Models of Health Behavior Essays

Models of Health Behavior Essays Models of Health Behavior Paper Models of Health Behavior Paper Note: If candidate uses incorrect formula: maximum 1/4 marks (for standard form) substitution into correct formula Note: If an error in subs and 4  ± ? 80 gets: and 6 states â€Å"no solution†: maximum 3/4 marks 4  ± 16 + 96 6 4  ± 112 = 6 2 ±2 7 = 3 = 2,43 or ? 1,10 112 If doesn’t conclude with â€Å"no solution†: maximum 2/4 marks 4  ± 112 or 6 decimal answer (4) Copyright reserved Mathematics/PI 3 NSC – Memorandum DBE/November 2011 OR 3x 2 ? 4 x = 8 3x 2 ? 4 x ? 8 = 0 x= = ? b  ± b ? 4ac 2a ? (? 4)  ± 2 (? 4)2 ? 4(3)(? 8) 2(3) Note: Penalise 1 mark for inaccurate rounding off to ANY number of decimal places if candidate gives decimal answers tandard form substitution into correct formula answer answer (4) = 2,43 or ? 1,10 1. 1. 3 4 x 2 + 1 ? 5x (4 x ? 1)(x ? 1) ? 0 + 0 1 4 x? OR 4 x 2 ? 5x + 1 ? 0 factors ? 0 1 + 1 4 1? ? OR ? ? ? ; ? ? [1; ? ) 4? ? 1 both critical 1 values of and 1 4 or OR ? answer (4) 1 or x ? 1 4 1 4 1 x OR 1 4 1 xNote: If candidate gives either of these correct graphical solutions but writes down the incorrect intervals or uses AND: max 3/4 marks NOTES: If a candidate gives an answer of 1 ? x ? If a candidate gives an answer of 1 then max 3/4 marks. 4 1 ? x ? 1 then max 2/4 marks. 4 1 If a candidate gives an answer of x ? nd x ? 1 then max 3/4 marks. 4 If the candidate leaves out the equality of the notation then penalty of 1 mark. 1 If a candidate gives an answer of x ? ; x ? 1 then max 3/4 marks. 4 1 If candidate gives x ? and/or x ? 1 , BREAKDOWN: max 2/4 marks. 4 If candidate gives : 0 ? 0 + + award 3/4 marks 1 1 4 Copyright reserved Please turn over Mathematics/PI 4 NSC – Memorandum DBE/November 2011 1. 2. 1 x 2 + 5 xy + 6 y 2 = 0 (x + 3 y )(x + 2 y ) = 0 x + 3y = 0 x = ? 3 y OR x = ? 3 y OR x + 2y = 0 x = ? 2 y x = ? 2 y Note: If a candidate gives x x ? = 3 or ? = 2 y y award 2/3 marks factors answers (3) Let k = y 2 x + 5 xy + 6 y 2 = 0 2 ?x? ?x? ? ? + 5? ? + 6 = 0 ? y? ? y? ? ? ? ? 2 k + 5k + 6 = 0 (k + 3)(k + 2) = 0 k = – 3 or k = – 2 x x = ? 3 or = ? 2 y y OR factors answers (3) x 2 + 5 xy + 6 y 2 = 0 x= x= ? 5 y  ± (5 y ) 2 ? 4(1)(6 y 2 ) 2(1) ? 5y  ± y2 2 ? 5y  ± y x= 2 x = ? 3 y x = ? 2 y or x x = ? 3 = ? 2 y y substitutes correctly into correct formula answers (3) OR x 2 + 5 xy + 6 y 2 = 0 ?5 x 2 + 5 xy + ? ?2 ? 5 ? y ? = ? 6 y 2 + ? ?2 ? 2 2 ? y? ? 2 5 ? 1 2 ? ? x + y? = y 2 ? 4 ? 5 1 x+ y= ± y 2 2 5 1 x=? y ± y 2 2 completing the square Copyright reservedPlease turn over Mathematics/PI x = ? 3 y x = ? 3 y x = ? y 5 NSC – Memorandum DBE/November 2011 answers (3) or x = ? 2 y OR Let k = x = ky x y x 2 + 5 xy + 6 y 2 = 0 (ky )2 + 5 y(ky ) + 6 y 2 = 0 k 2 y 2 + 5 y 2k + 6 y 2 = 0 y 2 k 2 + 5k + 6 = 0 ( (k ) 2 + 5k + 6 = 0 ) factors (k + 3)(k + 2) = 0 k = – 3 or k = – 2 x x = ? 3 or = ? 2 y y answers (3) Note: (x;y) = (0;0) is also a solution, but in this case OR x is undefined y Let y = 1 , x 2 + 5x + 6 = 0 (x + 2)(x + 3) = 0 x = ? 2 or x = ? 3 x x = ? 2 or = ? 3 y y x+ y =8 ? 3y + y = 8 ? 2y = 8 y = ? 