Technical Mathematics Grade 12 P1 2026 Memo (Eng)

hlayiso mock exams Practice Series: Grade 12 NSC Preparation WWW.HLAYISO.COM NATIONAL SENIOR CERTIFICATE GRADE 12 TECHNICAL MATHEMATICS P1 PREDICTION EXAMINATION 2026 MARKING GUIDELINES MARKS: 150 These marking guidelines consist of 13 pages. Copyright reserved Please turn over

Technical Mathematics/P1 2 Hlayiso/Prediction 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. • If a candidate has crossed out an attempt of a question and not redone the question, mark the crossed-out version. • Consistent accuracy (CA) applies in ALL aspects of the marking guidelines 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 (5 − 2x)(x + 3) = 0 ✓ x = 25 A x= 5 or x = −3 ✓ x = −3 A 2 (2) 1.1.2 3x2 − 2x − 7 =p 0 ✓ standard form A −(−2) ± (−2)2 − 4(3)(−7) ✓ SF CA x= √ 2(3) 2 ± 88 ✓ x ≈ 1,90 CA x= ✓ x ≈ −1,23 CA 6 x ≈ 1,90 or x ≈ −1,23 NPR (4) 1.1.3 x2 + x − 12 > 0 ✓ standard form & factors A (x + 4)(x − 3) > 0 ✓ both critical values CA Critical values: x = −4 and x = 3 x −4 3 ✓ correct notation CA x < −4 or x>3 OR x ∈ (−∞ ; −4) ∪ (3 ; ∞) (3) 1.2.1 y = 2x + 3 ✓ y = 2x + 3 A (1) Copyright reserved Please turn over

Technical Mathematics/P1 3 Hlayiso/Prediction 2026 NSC – Marking Guidelines 1.2.2 x2 + x(2x + 3) − (2x + 3)2 = −1 ✓ substitution CA x2 + 2x2 + 3x − 4x2 − 12x − 9 + 1 = 0 −x2 − 9x − 8 = 0 ✓ standard form CA x2 + 9x + 8 = 0 (x + 1)(x + 8) = 0 ✓ factors CA x = −1 or x = −8 ✓ both x-values CA y=1 y = −13 ✓ both y-values CA (5) 2Ek 1.3.1 v2 = ✓ v2 asqsubject A rm 2Ek 2Ek ✓v= m CA v= m AO: full marks (2) s 2(2,4 × 105 ) √ 1.3.2 v= = 320 ✓ substitution CA 1,5 × 103 ✓ answer CA v ≈ 17,89 m/s NPU NPR (2) 1.4 1101012 = 32 + 16 + 4 + 1 = 53 ✓ answer A (1) 1.5 53 + 27 = 80 ✓ 80 OR binary addition M 80 = 64 + 16 ✓ answer CA = 10100002 (2) [22] Copyright reserved Please turn over

Technical Mathematics/P1 4 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 2 2.1.1 2m + 6 = 0 m = −3 ✓ m = −3 A (1) 2.1.2 9 − 3m < 0 ✓ 9 − 3m < 0 A m>3 ✓m>3 CA (2) 2.2 2x2 − 8x + p + 4 = 0 ✓ standard form A Non-real roots: ∆ < 0 ✓∆<0 A (−8)2 − 4(2)(p + 4) < 0 ✓ SF CA 64 − 8p − 32 < 0 −8p < −32 p>4 ✓ p>4 CA (4) [7] Copyright reserved Please turn over

