# Mathematics Grade 12 P1 NSC November Prediction 2026 Memo

> Grade 12 Mathematics Paper 1 memorandum from the NSC November Prediction Exam 2026 (English), 10 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/grade-12-mathematics-2026-nsc-november-prediction-exam-memorandum-paper-1/d071a65e-74be-40a7-8d27-0249e0373d87
- Updated: 2026-10-29
- Subject: Mathematics
- Grade: Grade 12
- Language: English
- Document type: Memorandum
- Year: 2026
- Exam period: NSC November Prediction Exam
- Paper: 1
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 10

Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 12](https://www.hlayiso.com/library/grade-12) › [Mathematics](https://www.hlayiso.com/library/grade-12/mathematics) › [Mathematics Grade 12 P1 NSC November Prediction 2026 Memo](https://www.hlayiso.com/library/grade-12-mathematics-2026-nsc-november-prediction-exam-memorandum-paper-1/d071a65e-74be-40a7-8d27-0249e0373d87)

## About this document

Mathematics · Grade 12 · NSC November Prediction Exam · 2026 · English. Memorandum, 10 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

hlayiso mock exams
                           Practice Series:
                           Grade 12 NSC Preparation
                           WWW.HLAYISO.COM

                                NATIONAL
                           SENIOR CERTIFICATE

                                    GRADE 12

                                MATHEMATICS P1
                      PREDICTION EXAMINATION 2026
                            MARKING GUIDELINES

MARKS: 150

                     These marking guidelines consist of 10 pages.

Copyright reserved                                                   Please turn over

### Page 2

Mathematics/P1                                   2                             Hlayiso/Prediction 2026
                                       NSC – Marking Guidelines

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.

QUESTION 1

                     2
 1.1.1       x=          or   x = −4                                 ✓ x = 32
                     3
                                                                     ✓ x = −4                    (2)

 1.1.2       3x2 − 6x − 5 =√
                           0                 √                       ✓ standard form
                 −(−6) ± (−6)2 − 4(3)(−5) 6 ± 96                     ✓ substitution
             x=                          =
                             2(3)            6                       ✓ both answers              (3)
             x = 2,63 or x = −0,63

 1.1.3       x2 − 2x − 3 − 5 > 0
             x2 − 2x − 8 > 0                                         ✓ standard form
             (x − 4)(x + 2) > 0                                      ✓ critical values
             x < −2 or x > 4                                         ✓ x < −2 ✓ x > 4            (4)

                     √
 1.1.4       x − 4 = 2x + 7                                          ✓ isolate surd
             x2 − 8x + 16 = 2x + 7                                   ✓ squaring both sides
             x2 − 10x + 9 = 0
             (x − 9)(x − 1) = 0                                      ✓ factors
             x = 9 or x = 1 √
             Check x = 1: 1 − 9 = −2 ̸= 4 ∴ x = 9 only               ✓ rejection of x = 1        (4)

 1.1.5       (2x )2 − 6 · 2x − 16 = 0                                ✓ 2x+1 = 2 · 2x and
             Let k = 2x : k2 − 6k − 16 = 0                           4x = (2x )2
             (k − 8)(k + 2) = 0                                      ✓ standard form in k
             2x = 8 ⇒ x = 3 or 2x = −2 (no solution)                 ✓ factors
                                                                     ✓ x = 3 and rejection       (4)

Copyright reserved                                                                    Please turn over

### Page 3

Mathematics/P1                                  3                           Hlayiso/Prediction 2026
                                      NSC – Marking Guidelines

 1.2         Let length = x and breadth = y
             2x + 2y = 34 ⇒ x = 17 − y                              ✓ x + y = 17
             x2 + y2 = 132 = 169                                    ✓ x2 + y2 = 169
             (17 − y)2 + y2 = 169                                   ✓ substitution
             2y2 − 34y + 120 = 0 ⇒ y2 − 17y + 60 = 0                ✓ standard form
             (y − 12)(y − 5) = 0
             The garden is 12 m by 5 m                              ✓ 12 m and 5 m              (5)

