# Technical Mathematics Grade 11 P2 2026 Memo (Eng)

> Grade 11 Technical Mathematics Paper 2 memorandum from the November Mock Exam 2 2026 (English), 12 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-2/4494d07c-a4e9-4173-88f2-be51a52ddca5
- Updated: 2026-10-29
- Subject: Technical Mathematics
- Grade: Grade 11
- Language: English
- Document type: Memorandum
- Year: 2026
- Exam period: November Mock Exam 2
- Paper: 2
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 12

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## About this document

Technical Mathematics · Grade 11 · November Mock Exam 2 · 2026 · English. Memorandum, 12 pages. Read online or download the PDF.

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## Matching paper and memo

- [Grade 11 Technical Mathematics P2 November 2026 Mock 2](https://www.hlayiso.com/library/grade-11-technical-mathematics-2026-november-mock-exam-2-question-paper-paper-2/10aa5488-4c7d-4935-8857-d86b7652dd1f) — 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 P2
                     NOVEMBER EXAMINATION 2026 (MOCK)
                             MARKING GUIDELINES

MARKS: 150

                      These marking guidelines consist of 12 pages.

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### Page 2

Technical Mathematics/P2                        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
 S/R Statement/Reason                 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.
    • In QUESTIONS 6, 7 and 8 (Euclidean geometry) a statement without the correct reason earns
      no mark for the reason. Accept any valid alternative method.

QUESTION 1

                     3 − (−1) 4 1
 1.1         mAB =           = =                                      ✓ SF                       A
                     5 − (−3) 8 2
                                                                      ✓ answer                   A
                                                                                                (2)

                  √
 1.2         AB = (5 − (−3))2 + (3 − (−1))2                           ✓ SF
                                                                        √                       A
               √         √
             = 64 + 16 = 80                                           ✓ √ 80                   CA
                √
             = 4 5 units                                              ✓4 5                     CA
                                                                                               (3)
                (                 )
               −3 + 5 −1 + 3
 1.3         M       ;                = M(1 ; 1)                      ✓x=1                       A
                 2      2
                                                                      ✓y=1                       A
                                                                                                (2)

 1.4         m⊥ = −2 (mAB × m⊥ = −1)                                  ✓ m = −2                 CA
             y − 1 = −2(x − 1)                                        ✓ substitution of M      CA
             y = −2x + 3                                              ✓ answer                 CA
                                                                                               (3)

 1.5         Substitute x = −1: y = −2(−1) + 3 = 5, so C(−1 ; 5)      ✓ substitution             A
             lies on the line.                                        ✓ y = 5 / LHS = RHS        A
             The line is the perpendicular bisector of AB, so every
             point on it is equidistant from A and B: CA = CB.        ✓ CA = CB with reason      A
             ∴ △ABC is isosceles.                                                               (3)
                           √            √
                               2   2
             OR CA  √ = 2 +√6 = 40 and
             CB = 62 + 22 = 40, so CA = CB.

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### Page 3

Technical Mathematics/P2                        3                            Hlayiso/Mock 2026
                                      NSC – Marking Guidelines

                     5 − (−1)  6
 1.6         mAC =            = =3                               ✓ mAC = 3                     A
                    −1 − (−3) 2
             tan α = 3
                                                                 ✓ answer                    CA
             α = 71,57◦                                                                      (2)

                                        1
 1.7         Inclination of AB: tan β = , β = 26,57◦             ✓ β = 26,57◦                CA
                                        2
                                                                 ✓ method α − β               M
             CÂB = α − β = 71,57 − 26,57◦
                                     ◦
                                                                 ✓ answer                    CA
             = 45◦
                                                     √                                       (3)
             (Accept:
                   √   cosine rule
                                √  in △ABC with AC =   40,
             AB = 80, BC = 40.)

 1.8         CM ⊥ AB,
                   √ so CM is the perpendicular height.
                                            √
             CM = (1 − (−1))2 + (1 − 5)2 = 20                    ✓ SF     √                   A
                    1              1 √       √                   ✓ CM = 20                   CA
             Area = × AB ×CM = × 80 × 20                         ✓ substitution              CA
                    2              2
             = 20 square units                                   ✓ answer                    CA
             OR Area                                                                         (4)
               1                   1√ √
             = (AB)(AC) sinCÂB =     80 40 sin 45◦ = 20
               2                   2
                                             √
 1.9         BE 2 = (x − 5)2 + (−3 − 3)2 = (2 13)2               ✓ SF                         A
             (x − 5)2 + 36 = 52                                  ✓ (x − 5)2 = 16             CA
             (x − 5)2 = 16
                                                                 ✓ both values               CA
             x − 5 = ±4
             x = 9 or x = 1
                                                                 ✓ selection x = 1           CA
             Since x < 5: x = 1                                                              (4)