4 x = 12 factors answers (3) x+ y =8 ? 2y + y = 8 ? y =8 y = ? 8 x = 16 1. 2. 2 OR substitution x = – 3y subs x = ? 2 y values both x values correct (5) OR 8? y = ? 3 y 8 ? y = ? 3 y 8 = ? 2 y y = ? 4 x = 12 Copyright reserved 8? y = ? 2 OR y 8 ? y = ? 2 y 8 = ? y y = ? 8 x = 16 x=8–y substitution y values both correct x values (5) Please turn over Mathematics/PI 6 NSC – Memorandum DBE/November 2011 OR x+ y =8 y =8? x x = ? 3 OR 8? x x = ? 3(8 ? x ) x = ? 24 + 3 x ? 2 x = ? 24 x = 12 y = ? 4 OR y =8? x x = ? 2 8? x x = ? 2(8 ? x ) x = ? 16 + 2 x ? x = ? 16 x = 16 y = ? 8 substitution x values correct both y values correct (5) (x + 2 y )(x + 3 y ) = 0 x+ y =8 x =8? y ( y + 8)(2 y + 8) = 0 y = ? 8 or y = ? 4 x = 16 x = 12 x =8? y ubstitution y values correct both x values correct (5) OR x = 8? y x = 8? y 2 (8 ? y ) 2 + 5(8 ? y ) y + 6 y = 0 64 ? 16 y + y 2 + 40 y ? 5 y 2 + 6 y 2 = 0 2 y 2 + 24 y + 64 = 0 y 2 + 12 y + 32 = 0 ( y + 8)( y + 4) = 0 y = ? 8 or y = ? 4 x = 16 x = 12 OR substitution factors both y values correct both x values correct (5) Copyright reserved Please turn over Mathematics/PI 7 NSC – Memorandum DBE/November 2011 OR x =8? y (8 ? y ) 2 + 5(8 ? y ) y + 6 y = 0 2 2 2 2 x = 8? y substitution 64 ? 16 y + y + 40 y ? 5 y + 6 y = 0 2 y 2 + 24 y + 64 = 0 y 2 + 12 y + 32 = 0 ? 12  ± 12 ? 4(1)(32) y= 2(1) 2 ? 12  ± 16 2 y = ? 8 or y = ? x = 16 x = 12 = Note: If a candidate uses the formula and replaces x for y and then answers are swapped: maximum 4/5 marks substitutes into correct formula both y values correct both x values correct (5) OR y =8? x x 2 + 5 x(8 ? x ) + 6(8 ? x )2 = 0 x + 40 x ? 5 x + 6 64 ? 16 x + x 2 x 2 ? 56 x + 384 = 0 x 2 ? 28 x + 192 = 0 (x ? 16)(x ? 12) = 0 x = 16 x = 12 or y = ? 8 y = ? 4 2 2 y =8? x 2 ( )= 0 substitution factors both x values correct both y values correct (5) OR y =8? x x + 5 x(8 ? x ) + 6 (8 ? x ) = 0 2 2 y =8? x substitution x + 40 x ? 5 x 2 2 + 6(64 ? 16 x + x ) = 0 2 2 x 2 ? 56 x + 384 = 0 x 2 ? 28 x + 192 = 0 x= = ? ? 28)  ± 28  ± 416 2 (? 28)2 ? 4(1)(192) 2(1) x = 12 x = 16 or y = ? 4 y = ? 8 substitutes into correct formula both x values correct both correct y values (5) [19] Please turn over Copyright reserved Mathematics/PI 8 NSC – Memorandum DBE/November 2011 QUESTION 2 2. 1. 1 x ? 4 = 32 ? x 2 x = 36 x = 18 T2 ? T1 = T3 ? T2 Note: If answer only: award 2/2 marks answer (2) a + 2d = 32 and a = 4 OR a=4 a + 2d = 32 2d = 28 d = 14 x = 14 + 4 x = 18 OR Note: If candidate writes x? 4 32 ? x only (i. e. omits equality) : 0/2 marks answer (2) substitutes correctly into arithmetic mean 4 + 32 formula i. e. 2 answers (2) T2 T3 = T1 T2 x= + 32 = 18 2 2. 1. 2 x 32 = 4 x x 2 = 128 x =  ± 128 x =  ±8 2 OR a=4 x r= 4 ? x? ar 2 = 4? ? ? 4? ? x? 32 = 4? ? ? 4? x 2 = 128 OR x =  ±11,31 OR x =  ± 2 2 7 Note: If candidate 32 x only writes 4 x (i. e. omits e quality) : 0/2 marks x 2 = 128 both answers (surd or decimal or exponential form) (3) Note: If only x = 128 then penalty 1 mark 2 2 2 ? x? 32 = 4? ? ? 4? 2 x = 128 x =  ± 128 x =  ±8 2 or x =  ±11,31 or x =  ± 2 2 7 both answers (surd or decimal or exponential form) (3) substitutes correctly into geometric mean formula i. e.  ± 4? 32 both answers (surd or decimal or exponential form) (3) Please turn over OR x =  ± 4 ? 32 =  ± 128 or x =  ±8 2 or x =  ±11,31 or x =  ± 2 2 7 Copyright reserved Mathematics/PI 9 NSC – Memorandum DBE/November 2011 2. 2 13 P = ? 3k ? 5 a = 3? 4 or +3 3? 5 =3 k =1 1? 5 +3 2? 5 + + 3 13 ? 5 = 3 ? 4 + 3 ? 3 + 3- 2 + + 38 3 ? 4 313 ? 1 = 3 ? 1 = 9841,49 Note: Correct answer only: 1/4 marks only 1 81 ( ) r =3 subs into correct formula or 9841 40 797161 or 81 81 answer (4) OR 13 P = ? 3k ? 5 k =1 1? 5 =3 + 32 ? 5 + 33 ? 5 + + 313 ? 5 2. 3 2 S n = [2a + (n ? 1)d ] + [2a + (n ? 1)d ] + + [2a + (n ? 1)d ] + [2a + (n ? 1)d ] = n[2a + (n ? 1)d ] Sn = n [2a + (n ? 1)d ] 2 S n = [a + (n ? 1)d ] + [a + (n ? 2)d ] + [a + d ] + a = 3 ? 4 + 3? 3 + 3- 2 + + 38 1 1 1 = + + + + 6561 81 27 9 40 797161 or = 9841,49 or 9841 81 81 S n = a + [a + d ] + [a + 2d ] + + [a + (n ? 2 )d ] + [a + (n ? 1)d ] Note: If the candidate rounds off and gets 9841,46 (i. e. correct to one decimal place): DO NOT penalise for the rounding off. expand the sum 13 terms in expansion answer (4) writing out Sn â€Å"reversing† Sn expressing 2Sn grouping to get 2 S n = n[2a + (n ? 1)d ] (4) writing out Sn â€Å"reversing† Sn expressing 2Sn grouping to get 2S n = n[a + a + (n ? 1)d ] (4) OR S n = a + [a + d ] + [a + 2d ] + + (Tn ? d ) + Tn S n = Tn + (Tn ? d ) + [a + d ] + a = n[a + a + (n ? 1)d ] = [2a + (n ? 1)d ] Sn = n [2a + (n ? 1)d ] 2 2 S n = a + Tn + a + Tn + a + Tn + + a + Tn Note: If a candidate uses a circular argument (eg S n +1 = S n + Tn ): max 1/4 marks (for writing out Sn) Note: If a candidate uses a specific linear sequence, then NO marks. [13] Copyright reserved Please turn over Mathematics/PI 10 NSC – Memorandum DBE/November 2011 QUESTION 3 3. 1 21; 24 Note: If candidate writes T8 = 21 T7 = 24 : award 1/2 marks 21 24 (2) 3. 2 T2 k = 3. 2 k ? 1 and so T52 = 3. 2 26 ? 1 = 100663296 Note: If candidate writes out all 52 terms and gets correct answer: award 5/5 marks . 2 k ? 1 T52 6k ? 3 T51 T2k ? 1 = 3 + 6(k ? 1) = 6k ? 3 and so T51 = 6(26) ? 3 = 153 T52 ? T51 = 100663296 ? 153 = 100663143 answer Note: If candidate used k = 52: max 2/5 Note: if candidate interchanges order i. e. does T51 ? T52 : max 4/5 marks Note: writes out all 52 terms and subtracts T51 ? T52 : max 4/5 marks (5) OR Consider sequence P: 3 ; 6 ; 12 †¦ Pn = 3. 2 n? 1 P26 = 3. 2 26 ? 1 = 100663296 Consider sequence Q: 3 ; 9 ; 15 †¦ Qn = 6n ? 3 Q26 = 6(26) ? 3 = 153 T52 ? T51 = P26 ? Q26 = 100663296 ? 153 = 100663143 Pn = 3. 2 n? 1 P26 Qn = 6n ? 3 Q26 answer (5) Copyright reserved Please turn over Mathematics/PI 1 NSC – Memorandum DBE/November 2011 3. 3 For all n ? N , n = 2k or n = 2k ? 1 for some k ? N If n = 2k : Tn = T2k = 3. 2 k ? 1 If n = 2k ? 1 : Tn = T2k ? 1 = 6k ? 3 = 3(2k ? 1) factors 3. 2 k ? 1 Note: If a candidate only illustrates divisibility by 3 with a specific finite part of the sequence, not the general term: 0/2 marks factors 3(2k ? 1) (2) In either case, Tn has a factor of 3, so is divisible by 3. OR Pn = 3. 2 n ? 1 Which is a multiple of 3 Qn = 6 n ? 3 = 3(2n ? 1) Which is also a multiple of 3 Since Tn = Q2 k ? 