                 22023 (23 + 2) 8 + 2 10
 1.3         =                 =      =                             ✓ common factor 22023 in
                 22023 (22 − 1) 4 − 1   3
                                                                    numerator
                                                                    ✓ common factor in
                                                                    denominator
                                                                    ✓ 10
                                                                       3                     (3)

                                                                                               [25]

QUESTION 2

 2.1.1       Tn = 2 + (n − 1)(7) = 7n − 5                           ✓ Tn = 345
             7n − 5 = 345 ⇒ n = 50                                  ✓ n = 50                    (2)

                   50
 2.1.2       S50 =    [2 + 345] = 8 675                             ✓ S50
                    2
                   20                                               ✓ S20
             S20 =    [2(2) + 19(7)] = 1 370                        ✓ answer
                    2
              50                                                    OR: ✓ 142 and 30 terms
              ∑ (7n − 5) = S50 − S20 = 7 305                        ✓ substitution ✓ answer (3)
             n=21
                                         30
             OR: T21 = 142; 30 terms:       [142 + 345] = 7 305
                                          2

 2.1.3       Terms that are even numbers: 2 + 16 + 30 + . . .       ✓ a = 2 and d = 14
             (a = 2, d = 14)                                        ✓ last term 338
             Last even term ≤ 345: 338                              ✓ n = 25
             2 + (n − 1)(14) = 338 ⇒ n = 25                         ✓ answer                    (4)
                    25
             S25 =     [2 + 338] = 4 250
                     2

 2.2.1       First differences: 5 ; 9 ; 13; second difference = 4
             2a = 4 ⇒ a = 2                                         ✓a=2
             3a + b = 5 ⇒ b = −1                                    ✓ b = −1
             a+b+c = 3 ⇒ c = 2                                      ✓c=2                        (3)
             Tn = 2n2 − n + 2

Copyright reserved                                                                   Please turn over

### Page 4

Mathematics/P1                                   4                            Hlayiso/Prediction 2026
                                       NSC – Marking Guidelines

 2.2.2       First difference between Tn and Tn+1 :                   ✓ 4n + 1
             5 + (n − 1)(4) = 4n + 1                                  ✓ n = 99
             4n + 1 = 397 ⇒ n = 99                                    ✓ T99 and T100              (3)
             Between T99 and T100

                                                                                                 [15]

QUESTION 3

 3.1         a = 80; r = 0,9                                          ✓ r = 0,9
             T5 = 80(0,9)4 = 52,49 cm                                 ✓ answer                    (2)

 3.2         80(0,9)n−1 < 10                                          ✓ inequality
             (0,9)n−1 < 0,125
             n−1 >
                     log 0,125
                               = 19,74                                ✓ use of logs
                      log 0,9                                         ✓ 21st swing                (3)
             n > 20,74 ∴ the 21st swing
                     [            ]
                   80 1 − (0,9)15
 3.3         S15 =                  = 635,29 cm                       ✓ substitution
                       1 − 0,9
                                                                      ✓ answer                    (2)

                     80
 3.4         S∞ =          = 800 cm = 8 m                             ✓ S∞ formula
                   1 − 0,9
             Sn < S∞ for every n, so the total distance gets closer   ✓ 800 cm
             and closer to 8 m but never exceeds it.                  ✓ conclusion with reason (3)
             Thabo is NOT correct.

                                                                                                 [10]

Copyright reserved                                                                     Please turn over

### Page 5

Mathematics/P1                                        5                         Hlayiso/Prediction 2026
                                            NSC – Marking Guidelines

QUESTION 4

              1
 4.1            = b−1 ⇒ b = 3                                          ✓ substitution               (1)
              3

 4.2         x = 3y                                                    ✓ swop x and y
             y = log3 x                                                ✓ answer                     (2)

 4.3                      y                                            ✓ f : shape and (0; 1)
                              f
                                                                       ✓ g: shape and (1; 0)
                                                                       ✓ one other point on each
                                                                       graph, e.g. (1; 3) and (3; 1)(3)
                     (1; 3)
                                                g
                 (0; 1)
                                       (3; 1)       x
                              (1; 0)