Technical Mathematics/P1 5 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 3
1−x 52x+1 · 52 3.1.1 ✓ prime bases A 53 =5 2x+1+2−2x−3 ✓ adding/subtracting = 50 = 1 exponents CA ✓ answer CA (3) √ √ √ 3.1.2 = √ 25 × 2√ − 9 ×√2 + 4 × 2 √ √ √ = 5√2 − 3 2 + 2 2 ✓ 5 2−3 2+2 2 A =4 2 ✓ answer CA (2) log2 25 + log 10−2 3.1.3 = log 103 ✓ log2 32 = 5 A 5 + (−2) = ✓ log 0,01 = −2 A 3 =1 ✓ answer CA (3)   3.2 log3 (x + 1)(x − 5) = 3 ✓ product law A (x + 1)(x − 5) = 33 ✓ exponential form CA x2 − 4x − 32 = 0 ✓ standard form & factors (x − 8)(x + 4) = 0 CA x = 8 (x ̸= −4, since x − 5 > 0) ✓ answer with rejection CA (4) 3.3.1 z = (2 − 4) + (−3 + 1)i = −2 − 2i ✓ answer A (1) 3.3.2 Im ✓ point (−2 ; −2) in quadrant 3 CA −2 Re ✓ line from origin & labels CA −2 (2) z(−2 ; −2) Copyright reserved Please turn over

Technical Mathematics/P1 6 Hlayiso/Prediction 2026 NSC – Marking Guidelines p √ √ 3.3.3 r= (−2)2 + (−2)2 = 8 = 2 2 ≈ 2,83 ✓r CA −2 ✓ tan θ / ref. angle CA tan θ = = 1; reference angle = 45◦ −2 ◦ ◦ ◦ ✓ θ = 225◦ CA θ = 180 √ + 45 ◦= 225 ✓ polar form CA z = 2 2 cis ◦ √225 OR ◦ 2,83 cis 225 (4) (Accept 2 2 cis(−135 )) 3.4 (3 − 2i)2 − 4i = 9 − 12i + 4i2 − 4i ✓ expansion A = 9 − 12i − 4 − 4i ✓ i2 = −1 A = 5 − 16i m = 5 and n = −16 ✓ both values CA (3) [22] QUESTION 4 4.1.1 −4 ≤ x ≤ 4 OR x ∈ [−4 ; 4] ✓ critical values A ✓ notation A (2) 4.1.2 x=0 and y = −2 ✓x=0 A ✓ y = −2 A (2) 6 4.1.3 0 = − −2 ✓ g(x) = 0 A x 2x = −6 x = −3 (−3 ; 0) ✓ answer CA (2) Copyright reserved Please turn over

Technical Mathematics/P1 7 Hlayiso/Prediction 2026 NSC – Marking Guidelines 4.1.4 y f: ✓ intercepts ±4 and 4 A g ✓ semicircle shape (above 4 f x-axis) A g: ✓ both asymptotes CA ✓ x-intercept −3 CA −4 −3 4 x ✓ shape (2nd & 4th quad.) A (5) y = −2 g 4.1.5 −3 < x < 0 OR x ∈ (−3 ; 0) ✓ critical values CA ✓ notation CA (2) 4.2.1 h(x) = a(x + 1)(x − 5) ✓ a(x + 1)(x − 5) A Substitute C(0 ; 5): 5 = a(1)(−5) ✓ substitution of C A a = −1 ✓ a = −1 & simplification A h(x) = −(x2 − 4x − 5) h(x) = −x2 + 4x + 5 OR OR c = 5: h(x) = ax2 + bx + 5 ✓c=5 A A: a − b + 5 = 0; B: 25a + 5b + 5 = 0 ✓ both equations A Solving: a = −1 and b = 4 ✓ a = −1, b = 4 A h(x) = −x2 + 4x + 5 (3) b 4 4.2.2 x=− =− =2 ✓x=2 CA 2a 2(−1) −1 + 5 OR x = = 2 OR h′ (x) = −2x + 4 = 0, x = 2 ✓y=9 CA 2 y = −(2)2 + 4(2) + 5 = 9 (2) D(2 ; 9) 4.2.3 k(2) = −2 + 5 = 3 E(2 ; 3) ✓ yE = 3 CA DE = 9 − 3 = 6 units ✓ answer CA (2) Copyright reserved Please turn over