1 or Tn = P2 k for all n ? N , Tn is always divisible by 3 OR factors 3. 2 n ? 1 factors 3(2n ? 1) (2)The odd terms are odd multiples of 3 and the even terms are 3 times a power of 2. This means that all the terms are multiples of 3 and are therefore divisible by 3. odd multiples of 3 3 times a power of 2 (2) [9] Copyright reserved Please turn over Mathematics/PI 12 NSC – Memorandum DBE/November 2011 QUESTION 4 4. 1 The second, third, fourth and fift h terms are 1 ; – 6 ; T4 and – 14 First differences are: – 7 ; T4 + 6 ; – 14 – T4 So T4 + 6 + 7= – 14 – 2T4 – 6 T4 = – 11 d = – 11 + 6 + 7 = 2 or – 14 + 22 – 6 = 2 Note: Answer only (i. e. d = 2) with no working: 3 marks Note: Candidate gives T4 = ? 11 and d = 2 only: award 5/5 marks 7 T4 + 6 – 14 – T4 setting up equation T5 ? T2 = (T5 ? T4 ) + (T4 ? T3 ) + (T3 ? T2 ) answer (5) –7 –7+d – 7 + 2d setting up equation T5 ? T2 = (T5 ? T4 ) + (T4 ? T3 ) + (T3 ? T2 ) OR T2 1 -7 T3 -6 -7+d d T4 -7+2d d T5 -14 T5 ? T2 = (T5 ? T4 ) + (T4 ? T3 ) + (T3 ? T2 ) ? 15 = (? 7 + 2d ) + (? 7 + d ) + ? 7 ? 15 = ? 21 + 3d 6 = 3d d =2 Note: Candidate uses trial and error and shows this: award 5/5 marks answer (5) OR 4a + 2b + c = 1 9a + 3b + c = ? 6 5a + b = ? 7 25a + 5b + c = ? 14 16a + 2b = ? 8 10a + 2b = ? 14 6a = 6 a =1 d = 2a = 2 4a + 2b + c = 1 9a + 3b + c = ? 6 25a + 5b + c = ? 1 4 solved simultaneously answer (5) ORT1 1 – T1 T1 8 1 -7 T4+13 -6 T4+6 -20-2 T4 -14 T4 -14 T4 –7 T4 + 6 ? 14 ? T4 setting up equation answer (5) Please turn over T4 + 13 = ? 20 ? 2T4 3T4 = ? 33 T4 = ? 11 d = ? 11 + 13 d =2 Copyright reserved Mathematics/PI 13 NSC – Memorandum DBE/November 2011 OR T1 x 1–x -8+x T2 1 -7 y +13 T3 -6 y+6 20 – 2y T4 y -14 y T5 -14 –7 y+6 ? 14 ? y y + 13 = ? 20 ? 2 y 3 y = ? 33 y = ? 11 Second difference = y + 13 = ? 11 + 13 = 2 4. 2 T1 1 –6 -9 2 -7 Note: Answer only: award 2/2 marks Note: If incorrect d in 4. 1, 2/2 CA marks for T1 = d + 8 (since 1 ? T1 = ? 7 ? d ) setting up equation answer (5) method T1 = 10 (2) T1 = 10 OR =1 5a + b = ? 7 5(1) + b = ? 7 b = ? 12 a +b+c =1 4(1) + 2(? 12) + c = 1 c = 21 Tn = n 2 ? 12n + 21 T1 = (1) 2 ? 12(1) + 21 = 10 OR method T1 = 10 (2) T4 + 13 = ? 8 + T1 ? 11 + 13 = ? 8 + T1 T1 = 10 y + 13 = ? 8 + x method T1 = 10 (2) [7] OR ? 11 + 13 = ? 8 + x x = 10 Copyright res erved Please turn over Mathematics/PI 14 NSC – Memorandum DBE/November 2011 QUESTION 5 5. 1. 1 y = f (0) ? 6 = ? 1 0? 3 =1 (0 ; 1) OR ?6 ? 1 x? 3 ? 6 1= x? 3 x ? 3 = ? 6 0= x = ? 3 (? 3 ; 0) x = 0 and y = 1 5. 1. 2 Note: Mark 5. 1. 1 and 5. 1. 2 as a single question. If the intercepts are interchanged: max 3/5 marks y =1 x=0 (2) y=0 x ? 3 = ? 6 nswer (3) shape y 5. 1. 3 Note: The graph must tend towards the asymptotes in order to be awarded the shape mark (? 3; 0) (0 ; 1) 0 3 x y = ? 1 ?1 x=3 both intercepts correct horizontal asymptote vertical asymptote (4) Note: A candidate who draws only one ‘arm’ of the hyperbola loses the ‘shape’ mark i. e. max 3/4 marks 5. 1. 4 ? 3 lt; x lt; 3 OR (? 3; 3) OR ? 3 lt; x and x lt; 3 Note: if candidate writes ? 