 4.4         h(x) = 3x−1 − 3                                           ✓ equation of h
             3x−1 = 3 ⇒ x − 1 = 1 ⇒ x = 2                              ✓x=2                         (2)

 4.5         g(x) ≤ 0 for 0 < x ≤ 1 and g(x) ≥ 0 for x ≥ 1             ✓ critical values 1 and 2
             h(x) < 0 for x < 2 and h(x) ≥ 0 for x ≥ 2                 ✓0<x≤1
             0 < x ≤ 1 or x ≥ 2                                        ✓x≥2                      (3)

                                                                                                   [11]

QUESTION 5

 5.1         p=1                                                       ✓p
             q=3                                                       ✓q                           (2)

                  a
 5.2         1=      +3                                                ✓ substitution of (1 ; 1)
                 1+1
             a                                                         ✓ simplification          (2)
               = −2 ⇒ a = −4
             2

 5.3         y = x + c through (−1 ; 3): 3 = −1 + c ⇒ c = 4            ✓ substitution of (−1; 3)
             y = x+4                                                   ✓ answer                  (2)

Copyright reserved                                                                       Please turn over

### Page 6

Mathematics/P1                                  6                               Hlayiso/Prediction 2026
                                      NSC – Marking Guidelines

                  −4              1
 5.4         A:      +3 = 0 ⇒ x =                                       ✓ x = 31
                 x+1              3
                                1                                       ✓ x < −1
             x < −1 or 0 ≤ x ≤                                          ✓ 0 ≤ x ≤ 13               (3)
                                3

 5.5         Reflection in the line x = −1 (the vertical asymptote of   ✓ reflection
             f)                                                         ✓ in the line x = −1       (2)
                                      −4            4
             Check: f (−2 − x) =           +3 =         + 3 = g(x)
                                    −x − 1       x+1
             (Also accept: rotation through 90◦ about (−1 ; 3).)

                                                                                                  [11]

QUESTION 6

                      b       6
 6.1         x=−        =−         =3                                   ✓x=3
                     2a     2(−1)
                                                                        ✓ y = 16                   (2)
              f (3) = −9 + 18 + 7 = 16    D(3 ; 16)

 6.2         y ≤ 16    or   y ∈ (−∞ ; 16]                               ✓ answer                   (1)

 6.3         −x2 + 6x + 7 = 0 ⇒ (x − 7)(x + 1) = 0 A(−1 ; 0)            ✓ A(−1; 0) and m = 1
                   7−0                                                  ✓ equation                 (2)
             m=            = 1 g(x) = x + 1
                 6 − (−1)

 6.4         f (x) ≥ g(x) for −1 ≤ x ≤ 6; f increasing for x < 3        ✓ −1 ≤ x
             −1 ≤ x < 3 (accept −1 ≤ x ≤ 3)                             ✓x<3                       (2)

 6.5         MN = f (x) − g(x) = −x2 + 5x + 6                           ✓ MN expression
             x=− =−
                   b        5
                                 =
                                   5
                                        (OR:                            ✓ x = − 2a
                                                                                 b
                                                                                   / derivative = 0
                  2a      2(−1) 2                                       ✓x= 2 5

             dx (MN) = −2x + 5 = 0)
             d
                                                                        ✓ 12,25                   (4)
                        25 25          49
             MNmax = − + + 6 =            = 12,25 units
                         4     2        4

 6.6         Tangent is g moved up by the maximum length MN:            ✓ gradient 1 / + 49
                                                                                          4
             y = x+1+
                          49
                             = x+
                                   53                                   ✓ answer                   (2)
                           4       4
             OR: f ′ (x) = −2x + 6 = 1 ⇒ x = 52 , f ( 25 ) = 63
                                                              4,
                      53
             y = x+ 4

                                                                                                  [13]

Copyright reserved                                                                      Please turn over

### Page 7

Mathematics/P1                                  7                                Hlayiso/Prediction 2026
                                      NSC – Marking Guidelines

