# Technical Mathematics Grade 12 P1 2026 Memo (Eng)

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

- Page: https://www.hlayiso.com/library/grade-12-technical-mathematics-2026-nsc-november-mock-exam-memorandum-paper-1/aaed7b93-383b-46ce-a8e1-03102ac7c19b
- Updated: 2026-10-29
- Subject: Technical Mathematics
- Grade: Grade 12
- Language: English
- Document type: Memorandum
- Year: 2026
- Exam period: NSC November Mock Exam
- Paper: 1
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 12

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## About this document

Technical Mathematics · Grade 12 · NSC November Mock Exam · 2026 · English. Memorandum, 12 pages. Read online or download the PDF.

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

## Matching paper and memo

- [Grade 12 Technical Mathematics P1 November 2026 Mock](https://www.hlayiso.com/library/grade-12-technical-mathematics-2026-nsc-november-mock-exam-question-paper-paper-1/5d68d5eb-cb36-4bae-86ee-8de53405fff5) — QP · 2026 · Eng

## Document text

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### Page 1

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

                                NATIONAL
                           SENIOR CERTIFICATE

                                    GRADE 12

                        TECHNICAL MATHEMATICS P1
                          MOCK EXAMINATION 2026
                            MARKING GUIDELINES

  MARKS: 150

                     These marking guidelines consist of 12 pages.

Copyright reserved                                                   Please turn over

### Page 2

Technical Mathematics/P1                        2                                  Hlayiso/Mock 2026
                                      NSC – Marking Guidelines

MARKING CODES
 A   Accuracy                         CA     Consistent accuracy
 M   Method                           R      Rounding
 NPR No penalty for rounding          NPU    No penalty for units omitted
 S   Simplification                   SF     Substitution in correct formula
NOTE:

    • If a candidate answers a question TWICE, only mark the FIRST attempt.
    • Consistent accuracy (CA) applies in ALL aspects of the marking guidelines where indicated.
    • No penalty for rounding (NPR) for ALL questions, unless stated otherwise.
    • AO = answer only. Where AO is indicated, the stated marks are awarded for a correct answer
      without working.

QUESTION 1

 1.1.1       x(3x + 8) = 0                                            ✓x=0                        A
             x=0     or    x=−
                               8
                                     (x ≈ −2,67)                      ✓ x = − 38                  A
                               3                                                                 (2)

 1.1.2       2x2 − 6x − 4 =√
                           0 OR x2 − 3x − 2 = 0                       ✓ standard form            A
                 −(−3) ± (−3)2 − 4(1)(−2)                             ✓ SF                      CA
             x=
                     √       2(1)
                 3 ± 17                                               ✓ x ≈ 3,56                CA
             x=                                                       ✓ x ≈ −0,56               CA
                     2
             x ≈ 3,56 or x ≈ −0,56                                                              (4)

 1.1.3       x2 − 2x − 8 > 0                                          ✓ factors                  A
             (x − 4)(x + 2) > 0                                       ✓ both critical values    CA
             Critical values: x = 4 and x = −2

                                x
               −2          4
                                                                      ✓ correct notation        CA
             x < −2 or         x>4                                    AO: full marks
                                                                                                 (3)

 1.2.1       y = 5 − 2x                                               ✓ y = 5 − 2x                A
                                                                                                 (1)

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### Page 3

Technical Mathematics/P1                     3                             Hlayiso/Mock 2026
                                   NSC – Marking Guidelines

 1.2.2       x(5 − 2x) − 2 = 0                                ✓ substitution             CA
             5x − 2x2 − 2 = 0
             2x2 − 5x + 2 = 0                                 ✓ standard form            CA
             (2x − 1)(x − 2) = 0                              ✓ factors                  CA
             x=
                 1
                     or x = 2                                 ✓ both x-values            CA
                 2                                            ✓ both y-values            CA
             y=4          y=1
                                                                                         (5)

 1.3.1       V 2 =√PR                                         ✓ V 2 =√PR                  A
             V = PR                                           ✓ V = PR                   CA
                                                                                         (2)

