Technical Mathematics Grade 11 P1 November Mock 2026 Memo

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. Copyright reserved Please turn over

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 1 1.1.1 (3a2 b)0 + 8 3 = 1+2 ✓ 1 and 2 A =3 ✓ answer CA (2) 3 1 x2 · x2 √ = x 2 + 2 −(−1) 3 1 1 1.1.2 −1 ✓ x = x 2 / adding x exponents A = x3 ✓ answer CA (2) ( √ )2 ( √ )2 ( √ ) 1.1.3 2 5 − 1 = 2 5 − 2 2 5 (1) + 1 √ ✓ 20 √ A = 20 − 4 5 + 1 √ ✓ −4 5 A = 21 − 4 5 ✓ answer CA (3) √ √ √ 1.1.4 25 × 2 + 9 × 2 − 4 × 2 √ √ √ √ √ = 5 2+3 2−2 2 ✓ 5√2 and 3 2 A √ =6 2 ✓2 2 A ✓ answer CA (3) Copyright reserved Please turn over

Technical Mathematics/P1 3 Hlayiso/Mock 2026 NSC – Marking Guidelines log3 34 − log2 2−3 1.1.5 log 103 + log5 1 ✓4 A 4 − (−3) ✓ −3 A = 3+0 ✓3 A 7 = OR 2 13 ✓0 A 3 ✓ answer CA (5) 2x · 22 − 2x 1.2 ✓ exponent laws A 3 · 2x · 2−1 2x (4 − 1) ✓ common factor A = 3 x ✓ simplification CA 2 ·2 ✓ answer CA 3 · 2x (4) = 3 x 2 ·2 =2 32x · 2x+1 · 3x+1 1.3 ✓ 32x and 33x A 33x · 2x ✓ 6x+1 = 2x+1 · 3x+1 A = 2x+1−x · 32x+x+1−3x ✓ adding/subtracting = 21 · 31 exponents CA =6 ✓ answer CA (4) 1.4.1 X = 32 + 16 + 4 + 1 = 53 ✓ 53 A Y = 16 + 8 + 4 + 1 = 29 ✓ 29 A (2) 1.4.2 53 − 29 = 24 = 110002 ✓ answer CA (1) [26] Copyright reserved Please turn over

Technical Mathematics/P1 4 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 2 2.1.1 log 6 = log(2 × 3) ✓ product law A = log 2 + log 3 = a + b ✓ answer CA (2) 3 2.1.2 log 1,5 = log ✓ 23 A 2 = log 3 − log 2 ✓ quotient law A ✓ answer CA = b−a (3) ( ) 2.1.3 log 360 = log 22 × 32 × 10 ✓ factors A = 2 log 2 + 2 log 3 + log 10 ✓ product law CA = 2a + 2b + 1 ✓ power law CA ✓ answer (log 10 = 1) CA (4) 2.2.1 2x+1 = 2−4 ✓ 2−4 A x + 1 = −4 ✓ equating exponents CA x = −5 ✓ answer CA (3) 2.2.2 5x · 5 + 5x = 150 5x (5 + 1) = 150 ✓ common factor A 5x = 25 = 52 ✓ 5x = 52 CA x=2 ✓ answer CA (3) 2.2.3 2x − 1 = 32 ✓ exponential form A 2x = 10 ✓ 2x = 10 CA x=5 ✓ answer CA (3) [ ] 2.2.4 log x(x − 3) = 1 ✓ product law A x2 − 3x = 101 ✓ exponential form CA x2 − 3x − 10 = 0 ✓ standard form CA ✓ factors CA (x − 5)(x + 2) = 0 ✓ answer with rejection CA x = 5 or x ̸= −2 (log of a negative number is (5) undefined) Copyright reserved Please turn over