3 lt; x only: 1/2 marks Note: if candidate writes x lt; 3 only: 1/2 marks ?3 and 3 inequality OR interval notation (2) Copyright reserved Please turn over Mathematics/PI 15 NSC – Memorandum DBE/Novem ber 2011 5. 1. y= ?6 ? 1 ? 2? 3 1 = 5 1? 1 5 1 5 m= 0 ? (? 2) 2 = 5 formula substitution answer (4) OR m= = f (0) ? f (? 2) 0 ? (? 2) 1? 1 5 formula f (? 2) = 1 5 0+2 2 = 5 b lt; 0 since b lt; 0 and a lt; 0 2a y x substitution answer (4) y-intercept negative turning point on the x axis turning point on the left of the y axis maximum TP and quadratic shape 5. 2 x=? 0 (4) [19] Copyright reserved Please turn over Mathematics/PI 16 NSC – Memorandum y DBE/November 2011 QUESTION 6 f C(0 ; 4,5) g x O A B 6. 1 0 = 2x ? 8 8 = 2x 23 = 2 x x=3 A(3 ; 0) f (0) = 2 0 ? 8 = 1? 8 = ? 7 B(0 ; –7) Note: no CA marks Note: answer only: award 2/2 marks =0 answer for A x=0 answer for B (4) answer (1) 6. 2 6. 3 y = ? 8 OR y + 8 = 0 h( x ) = f ( 2 x ) + 8 = 22x ? 8 + 8 ( ) (2 2 x ? 8) answer of h( x ) = 4 x or 2 2 x (2) = 4 x or 2 2 x 6. 4 x = 4y y = log 4 x OR x = 22 y 2 y = log 2 x 1 y = log 2 x OR y = log 2 x 2 Note: answer only award 2/2 marks Note: candidate works out f -1 and gets y = l og 2 ( x + 8) award 1/2 marks log x OR y = log 4 switch x and y answer in the form y =†¦ (2) 6. 5 p ( x) = ? log 4 x OR p( x) = log 1 x 4 answer (1) OR p ( x) = log 4 OR 1 x OR 1 p( x) = ? log 2 x 2 y = ? log 2 x Copyright reserved Please turn over Mathematics/PI 17 NSC – Memorandum 5 DBE/November 2011 . 6 ? g (k ) ? ? g (k ) = g (0) + g (1) + g (2) + g (3) ? g (4) ? g (5) x = 3 is the axis of symmetry of g ? by symmetry g (2) = g (4) and g (1) = g (5) Answer = g (0) + g (3) = 4,5 + 0 = 4,5 OR k =0 k =4 3 = g (0) + g (1) + g (2) + g (3) ? g (4) ? g (5) g (2) = g (4) and g (1) = g (5) g (0) + g (3) answer (4) ? g (k ) ? ? g (k ) ? g (k ) = g (0) + g (1) + g (2) + g (3) ? g (k ) = g (4) + g (5) k =4 k =0 5 3 5 expansion k =0 3 k =4 x = 3 is the axis of symmetry of g ? by symmetry g (4) = g (2) g (5) = g (1) k =0 g (2) = g (4) and g (1) = g (5) ? g (k ) ? ? g (k ) k =4 3 5 = g (0) + g (3) = 4,5 + 0 = 4,5 OR g (0) + g (3) answer (4) ( x) = a( x ? 3) + 0 2 4,5 = a(0 ? 3) 2 + 0 4,5 = 9a 1 a= 2 1 2 g ( x) = ( x ? 3) 2 k =0 3 k =0 g ( x) = 1 (x ? 3)2 2 ? g (k ) ? ? g (k ) k =4 3 5 ? g (k ) = g (0) + g (1) + g (2) + g (3) = 4,5 + 2 + 0,5 + 0 =7 expansion Copyright reserved Please turn over Mathematics/PI 18 NSC – Memorandum DBE/November 2011 k =4 ? g (k ) = g (4) + g (5) = 0,5 + 2 = 2,5 ? g (k ) ? ? g (k ) k =4 3 5 5 k =0 = 7 ? 2,5 = 4,5 7 ? 2,5 answer (4) OR g ( x) = ax 2 + bx + c g (k ) = ak 2 + bk + c g (0) = c g (1) = a + b + c g (2) = 4a + 2b + c g (3) = 9a + 3b + c k =0 ? g (k ) = 14a + 6b + 4c 3 g (4) = 16a + 4b + c ? g (k ) = 41a + 9b + 2c 5 5 (5) = 25a + 9b + c k =4 3 k =0 ? g (k ) ? ? g (k ) = ? 27a ? 3b + 2c k =4 ? 27 a ? 3b + 2c g ( x) = a ( x ? 3)2 + 0 4,5 = a (0 ? 3) 2 + 0 4,5 = 9a 1 2 1 g ( x) = ( x ? 3)2 2 1 2 9 = x ? 3x + 2 2 a= k =0 ? g (k ) ? ? g (k ) = ? 27 a ? 3b + 2c k =4 3 5 g ( x) = 1 (x ? 3)2 2 ?9? ?1? = ? 27? ? ? 3(? 3) + 2? ? ? 2? ?2? = 4,5 answer (4) [14] Copyright reserved Please turn over Mathematics/PI 19 NSC – M emorandum DBE/November 2011 QUESTION 7 7. 1 A = P(1 ? i ) P n = P(1 ? 