QUESTION 7

                     √ = 680 000(1 − i)                                  ✓ substitution into
 7.1         301 719,61                 5

                       301 719,61                                        A = P(1 − i)n
             1−i = 5              = 0,85                                 ✓ 0,85
                         680 000
             i = 0,15 i.e. 15% p.a.                                      ✓ 15%                      (3)

 7.2         New taxi: 680 000(1,062)5 = R918 610,69                     ✓ R918 610,69
             Amount needed:                                              ✓ R616 891,08
             918 610,69 − 301
                            [ 719,61 = R616] 891,08                      ✓ substitution into Fv
                           x (1,008)60 − 1                               ✓ answer                   (4)
             616 891,08 =
                                0,008
             x = R8 050,90

                 0,114
 7.3.1       i=        = 0,0095; n = 240                                 ✓ i and n
                   12   [                 ]                              ✓ substitution into Pv
                       x 1 − (1,0095)−240
             950 000 =                                                   ✓ answer                   (3)
                              0,0095
             x = R10 065,70

 7.3.2       Balance after
                       [ 60 instalments    ](180 remaining):             ✓ n = 180
             10 065,70 1 − (1,0095)   −180
                                                                         ✓ R866 358,46
                                             = R866 358,46
                        0,0095                                           ✓ substitution with R12 500
                           12 500 [1 − (1,0095)−n ]                      ✓ n = 113,61 (logs)
             866 358,46 =
                                    0,0095
             (1,0095)−n = 0,34157 . . . ⇒ n = 113,61                     ✓ 66 months                (5)
             114 payments instead of 180
             Paid off 180 − 114 = 66 months sooner
             OR (balance):                    [               ]
                                   10 065,70   (1,0095)60 − 1
             950 000(1,0095)60 −                                =
                                              0,0095
             R866 358,83
             (Accept balances from R866 358,46 to R866 358,83.)

                                                                                                   [15]

QUESTION 8

 8.1         f (x + h) = (x + h)2 − 4(x + h) = x2 + 2xh + h2 − 4x − 4h   ✓ f (x + h)
             f (x + h) − f (x) = 2xh + h2 − 4h                           ✓ 2xh + h2 − 4h
                           2xh + h2 − 4h                                 ✓ formula with limit
             f ′ (x) = lim               = lim (2x + h − 4)
                       h→0       h          h→0                          ✓ common factor
             = 2x − 4                                                    ✓ answer                   (5)

Copyright reserved                                                                       Please turn over

### Page 8

Mathematics/P1                                            8                        Hlayiso/Prediction 2026
                                                NSC – Marking Guidelines

                     (x − 2)(x2 + 2x + 4)
 8.2.1       y=                           = x2 + 2x + 4                    ✓ factorising difference of
                             x−2
              dy                                                           cubes
              dx
                 = 2x + 2                                                  ✓ x2 + 2x + 4
                                                                           ✓ answer                   (3)
             (        )
              √     2 2
                         = x − 4x− 2 + 4x−2
                                   1
 8.2.2          x−                                                         ✓ expansion in index form
                    x
                                                                           ✓ 1 + 2x− 2
                                                                                     3

             Dx = 1 + 2x− 2 − 8x−3
                          3
                                                                           ✓ −8x−3                  (3)

                                                                                                     [11]

QUESTION 9

 9.1         f ′ (x) = k(x − 1)(x − 3)                                     ✓ form of f ′
             9 = k(−1)(−3) ⇒ k = 3                                         ✓k=3
             f ′ (x) = 3x2 − 12x + 9 = 3ax2 + 2bx + c                      ✓ equating coefficients    (3)
             ∴ a = 1 ; b = −6 ; c = 9
             OR: c = f ′ (0) = 9; f ′ (1) = 0: 3a + 2b = −9;
             f ′ (3) = 0: 27a + 6b = −9 ⇒ a = 1, b = −6

