# Technical Mathematics Grade 11 P1 2026 Memo (Eng)

> Grade 11 Technical Mathematics Paper 1 memorandum from the November Mock Exam 2 2026 (English), 11 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/grade-11-technical-mathematics-2026-november-mock-exam-2-memorandum-paper-1/27473a72-3ee0-409d-a405-80b606736a37
- Updated: 2026-10-29
- Subject: Technical Mathematics
- Grade: Grade 11
- Language: English
- Document type: Memorandum
- Year: 2026
- Exam period: November Mock Exam 2
- Paper: 1
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 11

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## About this document

Technical Mathematics · Grade 11 · November Mock Exam 2 · 2026 · English. Memorandum, 11 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 11 Technical Mathematics P1 November 2026 Mock 2](https://www.hlayiso.com/library/grade-11-technical-mathematics-2026-november-mock-exam-2-question-paper-paper-1/74f77288-302d-48bd-be7c-e97cdf4868f4) — QP · 2026 · Eng

## Document text

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### Page 1

Mock

                            hlayiso mock exams
                            Practice Series:
                            Grade 11 Exam Preparation
                            WWW.HLAYISO.COM

                                 NATIONAL
                            SENIOR CERTIFICATE

                                     GRADE 11

                         TECHNICAL MATHEMATICS P1
                     NOVEMBER EXAMINATION 2026 (MOCK)
                             MARKING GUIDELINES

MARKS: 150

                      These marking guidelines consist of 11 pages.

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### 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
 AO  Answer only
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.
    • Where a question says WITHOUT using a calculator, the working must be shown to earn full
      marks.

QUESTION 1
             (    )2
                8 3 ( 4 ) 34
 1.1.1               + 2
               27
               ( )2                                                    ✓ 94                     A
             =
                  2
                     + 23                                              ✓8                       A
                  3                                                    ✓ answer                CA
                4      76                                                                      (3)
             = +8 =          OR   8 49
                9       9
             (2 · 32 )x · (22 )x+1
 1.1.2
              (23 · 3)x · 3x−1
                                                                       ✓ numerator in primes    A
                2x · 32x · 22x+2                                       ✓ denominator in primes A
             = 3x x x−1
                 2 ·3 ·3                                               ✓ adding/subtracting
             = 2x+2x+2−3x · 32x−x−x+1                                  exponents              CA
             = 22 · 31 = 12                                            ✓ answer               CA
                                                                                               (4)

              (√    √ )2 ( √ )2
                3+2 3       3 3                                          √      √
 1.1.3            √       = √                                          ✓ 12 = 2 3               A
                 3 3        3 3                                        ✓ 27                    CA
                27     9    √
             = √ = √ =3 3                                              ✓ answer                CA
               3 3      3    √                                                                 (3)
                  √         9 3
             OR     27 OR
                              3

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### Page 3

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

              log2 2 2 + log3 3−3
                         3

 1.1.4                         1                                  ✓ 23                        A
             log 10−2 × log4 4 2                                  ✓ −3                        A
                3
                  + (−3)                                          ✓ −2                        A
             = 2
               (−2) × 12                                          ✓ 12                        A
                                                                  ✓ answer                   CA
               − 32   3
             =           =                                                                   (5)
                 −1          2
                     log 36 − log 4 − log 0,9
 1.2         LHS =                  1                             ✓ log 36 (power law)          A
                             log 10 2
                   (         )
               log
                        36                                        ✓ quotient law                A
                     4 × 0,9                                      ✓ 21                          A
             =     1
                   2 log 10
               log 10 1                                           ✓ log 10 = 1               CA
             = 1 = 1                                              ✓ answer = 2               CA
                     2           2
                                                                                             (5)
             = 2 = RHS

 1.3.1           1101                                             ✓ partial products / method
                                                                  M
               × 101

                 1101

                 0000

               1101                                               ✓ answer                   CA
                                                                                             (2)
             1000001

             11012 × 1012 = 10000012

 1.3.2       1 × 26 + 1 × 20 = 64 + 1 = 65                        ✓ expansion in powers of 2
                                                                  CA
                                                                  ✓ answer                CA
                                                                                           (2)

 1.4         r = 0,25 mm = 0,000 25 m = 2,5 × 10−4 m              ✓ r in metres & SF            A
                   (          )2
             A = π 2,5 × 10−4
                                                                  ✓ answer in sci. notation CA
             A = 1,96 × 10−7 m2
                                                                  NPU
                                                                                            (2)

                                                                                             [26]