0,07 ) 2 1 = 0,93n 2 1 log = n log 0,93 2 1 log 2 n= log 0,93 = 9,55 years n OR A = P(1 ? i )n P = P(1 ? 0,07 )n 2 1 = 0,93 n 2 1 log 0,93 = n 2 n = 9,55 yearsP 2 subs into correct formula A= log answer (4) Note: If candidate uses incorrect formula: max 1/4 marks P for A = 2 Note: If candidate interchanges A and P A i. e. uses P = : max 2/4 marks 2 Copyright reserved Please turn over Mathematics/PI 20 NSC – Memorandum DBE/November 2011 7. 2 Radesh: A = P(1 + in ) = 8 550 Bonus = 0,05 ? 6 000 = 300 = 6 000(1 + 0,085 ? 5) A = 6 000 + 8,5% of 6000 ? 5 OR = 6000 + 510 ? 5 = 6000 + 2550 = 8 550 8 550 Received = 8 550 + 300 = R 8 850 Thandi: n A = P(1 + i ) R8 850 ? 0,08 ? = 6 000? 1 + ? 4 ? ? = R 8 915,68 20 n = 20 0,08 i= 4 answer choice made (6) 0,15 1 or or 0,0125 12 80 n = 18 n = 18 7. 3Thandis investment is bigger. Fv = initial deposit with interest + annuity ? ? ? 0,15 ? 18 ? ?1 + ? ? 1? 18 ? 12 ? ? 0,15 ? = 1 000? 1 + ? + 700? ? 0,15 12 ? ? ? ? ? ? 12 ? ? = 1 250,58 + 14 032,33 = R15 282,91 i= OR ? 0,15 ? 1 000? 1 + ? 12 ? ? ? ? 0,15 ? 18 ? ? ? 1 + ? ?1? ? 12 ? 700? ? 0,15 ? ? ? ? 12 ? ? answer (6) 18 Fv = initial deposit with interest + annuity ?18 ? ? ? ? 1 ? ?1 + 0,15 ? ? ? 18 18 ? ? 0. 15 ? 0,15 ? 12 ? ? = 1 000? 1 + ? ? + 700? 1 + 0,15 12 ? 12 ? ? ? ? ? 12 ? ? 0,15 1 or or 0,0125 12 80 n = 18 n = 18 i= 0. 15 ? ? = 1 250,58 + 11220,68? 1 + ? 12 ? ? = 1 250,58 + 14 032,33 = R15 282,91 Copyright reserved 8 ? 0,15 ? 1 000? 1 + ? 12 ? ? ?18 ? ? ? 1 ? ?1 + 0,15 ? ? ? ? 12 ? 700? 0,15 ? ? 12 ? 18 ? ? 18 0. 15 ? ? 1 + 12 ? ? ? answer (6) Please turn over Mathematics/PI 21 NSC – Memorandum DBE/November 2011 OR ? ? 0,15 ? 19 ? ? ? 1 + ? ?1? 18 ? 12 ? ? 0,15 ? Fv = 300? 1 + ? + 700? ? 0,15 12 ? ? ? ? ? ? 12 ? ? = 375,17 + 14 907,74 = R15 282,91 0,15 1 or or 0,0125 80 12 n = 19 (corresponding to 700) n = 18 (corresponding to 300) i= ? 0,15 ? 300? 1 + ? 12 ? ? ? ? 0,15 ? 19 ? ? ? 1 + ? ?1? 12 ? ? 700? ? 0,15 ? ? ? ? 12 ? ? answer (6) [16] 18 QUESTION 8 8. 1 f ? ( x ) = lim = lim f (x + h ) ? f (x ) h h;0 ? 4( x + h )2 ? 4 x 2 h h;0 ( ) Note: Incorrect notation: formula substitution expansion = lim = lim ? 4 x 2 + 2 xh + h 2 + 4 x 2 h h;0 ? 4 x 2 ? 8 xh ? 4h 2 + 4 x 2 h h;0 ( ) no lim written: penalty 2 marks lim written before equals sign: penalty 1 mark Note: A candidate who gives –8x only: 0/5 marks Note: A candidate who omits brackets in the line lim (? 8 x ? 4h ) : h ;0 ? 8 xh ? 4h 2 = lim h h;0 h(? 8 x ? 4h ) = lim h h;0 = lim (? 8 x ? 4h ) h;0 ? 8 x ? 4h answer (5) = ? 8 x NO penalty OR Copyright reserved Please turn over Mathematics/PI 22 NSC – Memorandum DBE/November 2011 f ( x ) = ? 4 x 2 f ( x + h) = ? 4( x + h) 2 = ? 4 x 2 ? xh ? 4h 2 f ( x + h) ? f ( x) = ? 8 xh ? 4h 2 ? 8 xh ? 4h 2 f ? ( x) = lim h h;0 h ( ? 8 x ? 4 h ) = lim h h;0 = lim (? 8 x ? 4h) h;0 substitution expansion formula ? 8 x ? 4h = ? 8 x 8. 2. 1 y= 3 x2 ? 2x 2 3 1 = x ? 1 ? x 2 2 2 answer (5) 3 ? 1 x 2 ? Note: Incorrect notation in 8. 2. 1 and/or 8. 2. 2: Penalise 1 mark 3 dy = ? x ? 2 ? x 2 dx 3 =? 2 ? x 2x 8. 2. 2 f ( x) = (7 x + 1) 2 3 ? 