                 √          √
 1.3.2       V = (60)(240) = 14 400                           ✓ substitution             CA
             V = 120 V                                        ✓ answer                   CA
                                                              NPU
                                                                                          (2)

 1.4         45 = 1011012                                     ✓ answer                     A
                                                                                          (1)

 1.5         1011012 + 11102                                  ✓ 59 OR binary addition M
             = 45 + 14 = 59                                   ✓ answer               CA
             = 1110112                                                                (2)

                                                                                         [22]

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### Page 4

Technical Mathematics/P1                        4                             Hlayiso/Mock 2026
                                      NSC – Marking Guidelines

QUESTION 2

 2.1.1       ∆ = b2 − 4ac = (−3)2 − 4(2)(5)                      ✓ SF                       A
             ∆ = −31                                             ✓ answer                  CA
                                                                                           (2)

 2.1.2       Roots are non-real   (∆ < 0)                        ✓ non-real                CA
                                                                                           (1)

 2.2         For equal roots: ∆ = 0                              ✓∆=0                        A
             (−p)2 − 4(2)(8) = 0                                 ✓ SF                        A
             p2 = 64
             p = ±8                                              ✓ p = ±8                  CA
                                                                                           (3)

 2.3.1       p+1 = 0
             p = −1                                              ✓ p = −1                    A
                                                                                            (1)

 2.3.2       2p − 4 < 0                                          ✓ p<2                       A
             p < 2 (and p ̸= −1)                                                            (1)

                                                                                            [8]

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Technical Mathematics/P1                        5                              Hlayiso/Mock 2026
                                      NSC – Marking Guidelines

QUESTION 3

             √
                                                                 ✓ 2x2
             3
 3.1.1           8x6 = 2x2                                                                    A
                                                                                             (1)

             2x+2 · 22x
 3.1.2                                                           ✓ prime bases                A
              23x · 21
             = 2x+2+2x−3x−1                                      ✓ adding/subtracting
             = 21 = 2                                            exponents                  CA
                                                                 ✓ answer                   CA
                                                                                            (3)

                                            √
 3.1.3       log 8 + log 125 − log3 9 + log2 2
                                               1
             = log(8 × 125) − log3 32 + log2 2 2                 ✓ log 1 000                  A
             = log 1 000 − 2 + 21                                ✓2                           A
             = 3 − 2 + 21                                        ✓ 21                         A
             = 32 OR 1,5
                                                                 ✓ answer                   CA
                                                                                            (4)

                  [             ]
 3.2         log2 (x + 2)(x − 2) = 5                             ✓ product law            A
             x 2 − 4 = 25                                        ✓ exponential form      CA
             x2 = 36                                             ✓ x2 = 36               CA
             x = 6 (x ̸= −6, since x − 2 > 0)                    ✓ answer with rejection CA
                                                                                         (4)

 3.3         3x · 3 − 3x = 54
             3x (3 − 1) = 54                                     ✓ common factor             A
             3x = 27 = 33                                        ✓ 3x = 33                  CA
             x=3                                                 ✓ answer                   CA
                                                                 AO: full marks
                                                                                             (3)

 3.4.1       z = (1 + 2i)(i − 2)
             = i − 2 + 2i2 − 4i                                  ✓ expansion                  A
             = −2 − 3i − 2
             = −4 − 3i                                           ✓ answer                   CA
                                                                                            (2)

 3.4.2       z̄ = −4 + 3i                                        ✓ answer                   CA
                                                                                            (1)

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Technical Mathematics/P1                             6                            Hlayiso/Mock 2026
                                           NSC – Marking Guidelines

 3.4.3                               Im                               ✓ point (−4 ; −3) in
                                                                      quadrant 3                CA
                           −4
                                              Re
                                                                      ✓ line from origin/labels CA
                                                                                                (2)