Technical Mathematics/P1 8 Hlayiso/Prediction 2026 NSC – Marking Guidelines 4.2.4 0≤x≤5 OR x ∈ [0 ; 5] ✓ critical values A ✓ notation A (2) 4.3 y ✓ asymptote y = −4 A ✓ intercepts −3 and 2 A p ✓ increasing shape A (3) x 2 −3 y = −4 (p(x) = 2x − 4) [25] QUESTION 5   0,108 12 5.1 ieff = 1 + −1 ✓ 0,108 12 A 12 ✓ SF A = 0,113509 . . . ieff ≈ 11,35% ✓ answer CA (3) 5.2.1 A = P(1 − i)n ✓ SF A A = 185 000(1 − 0,18)3 A = R102 003,08 ✓ answer CA (2) 5.2.2 40 000 = 185 000(0,82)n ✓ SF A (0,82)n = 40 000 = 0,216216 . . . ✓ simplification CA 185 000 ✓ use of logs CA log 0,2162 n = log0,82 (0,216216 . . .) OR n = log 0,82 ✓ 8 years CA n ≈ 7,72 (After 7 years the value is The compressor will be replaced after 8 years. R46 117,81; after 8 years R37 816,61.) (4) Copyright reserved Please turn over

Technical Mathematics/P1 9 Hlayiso/Prediction 2026 NSC – Marking Guidelines   0,072 24 5.3 A2 = 50 000 1 + = R57 719,36 ✓ i = 0,072 12 ; n = 24 A 12   ✓i= 4 0,084 A 0,084 4 A3 = 57 719,36 1 + = R62 722,66 ✓ −8 000 with n = 4 (or 4   n = 12) CA 0,084 4 ✓ +10 000 with n = 8 CA A4 = (62 722,66 − 8 000) 1 + = R59 466,20 4   0,084 8 ✓ R82 031,28 CA A6 = (59 466,20 + 10 000) 1 + 4 A6 = R82 031,28 OR A6 = 50 000(1,006)24 (1,021)16 − 8 000(1,021)12 + ✓ conclusion CA 10 000(1,021)8 Accept answers in the range = R82 031,28 R82 031 – R82 032 (NPR for R82 031,28 > R80 000, so Lerato WILL have enough intermediate values, e.g. money to pay the deposit. R62 722,66 or R62 722,67) (6) [15] QUESTION 6 6.1 f (x + h) = 8 − 3(x + h) = 8 − 3x − 3h ✓ f (x + h) A f (x + h) − f (x) ✓ formula A f ′ (x) = lim h→0 h ✓ SF CA 8 − 3x − 3h − (8 − 3x) ✓ −3h CA = lim h h→0 h −3h ✓ answer CA = lim h→0 h Penalty of 1 mark for = lim (−3) h→0 incorrect notation (no lim). f ′ (x) = −3 AO: 0 marks (5) 6.2.1 y = 2x5 − 3x−1 + 7 ✓ −3x−1 A dy = 10x4 + 3x−2 OR 3 10x4 + 2 ✓ 10x4 A dx x ✓ +3x−2 (and 0 for 7) CA (3) x3 − 4x x2 2 6.2.2 = −2 ✓ x2 A  2x 2 2 ✓ −2 A x Dx −2 = x ✓ answer CA 2 (3) Copyright reserved Please turn over

Technical Mathematics/P1 10 Hlayiso/Prediction 2026 NSC – Marking Guidelines 2 6.2.3 f (x) = 4x 3 ✓ exponential form A 8 1 f ′ (x) = x− 3 OR 8 √ ✓ answer CA 3 33x (2) 6.3.1 g(−1) = −1 − 2 + 3 = 0 g(2) = 8 − 8 + 3 = 3 ✓ g(−1) and g(2) A g(2) − g(−1) 3 − 0 ✓ SF CA Average gradient = = 2 − (−1) 3 =1 ✓ answer CA (3) 6.3.2 g′ (x) = 3x2 − 4x ✓ g′ (x) A m = g′ (2) = 3(2)2 − 4(2) = 4 ✓m=4 CA Point: (2 ; 3) y − 3 = 4(x − 2) ✓ substitution of m and y = 4x − 5 (2 ; 3) CA OR 3 = 4(2) + c ✓ equation CA c = −5 (4) y = 4x − 5 [20] Copyright reserved Please turn over

