# Mathematics Grade 12 Paper 1 Challenge Paper 2026 Memo

> Grade 12 Mathematics Paper 1 memorandum from the Challenge Paper 2026, 11 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-1/d6b0a2dc-62d0-491b-91d1-fbb5f247ce89
- Updated: 2026-10-21
- Subject: Mathematics
- Grade: Grade 12
- Document type: Memorandum
- Year: 2026
- Exam period: Challenge Paper
- Paper: 1
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 11

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## About this document

Mathematics · Grade 12 · Hlayiso Challenge Paper · 2026. A deliberately harder practice Paper 1 (memorandum), 11 pages. Read online or download the PDF.

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### Page 1

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hlayiso mock exams
Practice Series:
Grade 12 NSC Preparation
WWW.HLAYISO.COM
NATIONAL
SENIOR CERTIFICATE
GRADE 12
MATHEMATICS P1
CHALLENGE PAPER 2026
MARKING GUIDELINES
MARKS: 150
These marking guidelines consist of 11 pages.

### Page 2

Mathematics/P1
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.
QUESTION 1
1.1.1
x4 +12 = 7x2
(x̸ = 0)
x4 −7x2 +12 = 0
(x2 −3)(x2 −4) = 0
x = ±
√
3
or
x = ±2
✓multiply by x2
✓standard form
✓factors
✓all four answers
(answers only: max 2/4)
(4)
1.1.2
3·3x + 3
3x = 10
Let k = 3x: 3k2 −10k +3 = 0
(3k −1)(k −3) = 0
3x = 1
3 ⇒x = −1
or
3x = 3 ⇒x = 1
✓3·3x and 3·3−x
✓standard form
✓factors
✓x = ±1
(4)
1.2
12+x−x2 ≥0 ⇒x2 −x−12 ≤0
(x−4)(x+3) ≤0 ⇒−3 ≤x ≤4
and x̸ = 2
−3 ≤x ≤4, x̸ = 2
(OR x ∈[−3 ; 2)∪(2 ; 4])
✓12+x−x2 ≥0
✓−3 ≤x ≤4
✓x̸ = 2
(3)
1.3
2x ·22y = 25 ⇒x+2y = 5 ⇒x = 5−2y
(5−2y)2 +2y2 = 11
6y2 −20y+14 = 0 ⇒3y2 −10y+7 = 0
(3y−7)(y−1) = 0
y = 1 and x = 3
or
y = 7
3 and x = 1
3
✓x+2y = 5
✓substitution
✓standard form
✓y-values
✓x-values
(5)
1.4.1
∆= [−2(k +1)]2 −4(k −1)(k +3)
= 4(k2 +2k +1)−4(k2 +2k −3) = 16 = 42
∆is a perfect square (independent of k) and the
coefficients are rational, so the roots are rational for
every rational k̸ = 1.
✓substitution into ∆
✓∆= 16
✓conclusion: perfect square
(3)
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### Page 3