 9.2          f (x) = x3 − 6x2 + 9x − 4                                    ✓ d = −4
              f (1) = 1 − 6 + 9 − 4 = 0   (1 ; 0)                          ✓ (1; 0)
              f (3) = 27 − 54 + 27 − 4 = −4     (3 ; −4)                   ✓ (3; −4)                  (3)

 9.3                 y                                                     ✓ x-intercepts 1 and 4
                                                                           ✓ y-intercept −4
                                                                           ✓ turning points
                                                                           ✓ shape                    (4)
                         (1; 0)             4
                                                   x

              −4
                                  (3; −4)
             (1; 0) is a turning point on the x-axis
             ⇒ f (x) = (x − 1)2 (x − 4)

 9.4         f ′′ (x) = 6x − 12 = 0 ⇒ x = 2                                ✓x=2
             f (2) = 8 − 24 + 18 − 4 = −2 (2 ; −2)                         ✓ y = −2                   (2)
             (OR: x = 1+3  2 = 2, midway between turning points)

Copyright reserved                                                                         Please turn over

### Page 9

Mathematics/P1                                   9                         Hlayiso/Prediction 2026
                                       NSC – Marking Guidelines

 9.5         m = f ′ (2) = 12 − 24 + 9 = −3                       ✓ m = −3
             y + 2 = −3(x − 2) ⇒ y = −3x + 4                      ✓ equation
             tan θ = −3 ⇒ θ = 180◦ − 71,57◦ = 108,43◦             ✓ tan θ = −3
                                                                  ✓ 108,43◦                  (4)

 9.6         −4 < k < 0                                           ✓ −4 and 0
                                                                  ✓ strict inequality        (2)

                                                                                            [18]

QUESTION 10

                                    12 − h r
 10.1        By similar triangles:         =                      ✓ ratio from similar triangles
                                      12     6
             12 − h = 2r ⇒ h = 12 − 2r                            ✓ h = 12 − 2r
             V = π r2 h = π r2 (12 − 2r) = 12π r2 − 2π r3         ✓ substitution into V = π r2 h
                                                                  (3)

 10.2        V ′ (r) = 24π r − 6π r2 = 0                          ✓ V ′ (r) = 0
             6π r(4 − r) = 0 ⇒ r = 4 (r ̸= 0)                     ✓r=4
             V (4) = 12π (16) − 2π (64) = 64π ≈ 201,06 cm3        ✓ 64π or 201,06 cm3        (3)

                                                                                             [6]

Copyright reserved                                                                Please turn over

### Page 10

Mathematics/P1                                         10                        Hlayiso/Prediction 2026
                                             NSC – Marking Guidelines

QUESTION 11

                                         60   40   1
 11.1.1      P(T and I) = P(T ) × P(I) =    ×    =                      ✓ product rule
                                         120 120 6
                          1                                             ✓ substitution               (2)
             n(T and I) = × 120 = 20
                          6

 11.1.2                                  S                              ✓ 8 and 10
                     T              W
                                                                        ✓ equation in y
                     24       16    32                                  ✓ y = 16
                              12                                        ✓ completed Venn diagram
                          8        10                                   (24 ; 32 ; 8 outside)  (4)
                       10
              8         I
             T and I only: 20 − 12 = 8; only I:
             40 − 12 − 8 − 10 = 10
             Let T and W only = y: only T = 40 − y, only W
             = 48 − y
             8 + (40 − y) + 10 + (48 − y) + 8 + y + 10 + 12 = 120
             136 − y = 120 ⇒ y = 16

 11.1.3      Exactly one: 24 + 32 + 10 = 66                             ✓ 66
             P=
                  66
                      =
                        11
                           = 0,55                                       ✓ answer                     (2)
                 120 20

 11.2.1      7 × 6 × 5 × 4 × 3 = 2 520                                  ✓ answer                     (1)

 11.2.2      First digit 5 or 7: 2 × (5 × 4 × 3) × 3 = 360              ✓ case first digit 5 or 7
             First digit 6: 1 × (5 × 4 × 3) × 2 = 120                   ✓ case first digit 6
             Total: 360 + 120 = 480                                     ✓ 480                        (3)