2 x 2 (3) ?x = 49 x 2 + 14 x + 1 f ? ( x) = 98 x + 14 f ? (1) = 98(1) + 14 = 112 multiplication 98 x 14 answer (4) OR f ( x) = (7 x + 1) 2 f ? ( x) = 2(7 x + 1)(7 ) By the chain rule . f ? ( x) = 98 x + 14 f ? (1) = 98(1) + 14 = 112 hain rule answer (4) [12] Copyright reserved Please turn over Mathematics/PI 23 NSC – Memorandum DBE/November 2011 QUESTION 9 9. 1 f ( x ) = ? 2 x 3 + ax 2 + bx + c f ? ( x ) = ? 6 x 2 + 2ax + b = ? 6( x ? 5)( x ? 2) = ? 6 x 2 ? 7 x + 10 ( ) = ? 6 x 2 + 42 x ? 60 2a = 42 a = 21 b = ? 60 f (5) = ? 2(5) + 21(5) ? 60(5) + c 3 2 Note: A candidate who substitutes the values of a, b and c and then checks (by substitution) that T (2; ? 9 ) and S (5;18) lie on the curve: award max 2/7 marks f ? ( x ) = ? 6 x 2 + 2ax + b ? 6( x ? 5)( x ? 2 ) b= –60 2a = 42 f (2) = ? 2(2) + 21(2) ? 60(2) + c OR ? 9 = ? 52 + c c = 43 3 2 18 = ? 5 + c c = 43 subs (5 ; 18) or (2 ; -9) c = 43 (7) Note: A candidate who substitutes the values of a, b and c into the a = 21 ; b = ? 60 ; c = 43 function i. e. gets f ( x) = ? 2 x 3 ? 21x 2 ? 60 x + 43 and then shows by substitution that T (2; ? 9 ) and S (5;18) are on the curve and works out the derivative i. e. gets f ? ( x ) = ? 6 x 2 ? 42 x ? 60 and shows (by substitution into the derivative) that the turning points are at x = 2 and x = 5 (assuming what s/he sets out to prove and proving what is given): award max 4/7 marks as follows: x = 2 from f ? ( x ) = 0 OR subs x = 2 into the derivative and gets 0 x = 5 from f ? x ) = 0 OR subs x = 5 into the derivative and gets 0 substitution of x = 2 in f and gets – 9 substitution of x = 5 in f and gets 18 f ? ( x ) = ? 6 x 2 + 2ax + b f ? (2) = 0 f ? (5) = 0 6a = 126 OR f ? ( x ) = ? 6 x 2 + 2ax + b f ? (2) = ? 6(2) + 2a (2) + b 0 = ? 24 + 4a + b 2 b = 24 ? 4a f ? (5) = ? 6(5) 2 + 2a (5) + b 0 = ? 150 + 10a + b 0 = ? 150 + 10a + (24 ? 4a) 0 = ? 126 + 6a 6a = 126 a = 21 b = ? 60 f (5) = ? 2(5) + 21(5) ? 60(5) + c 3 2 Note: If derivative equal to zero is not written: penalize once only b = – 60 f (2) = ? 2(2) + 21(2) ? 60(2) + c 3 2 subs (5 ; 18) or (2 ; -9) c = 43 (7) Please turn over 18 = ? 25 + c c = 43Copyright reserved a = 21 ; b = ? 60 ; c = 43 OR ? 9 = ? 52 + c c = 43 Mathematics/PI 24 NSC – Memorandum DBE/November 2011 OR f (2) = ? 9 i. e. ? 16 + 4a + 2b + c = ? 9 4a + 2b + c = 7 f (5) = 18 i. e. ? 250 + 25a + 5b + c = 18 25a + 5b + c = 268 21a + 3b = 261 f ? (5) = 0 f ? ( x ) = ? 6 x 2 + 2ax + b and f ? (2 ) = 0 OR 4a + b = 24 10a + b = 150 ? 16 + 4a + 2b + c = ? 9 and ? 250 + 25a + 5b + c = 18 f ? ( x ) = ? 6 x 2 + 2ax + b f ? (2) = 0 or f ? (5) = 0 12a + 3b = 72 9a = 189 189 9 a = 21 a= 12(21) + 3b = 72 3b = ? 180 b = ? 60 4(21) + 2(? 60 ) + c = 7 c = 43 4a + 2b + c = 7 OR 30a + 3b = 450 9a = 189 189 a= 9 a = 21 9a = 189 b = – 60 5a + 5b + c = 268 c = 43 subs (5 ; 18) or (2 ; -9) c = 43 (7) subs f ? (1) m tan = ? 24 f(1) = 2 f ? ( x) = ? 6 x + 42 x ? 60 2 OR 25(21) + 5(? 60 ) + c = 268 9. 2 f ? ( x) = ? 6 x + 42 x ? 60 2 m tan = ? 6(1) + 42(1) ? 60 2 = ? 24 3 2 f (1) = ? 2(1) + 21(1) ? 60(1) + 43 =2 Point of contact is (1 ; 2) y ? 2 = ? 24( x ? 1) y = ? 24 x + 26 9. 3 f ? ( x) = ? 6 x + 42 x ? 60 f ( x) = ? 12 x + 42 2 OR y = ? 24 x + c 2 = ? 