Mathematics/P1
3
NSC – Marking Guidelines
Hlayiso/Challenge 2026
1.4.2
x = 2(k +1)±4
2(k −1)
x = 1
or
x = k +3
k −1 = 1+
4
k −1
k −1 must be a factor of 4: k −1 ∈{±1 ; ±2 ; ±4}
k ∈{−3 ; −1 ; 0 ; 2 ; 3 ; 5}
✓roots 1 and k+3
k−1
✓1+
4
k−1 / factors of 4
✓all six values
(3)
1.5
xy = (
√
3)2 −(
√
2)2 = 1
x2 +y2 = (5+2
√
6)+(5−2
√
6) = 10
x
y + y
x = x2 +y2
xy
= 10
1 = 10
✓xy = 1
✓x2 +y2 = 10
✓answer 10
(3)
[25]
QUESTION 2
2.1.1
First differences: p+7 ; 5−p ; q−5 ; 33−q
Second differences equal:
(5−p)−(p+7) = (q−5)−(5−p) ⇒q = 8−3p
(q−5)−(5−p) = (33−q)−(q−5) ⇒p+3q = 48
p+3(8−3p) = 48 ⇒p = −3; q = 8+9 = 17
OR: Tn = an2 +bn+c with T1 = −7, T3 = 5, T5 = 33:
8a+2b = 12; 16a+2b = 28 ⇒a = 2, b = −2,
c = −7
p = T2 = −3; q = T4 = 17
✓first differences
✓first equation
✓second equation
✓solving
OR: ✓3 equations
✓✓a, b, c
✓T2 and T4
(4)
2.1.2
Second difference: (5−p)−(p+7) = 8−4 = 4
⇒2a = 4, a = 2
3a+b = T2 −T1 = 4 ⇒b = −2
a+b+c = −7 ⇒c = −7
Tn = 2n2 −2n−7
✓a = 2
✓b = −2
✓c = −7
(3)
2.1.3
Tn = 2n(n−1)−7
2n(n−1) is even, so Tn = even −odd = odd for every
n.
2 026 is even, so it is not a term.
OR: 2n2 −2n−7 = 2026 ⇒2n2 −2n−2033 = 0
n = 2±
√
16268
4
= 32,39 or −31,39; n /∈N
✓2n(n−1) even / equation
✓conclusion: Tn odd / n /∈N
(2)
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### Page 4

Mathematics/P1
4
NSC – Marking Guidelines
Hlayiso/Challenge 2026
2.2
log2
(8r
2
)
= log2
(
23r−1)
= 3r −1
k
∑
r=1
(3r −1) = 2+5+8+... (a = 2, d = 3)
k
2 [2(2)+(k −1)(3)] = 100
3k2 +k −200 = 0
(3k +25)(k −8) = 0
k = 8
(k̸ = −25
3 )
✓3r −1
✓a = 2 and d = 3
✓substitution into Sn = 100
✓k = 8 (with rejection)
(4)
[13]
QUESTION 3
3.1.1
T1 = a; T3 = a+2d; T7 = a+6d
a+2d
a
= a+6d
a+2d ⇒(a+2d)2 = a(a+6d)
a2 +4ad +4d2 = a2 +6ad
4d2 = 2ad ⇒2d(2d −a) = 0
d̸ = 0 ∴a = 2d
✓a+2d and a+6d
✓equal ratios
✓simplification with d̸ = 0
(3)
3.1.2
S10 = 10
2 [2(2d)+9d] = 65d = 260
d = 4; a = 8
Geometric: 8 ; 16 ; 32 ; ... r = 2
T5 = 8(2)4 = 128
✓65d = 260
✓a = 8, d = 4 and r = 2
✓128
(3)
3.2.1
r = 1−2x
−1 < 1−2x < 1
−2 < −2x < 0 ⇒0 < x < 1
✓−1 < 1−2x < 1
✓0 < x < 1
(2)
3.2.2
12
1−r = 20 ⇒r = 0,4
S∞−Sn = 20−12(1−0,4n)
1−0,4
= 20(0,4)n
20(0,4)n < 0,01 ⇒(0,4)n < 0,0005
n > log0,0005
log0,4
= 8,30
(inequality reverses)
At least 9 terms
(Note: x = 0,3, but x is not needed.)
✓r = 0,4
✓20(0,4)n / S∞−Sn
✓use of logs
✓9 terms
(4)
[12]
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### Page 5

