MATHEMATICS PAPER 1 MEMORANDUM GRADE 12 - 2018 SEPTEMBER PREPARATORY EXAM PAPERS AND MEMOS

Monday, 06 September 2021 07:30

MATHEMATICS PAPER 1 MEMORANDUM GRADE 12 - 2018 SEPTEMBER PREPARATORY EXAM PAPERS AND MEMOS

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MATHEMATICS PAPER 1
NATIONAL SENIOR CERTIFICATE
GRADE 12
MEMORANDUM
SEPTEMBER 2018

NOTE:

If a candidate answers a question TWICE, mark the FIRST attempt ONLY.
Consistent accuracy applies in ALL aspects of the marking guideline.
If a candidate crossed out an attempt of a question and did not redo the question, mark the crossed-out attempt.
The mark for substitution is awarded for substitution into the correct formula.

QUESTION 1

1.1.1	½ 𝑥² − 𝑥 − 4 = 0 𝑥2 − 2𝑥 − 8 = 0 (𝑥 − 4)(𝑥 + 2) = 0 𝑥 = 4 or/of 𝑥 = −2	✓✓ factors ✓𝑥-values (3)
1.1.2	-3(x² + 3x) + 7 = 0 −3𝑥² − 9𝑥 + 7 = 0 𝑥 = −𝑏 ± √𝑏² − 4𝑎𝑐 2𝑎 𝑥 = −(−9) ± √(−9)² − 4(1)(7) 2(−3) 𝑥 = 0,64 or/of 𝑥 = −3,64	✓ standard form ✓ substitution ✓✓ 𝑥-values (4)
1.1.3	2𝑥² − 3𝑥 < 0 𝑥(2𝑥 − 3) < 0 0 < 𝑥 <³/₂	✓ 𝑥(2𝑥 − 3) ✓ critical values/kritiese waardes ✓ ✓ 0 < 𝑥 <³/₂ (4)
1.2	𝑥−2𝑦=3 …………………..(1) 4𝑥²−3=−6𝑦+5𝑥𝑦……….(2) from (1) 𝑥=2𝑦+3 sub into (2) 2(2𝑦+3)²−3=−6𝑦+5𝑦(2𝑦+3) 4(4𝑦²+12𝑦+9)−3=−6𝑦+10𝑦²+15𝑦 16𝑦²+48𝑦+36−3=10𝑦²+9𝑦 6𝑦2+39𝑦+33=0 2𝑦²+13𝑦+11=0 (𝑦+1)(2𝑦+11)=0 𝑦=−1 or 𝑦=−¹¹/₂ 𝑥=1 or 𝑥=−8	✓ 𝑥=2𝑦+3 ✓ substitution ✓ standard form ✓ factors ✓ y-values s ✓ 𝑥-values (6)
1.3	2𝑥²−(𝑘−1)𝑥+𝑘−3=0 Δ =𝑏²−4𝑎𝑐 Δ =[−(𝑘−1)]²−4(2)(𝑘−3) Δ = 𝑘²−2𝑘+1−8𝑘+24 Δ = 𝑘²−10𝑘+25 Δ = (𝑘−5)² (𝑘−5)2≥0 ∴Roots are Real	✓ Δ =[−(𝑘−1)]²−4(2)(𝑘−3) ✓ Δ = 𝑘²−2𝑘+1−8𝑘+24 ✓ Δ = 𝑘²−10𝑘+25 ✓ Δ = (𝑘−5)² ✓ (𝑘−5)2≥0 (5)
1.4.1	3^2𝑚= 3𝑝 3−𝑝 3^2𝑚=3(1,5) 3−1,5 3^2𝑚=3 ∴2𝑚=1 ∴ 𝑚=½	✓ sub. of 𝑝=1,5 ✓ answer (2)
1.4.2	32𝑚= 3𝑝 3−𝑝 3²⁽⁰⁾= 3𝑝 3−𝑝 1= 3𝑝 3−𝑝 3𝑝=3−𝑝 4𝑝=3 ∴ 𝑝=³/₄	✓ sub. of 𝑚=0 ✓ answer (2) [26]

