✓a = 2 and expansion
OR:
✓P ✓any two equations
✓third equation ✓solving
(4)
5.3.1
f(x) > −18
Turning point of f: x = −−4
2(2) = 1;
f(1) = 2−4−16 = −18
f(x) > −18 everywhere except at the turning point
x ∈R, x̸ = 1
OR: 2x2 −4x+2 > 0 ⇒2(x−1)2 > 0 ⇒x̸ = 1
✓f(x) > −18
✓turning point (1;−18)
✓x ∈R, x̸ = 1
OR:
✓✓2(x−1)2 > 0 ✓
answer
(3)
5.3.2
g(x)·g−1(x) = 0 ⇒g(x) = 0 or g−1(x) = 0
g(x) = 0: x = −2
g−1: x = 2y+4 ⇒g−1(x) = x
2 −2
g−1(x) = 0: x = 4
x = −2
or
x = 4
(The x-intercept of g−1 is the image of the y-intercept
(0;4) of g.)
✓x = −2
✓g−1(x) = x
2 −2 or (4;0)
✓x = 4
(3)
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QUESTION 6
6.1
x = −2
✓x = −2
(1)
6.2
y ∈R, y̸ = 3
or
(−∞; 3)∪(3; ∞)
✓answer
(1)
6.3
0 =
4
x+2 +3
3(x+2) = −4 ⇒x = −10
3
B
(
−10
3 ; 0
)
✓y = 0
✓simplification
✓B
(
−10
3 ;0
)
(3)
6.4
4
x+2 +3 = −4x+k
4+3(x+2) = (−4x+k)(x+2)
4x2 +(11−k)x+10−2k = 0
For a tangent: ∆= 0
(11−k)2 −4(4)(10−2k) = 0
k2 +10k −39 = 0
(k +13)(k −3) = 0
k = 3
or
k = −13
OR (calculus): f ′(x) = −4(x+2)−2 = −4
(x+2)2 = 1 ⇒x = −1 or x = −3
f(−1) = 7; f(−3) = −1
7 = −4(−1)+k ⇒k = 3;
−1 = −4(−3)+k ⇒k = −13
✓equating
✓standard form
✓∆= 0
✓substitution
✓k2 +10k −39 = 0
✓both values of k
OR:
✓f ′(x) ✓f ′(x) = −4
✓both x-values
✓both y-values
✓substitution ✓both k
(6)
6.5
Lines of symmetry pass through C(−2;3) with m = ±1
Negative gradient: y−3 = −(x+2) ⇒y = −x+1
Reflection in y = −x+1: (x;y) →(1−y; 1−x)
D
(
1−0; 1+ 10
3
)
= D
(
1; 13
3
)
BD =
√(
1+ 10
3
)2 +
(13
3 −0
)2 = 13
√
2
3
≈6,13 units
OR: perpendicular through B: y = x+ 10
3 ; meets
y = −x+1 at M
(
−7
6; 13
6
)
BD = 2BM = 2
√(13
6
)2 +
(13
6
)2 = 13
√
2
3
≈6,13
✓y = −x+1
✓xD = 1
✓yD = 13
3
✓substitution into distance
formula
✓6,13 units
OR:
✓y = −x+1 ✓
perpendicular line
✓M ✓BD = 2BM
✓6,13
Note: CA from 6.3.
(5)
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UNOFFICIAL
Mathematics/P1
9
NSC – Unofficial Marking Guidelines
Hlayiso/November 2026
QUESTION 7
7.1
197700 = 465000(1−0,115)n
(0,885)n = 197700
465000
n = log0,885
(197700
465000
)
= 7,00
After 7 years
✓substitution into
A = P(1−i)n
✓use of logs
✓7 years
(3)
7.2
A = 10000
(
1+ 0,065
2
)6 (
1+ 0,07
4
)8
After 3 years: 10000(1,0325)6 = R12115,47
After 5 years: 12115,47(1,0175)8 = R13919,25
✓i = 0,065
2
and n = 6
✓i = 0,07
4
and n = 8
✓product of the two growth
factors / using R12 115,47
✓R13 919,25
(4)
7.3.1
n = 44×12 = 528; i = 0,075
12
F =
800
[(
1+ 0,075
12
)528
−1
]
0,075
12
F = R3306959,91
✓n = 528
✓i = 0,075
12
✓substitution into F
✓R3 306 959,91
(4)
7.3.2
i = 0,06
12 = 0,005; n = 240
3306959,91 = x
[
1−(1,005)−240]
0,005
x = R23692,09
✓i = 0,005 and n = 240
✓substitution into P with
7.3.1 as P
✓R23 692,09
Note: CA from 7.3.1.
(3)
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UNOFFICIAL
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10
NSC – Unofficial Marking Guidelines
Hlayiso/November 2026
QUESTION 8
8.1
f(x+h) = −(x+h)2+(x+h) = −x2−2xh−h2+x+h
f(x+h)−f(x) = −2xh−h2 +h
f ′(x) = lim
h→0
−2xh−h2 +h
h
= lim
h→0
h(−2x−h+1)
h
= lim
h→0(−2x−h+1)
= −2x+1
✓f(x+h)
✓−2xh−h2 +h
✓formula with limit
✓common factor
✓−2x+1
Note: Incorrect notation (e.g. no
“lim” or “=”): penalise 1 mark.
Answer only: 0/5.
(5)
8.2.1
g(x) = 3x
1
3 −3x2
g′(x) = x−2
3 −6x
✓3x
1
3
✓x−2
3
✓−6x
(3)
8.2.2
y = x(4x2 −1)
2x−1
= x(2x−1)(2x+1)
2x−1
y = 2x2 +x
dy
dx = 4x+1
✓factorisation
✓2x2 +x
✓4x+1
Note: Quotient rule (not in
CAPS) leading to 4x+1: full
marks.