Mathematics/P1
5
NSC – Marking Guidelines
Hlayiso/Challenge 2026
QUESTION 4
4.1
f(x) = 2(x−1)+3
x−1
= 2(x−1)
x−1
+
3
x−1 = 2+
3
x−1
OR:
3
x−1 +2 = 3+2(x−1)
x−1
= 2x+1
x−1
✓2x+1 = 2(x−1)+3 /
common denominator
✓splitting the fraction
(2)
4.2
x = 1
y = 2
✓x = 1
✓y = 2
(2)
4.3
mx = 2x+1
x−1 ⇒mx2 −mx = 2x+1
mx2 −(m+2)x−1 = 0
No intersection: ∆< 0
(m+2)2 +4m < 0 ⇒m2 +8m+4 < 0
Critical values: m = −8±
√
48
2
= −4±2
√
3
−4−2
√
3 < m < −4+2
√
3
(−7,46 < m < −0,54)
(For m = 0 the line y = 0 cuts f at x = −1
2; x = 1 is
never a root.)
✓standard form
✓∆< 0
✓m2 +8m+4 < 0
✓answer
(4)
4.4
g(x) = −f(−x): reflection in the y-axis followed by
reflection in the x-axis
(i.e. rotation through 180◦about the origin)
OR: translation 2 units to the left and 4 units down
(The two reflections may be in either order.)
✓✓correct description
(OR ✓2 left ✓4 down) (2)
[10]
QUESTION 5
5.1
O(0 ; 0): 2p +q = 0
…(1)
B(2 ; 12): 2p+2 +q = 12 ⇒4·2p +q = 12
…(2)
(2)−(1): 3·2p = 12 ⇒2p = 4 ⇒p = 2
q = −2p = −4
✓equation (1)
✓equation (2)
✓2p = 4
✓p = 2 and q = −4
(4)
5.2
h : y = 2x+2 −4
h−1 : x = 2y+2 −4
2y+2 = x+4 ⇒y+2 = log2(x+4)
y = log2(x+4)−2
✓swop x and y
✓log2(x+4)
✓answer
(3)
5.3
x > −4
(OR x ∈(−4 ; ∞))
✓answer
(1)
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### Page 6

Mathematics/P1
6
NSC – Marking Guidelines
Hlayiso/Challenge 2026
5.4
log2(x+4)−2 ≤2 ⇒log2(x+4) ≤4
x+4 ≤16 ⇒x ≤12
Domain: x > −4 ∴−4 < x ≤12
(OR: B(2;12) on h ⇒(12;2) on h−1, and h−1 is
increasing.)
✓x+4 ≤16 / point (12;2)
✓x ≤12
✓−4 < x
(3)
[11]
QUESTION 6
6.1
Axis of symmetry: x = −1+5
2
= 2
D on y = 3x−15: y = 3(2)−15 = −9; D(2 ; −9)
f(2) = a(3)(−3) = −9a = −9 ⇒a = 1
✓x = 2
✓y = −9
✓substitution: −9a = −9(3)
6.2
f(x) = x2 −4x−5
f ′(x) = 2x−4 = 2 ⇒x = 3
f(3) = 9−12−5 = −8; −8 = 2(3)+k ⇒k = −14
OR: x2 −4x−5 = 2x+k ⇒x2 −6x−(5+k) = 0
∆= 36+4(5+k) = 0 ⇒k = −14
✓f ′(x) = 2
✓point (3 ; −8)
✓k = −14
OR: ✓equating ✓∆= 0 ✓
k
(3)
6.3
Two distinct points: k > −14 (line above the tangent)
The line y = 2x+k passes through (0 ; −5), the
y-intercept of f, when k = −5; then the points of
intersection have x = 0 and x = 6.
For both x-values to be positive the line must lie below
this position: k < −5
−14 < k < −5
OR: x2 −6x−(5+k) = 0 ⇒x = 3±
√
14+k
Distinct: 14+k > 0; smaller root positive:
√
14+k < 3 ⇒k < −5
✓k > −14
✓k < −5 (with reason)
✓answer
(3)
6.4
x-intercepts of g: x = −1+t and x = 5+t
Both positive: −1+t > 0 ⇒t > 1
✓−1+t
✓t > 1
(2)
6.5
f(x) > 0 for x < −1 or x > 5; f(x) < 0 for −1 < x < 5
f ′(x) = 2x−4: f ′(x) > 0 for x > 2; f ′(x) < 0 for
x < 2
Same sign: −1 < x < 2
or
x > 5
✓critical values −1; 2; 5
✓−1 < x < 2
✓x > 5
(3)
[14]
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### Page 7