                                                                                             [26]

QUESTION 2

 2.1.1       tan(2(48,6◦ ) − 117,3◦ ) = tan(−20,1◦ )             ✓ −20,1◦                     A
             = −0,37                                             ✓ answer                    CA
                                                                                             (2)

             sec 117,3◦         1
 2.1.2                ◦
                        =              × tan2 48,6◦              ✓ reciprocals                 A
                2
             cot 48,6      cos 117,3◦
                                                                 ✓ substitution                A
             = (−2,1803 . . .) × (1,2865 . . .)
             = −2,81                                             ✓ answer                    CA
                                                                                             (3)

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### Page 4

Technical Mathematics/P2                                 4                            Hlayiso/Mock 2026
                                               NSC – Marking Guidelines

                              8
 2.2.1       cos θ =            and sin θ < 0: θ in the 4th quadrant      ✓ sketch in 4th quadrant     A
                             17
                y

                         8
                                    x

                    17        −15                                         ✓ y = −15                  CA
                                                                          ✓ answer                   CA
                                                                                                     (3)
             y2 = 172 − 82 = 225, y = −15
                        15
             tan θ = −
                         8
                                   (     )     (     )
                                      15          15
 2.2.2       8 tan θ − 17 sin θ = 8 −      − 17 −                         ✓ tan θ substituted        CA
                                       8          17
                                                                          ✓ sin θ = − 17
                                                                                      15
                                                                                                     CA
             = −15 + 15
                                                                          ✓ answer                   CA
             =0                                                                                      (3)

                                                        5
 2.3         4 cot θ = −5               ⇒   cot θ = −                     ✓ cot θ = − 54              A
                                                        4
                      4                                                   ✓ tan θ = −0,8             CA
             tan θ = − = −0,8
                      5
             Reference angle = 38,66◦                                     ✓ 141,3◦                   CA
             θ = 180◦ − 38,66◦ = 141,3◦                                   ✓ 321,3◦                   CA
                                                                                                     (4)
             or θ = 360◦ − 38,66◦ = 321,3◦

                                                                                                     [15]

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### Page 5

Technical Mathematics/P2                                              5                                  Hlayiso/Mock 2026
                                                            NSC – Marking Guidelines

QUESTION 3

               (− sin x)(tan x)
 3.1         =                                                                             ✓ − sin x                     A
                 (− cos x)(1)
                                                                                           ✓ tan x                       A
                sin x
             =        · tan x                                                              ✓ − cos x                     A
               cos x                                                                       ✓ tan 225◦ = 1                A
             = tan x · tan x
                                                                                           ✓ cos
                                                                                              sin x
                                                                                                  x = tan x             CA
             = tan2 x
                                                                                           ✓ answer                     CA
                                                                                                                        (6)

                       cos x      sin x
 3.2         LHS =             −                                                           ✓ tan x = cos
                                                                                                     sin x
                                                                                                                          A
                     1 − sin x cos x                                                                     x
                                                                                           ✓ common denominator           A
               cos2 x − sin x(1 − sin x)
             =
                    cos x(1 − sin x)                                                       ✓ sin2 x + cos2 x = 1          A
               cos x + sin2 x − sin x
                   2
             =
                  cos x(1 − sin x)                                                         ✓ simplification to cos1 x   CA
                   1 − sin x                                                                                            (4)
             =
               cos x(1 − sin x)
                 1
             =       = sec x = RHS
               cos x

                                                                                                                        [10]

QUESTION 4

 4.1               y                                                                       f:
                       f
                                                                                           ✓ x-intercepts 90◦ , 270◦ A
               2
                                                                                           ✓ end points and turning
                                 (120◦ ; 1)                                                point                        A
                                                                                           ✓ shape                      A
                           g
                                                                                       x   g:
                               90◦               180◦           270◦            360◦
                                                                                           ✓ x-intercepts 30◦ , 210◦ A
                                                                  (300◦ ; −1)              ✓ turning points (120◦ ; 1),
              −2
                                                                                           (300◦ ; −1)                  A
                                              (180◦ ; −2)                                  ✓ end points (0◦ ; − 12 ),
                                                                                           (360◦ ; − 21 ) and shape     A
             f : (0◦ ; 2), (90◦ ; 0), (180◦ ; −2), (270◦ ; 0), (360◦ ; 2)                                              (6)
             g: (0◦ ; − 12 ), (30◦ ; 0), (120◦ ; 1), (210◦ ; 0), (300◦ ; −1),
             (360◦ ; − 12 )