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### Page 4

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

QUESTION 2

                      5x · 5 − 5x · 5−1
 2.1         LHS =                                                   ✓ exponent laws            A
                  (        ) 2 · 5x
                                                                     ✓ common factor 5x         A
               5x 5 − 15
             =                                                       ✓ 24
                                                                        5                      CA
                  2 · 5x                                             ✓ simplification to 2,4   CA
               24
                       24                                                                      (4)
             = 5 =
                2      10
             = 2,4 = RHS

               3                                                         3
 2.2.1       x 2 = 27                                                ✓ x 2 = 27                 A
                     ( )2
                     2
                                                                     ✓ raising to power 32     CA
             x = 27 = 33 3
                     3

             x=9                                                     ✓ answer                  CA
                                                                                               (3)

 2.2.2       Let k = 2x : k2 − 6k + 8 = 0                            ✓ quadratic in 2x          A
             (k − 2)(k − 4) = 0                                      ✓ factors                 CA
             2x = 2 or 2x = 4 = 22                                   ✓ 2x = 2 and 2x = 22      CA
             x = 1 or x = 2                                          ✓x=1                      CA
                                                                     ✓x=2                      CA
                                                                                               (5)

             √
 2.2.3         x+7 = x+1                                             ✓ isolating the surd       A
             x + 7 = x2 + 2x + 1                                     ✓ squaring both sides      A
             x2 + x − 6 = 0                                          ✓ standard form           CA
                                                                     ✓ factors                 CA
             (x + 3)(x − 2) = 0
             x = 2 or x = −3
                             √
             Check x = −3: 4 = 2 ̸= −2 ∴ x ̸= −3                     ✓ x = 2 with rejection of −3
             x=2                                                     CA
                                                                                               (5)

                 [               ]
 2.2.4       log2 (x + 2)(x − 5) = 3                                 ✓ product law              A
             (x + 2)(x − 5) = 23                                     ✓ exponential form        CA
             x2 − 3x − 10 − 8 = 0                                    ✓ expansion               CA
                                                                     ✓ standard form           CA
             x2 − 3x − 18 = 0
                                                                     ✓ factors                 CA
             (x − 6)(x + 3) = 0                                      ✓ x = 6 with rejection    CA
             x = 6 (x ̸= −3: log of a negative number is                                       (6)
             undefined)

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### Page 5

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

                               log 40
 2.3         x − 1 = log5 40 =                                      ✓ log form                  A
                                log 5
                                                                    ✓ answer                   CA
             x = 1 + 2,29 = 3,29
                                                                                               (2)
                  √
             T       L           T2     L
 2.4.1          =          ⇒          =                             ✓ squaring correctly        A
             2π      g           4π 2   g
                   2
                                                                    ✓ answer                   CA
                gT                                                                             (2)
             L= 2
                4π
                 9,8 × (1,8)2
 2.4.2       L=                                                     ✓ substitution             CA
                     4π 2
             L = 0,80 m                                             ✓ answer                   CA
                                                                    NPU
                                                                                                (2)

                                                                                               [29]

QUESTION 3

 3.1.1       2x2 + x − 15 = 0                                       ✓ standard form             A
             (2x − 5)(x + 3) = 0                                    ✓ factors                  CA
                 5                                                  ✓ both answers             CA
             x=      or x = −3                                                                 (3)
                 2

 3.1.2       × x: 2x2 − 3 = 4x                                      ✓ standard form              A
             2x2 − 4x − 3 =√
                           0                 √
                 −(−4) ± (−4)2 − 4(2)(−3) 4 ± 40                    ✓ SF                       CA
             x=                          =                          ✓ x ≈ 2,58                 CA
                             2(2)            4
             x ≈ 2,58 or x ≈ −0,58                                  ✓ x ≈ −0,58                CA
                                                                                               (4)

 3.1.3       x2 − x − 6 > 0                                         ✓ standard form & factors A
             (x − 3)(x + 2) > 0                                     ✓ critical values       CA
             Critical values: x = −2 and x = 3                      ✓ notation              CA
             x < −2 or x > 3
                                                                    ✓ number line (open dots)
                                                                    CA
                     −2              3
                                                                                                (4)

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### Page 6

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

 3.2         y = 5 − 3x                                             ✓ y = 5 − 3x                A
             x2 + x(5 − 3x) + (5 − 3x)2 = 7                         ✓ substitution             CA
             x2 + 5x − 3x2 + 25 − 30x + 9x2 − 7 = 0
                                                                    ✓ standard form            CA
             7x2 − 25x + 18 = 0
                                                                    ✓ factors                  CA
             (7x − 18)(x − 1) = 0
                                                                    ✓ both x-values            CA
                 18
             x=        or x = 1                                     ✓ both y-values            CA
                  7                                                                            (6)
                    19
             y=−         or y = 2
                     7