Mathematics/P1
7
NSC – Marking Guidelines
Hlayiso/Challenge 2026
QUESTION 7
7.1
After 3 years: A = 50000
(
1+ 0,08
4
)12
= R63412,09
100000 = 63412,09
(
1+ 0,09
12
)m
(1,0075)m = 1,57698...
m = log1,57698...
log1,0075
= 60,96
61 months after the rate change
Total: 36+61 = 97 months
✓i = 0,02 and n = 12
✓R63 412,09
✓equation with i = 0,0075
✓m = 60,96
✓97 months
(5)
7.2.1
i = 0,12
12 = 0,01
Balance after 2 months: 420000(1,01)2 = R428442
428442 = x
[
1−(1,01)−72]
0,01
x = R8376,12
OR: 420000 = x
[
1−(1,01)−72]
0,01
(1,01)−2
✓i = 0,01 and n = 72
✓420000(1,01)2
✓substitution into Pv
✓x = 8376,12 (must be
shown)
(4)
7.2.2
Remaining instalments: 72−30 = 42
Balance = 8376,12
[
1−(1,01)−42]
0,01
= R286112,41
(Accept the future-value method:
428442(1,01)30 −8376,12
[
(1,01)30 −1
]
0,01
=
R286112,66.)
New balance: 286112,41−60000 = R226112,41
226112,41 = 8376,12[1−(1,01)−n]
0,01
(1,01)−n = 0,73005... ⇒n = −log0,73005...
log1,01
=
31,62
She will pay 31 more full instalments (and a smaller
final payment).
✓n = 42
✓substitution
✓R286 112,41
✓subtract R60 000
✓use of logs: n = 31,62
✓31 instalments
(6)
[15]
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### Page 8

Mathematics/P1
8
NSC – Marking Guidelines
Hlayiso/Challenge 2026
QUESTION 8
8.1
f(x+h) = 2(x+h)2 −
1
x+h = 2x2 +4xh+2h2 −
1
x+h
f(x+h)−f(x) = 4xh+2h2 −
1
x+h + 1
x =
4xh+2h2 +
h
x(x+h)
f ′(x) = lim
h→0
4xh+2h2 +
h
x(x+h)
h
=
lim
h→0
(
4x+2h+
1
x(x+h)
)
= 4x+ 1
x2
✓f(x+h)
✓
h
x(x+h)
✓formula with limit
✓dividing by h
✓answer
(Incorrect notation: −1)
(5)
8.2
(x+3)2 −9 = x2 +6x
y = x2 +6x
x
1
3
= x
5
3 +6x
2
3
dy
dx = 5
3x
2
3 +4x−1
3
✓x2 +6x
✓x
5
3 +6x
2
3
✓answer
(3)
8.3
dy
dx = 3x2 −3; mtangent = 3p2 −3
Tangent: y−(p3 −3p) = (3p2 −3)(x−p)
Through (0 ; 2): 2−p3 +3p = −3p3 +3p
2p3 = −2 ⇒p3 = −1 ⇒p = −1
P(−1 ; 2)
(Check: tangent at P is y = 2, which passes through
(0;2).)
OR:
m = p3 −3p−2
p−0
= 3p2 −3 ⇒p3 −3p−2 = 3p3 −3p
✓3p2 −3
✓substitution of (0;2) into
tangent / gradient equation
✓p3 = −1
✓P(−1 ; 2)
(4)
[12]
QUESTION 9
9.1
f ′(x) = 3x2 +2px+q
f ′(2) = 0: 12+4p+q = 0 ⇒4p+q = −12
…(1)
f(2) = 0: 8+4p+2q+16 = 0 ⇒2p+q = −12
…(2)
(1)−(2): 2p = 0 ⇒p = 0; q = −12
OR: f(x) = (x−2)2(x−c);
f(0) = −4c = 16 ⇒c = −4
f(x) = (x−2)2(x+4) = x3 −12x+16
✓f ′(x)
✓f ′(2) = 0
✓f(2) = 0
✓solving simultaneously
OR: ✓form ✓c = −4
✓✓expansion and
comparing
(4)
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### Page 9