QUESTION 2

2.1	2𝑥+2= 𝑥−1 7𝑥+1 2𝑥+2 (2𝑥+2)²=(7𝑥+1)(𝑥−1) 4𝑥2+8𝑥+4=7𝑥²−6𝑥−1 3𝑥²−14𝑥−5=0 (3𝑥+1)(𝑥−5)=0 𝑥=−¹/₃ or 𝑥=5	✓ 𝑇2 = 𝑇3 𝑇1 𝑇2 ✓ standard form ✓ factors ✓ 𝑥=−¹/₃ ✓ 𝑥=5 (5)
2.2	25 ;20 ;16;…. 𝑎=25 ; 𝑟 =⁴/₅ 𝑆∞= 𝑎 1−𝑟 𝑆∞= 25 1−⁴/₅ 𝑆∞=125𝑚	✓ 𝑟=45 ✓ 𝑆∞=𝑎1−𝑟 ✓ substitution ✓𝑆∞=125 𝑚 (4)
2.3	𝑎=2 𝑙=29 𝑆𝑛=155 𝑆𝑛=^𝑛/₂(𝑎+𝑙) 155=^𝑛/₂(2+29) 𝑛=10 29=2+(10−1)𝑑 9𝑑=27 𝑑=3	✓ sum formula of AS ✓sub. 𝑎 and 𝑙 ✓𝑛=10 ✓ sub.n 𝑛=10 ✓ 𝑑=3 (5)

[14]

QUESTION 3

2𝑎=4 𝑎=2
24=2+𝑏+𝑐 ............(1)
24=50+5𝑏+𝑐 ........(2)
4𝑏=−48 (2) – (1)
𝑏=−12
24=2−12+𝑐
𝑐=34
𝑇𝑛=2𝑛²−12𝑛+34
See alternative answers

✓ 2𝑎=4
✓ 𝑎=2
✓sub. into 𝑇1
✓sub. into 𝑇5
✓method of solving
✓ 𝑏=−12
✓ sub.of 𝑏
✓ 𝑐=34
(8)

[8]

QUESTION 4

4.1	36=𝑘² ∴𝑘=6	✓ sub of point ✓ 𝑘=6 (2)
4.2	𝑦=6𝑥 𝑥=6𝑦 𝑦=log6𝑥	✓ swop of 𝑥 and 𝑦. ✓ 𝑦=log₆𝑥 (2)
4.3	0<𝑥≤1	✓✓ answer (2)
4.4	𝑦>2	✓✓ answer (2)

[8]

QUESTION 5

5.1	𝐵(5;0)	✓✓ answer (2)
5.2	𝑦=𝑎(𝑥−𝑝)²+𝑞 𝑦=𝑎(𝑥−³/₂)²+⁴⁹/₄ 0=𝑎(5−³/₂)²+⁴⁹/₄ 𝑎=−1 𝑦=−1(𝑥−³/₂)²+⁴⁹/₄ 𝑦=−1(𝑥²−3𝑥+⁹/₄)+⁴⁹/₄ 𝑦=−𝑥²+3𝑥+10 OR 𝑦=𝑎(𝑥−5)(𝑥+2) Inspection ⁴⁹/₄=𝑎(³/₂−5)(³/₂+2) 𝑎=−1 𝑦=−1(𝑥−5)(𝑥+2) 𝑦=−𝑥²+3𝑥+10	✓ sub of turning point ✓ sub of point B ✓ 𝑎=−1 ✓𝑦=−𝑥²+3𝑥+10 (4) ✓ sub of 𝑥-intercepts ✓sub of turning point ✓ 𝑎=−1 ✓𝑦=−𝑥²+3𝑥+10 (4)
5.3	−𝑥²+3𝑥+10=−𝑥+5 𝑥²−4𝑥−5=0 (𝑥−5)(𝑥+1)=0 𝑥=5 or 𝑥=−1 𝑆(−1;6)	✓𝑓(𝑥)=𝑔(𝑥) ✓ standard form ✓ factors ✓𝑆(−1;6) (4)
5.4.1	−1≤𝑥≤5	✓✓ answer (2)
5.4.2	−𝑥²+3𝑥−2,25<0 −𝑥²+3𝑥+10<2,25+10 ∴ 𝑓(𝑥)<12,25 𝑥∈𝑅 ;𝑥≠1,5	✓ 𝑓(𝑥)<12,25 ✓✓ 𝑥∈𝑅 ;𝑥≠1,5 accuracy (3)