 11.2.3      Positions for the pair: 4 adjacent pairs of positions; 2   ✓ 4×2
             orders (12 or 21)                                          ✓ 480
             Remaining 3 positions: 5 × 4 × 3 = 60                      ✓ answer                     (3)
             4 × 2 × 60 = 480
                   480      4
             P=         =      ≈ 0,19
                  2 520 21

                                                                                                    [15]

                                                                                          TOTAL: 150

Copyright reserved                                                                       Please turn over

## Related documents

- [MATHS P1 GR12 MEMO SEPT 2021 English IsiXhosa](https://www.hlayiso.com/library/maths-p1-gr12-memo-sept-2021-english-isixhosa/6e14ca2d-44f7-417b-b314-e40962f91676) — Mathematics · Memo · 2021 · Eng
- [MATHS P1 GR12 QP JUNE 2023 English](https://www.hlayiso.com/library/maths-p1-gr12-qp-june-2023-english/245dbccd-9e3b-4add-924a-72f6c39ee078) — Mathematics · QP · 2023 · Eng
- [Mathematics P1 (English) QP](https://www.hlayiso.com/library/mathematics-p1-english-qp/11462bfe-d0db-4e5e-8cc4-7aa8f32251ac) — Mathematics · QP · 2023 · Eng
- [Gr12 Mathematics P1 (English) SEP 2022 Question Paper](https://www.hlayiso.com/library/gr12-mathematics-p1-english-sep-2022-question-paper/14941e0e-8578-4c8c-ba9e-f87f93e33e32) — Mathematics · QP · 2022 · Eng
- [MATHS P1 QP GR12 SEPT 2021 English](https://www.hlayiso.com/library/maths-p1-qp-gr12-sept-2021-english/12f7db49-5bd7-414f-af58-65ed09d4c3e4) — Mathematics · QP · 2021 · Eng
- [Gr12 Mathematics P1 (English) 2021 Preparatory Question Paper](https://www.hlayiso.com/library/gr12-mathematics-p1-english-2021-preparatory-question-paper/37860cd4-80b8-4144-af87-6a1797b9bda3) — Mathematics · QP · 2021 · Eng

## More from Grade 12 Mathematics

- [Maths Grade 12 NSC P2 QP September 2025 Gauteng](https://www.hlayiso.com/library/maths-grade-12-nsc-p2-qp-september-2025-gauteng/1d6bf7fd-d288-4a01-99f7-2c87e247760c) — QP · 2025
- [Maths Grade 12 NSC P2 QP September 2025 Eng Northern Cape](https://www.hlayiso.com/library/maths-grade-12-nsc-p2-qp-september-2025-eng-northern-cape/4ede8207-ca82-4419-bae1-fae734498cc7) — QP · 2025 · Eng
- [MATHS P2 ANSWER BOOK GR12 JUNE 2024 Afr+English](https://www.hlayiso.com/library/maths-p2-answer-book-gr12-june-2024-afr-english/4a71e975-4ac4-4f55-8e23-0b8155a5b0c6) — Addendum · 2024 · Eng
- [MATHS P2 GR12 QP JUNE 2024 English](https://www.hlayiso.com/library/maths-p2-gr12-qp-june-2024-english/f19338aa-ae27-4fb7-9f1b-e60b2b91eb8c) — QP · 2024 · Eng
- [FS Gr 12 Sept 2024 Afr en Eng P1](https://www.hlayiso.com/library/fs-gr-12-sept-2024-afr-en-eng-p1/4d7be3dc-5f53-4da3-a833-8fa9c4d73be1) — QP · 2024 · Eng
- [G12 Maths P2 Sept Free State 2024](https://www.hlayiso.com/library/g12-maths-p2-sept-free-state-2024/b81ee450-da0b-40b1-84a9-e4089f8e3d30) — QP · 2024

## Collection

- [Grade 12 past papers](https://www.hlayiso.com/grade-12/past-papers)

---

Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. When citing, link the page URL above, not a PDF or /api/ URL.