(3)
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UNOFFICIAL
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11
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QUESTION 9
9.1.1
f(1) = −1+6−9+4 = 0 ∴(x−1) is a factor
✓f(1) = 0
(1)
9.1.2
f(x) = (x−1)(−x2 +5x−4)
= −(x−1)(x−1)(x−4) = −(x−1)2(x−4)
(1; 0) and (4; 0)
✓quadratic factor
✓factorised fully
✓both x-intercepts
(3)
9.1.3
f ′(x) = −3x2 +12x−9 = 0
x2 −4x+3 = 0 ⇒(x−1)(x−3) = 0
x = 1
or
x = 3
f(1) = 0 (local minimum);
f(3) = −27+54−27+4 = 4 (local maximum)
0 < k < 4
✓f ′(x)
✓f ′(x) = 0
✓x = 1 and x = 3
✓y-values 0 and 4
✓0 < k < 4
(5)
9.1.4
g′′(x) = f ′′(−x); f ′′(x) = −6x+12 < 0 for x > 2 and
> 0 for x < 2
g is f reflected in the y-axis (and shifted up 2): the
point of inflection moves from x = 2 to x = −2 and the
concave-down part (x > 2) moves to x < −2
g′′(x) = 6x+12 > 0 ⇒g is concave up for x > −2
✓critical value −2
✓x > −2
Note: Vertical shift does not
affect concavity.
(2)
9.2
f ′(x) = 9x2 +2ax+b; f ′′(x) = 18x+2a
f ′′(2) = 0: 36+2a = 0 ⇒a = −18
Tangent y = 3 is horizontal: f ′(2) = 0
36+4(−18)+b = 0 ⇒b = 36
f(2) = 3: 3(8)+(−18)(4)+36(2)+c = 3
24−72+72+c = 3 ⇒c = −21
✓f ′(x) ✓f ′′(x)
✓a = −18
✓b = 36
✓f(2) = 3
✓c = −21
(6)
[17]
QUESTION 10
10.1
Total surface area (open top):
(2x)(x)+2(2x)(h)+2(x)(h) = 300
2x2 +6xh = 300
h = 300−2x2
6x
= 150−x2
3x
(
= 50
x −x
3
)
✓surface area = 300
✓h in terms of x
(2)
10.2
V = (2x)(x)(h) = 2x2 · 150−x2
3x
= 2x(150−x2)
3
= 100x−2
3x3
✓V = 2x2h
✓substitution of h
(2)
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10.3
V ′(x) = 100−2x2 = 0
x2 = 50 ⇒x = 5
√
2 ≈7,07 (x > 0)
V ′′(x) = −4x < 0 ∴maximum
Vmax = 100(5
√
2)−2
3(5
√
2)3 = 1000
√
2
3
≈
471,40 cm3
✓V ′(x)
✓V ′(x) = 0
✓x = 5
√
2 or 7,07
✓471,40 cm3
Note: Using x = 7,07 gives
471,40; accept 471,4. CA from
x.
(4)
[8]
QUESTION 11
11.1.1
No. P(A and B) = 0,3̸ = 0
✓No
✓reason P(A and B)̸ = 0 (2)
11.1.2
P(only B) = 1−0,16−0,3−0,28 = 0,26
P(B) = 0,26+0,3 = 0,56
OR: P(A or B) = 1−0,28 = 0,72; P(A) = 0,46
0,72 = 0,46+P(B)−0,3 ⇒P(B) = 0,56
✓method
✓0,26
✓0,56
OR:
✓0,72 and 0,46 ✓
substitution
✓0,56
(3)
11.2.1
(49−x)+4+2+8+x+5+(60−x)+14 = 100
142−x = 100 ⇒x = 42
✓sum of all regions = 100
(1)
11.2.2
n(M) = (49−42)+4+8+42 = 61 ⇒P(M) = 0,61
n(L) = 42+8+5+18 = 73 ⇒P(L) = 0,73
P(M)×P(L) = 0,61×0,73 = 0,4453
P(M and L) = 42+8
100
= 0,5
P(M and L)̸ = P(M)×P(L)
∴taking Mathematics is NOT independent of taking
Life Sciences
✓P(M) = 0,61
✓P(L) = 0,73
✓0,4453
✓P(M and L) = 0,5
✓conclusion (not
independent)
Note: Conclusion mark only
with comparison of P(M and L)
and P(M)×P(L).
(5)
11.3.1
3+2+4 = 9 bottles: 9! = 362880
✓362880
(1)
Copyright reserved
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### Page 13

UNOFFICIAL
Mathematics/P1
13
NSC – Unofficial Marking Guidelines
Hlayiso/November 2026
11.3.2
Let n = number of bottles remaining.
Energy drinks at both ends: 2×1 ways; the other
(n−2) bottles: (n−2)! ways
P = 2×(n−2)!
n!
=
2
n(n−1) = 1
21
n(n−1) = 42 ⇒n2 −n−42 = 0
(n−7)(n+6) = 0 ⇒n = 7 (n > 0)
Bottles of juice that broke: 9−7 = 2
OR: P = 2
n ×
1
n−1 = 1
21 ⇒n = 7
Check: 2×5!
7!
= 240
5040 = 1
21
✓2×(n−2)!
✓n! as total
✓equation = 1
21
✓n = 7
✓2 bottles
Note: Trial and error that tests
the number broken with full
calculations (e.g. 2×5!
7! = 1
21) and
concludes 2: full marks. Answer
only: 1/5.
(5)
[17]
TOTAL: 150
Copyright reserved
Please turn over

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