                                       −3
              (−4 ; −3) OR z

                  √
 3.4.4       r=     (−4)2 + (−3)2 = 5                                 ✓r=5                        CA
                      −3                                              ✓ tan θ                     CA
             tan θ =
                      −4
             reference angle = 36,87◦                                 ✓ θ = 216,87◦               CA
             θ = 180◦ + 36,87◦ = 216,87◦                              ✓ polar form                CA
             z = 5 cis 216,87◦ OR 5 cis(−143,13◦ )                                                (4)

                                                                                                  [24]

QUESTION 4

 4.1.1       y = −3      OR     (0 ; −3)                              ✓ answer                      A
                                                                                                   (1)

 4.1.2       x2 − 2x − 3 = 0                                          ✓ factors                     A
             (x − 3)(x + 1) = 0
             x = 3 or x = −1                                          ✓ both x-values             CA
                                                                                                  (2)

                    b       −2
 4.1.3       x=−      =−         =1                                   ✓x=1                         A
                   2a      2(1)
                                                                      ✓ substitution              CA
             y = (1)2 − 2(1) − 3 = −4 OR
                                                                      ✓ y = −4                    CA
                 4ac − b2 4(1)(−3) − 4
             y=           =                                                                       (3)
                    4a             4
             Turning point: (1 ; −4)

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Technical Mathematics/P1                                       7                             Hlayiso/Mock 2026
                                                     NSC – Marking Guidelines

 4.1.4                   y                   g                                   f:
                                                                                ✓ x-intercepts             CA
                                                                                ✓ y-intercept              CA
                                         f                                      ✓ turning point & shape CA
                                                                                g:
                     1                                                          ✓ x-intercept −1            A
                −1
                                                 x                              ✓ y-intercept 1 & straight
                                     3
                                                                                line                        A
                                                                                                           (5)
                             −3

                         (1 ; −4)

 4.1.5       y ≥ −4          OR     y ∈ [−4 ; ∞)                                ✓ answer                  CA
                                                                                                          (1)

 4.1.6       −1 < x < 3             OR       x ∈ (−1 ; 3)                       ✓ critical values         CA
                                                                                ✓ notation                CA
                                                                                                          (2)

 4.2.1       x=0         and y = −2                                             ✓x=0                        A
                                                                                ✓ y = −2                    A
                                                                                                           (2)

                    a
 4.2.2       h(x) = − 2
                    x
                  a                                                             ✓ substitution of P        A
             2=      −2
                 −1                                                             ✓ answer                  CA
             a = −4                                                                                       (2)

 4.2.3       k(x) = bx − 2
             7 = b2 − 2                                                         ✓ substitution of R         A
             b2 = 9
             b = 3 (b > 0)                                                      ✓ answer                  CA
                                                                                                          (2)

                 −4
 4.2.4       0=     −2                                                          ✓ h(x) = 0                CA
                  x
             2x = −4
             x = −2    B(−2 ; 0)                                                ✓ answer                  CA
                                                                                                          (2)

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### Page 8

Technical Mathematics/P1                          8                            Hlayiso/Mock 2026
                                        NSC – Marking Guidelines

 4.2.5       y > −2    OR     y ∈ (−2 ; ∞)                         ✓ answer                   A
                                                                                             (1)

 4.2.6       x ∈ R , x ̸= 0                                        ✓ answer                   A
                                                                                             (1)

                                                                                            [24]

QUESTION 5

 5.1         A = P(1 − i)n                                         ✓ formula                 A
             A = 480 000(1 − 0,12)5                                ✓ SF                      A
             A = R253 311,32                                       ✓ answer                 CA
                                                                                            (3)
                    (         )
                         0,096 12
 5.2         ieff = 1 +           −1                               ✓ 0,096
                                                                      12                      A
                           12
             = 0,100338 . . .                                      ✓ SF                       A
             ieff = 10,03%
                                                                   ✓ answer                 CA
                                                                                            (3)