 3.3.1       2π rh = A − 2π r2                                      ✓ isolating 2π rh           A
                  A − 2π r2              A                          ✓ answer                   CA
             h=              OR    h=        −r
                    2π r                2π r                                                   (2)

                 4,5 − 2π (0,4)2
 3.3.2       h=                                                     ✓ SF                       CA
                     2π (0,4)
                                                                    ✓ simplification           CA
                 3,4947 . . .
             h=                                                     ✓ answer                   CA
                 2,5133 . . .                                                                  (3)
             h = 1,39 m

 3.3.3       2π r2 + 2π r(1,2) = 6                                  ✓ substitution of h and A A
             2π r2 + 2,4π r − 6 = 0                                 ✓ standard form          CA
                            √
                  −2,4π ± (2,4π )2 − 4(2π )(−6)                     ✓ SF                     CA
             r=                                                     ✓ r = 0,55 m with rejection
                                2(2π )
             r = 0,55 m (r ̸= −1,75; a radius cannot be negative)   CA
                                                                                              (4)

                                                                                               [26]

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### Page 7

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

QUESTION 4

 4.1         4x2 − 12x + 9 = 0                                        ✓ standard form & SF         A
             ∆ = (−12)2 − 4(4)(9) = 144 − 144
             ∆=0                                                      ✓∆=0                       CA
             Roots are real, rational and equal.                      ✓ nature                   CA
                                                                                                 (3)

 4.2         ∆ ≥ 0: (−6)2 − 4(k)(3) ≥ 0                               ✓∆≥0                        A
             36 − 12k ≥ 0                                             ✓ substitution              A
             k ≤ 3 , k ̸= 0                                           ✓k≤3                       CA

                                                                      (no penalty if k ̸= 0 omitted)
                                                                      (3)

                 (           )2
 4.3         ∆ = − (p + 1) − 4(1)(p)                                  ✓ ∆ = p2 − 2p + 1          A
             = p2 − 2p + 1                                            ✓ (p − 1) : ≥ 0 and perfect
                                                                               2

             = (p − 1)2 ≥ 0 for all real p, and is a perfect square   square                     A
                                                                                                (2)
             ∴ roots are real and rational

                                                                                                  [8]

QUESTION 5

 5.1         x=0      and y = 1                                       ✓x=0                         A
                                                                      ✓y=1                         A
                                                                                                  (2)

                  3
 5.2         0 = − +1                                                 ✓ f (x) = 0                 A
                  x
             x=3    (3 ; 0)                                           ✓ answer                   CA
                                                                                                 (2)

 5.3         y-int: g(0) = 1 − 2 = −1       (0 ; −1)                  ✓ y-intercept                A
                    ( )x
             x-int: 12 = 2                                            ✓ g(x) = 0                   A
             2−x = 21
                                                                      ✓ x-intercept              CA
             x = −1       (−1 ; 0)
                                                                                                 (3)

 5.4         y = −2                                                   ✓ answer                     A
                                                                                                  (1)

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### Page 8

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

 5.5                             y                                     f:
                                                                       ✓ asymptotes x = 0, y = 1
                                                                       CA
                      g                                                ✓ x-intercept (3 ; 0)    CA
                                                                       ✓ shape (branches in 2nd &
                f
                                                                       4th quadrants of asymptotes)
                                                       y=1
                                                        x              A
                           −1               3      f                   g:
                                     −1
                                                       y = −2          ✓ asymptote y = −2       CA
                                                                       ✓ both intercepts        CA
                                                                       ✓ decreasing shape         A
                                                                                                 (6)

 5.6         Centre of symmetry (0 ; 1)                                ✓ m = −1                   A
             y = −x + 1                                                ✓c=1                      CA
                                                                                                 (2)

 5.7         y > −2       OR     y ∈ (−2 ; ∞)                          ✓ answer                  CA
                                                                                                 (1)

 5.8         0<x≤3          OR       x ∈ (0 ; 3]                       ✓ critical values         CA
                                                                       ✓ notation                CA
                                                                                                 (2)

                    ( )−x
 5.9         h(x) = 12    −2                                           ✓ answer                  CA
             h(x) = 2 − 2
                     x                                                                           (1)

                                                                                                 [20]

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### Page 9

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

QUESTION 6

 6.1         r2 = (−3)2 + (−4)2 = 25                             ✓ r2 = 25 (r = 5)          A
                     √
             g(x) = − 25 − x2                                    ✓ equation                CA
                                                                                           (2)

 6.2         A(−5 ; 0),    B(5 ; 0), C(0 ; −5)                   ✓ A and B                 CA
                                                                 ✓C                        CA
                                                                                           (2)

                  0 − (−5)
 6.3         m=            =1                                    ✓ gradient                CA
                    5−0
                                                                 ✓ equation                CA
             h(x) = x − 5
                                                                                           (2)