Mathematics/P1
9
NSC – Marking Guidelines
Hlayiso/Challenge 2026
9.2
f(x) = x3 −12x+16
f ′(x) = 3x2 −12 = 0 ⇒x = ±2
f(−2) = −8+24+16 = 32; f(2) = 0
(−2 ; 32) and (2 ; 0)
✓f ′(x) = 0
✓x = ±2
✓y-values
(3)
9.3
x
y
−4
2
16
(−2;32)
f
f(x) = (x−2)2(x+4): x-intercepts −4 and 2
✓x-intercepts −4 and 2 and
y-intercept 16
✓turning points
✓shape
(3)
9.4
f(x) = k: horizontal line y = k cuts f.
For 0 < k < 16 the line cuts f once for x < −2, once
for 0 < x < 2 and once for x > 2.
For k = 16 one root is x = 0 (not positive); for k = 0
the positive root is repeated.
0 < k < 16
✓k > 0
✓k < 16 (reference to
y-intercept)
✓strict inequalities
(3)
9.5
f is decreasing for −2 < x < 2
g is f shifted 1 unit right: −1 < x < 3
(Accept −1 ≤x ≤3.)
✓−2 < x < 2 /
g′(x) = 3(x−1)2 −12
✓−1 < x < 3
(2)
[15]
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### Page 10

Mathematics/P1
10
NSC – Marking Guidelines
Hlayiso/Challenge 2026
QUESTION 10
10.1
πr2h = 500π ⇒h = 500
r2
C = 0,04
(
2πr2)
+0,02(2πrh)
= 0,08πr2 +0,04πr · 500
r2 = 0,08πr2 + 20π
r
✓h = 500
r2
✓cost of top/bottom and side
✓substitution of h
(3)
10.2
C′(r) = 0,16πr −20π
r2 = 0
0,16r3 = 20 ⇒r3 = 125 ⇒r = 5 cm
(C′′(r) = 0,16π + 40π
r3 > 0, so this is a minimum.)
C(5) = 0,08π(25)+ 20π
5
= 2π +4π = 6π
Minimum cost = R18,85
✓C′(r)
✓C′(r) = 0
✓r = 5
✓substitution r = 5
✓R18,85
(5)
[8]
QUESTION 11
11.1.1
Independent: P(A and B) = x(x+0,2)
0,76 = x+(x+0,2)−x(x+0,2)
x2 −1,8x+0,56 = 0 ⇒(x−0,4)(x−1,4) = 0
(OR 25x2 −45x+14 = 0 ⇒(5x−2)(5x−7) = 0)
x = 0,4
(x = 1,4 > 1, rejected)
✓P(A and B) = x(x+0,2)
✓substitution into the
addition rule
✓standard form
✓x = 0,4 with rejection
(4)
11.1.2
P(A and B) = 0,4×0,6 = 0,24
P(exactly one) = 0,76−0,24 = 0,52
OR: 0,4(0,4)+0,6(0,6) = 0,52
✓0,24 / method
✓0,52
(2)
11.2.1
Treat the 3 girls as one unit: 5 units, 5! arrangements
Girls among themselves: 3!
5!×3! = 720
✓5!
✓3!
✓720
(3)
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### Page 11

Mathematics/P1
11
NSC – Marking Guidelines
Hlayiso/Challenge 2026
11.2.2
Seat the 4 boys: 4! = 24
This creates 5 gaps (including the two ends); girls in 3
different gaps: 5×4×3 = 60
n(E) = 24×60 = 1440; n(S) = 7! = 5040
P = 1440
5040 = 2
7 ≈0,29
OR: girls’ seat sets with no two adjacent:
{1,3,5},{1,3,6},{1,3,7},{1,4,6},{1,4,7},
{1,5,7},{2,4,6},{2,4,7},{2,5,7},{3,5,7}: 10 sets;
10×3!×4! = 1440
✓4!
✓5×4×3
✓7!
✓2
7
(4)
11.2.3
Together: 2×6! = 1440
P(not together) = 1−1440
5040 = 5
7 ≈0,71
✓2×6!
✓5
7
(2)
[15]
TOTAL: 150
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