Technical Mathematics/P1 5 Hlayiso/Mock 2026 NSC – Marking Guidelines log 20 2.3 x = log3 20 = ✓ log form A log 3 ✓ answer CA x ≈ 2,73 (2) 2E 2.4.1 v2 = ✓ v2 = 2E m A √m ✓ answer CA 2E v= (2) m √ 2(1 250) √ 2.4.2 v= = 25 ✓ substitution CA 100 ✓ answer CA v = 5 m/s NPU (2) [29] QUESTION 3 3.1.1 2x − 3 = 0 or x + 4 = 0 x= 3 or x = −4 ✓ x = 23 A 2 ✓ x = −4 A (2) √ −(−1) ± (−1)2 − 4(3)(−5) 3.1.2 x= ✓ SF √ A 2(3) √ ✓ 61 CA 1 ± 61 x= 6 ✓ x ≈ 1,47 CA x ≈ 1,47 or x ≈ −1,14 ✓ x ≈ −1,14 CA (4) 3.1.3 (x + 5)(x − 3) ≤ 0 ✓ factors A Critical values: x = −5 and x = 3 ✓ critical values CA −5 3 x − ✓ correct notation CA −5 ≤ x ≤ 3 OR x ∈ [−5 ; 3] AO: full marks (3) Copyright reserved Please turn over

Technical Mathematics/P1 6 Hlayiso/Mock 2026 NSC – Marking Guidelines 3.1.4 × 2x: x2 − 8 = 2x ✓ multiplying by LCD A x2 − 2x − 8 = 0 ✓ standard form CA (x − 4)(x + 2) = 0 ✓ factors CA ✓ both answers CA x = 4 or x = −2 (4) 3.2 y = 2x + 1 ✓ y = 2x + 1 A x2 + (2x + 1)2 = 13 ✓ substitution CA x2 + 4x2 + 4x + 1 − 13 = 0 5x2 + 4x − 12 = 0 ✓ standard form CA ✓ factors CA (5x − 6)(x + 2) = 0 ✓ both x-values CA 6 x= or x = −2 ✓ both y-values CA 5 (6) 17 y= y = −3 5 3.3.1 A = w(w + 5) OR A = w2 + 5w ✓ answer A (1) 3.3.2 w2 + 5w = 84 ✓ equation CA w2 + 5w − 84 = 0 (w + 12)(w − 7) = 0 ✓ factors CA ✓ w = 7 with rejection CA w = 7 (w ̸= −12, a length cannot be negative) ✓ length = 12 cm CA Width = 7 cm and length = 12 cm (4) A 3.3.3 r2 = ✓ r2 = πA A √π ✓ answer CA A r= (2) π [26] Copyright reserved Please turn over

Technical Mathematics/P1 7 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 4 4.1 ∆ = b2 − 4ac = (−7)2 − 4(2)(3) ✓ SF A ∆ = 25 ✓ ∆ = 25 CA Roots are real, rational and unequal. ✓ nature CA (∆ > 0 and a perfect square) (3) 4.2.1 k = −3 ✓ answer A (1) 4.2.2 k−2 < 0 ✓k<2 A k < 2 (k ̸= −3) (1) 4.2.3 k−2 = 0 ✓ answer A k=2 (1) 4.3 ∆ = 0: p2 − 4(1)(9) = 0 ✓ ∆ = 0 with substitution A p2 = 36 p = ±6 ✓ answer CA (2) [8] QUESTION 5 5.1 x=0 and y = −2 ✓x=0 A ✓ y = −2 A (2) 4 5.2 0= −2 ✓ f (x) = 0 A x 2x = 4 ✓ answer CA x=2 (2 ; 0) (2) √ 5.3 0 = 16 − x2 ✓ g(x) = 0 A x2 = 16 x = ±4 (−4 ; 0) and (4 ; 0) ✓ answer CA (2) Copyright reserved Please turn over