 4.2         360◦                                                                          ✓ answer                       A
                                                                                                                         (1)

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### Page 6

Technical Mathematics/P2                         6                              Hlayiso/Mock 2026
                                       NSC – Marking Guidelines

 4.3         −2 ≤ y ≤ 2    OR     y ∈ [−2 ; 2]                    ✓ critical values            A
                                                                  ✓ notation                   A
                                                                                              (2)

 4.4         30◦ < x < 90◦ or 210◦ < x < 270◦                     ✓ 30◦ < x < 90◦           CA
             OR x ∈ (30◦ ; 90◦ ) ∪ (210◦ ; 270◦ )                 ✓ 210◦ < x < 270◦         CA
                                                                  ✓ notation (open intervals)
                                                                  CA
                                                                                            (3)

                                                                                             [12]

QUESTION 5

 5.1         PR2 = PQ2 + QR2 − 2(PQ)(QR) cos Q̂                   ✓ cosine rule                A
             PR2 = 4,22 + 5,62 − 2(4,2)(5,6) cos 105◦             ✓ SF                         A
             PR2 = 61,175 . . .
                                                                  ✓ answer                   CA
             PR = 7,82 m
                                                                                             (3)

             sin QP̂R sin 105◦
 5.2                  =                                           ✓ sine rule                CA
                5,6       7,82
                                                                  ✓ SF                       CA
                        5,6 sin 105◦
             sin QP̂R =              = 0,6916 . . .               ✓ answer                   CA
                            7,82                                                             (3)
             QP̂R = 43,76◦

                    1
 5.3         Area = (4,2)(5,6) sin 105◦                           ✓ SF                        A
                    2
                                                                  ✓ answer                   CA
             = 11,36 m2
                                                                                             (2)

                                    QS
 5.4         In △PQS: sin QP̂S =                                  ✓ trig ratio / area method CA
                                    PQ
                                                                  ✓ SF                       CA
             QS = 4,2 sin 43,76◦
                                                                  ✓ answer                   CA
             QS = 2,90 m                                                                     (3)
                  1                         2(11,36)
             OR     (PR)(QS) = 11,36 ⇒ QS =          = 2,90
                  2                           7,82
             m

                                                                                             [11]

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### Page 7

Technical Mathematics/P2                        7                             Hlayiso/Mock 2026
                                      NSC – Marking Guidelines

QUESTION 6

 6.1         …the interior opposite angle.                       ✓ answer                    A
                                                                                            (1)

                     1
 6.2.1       AD̂C =    × 124◦ = 62◦                              ✓S                          A
                     2
             (∠ at centre = 2 × ∠ at circumference)              ✓R                          A
                                                                                            (2)

 6.2.2       AB̂C = 180◦ − 62◦ = 118◦                            ✓S                        CA
             (opp ∠s of cyclic quad)                             ✓R                         A
                                                                                           (2)

 6.2.3       OA = OC (radii)                                     ✓ S/R radii / isosceles     A
             OÂC = OĈA (∠s opp equal sides)                    ✓ R sum of ∠s in △          A
                    180◦ − 124◦                                  ✓ answer                    A
             OÂC =             = 28◦ (sum of ∠s in △)                                      (3)
                         2

 6.2.4       BĈE = DÂB = 95◦                                   ✓S                          A
             (ext ∠ of cyclic quad)                              ✓R                          A
                                                                                            (2)

                             180◦ − 62◦
 6.2.5       DÂC = DĈA =              = 59◦                    ✓ S/R DÂC = 59◦          CA
                                  2
             (∠s opp equal sides; sum of ∠s in △)                ✓ answer                  CA
                                                                                           (2)
             BÂC = 95◦ − 59◦ = 36◦
             OR AD = DC ⇒ AĈD = DÂC (equal chords subtend
             equal ∠s), then same working. Accept any valid
             method.

                                                                                           [12]

QUESTION 7

 7.1         …radius (diameter) …                                ✓ answer                    A
                                                                                            (1)

 7.2.1       AB̂C = CÂT = 32◦                                   ✓S                          A
             (tan-chord theorem)                                 ✓R                          A
                                                                                            (2)

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### Page 8

Technical Mathematics/P2                       8                            Hlayiso/Mock 2026
                                     NSC – Marking Guidelines

 7.2.2       BÂC = 90◦                                          ✓S                          A
             (∠ in semicircle)                                   ✓R                          A
                                                                                            (2)