[15]

QUESTION 6

6.1		✓ asymptote ✓ 𝑥-intercept ✓ shape ✓ other point (4)
6.2	𝑓(𝑥)=2𝑥−1−1 𝑓′(𝑥)=−2𝑥⁻²	✓ 𝑓′(𝑥)=−2𝑥⁻²−1 ✓ 𝑓′(𝑥)=−2𝑥⁻² (2)
6.3	ℎ(𝑥)=−𝑥−1	✓✓ answer (2)
6.4	𝑓′(𝑥)= −2 and ℎ(𝑥)=−𝑥−1 𝑥² -2 =−1 𝑥² 𝑥²=2 𝑥=√2 ; 𝑥>0 𝑓(√2)= 2 −1 √2 Equation of tangent: 𝑦=−𝑥+𝑐 2 −1=−√2+𝑐 √2 𝑐=−1+2√2 ∴𝑘=2√2	✓ setting up of equation ✓ 𝑥=√2 ✓ ²/_√2−1 ✓sub of (√2; ²/_√2−1) ✓ 𝑘=2√2 (5)

[13]

QUESTION 7

7.1	2000(1+ 8 )¹²=2000(1+ 𝑟 )² 1200 200 √(1+ 8 )¹²=(1+ 𝑟 ) 1200 200 𝑟=8,13%	✓ 8 and 𝑟 1200 200 ✓ 𝑛=12 and 𝑛=2 ✓ 𝑟=8,13% (3)
7.2	𝐴=𝑃(1−𝑖)^𝑛 4 500=9 500(1−7,7%)^𝑛 𝑛= log 4500 9500 log(1−7,7%) 𝑛≈9,325 It will take 10 years	✓ correct formula ✓ sub. of A and P ✓ use of logs ✓𝑛≈9,325 ✓ 10 years (5)
7.3.1	⁷⁵/₁₀₀ ×170 500=𝑅127 875 OR 25 ×170 500=42 625 100 Loan =170 500−42 625 Loan = R127 875	✓✓ answer (2) OR ✓ R 42 625 ✓ answer (2)
7.3.2	127 875=𝑥[1−(1+ 13,2 )⁻⁶⁰] 1200 13.2 1200 𝑥=𝑅 2 922,66	✓ 13.2 1200 ✓ 𝑛=60 ✓ sub of i, n and 127 875 into correct formula ✓✓ answer (5)

[15]