Technical Mathematics/P1 8 Hlayiso/Mock 2026 NSC – Marking Guidelines 5.4 y = 4 OR (0 ; 4) ✓ answer A (1) 5.5 y f: ✓ asymptotes CA ✓ x-intercept (2 ; 0) CA 4 ✓ shape (both branches) A g g: ✓ x-intercepts CA f x ✓ y-intercept CA −4 2 4 ✓ semicircle above x-axis A y = −2 (6) 5.6 −4 ≤ x ≤ 4 OR x ∈ [−4 ; 4] ✓ answer CA (1) 5.7 0 ≤ y ≤ 4 OR y ∈ [0 ; 4] ✓ answer CA (1) 5.8 y ∈ R , y ̸= −2 ✓ answer CA (1) 5.9 0<x<2 OR x ∈ (0 ; 2) ✓ critical values CA ✓ notation CA (2) 4 5.10 −2 = 2 ✓ equation A x 4 =4 ✓ answer CA x x=1 (2) [20] Copyright reserved Please turn over

Technical Mathematics/P1 9 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 6 6.1 y = −9 ✓ answer A (1) 6.2 −6 = b1 − 9 ✓ substitution of P A b=3 ✓ answer CA (2) 6.3 y = 30 − 9 = 1 − 9 = −8 ✓ substitution x = 0 CA C(0 ; −8) ✓ answer CA (2) 6.4 0 = 3x − 9 ✓ f (x) = 0 CA 3x = 9 = 32 x=2 E(2 ; 0) ✓ answer CA (2) 6.5 y > −9 OR y ∈ (−9 ; ∞) ✓ answer CA (1) 6.6 g(x) = − (3x − 9) ✓ − f (x) CA g(x) = −3x + 9 ✓ answer CA (2) 6.7 x<2 OR x ∈ (−∞ ; 2) ✓ answer CA (1) [11] Copyright reserved Please turn over

Technical Mathematics/P1 10 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 7 7.1 −x2 + 2x + 8 = 0 ✓ f (x) = 0 A x2 − 2x − 8 = 0 (x − 4)(x + 2) = 0 ✓ factors CA A(−2 ; 0) and D(4 ; 0) ✓ both coordinates CA (3) b 2 7.2 x=− =− =1 ✓x=1 A 2a 2(−1) ✓ substitution CA y = −(1)2 + 2(1) + 8 = 9 ✓y=9 CA C(1 ; 9) (3) 7.3 −x2 + 2x + 8 = x + 2 ✓ equating f = g A x2 − x − 6 = 0 (x − 3)(x + 2) = 0 ✓x=3 CA x = 3 (x = −2 at A) ✓y=5 CA y = 3+2 = 5 B(3 ; 5) (3) 7.4 −2 ≤ x ≤ 3 OR x ∈ [−2 ; 3] ✓ critical values CA ✓ notation CA (2) 7.5 y ≤ 9 OR y ∈ (−∞ ; 9] ✓ answer CA (1) 7.6 (1 ; 6) ✓ answer CA (1) [13] Copyright reserved Please turn over

Technical Mathematics/P1 11 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 8 8.1 A = P(1 + ni) ✓ formula A ( ) A = 12 000 1 + 4 × 0,075 ✓ SF A A = R15 600 ✓ answer CA (3) ( ) 0,09 4 8.2 ieff = 1 + −1 ✓ 0,09 4 A 4 ✓ SF A = 0,093083 . . . ieff = 9,31% ✓ answer CA (3) 8.3 A = P(1 − i)n ✓ formula A 210 000 = P(1 − 0,12)5 ✓ SF A 210 000 P= ✓ answer CA (0,88)5 P = R397 929,31 (3) ( ) 0,08 24 8.4.1 A = 40 000 1 + ✓ 0,08 12 A 12 ✓ n = 24 & SF A A = R46 915,52 ✓ answer CA (3) 8.4.2 Amount after withdrawal: ✓ subtracting R10 000 CA 46 915,52 − 10 000 = R36 915,52 ✓ 0,09 A ( ) 2 0,09 6 ✓n=6 A A = 36 915,52 1 + ✓ SF CA 2 A = R48 073,61 ✓ answer CA (5) OR A= ( )24 ( )6 ( )6 40 000 1 + 0,08 12 1 + 0,09 2 − 10 000 1 + 0,09 2 A = R48 073,61 [17] TOTAL: 150 Copyright reserved Please turn over