 7.2.3       OÂT = 90◦ (tangent ⊥ radius)                       ✓ S/R OÂT = 90◦            A
             AÔT = 2 × AB̂C = 64◦ (∠ at centre = 2 × ∠ at       ✓ S/R AÔT = 64◦           CA
             circumference)                                      ✓ answer                   CA
             T̂ = 180◦ − 90◦ − 64◦ = 26◦                                                    (3)
             OR In △TAB: T ÂB = 32◦ + 90◦ = 122◦ ;
             T̂ = 180◦ − 122◦ − 32◦ = 26◦

 7.2.4       AĈB = 180◦ − 90◦ − 32◦ = 58◦ (sum of ∠s in △)      ✓S                         CA
             OR AD̂B = 90◦ − 32◦ = 58◦ (CA)                      ✓R                          A
             AD̂B = AĈB = 58◦ (∠s in same segment)                                         (2)

 7.3         In △TAC and △T BA:
             T̂ = T̂ (common)                                    ✓S                          A
             T ÂC = T B̂A (= 32◦ ) (tan-chord theorem)          ✓ S/R                       A
             T ĈA = T ÂB (sum of ∠s in △)                      ✓ S/R                       A
                                                                 ✓ conclusion with reason    A
             ∴ △TAC ||| △T BA (∠∠∠)
                                                                                            (4)

 7.4.1       OM̂K = 90◦ and ON̂K = 90◦ (tangent ⊥ radius)        ✓ S/R                       A
             OM̂K + ON̂K = 180◦                                  ✓ sum = 180◦                A
             ∴ KMON is a cyclic quad (opp ∠s supplementary /     ✓R                          A
             converse opp ∠s of cyclic quad)                                                (3)

 7.4.2       M ÔN = 180◦ − 64◦ = 116◦ (opp ∠s of cyclic quad    ✓ S/R M ÔN = 116◦          A
             KMON)
             Reflex M ÔN = 360◦ − 116◦ = 244◦                   ✓R                          A
                      1                                          ✓ answer                   CA
             M P̂N = (244◦ ) = 122◦ (∠ at centre = 2 × ∠ at
                      2                                                                     (3)
             circumference)
             OR KM = KN (tangents from same point),
             K M̂N = 58◦ ; angle in the major segment = 58◦
             (tan-chord); M P̂N = 180◦ − 58◦ = 122◦ (opp ∠s of
             cyclic quad)

                                                                                          [20]

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### Page 9

Technical Mathematics/P2                       9                            Hlayiso/Mock 2026
                                     NSC – Marking Guidelines

QUESTION 8

 8.1         …bisects the chord.                                ✓ answer                     A
                                                                                            (1)

                     1
 8.2.1       AF = AB = 24 cm                                    ✓S                           A
                     2
             (line from centre ⊥ to chord)                      ✓R                           A
                                                                                            (2)

 8.2.2       OE = 22 − x                                        ✓ answer                     A
                                                                                            (1)

 8.2.3       OA2 = 242 + x2 (Pythagoras)                        ✓ OA2                        A
             CE = 20 cm (line from centre ⊥ to chord)
             OC2 = 202 + (22 − x)2                              ✓ OC2                      CA
             OA = OC (radii): 576 + x2 = 400 + 484 − 44x + x2   ✓ equating (radii)          A
             44x = 308
                                                                ✓ simplification to
             x=7                                                44x = 308                  CA
                                                                                           (4)

 8.2.4       r2 = 242 + 72 = 625                                ✓ substitution             CA
             r = 25 cm                                          ✓ answer                   CA
                                                                                           (2)

                                                                                           [10]

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### Page 10

Technical Mathematics/P2                         10                             Hlayiso/Mock 2026
                                       NSC – Marking Guidelines

QUESTION 9

                           72 × 1 000
 9.1.1       72 km/h =                m/s = 20 m/s                  ✓ conversion               A
                             3 600
                                                                    ✓ answer                   A
                                                                                              (2)

 9.1.2       v = π Dn                                               ✓ formula                 A
             20 = π (0,64)n                                         ✓ SF                     CA
             n = 9,95 r/s                                           ✓ answer                 CA
                                                                                             (3)

 9.1.3       ω = 2π n = 2π (9,947 . . .)                            ✓ SF                     CA
             ω = 62,50 rad/s                                        ✓ answer                 CA
                       v     20                                                              (2)
             OR ω = =            = 62,5 rad/s
                       r 0,32

 9.1.4       5 minutes = 300 s                                      ✓ ×300                   CA
             Revolutions = 9,947 . . . × 300 = 2 984,2 . . .        ✓ answer                 CA
             2 984 complete revolutions (accept 2 985 if n = 9,95                            (2)
             was used; CA from candidate’s n in 9.1.2, NPR)

                     r2 θ
 9.2.1       Area =                                                 ✓ formula                 A
                      2
                                                                    ✓ SF                      A
                    32 θ
             6,75 =                                                 ✓ answer                 CA
                      2                                                                      (3)
             θ = 1,5 rad