QUESTION 8

8.1	𝑓(𝑥)=𝑥−2𝑥² 𝑓′(𝑥)=lim 𝑓(𝑥+ℎ)−𝑓(𝑥) ℎ→0 ℎ = lim (𝑥+ℎ)−2(𝑥+ℎ)² −(𝑥−2𝑥2 ) ℎ→0 ℎ =lim (𝑥+ℎ)−2(𝑥²+2𝑥ℎ+ℎ²)−𝑥+2𝑥² ℎ→0 ℎ = lim 𝑥+ℎ−2𝑥²−4𝑥ℎ−2ℎ²−𝑥+2𝑥² ℎ→0 ℎ =lim −4𝑥ℎ−2ℎ²+ℎ ℎ→0 ℎ =lim ℎ(−4𝑥−2ℎ+1) ℎ→0 ℎ =lim (−4𝑥−2ℎ+1) ℎ→0 = −4𝑥+1 Answer ONLY: 0 marks Penalise 1 mark for incorrect use of formula. Must show 𝑓′(𝑥).	✓ formula ✓ substitution of (𝑥+ℎ) ✓ simplification ✓ simplification to (−4𝑥ℎ−2ℎ²+ℎ) ✓ common factor ✓ answer (6)
8.2.1	𝑦=¹/₉𝑥−3+9𝑥 𝑑𝑦 =−¹/₃𝑥−4+9 𝑑𝑥 Penalise 1 mark for incorrect notation.	✓−¹/₃𝑥⁻⁴ ✓ 9 (2)
8.2.2	𝑦= −1 +𝑥3 2𝑥√𝑥 𝑦= −1 + 𝑥³ 2𝑥.𝑥^½ 𝑦=−½𝑥−^3/^₂+𝑥³ 𝑑𝑦 =³/₄𝑥−^5/₄+3𝑥² 𝑑𝑥	✓ √𝑥=𝑥^½ ✓ −12𝑥−32 ✓ 34𝑥−54 ✓ 3𝑥2 (4)

[12]

QUESTION 9

9.1	ℎ(𝑥)= 𝑥³−9𝑥²+23𝑥−15. ℎ′(𝑥)=3𝑥²−18𝑥+23 𝑥=−𝑏±√𝑏²−4𝑎𝑐 2𝑎 𝑥=−(−18)±√(−18)²−4(3)(23) 2(3) 𝑥=4,15 or 𝑥=1,85 𝑥=1,85 at C	✓ ℎ′(𝑥)=3𝑥²−18𝑥+23 ✓ sub into formula ✓ both 𝑥 values ✓ 𝑥=1,85 (4)
9.2	ℎ(𝑥)= 𝑥³−9𝑥²+23𝑥−15. ℎ(𝑥)=(𝑥−1)(𝑥²−8𝑥+15) ℎ(𝑥)=(𝑥−1)(𝑥−3)(𝑥−5) ∴𝐹(5;0)	✓(𝑥−1)(𝑥²−8𝑥+15) ✓(𝑥−1)(𝑥−3)(𝑥−5) ✓✓𝐹(5;0) (4)
9.3	ℎ(𝑥)= 𝑥³−9𝑥²+23𝑥−15. ℎ′(𝑥)=3𝑥²−18𝑥+23 ℎ′′(𝑥)=6𝑥−18 6𝑥−18=0 𝑥=3 ∴𝑘=3 𝑥=4,15+1,85 2 𝑥=⁶/₂ 𝑥=3	✓ ℎ′′(𝑥)=6𝑥−18 ✓ 6𝑥−18=0 ✓∴𝑘=3 (3)
9.4	ℎ′(𝑥)=3𝑥²−18𝑥+23 ℎ′(3)=3(3)²−18(3)+23 ℎ′(3)=−4 𝑦=−4𝑥+𝑐 0=−4(3)+𝑐 𝑐=12 𝑦=−4𝑥+12	✓ ℎ′(3)=−4 ✓ sub of point D ✓𝑦=−4𝑥+12 (3)

[14]

QUESTION 10

10.1	𝑃=𝑥(50−½𝑥)−(¼𝑥²+35𝑥+25) 𝑃=50𝑥−½𝑥²−¼𝑥²−35𝑥−25 𝑃=−³/₄𝑥²+15𝑥−25	✓ 𝑥(50−½ 𝑥) ✓ subtracting total cost (2)
10.2	𝑑𝑃 =−³/₂𝑥+15 𝑑𝑥 −³/₂𝑥+15=0 𝑥=10	✓ 𝑑𝑃 =−³/₂𝑥+15 𝑑𝑥 ✓−³/₂ 𝑥+15=0 ✓ 𝑥=10 (3)
10.3	𝐶=¼𝑥²+35𝑥+25 𝑥 𝐶=¼𝑥+35+25𝑥⁻¹ 𝑑𝐶 =¼−25𝑥⁻² 𝑑𝑥 ¼ −25𝑥⁻²=0 25 = ¼ 𝑥² 𝑥²=100 𝑥=10 ∴Minimum	✓𝐶= ¼𝑥²+35𝑥+25 𝑥 ✓𝐶=¼𝑥+35+25𝑥⁻¹ ✓ 𝑑𝐶 =¼ −25𝑥⁻² 𝑑𝑥 ✓ ¼ −25𝑥⁻²=0 ✓ 𝑥=10 (5)

