Technical Mathematics Grade 11 P1 2026 Memo (Eng)

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. 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 ( )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 Copyright reserved Please turn over

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] Copyright reserved Please turn over

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) Copyright reserved Please turn over

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) Copyright reserved Please turn over

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] Copyright reserved Please turn over

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) Copyright reserved Please turn over

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] Copyright reserved Please turn over

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] Copyright reserved Please turn over

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] Copyright reserved Please turn over

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 Copyright reserved