 6.4         −5 ≤ y ≤ 0    OR    y ∈ [−5 ; 0]                    ✓ answer                  CA
                                                                                           (1)

 6.5         −5 ≤ x ≤ 0    or   x=5                              ✓ −5 ≤ x ≤ 0              CA
                                                                 ✓x=5                      CA
                                                                                           (2)

                 √
 6.6         BC = (5 − 0)2 + (0 − (−5))2                         ✓ SF                      CA
               √     √
             = 50 = 5 2 ≈ 7,07 units                             ✓ answer                  CA
                                                                                           (2)

                                                                                           [11]

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### Page 10

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

QUESTION 7

 7.1         f (x) = a(x − 2)2 − 9                                ✓ substitution of turning
             Substitute C(0 ; −5): −5 = a(0 − 2)2 − 9             point                      A
             4a = 4 ⇒ a = 1                                       ✓a=1                      CA
             f (x) = (x − 2)2 − 9 = x2 − 4x − 5
                                                                  ✓ b = −4 and c = −5       CA
             a = 1; b = −4; c = −5
                                                                                            (3)
             OR
                          b
             c = −5; − = 2 ⇒ b = −4a
                         2a
             −9 = 4a + 2b − 5 ⇒ −4 = 4a − 8a ⇒ a = 1; b = −4

 7.2         x2 − 4x − 5 = 0                                      ✓ f (x) = 0               CA
             (x − 5)(x + 1) = 0                                   ✓ factors                 CA
             A(−1 ; 0) and D(5 ; 0)                               ✓ both coordinates        CA
                                                                                            (3)

 7.3         x2 − 4x − 5 = 2x − 10                                ✓ equating f = g          CA
             x2 − 6x + 5 = 0
             (x − 1)(x − 5) = 0                                   ✓x=1                      CA
             x = 1 (x = 5 at D)
                                                                  ✓ y = −8                  CA
             y = 2(1) − 10 = −8        E(1 ; −8)
                                                                                            (3)

 7.4         1<x<5         OR   x ∈ (1 ; 5)                       ✓ critical values         CA
                                                                  ✓ notation                CA
                                                                                            (2)

 7.5         Minimum value of f is −9.                            ✓ −9 + k > 0 (or ∆ < 0:
             −9 + k > 0                                           16 − 4(−5 + k) < 0)     CA
             k>9                                                  ✓ answer                CA
                                                                                          (2)

                                                                                            [13]

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### Page 11

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

QUESTION 8

 8.1.1       10% × 8 400 = R840                                    ✓ answer                    A
                                                                                              (1)

 8.1.2       Loan = 8 400 − 840 = R7 560                           ✓ R7 560                  CA
                      (             )
             A = 7 560 1 + 2 × 0,125 = R9 450                      ✓ SF                      CA
                          9 450                                    ✓ answer                  CA
             Instalment =       = R393,75                                                    (3)
                            24

 8.2         A = P(1 − i)n                                         ✓ SF                        A
             18 000 = 45 000(1 − i)6
                     √                                             ✓ simplification/root     CA
             1 − i = 6 0,4 = 0,8583 . . .
             i = 0,1416 . . .
                                                                   ✓ answer                  CA
             i = 14,16%
                                                                                             (3)
                             (        )
                                 0,075 n
 8.3         35 000 = 20 000 1 +                                   ✓ SF                        A
                                   12
             1,75 = (1,00625) n
                                                                   ✓ using logs              CA
                   log 1,75                                        ✓ answer in years         CA
             n=              = 89,82 months
                 log 1,00625
             ≈ 7,48 years                                          (accept 90 months / 7,5
                                                                   years)                     (3)
                   (         )
                        0,096 12
 8.4.1       ieff = 1 +          −1                                ✓ SF                       A
                         12
                                                                   ✓ answer                  CA
             = 10,03%
                                                                                             (2)

                                  (         )36
                                      0,096
 8.4.2       End of year 3: 60 000 1 + 12       = R79 933,79       ✓ year 3 amount            A
             After deposit: R94 933,79                             ✓ adding R15 000          CA
                                     (        )4                   ✓ 0,1
                                                                      4 and n = 4, n = 8      A
             End of year 4: 94 933,79 1 + 0,1
                                           4     = R104 789,14     ✓ subtracting R8 000      CA
             After withdrawal: R96 789,14                          ✓ answer                  CA
                                     (        )8                                             (5)
             End of year 6: 96 789,14 1 + 0,1
                                           4     = R117 928,17
             OR
             A = 60 000(1,008)36 (1,025)12 + 15 000(1,025)12
                 − 8 000(1,025)8 = R117 928,17

                                                                                             [17]

                                                                                   TOTAL: 150

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