[10]

QUESTION 11

11.1.1	𝑃(𝑀)=1200 1600 𝑃(𝑀)=³/₄ or 0,75	✓ answer(1)
11.1.2	𝑃(𝐹𝑎𝑖𝑙)= 200 1600 𝑃(𝐹𝑎𝑖𝑙)=¹/₈	✓ answer(1)
11.1.3	𝑃(𝑀)×𝑃(𝐹)=³/₄×¹/₈ = 3 32 3 = 𝐴 32 1600 𝐴=150	✓ ³/₄×¹/₈ ✓ 3 32 ✓ 3 = 𝐴 32 1600 (3)
11.1.4	𝐵=1050 𝐶=50 𝐷=350	✓𝐵=1050 ✓𝐶=50 ✓𝐷=350 (3)
11.1.5	𝑃(𝐹/𝐹)= 50 1600 𝑃(𝐹/𝐹)= 1 32	✓50 ✓ 1600 (2)
11.2.1	9!=362880	✓✓ answer (2)
11.2.2	4!×5!×6 =17 280 OR 6!×4!=17280	✓ 4!×5! ✓ ×6 ✓ 17280 (3)

[15]

TOTAL:150

ALTERNATIVE ANSWERS

1.1.1	12𝑥²−𝑥−4=0 (½𝑥+1)(𝑥−4)=0 𝑥=−2 or 𝑥=4 OR 12𝑥²−𝑥−4=0 (½ 𝑥−2)(𝑥+2)=0 𝑥=4 or 𝑥=−2 OR 𝑥=−𝑏±√𝑏²−4𝑎𝑐 2𝑎 𝑥=−(−1)±√(−1)²−4(½)(−4) 2(½) 𝑥=4 or 𝑥=−2	✓✓ factors ✓ 𝑥-values (3) ✓✓ factors ✓ 𝑥-values (3) ✓✓sub into formula ✓ 𝑥-values
3.1	2𝑎=4 𝑎=2 𝑇₃=axis of symmetry 𝑇_𝑛=𝑎(𝑛+𝑝)²+𝑞 𝑇_𝑛=2(𝑛+𝑝)²+𝑞 𝑇_𝑛=2(𝑛−3)²+𝑞 24=2(1−3)²+𝑞 𝑞=16 𝑇_𝑛=2(𝑛−3)²+16 𝑇_𝑛=2(𝑛²−6𝑛+9)+16 𝑇_𝑛=2𝑛²−12𝑛+34	✓ 2𝑎=4 ✓ 𝑎=2 ✓ 𝑇_𝑛=2(𝑛+𝑝)²+𝑞 ✓ 𝑇_𝑛=2(𝑛−3)²+𝑞 ✓ 24=2(1−3)²+𝑞 ✓ 𝑞=16 ✓ 𝑇_𝑛=2(𝑛²−6𝑛+9)+16 ✓ 𝑇_𝑛=2𝑛²−12𝑛+34 (8)
5.4.2	−𝑥²+3𝑥−⁹/₄<0 𝑥²−3𝑥+⁹/₄>0 4𝑥²−12𝑥+9>0 (2𝑥−3)(2𝑥−3)>0 ∴𝑥∈𝑅 ;𝑥≠³/₂	✓ factors ✓✓ answer (3)

Last modified on Tuesday, 07 September 2021 11:02

Published in Grade 12 September 2018 Preparatory Examinations

MATHEMATICS PAPER 1 MEMORANDUM GRADE 12 - 2018 SEPTEMBER PREPARATORY EXAM PAPERS AND MEMOS

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