 5.3         A = P(1 + i)n
             20 000 = 12 000(1 + i)6                               ✓ SF                      A
             (1 + i)6 =
                        5                                          ✓ simplification         CA
                     √3                                            ✓ 6th root               CA
             1 + i = 6 53 = 1,08887 . . .
             i = 0,0889                                            ✓ answer                 CA
             r = 8,89% per annum                                                            (4)

                        (          )
                              0,08 8
 5.4         A2 = 25 000 1 +           = R29 291,48                ✓ i = 0,08
                                                                          4 ;n=8              A
                                4
                           (          )                                  0,09
                                                                   ✓ i = 12 ; n = 12          A
                                  0,09 12
             A3 = 29 291,48 1 +           = R32 039,23             ✓ +5 000                 CA
                                   12                              ✓ n = 24
                                      (         )                                           CA
                                           0,09 24                 ✓ answer                 CA
             A5 = (32 039,23 + 5 000) 1 +
                                            12                     Accept answers in the range
             A5 = R44 314,23                                       R44 314 – R44 315
             OR                                                                              (5)
             A = 25 000(1,02)8 (1,0075)36 + 5 000(1,0075)24
             = R44 314,23

                                                                                            [15]

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### Page 9

Technical Mathematics/P1                          9                            Hlayiso/Mock 2026
                                        NSC – Marking Guidelines

QUESTION 6

 6.1         f (x + h) = 2(x + h)2 − 3 = 2x2 + 4xh + 2h2 − 3       ✓ f (x + h) expanded        A
             f (x + h) − f (x) = 4xh + 2h2                         ✓ 4xh + 2h2                CA
                           4xh + 2h2                               ✓ formula with limit
             f ′ (x) = lim
                       h→0     h                                   notation                    A
             = lim (4x + 2h)                                       ✓ 4x + 2h                  CA
                h→0
              f ′ (x) = 4x                                         ✓ answer                   CA
                                                                                              (5)

              dy
 6.2.1           = 20x3 − 3                                        ✓ 20x3                       A
              dx
                                                                   ✓ −3                         A
                                                                                               (2)

              f (x) = 6x−2 − x 3
                                1
 6.2.2                                                             ✓ exponential form          A
                                1 2
              f ′ (x) = −12x−3 − x− 3                              ✓ −12x−3                   CA
                                                                   ✓ − 31 x− 3
                                                                             2
                                3                                                             CA
                                                                                              (3)

             x2 − 9 (x − 3)(x + 3)
 6.2.3              =              = x−3                           ✓ factorising               A
             x+3            x+3
             Dx [x − 3] = 1                                        ✓ x−3                      CA
                                                                   ✓ answer                   CA
                                                                                              (3)

 6.3.1       g′ (x) = 3x2 − 4                                      ✓ answer                     A
                                                                                               (1)

 6.3.2       g′ (1) = 3(1)2 − 4 = −1                               ✓ answer                   CA
                                                                                              (1)

 6.3.3       g(1) = (1)3 − 4(1) = −3                               ✓ y-value                   A
             y − (−3) = −1(x − 1)                                  ✓ substitution             CA
             y = −x − 2                                            ✓ equation                 CA
                                                                                              (3)

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### Page 10

Technical Mathematics/P1                        10                               Hlayiso/Mock 2026
                                      NSC – Marking Guidelines

 6.3.4       3x2 − 4 = 8                                         ✓ g′ (x) = 8                 CA
             x2 = 4
             x = 2 or x = −2                                     ✓ both values                CA
                                                                                              (2)

                                                                                              [20]

QUESTION 7

 7.1         C(0 ; 27)                                           ✓ answer                       A
                                                                                               (1)

 7.2         f (−3) = (−3)3 − 3(−3)2 − 9(−3) + 27                ✓ substitution of −3           A
             = −27 − 27 + 27 + 27 = 0                            ✓ = 0 with conclusion          A
             ∴ (x + 3) is a factor of f (x)                                                    (2)

 7.3         f (x) = (x + 3)(x2 − 6x + 9)                        ✓ quadratic factor           CA
             = (x + 3)(x − 3)2                                   ✓ (x − 3)2                   CA
             x = −3 or x = 3
             A(−3 ; 0) and B(3 ; 0)                              ✓ both coordinates           CA
                                                                                              (3)