Technical Mathematics/P1 11 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 7 7.1 f (4) = 0 ✓ f (4) = 0 A (4)3 − 12(4) + k = 0 ✓ substitution & 64 − 48 + k = 0 simplification A k = −16 (2) 7.2 y = −16 OR (0 ; −16) ✓ answer A (1) 7.3 f (x) = (x − 4)(x2 + 4x + 4) ✓ quadratic factor CA (by inspection, OR synthetic/long division by x − 4: ✓ (x + 2)2 CA coefficients 1 0 −12 −16 → 1 4 4, remainder 0) = (x − 4)(x + 2)2 ✓ both x-intercepts CA x = 4 or x = −2 (3) (4 ; 0) and (−2 ; 0) 7.4 f ′ (x) = 3x2 − 12 = 0 ✓ f ′ (x) A x2 = 4 ✓ f ′ (x) = 0 CA x = 2 or x = −2 ✓ both x-values CA f (2) = 8 − 24 − 16 = −32 f (−2) = −8 + 24 − 16 = 0 ✓ y = −32 CA Turning points: (2 ; −32) and (−2 ; 0) ✓y=0 CA (5) 7.5 y ✓ x-intercepts CA f ✓ y-intercept CA ✓ turning points CA (−2 ; 0) ✓ shape A x (4) 4 −16 (2 ; −32) 7.6 −2 < x < 2 OR x ∈ (−2 ; 2) ✓ critical values CA ✓ notation CA (2) [17] Copyright reserved Please turn over

Technical Mathematics/P1 12 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 8 8.1 2(x · x + x · h + x · h) = 600 ✓ SF in surface area = 600 A 2x2 + 4xh = 600 4xh = 600 − 2x2 ✓ h as subject A 600 − 2x2 300 − x2 (2) h= = 4x 2x   300 − x2 8.2 V = x·x·h = x 2 ✓ substitution of h A 2x ✓ simplification A x(300 − x ) 2 = (2) 2 1 V (x) = 150x − x3 2 3 8.3 V ′ (x) = 150 − x2 = 0 ✓ V ′ (x) A 2 x2 = 100 ✓ V ′ (x) = 0 CA x = 10 (x > 0) ✓ x = 10 CA 300 − (10)2 h= = 10 ✓ h = 10 CA 2(10) Dimensions: 10 cm × 10 cm × 10 cm (4) (Maximum volume = 1 000 cm3 ; not required) [8] QUESTION 9 3x 9.1.1 = 6 ln x − +C ✓ 6 ln x A ln 3 x ✓ − ln3 3 A ✓ +C A (3) Z   3 3 9.1.2 x − 4x dx 2 ✓ x 2 − 4x A 5 2 5 ✓ 52 x 2 CA = x 2 − 2x2 +C ✓ −2x2 +C CA 5 (3) 9.2.1 x2 − 4x = 0 ✓ factors A x(x − 4) = 0 x = 0 and x = 4 ✓ both values CA (2) Copyright reserved Please turn over

Technical Mathematics/P1 13 Hlayiso/Prediction 2026 NSC – Marking Guidelines Z 6  3 6
x 9.2.2 x − 4x dx = 2 − 2x 2 ✓ integration A 0 3 0 = (72 − 72) − 0 = 0 ✓ = 0 shown CA (2) 9.2.3 Region A lies BELOW the x-axis, so its integral is ✓ reason (A below x-axis) A negative and cancels the integral of B. Area cannot be (1) negative, so Thabo’s claim is NOT valid. Z 4  
64 32 9.2.4 x − 4x dx = 2 − 32 − 0 = − 3 3 0 ✓ Area A = 32 3 CA 32 32 Area A = − = 3 3 ✓ Area B = 32 Z 6
3 CA Area B = x2 − 4x dx ✓ total CA 4 (3)
32 = (72 − 72) − 64 3 − 32 = 3 32 32 64 Total area = + = ≈ 21,33 square units 3 3 3 [14] TOTAL: 150 Copyright reserved