24(1) + c c = 26 y = ? 24 x + 26 y ? 2 = ? 24( x ? 1) OR y = ? 24 x + 26 f ( x ) = ? 12 x + 42 (5) 0 = ? 12 x + 42 x= 7 2 x= 7 2 (2) OR 2+5 x= 2 Please turn over Copyright reserved Mathematics/PI 25 NSC – MemorandumDBE/November 2011 2+5 2 7 x= 2 OR ? 21 x= 3(? 2 ) 7 = 2 x= QUESTION 10 y x= 7 2 (2) x= ? 21 3(? 2 ) 7 x= 2 (2) [14] ?4 0 1 x y = f /(x) 10. 1 x-value of turning point: ? 4 +1 x= 2 3 =? 2 3 ? 3 ? ?x ;gt; ? OR ? x ? ? ? ; ? ? 2 ? 2 ? f has a local minimum at x = ? 4 because: (1; y) f ? 4 xgt;? 3 ? 3 ? OR ? ? ; ? ? 2 ? 2 ? (1) 10. 2 x=†“4 graph (3) f 1 –4 OR f ( x) lt; 0 for x lt; ? 4 , so f is decreasing for x lt; ? 4 . f / ( x) gt; 0 for ? 4 lt; x lt; 1 , so f is increasing for ? 4 lt; x lt; 1 . / i. e. –4 OR Copyright reserved ? f has a local minimum at x = ? 4 x=–4 f / ( x) lt; 0 for x lt; ? 4 f / ( x) gt; 0 for ? lt; x 0 so graph is concave up at x = – 4, so f has a local minimum at x = – 4. x=–4 gradient negative for x lt; ? 4 gradient positive for ? 4 lt; x lt; 1 (3) f ? (? 4) = 0 f (? 4) gt; 0 x=–4 (3) [4] QUESTION 11 11. 1 11. 2 V (0) = 100 ? 4(0) = 100 litres Rate in – rate out = 5 – k l / min answer (1) 5–k –4 units stated once 5 ? k = ? 4 k=9 V ? (t ) = ? 4 l / min 11. 3 5 ? k = ? 4 k = 9 l / min OR (3) (2) Note: Answer only: award 2/2 marks Volume at any time t = initial volume + incoming total – outgoing total 100 + 5t ? kt = 100 ? 4t 5t ? kt = ? 4t 9t ? kt = 0 t (9 ? k ) = 0 At 1 minute from start, t = 1, 9 â€⠀œ k = 0, so k = 9 OR 00 + 5t ? kt = 100 ? 4t k=9 (2) dV = ? 4 , the volume of water in the tank is decreasing by 4 dt litres every minute. So k is greater than 5 by 4, that is, k = 9. Since k=9 (2) [6] Copyright reserved Please turn over Mathematics/PI 27 NSC – Memorandum DBE/November 2011 QUESTION 12 Note: If the wrong inequality 50x + 25y ? 500 is used, candidate wrongly says that there are more learners than available seats. Maximum of 10 marks. 12. 1 x, y ? N x + y ? 15 50 x + 25 y ? 500 y? 8 y ? ? x + 15 OR y ? ?2 x + 20 y? 8 Note: for the inequality’s marks to be awarded, the LHS and the RHS must be correctNote: If candidate gives 50 x + 25 y = 500 : max 5/6 marks x + y ? 15 y ? 8 50 x + 25 y ? 500 (6) 12. 2 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 y x + y ? 15 50 x + 25 y ? 500 y? 8 feasible region (4) Blue buses x 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Red buses 12. 3 12. 4. 1 C = 600 x + 300 y (6 ; 8) ; (7 ; 6) ; (8 ; 4) ; (9 ; 2) and (10 ; 0) NOTE: The gradient of the search line is m = ? answer 2 1 (1) 3 marks for all correct solutions 2 marks if only 3 or 4 correct solutions 1 mark if only 1 or 2 correct solutions (3) subs answer (2) 2. 4. 2 12. 5 C = 6(600) + 8(300) = R 6 000 or C = 7(600) + 6(300) = R 6 000 or C = 8(600) + 4(300) = R 6 000 or C = 9(600) + 2(300) = R 6 000 or C = 10(600) + 0(300) = R 6 000 8 red ; 4 blue answer (1) [17] TOTAL: 150 Copyright reserved Please turn over Mathematics/P1 28 NSC – Memorandum DBE/November 2011 QUESTION 12. 2 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 y Blue Buses Red Buses 1 2 3 4 5 6 7 8 x 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Copyright reserved