                     180◦                                                 ◦
 9.2.2       1,5 ×        = 85,94◦                                  ✓ × 180
                                                                         π                   CA
                      π
                                                                    ✓ answer                 CA
                                                                                             (2)

 9.2.3       s = rθ = 3 × 1,5                                       ✓ SF                     CA
                                       rs       3s                  ✓ answer                 CA
             s = 4,5 m OR        Area = : 6,75 = , s = 4,5 m
                                       2        2                                            (2)

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### Page 11

Technical Mathematics/P2                       11                           Hlayiso/Mock 2026
                                     NSC – Marking Guidelines

 9.3         4h2 − 4dh + x2 = 0                                 ✓ SF                        A
             4h2 − 4(2)h + (1,6)2 = 0
             4h2 − 8h + 2,56 = 0 (h2 − 2h + 0,64 = 0)           ✓ standard form         CA
                                                                ✓ factors / formula     CA
             (h − 0,4)(h − 1,6) = 0
                                                                ✓ h = 0,4 or h = 1,6    CA
             h = 0,4 m (minor segment) or h = 1,6 m (major
                                                                ✓ selecting h = 1,6 m (major
             segment)
                                                                segment)                CA
             Height of the tunnel h = 1,6 m                                              (5)
             OR radius = 1 m ✓A; half-chord√   = 0,8 m ✓A;
             distance from centre to KL = 1 − 0,82 ✓A = 0,6 m
                                             2
             ✓CA;
             h = 1 + 0,6 = 1,6 m ✓CA

 9.4.1       V = l × b × h = 1,2 × 0,8 × 0,75                   ✓ SF                       A
             V = 0,72 m3                                        ✓ 0,72 m3                 CA
             Capacity = 0,72 × 1 000 = 720 litres               ✓ answer in litres        CA
                                                                                          (3)

                      [                                     ]
 9.4.2       TSA = 2 (1,2)(0,8) + (1,2)(0,75) + (0,8)(0,75)     ✓ SF                       A
             = 2(0,96 + 0,9 + 0,6) = 4,92 m2                    ✓ 4,92 m2                 CA
             For 6 tanks: 6 × 4,92 = 29,52 m2                   ✓ ×6                      CA
                                                                ✓ ×R185,50                CA
             Cost = 29,52 × R185,50
                                                                ✓ answer                  CA
             = R5 475,96                                                                  (5)

                                                                                          [29]

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### Page 12

Technical Mathematics/P2                        12                                 Hlayiso/Mock 2026
                                      NSC – Marking Guidelines

QUESTION 10

                     36
 10.1        a=         =6m                                           ✓ answer                    A
                      6
                                                                                                 (1)

                                            8+y           y+9
 10.2        m1 = 6,75; m2 = 7,75; m3 =           ; m4 =       ;      ✓ numerical mid-ordinates A
                                              2            2
             m5 = 8,75; m6 = 7,75                                     ✓ m3 and m4 in terms of y A
             AT = a(m1 + m2 + . . . + m6 )
                     (                                            )   ✓ SF = 294                CA
                                      8+y y+9
             294 = 6 6,75 + 7,75 +         +        + 8,75 + 7,75
                                        2       2
             294 = 6(39,5 + y)
                                                                      ✓ answer                  CA
             49 = 39,5 + y                                                                      (4)
             y = 9,5 m     (                               )
                             6+7
             OR 294 = 6             + 7,5 + 8 + y + 9 + 8,5 =
                               2
             6(39,5 + y), y = 9,5 m

                                                                                                 [5]

                                                                                       TOTAL: 150

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- [TECH MATHS G11 LB m01](https://www.hlayiso.com/library/tech-maths-g11-lb-m01/4c7c2a64-b0c2-4a60-93b7-19462c110221) — Textbook
- [TECH MATHS G11 LB m05 AFR2](https://www.hlayiso.com/library/tech-maths-g11-lb-m05-afr2/48842861-f8f3-4f31-8902-ea599124a6d0) — Textbook
- [TECH MATHS P1 QP GR11 NOV2023 English](https://www.hlayiso.com/library/tech-maths-p1-qp-gr11-nov2023-english/23847445-90bc-4112-a3f4-acf5c0c9f3e3) — QP · 2023 · Eng

## Collection

- [Grade 11 past papers](https://www.hlayiso.com/grade-11/past-papers)

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