 7.4         f ′ (x) = 3x2 − 6x − 9 = 0                          ✓ f ′ (x)                     A
             x2 − 2x − 3 = 0                                     ✓ f ′ (x) = 0                CA
             (x − 3)(x + 1) = 0                                  ✓ factors                    CA
             x = −1 or x = 3                                     ✓ x-values                   CA
             f (−1) = −1 − 3 + 9 + 27 = 32
             D(−1 ; 32) and B(3 ; 0)                             ✓ y = 32                     CA
                                                                                              (5)

 7.5         −1 < x < 3    OR     x ∈ (−1 ; 3)                   ✓ critical values            CA
                                                                 ✓ notation                   CA
                                                                                              (2)

 7.6          f ′ (0) = 3(0)2 − 6(0) − 9 = −9                    ✓ answer                     CA
                                                                                              (1)

                                                                                              [14]

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### Page 11

Technical Mathematics/P1                            11                               Hlayiso/Mock 2026
                                          NSC – Marking Guidelines

QUESTION 8

 8.1         V (0) = 200 kilolitres                                  ✓ answer                      A
                                                                                                  (1)

 8.2         V ′ (t) = −3t 2 + 12t + 15                              ✓ −3t 2 + 12t                 A
                                                                     ✓ 15                          A
                                                                                                  (2)

 8.3         V ′ (2) = −3(2)2 + 12(2) + 15                           ✓ substitution              CA
             = 27 kilolitres per hour                                ✓ answer                    CA
                                                                                                 (2)

 8.4         −3t 2 + 12t + 15 = 0                                    ✓ V ′ (t) = 0               CA
             t 2 − 4t − 5 = 0
             (t − 5)(t + 1) = 0
             t = 5 (t ̸= −1)                                         ✓t =5                       CA
             V (5) = −(5)3 + 6(5)2 + 15(5) + 200
             V (5) = 300 kilolitres                                  ✓ answer                    CA
                                                                                                 (3)

                                                                                                  [8]

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### Page 12

Technical Mathematics/P1                         12                            Hlayiso/Mock 2026
                                       NSC – Marking Guidelines

QUESTION 9
             ∫
                 3
 9.1.1             dx = 3 ln x +C                                 ✓ 3 ln x                     A
                 x
                                                                  ✓ +C                         A
                                                                                              (2)

             ∫
 9.1.2           (4x2 − 4x + 1) dx                                ✓ expansion                 A
                                                                        3
                 4x3                                              ✓ 4x3                      CA
             =       − 2x2 + x +C                                 ✓ −2x2 + x +C              CA
                  3
                                                                                             (3)

             ∫ (              )
                     1
 9.1.3             x +2
                     2    x
                                  dx                              ✓ exponential form          A
                                                                         3
              2 3        2x                                       ✓ 32 x 2                   CA
             = x2 +      +C                                            x
              3     ln 2                                          ✓ ln2 2 +C                  A
                                                                                             (3)

 9.2.1       −x2 + 2x + 3 = 0                                     ✓ factors                    A
             x2 − 2x − 3 = 0
             (x − 3)(x + 1) = 0
             x = −1 and x = 3                                     ✓ both values              CA
                                                                                             (2)

                    ∫ 3(              )
 9.2.2       Area =       −x2 + 2x + 3 dx                         ✓ integral with limits     CA
               [ 3 −1         ]3                                  ✓ integration              CA
                  x                                               ✓ substitution of limits   CA
             = − + x + 3x
                        2

               ( 3           ) −1(         )                      ✓ simplification           CA
                   27
             = − +9+9 −
                                    1
                                      +1−3                        ✓ answer                   CA
                  (3 )              3                                                        (5)
                      5
             = 9− −
                      3
               32
             =    ≈ 10,67 square units
                3

                                                                                             [15]

                                                                                  TOTAL:      150

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