Copyright reserved Please turn over MARKS: 150 This memorandum consists of 23 pages. MATHEMATICS P1 NOVEMBER 2008 MEMORANDUM NATIONAL SENIOR CERTIFICATE GRADE 12 Mathematics/P1 2 DoE/November 2008 NSC – Memorandum Copyright reserved Please turn over • Continued Accuracy will apply as a general rule. • If a candidate does a question twice and does not delete either, only mark the FIRST attempt. • If a candidate does a question, crosses it out and does not re-do it, mark the deleted attempt. QUESTION 1 1.1.1 4 5 2 − = x x 0 ) 1 )( 4 ( 0 4 5 2 = − − = + − x x x x x = 4 or x = 1 standard form = 0 factorisation both answers (3) OR By the formula standard form = 0 substitution both answers 1.1.2 0 3 3 3 3 3 ) 3 (2 2 = − − − = − − = − x x x x x x 2 21 3 ) 1 ( 2 ) 3 )( 1 ( 4 ) 3 ( 3 2 ± = − − − ± = xx 79 , 3 = x or 79 , 0 − = x OR 0 3 3 3 3 3 ) 3 ( 2 2 = + + − − = − − = − x x x x x x 2 21 3 ) 1 ( 2 ) 3 )( 1 ( 4 ) 3 ( 3 2 −± − = − − − ± − = xx 79 , 3 = x or 79 , 0 − = x simplification standard form substitution into formula answers (5) simplification standard form substitution into formula answers (5) Note: If negative discriminant: max 2 /5 – 1 for not equal to zero in this question only. If = 0 appears once in this question then full marks – 1 for inaccurate rounding off for both answers. – 1 for inaccurate rounding off for both answers.Mathematics/P1 3 DoE/November 2008 NSC – Memorandum Copyright reserved Please turn over 1.1.3 0 ) 1 )( 3 2 ( 0 3 2 0 3 2 2 3 2 2 2 > − + > − + < + − − < − x x x x x x x x 23 − < x or 1 > x OR ) ; 1 ( ) 23 ; ( ∞ ∪ − −∞ ∈ x OR ) 1 )( 3 2 ( 0 3 2 0 2 3 2 2 − + < − + < < − x x x x x x 23 − < x or 1 > x standard form factorisation OR /∪ 23 − < x 1 > x (5) standard form factorisation OR /∪ 23 − < x 1 > x (5) 23 − 1 + + + + Note: 4 /5 Inaccurate inequality in the beginning 2 /5 If final answer does not have inequality signs (ie. question has been changed to an equation) 4 /5 If the candidate has used AND or ∩ instead of OR or ∪ If Answer is 1 23 0 ) 1 )( 3 2 ( < < − > − + x x x then: 2 /5 Mathematics/P1 4 DoE/November 2008 NSC – Memorandum Copyright reserved Please turn over 1.2 x y 2 3 − = 0 ) 3 )( 2 3 ( 0 6 11 3 4 12 9 2 3 ) 2 3 ( ) 2 3 ( 2 2 2 2 2 = − − = + − + − = + − + − = + − + x x x x x x x x x x x x x 32 = x or 3 = x ∴ 35 = y ∴ 3 − = y OR x y 2 3 − = substitution simplification of 2 ) 2 3 ( x − standard form factorisation both x values y values (8) 2 3 y x − = ) 3 )( 5 3 ( 0 15 4 3 0 4 2 6 4 6 9 2 3 4 6 9 2 3 2 3 2 2 2 2 2 2 2 + − = − + = = − + + + − = − + + + − = − + + ⎟⎠ ⎞ ⎜⎝ ⎛ − y y y y y y y y y y y y y y y y y y 35 = y or 3 − = y ∴ 32 = x ∴ 3 = x OR x y 2 3 − = 0 ) 1 )( ( 0 ) ( ) )( ( 0 2 2 = + − + = + + − + = + + − y x y x y x y x y x y x y x 3 3 2 3 − == − = − − = yx x x x y or 3532 1 2 3 1 == + = − + = yx x x x y OR 2 3 y x − = substitution simplification of 2 2 3 ⎟⎠ ⎞ ⎜⎝ ⎛ − y standard form factorisation both y values x values (8) y = 3 – 2x common factor common bracket y = – x 3 – 2x = – x both x-values y-values (8)Mathematics/P1 5 DoE/November 2008 NSC – Memorandum Copyright reserved Please turn over 2 3 y x − = 0 ) 1 )( ( 0 ) ( ) )( ( 0 2 2 = + − + = + + − + = + + − y x y x y x y x y x y x y x 333 2 2 3 = − = + − = − − = − = xy y y y y x y or 32355 3 2 3 2 1 2 3 1 === + − = + − = + = xyy y y y y x y 2 3 y x − = common factor common bracket y = – x 2 3 y y − − = both y-values x-values (8) 1.3 ( )( ) ( ) 2 2 2 2 24 2 + = − − + = −− x x x x x x Therefore when x = 999 999 999 999, the value is 999 999 999 999 +2 = 1 000 000 000 001. OR ( )( ) ( ) 2 2 2 2 24 2 + = − − + = −− x x x x x x 999 999 999 999 = 1012 – 1 x + 2 = 999 999 999 999 + 2 = 1012 + 1 factorisation simplification answer (3) Note: If candidate has substituted directly, 0/3 (answer would be 1 × 1012 by substitution) Answer only: 2 /3 Correct answer but incorrect mathematics 0 /3 1.4 1 1 1 1 4 4 4 > + = + x x x since 0 14 > x 4 4 1 x x + ∴ can never be equal to ½ OR 21 1 2 2 4 4 4 − = = + x x x Which has no real solution since 0 14 > x for all R x ∈ –{0} OR inequality conclusion (2) equation conclusion (2) Mathematics/P1 6 DoE/November 2008 NSC – Memorandum Copyright reserved Please turn over 08 ) 2 )( 1 ( 4 0 4 0 2 0 0 22 22 2 4 4 4 4 < − = − = − = + + = + = +ac b x x x x x ∴ no real roots calculation ∆ < 0 or ∆ = – 8 (2) 2 2 2 4 4 4 − = ∴ = + x x x Which has no real solution since 0 4 ≥ x for all R x ∈ equation conclusion (2) [26] QUESTION 2 2.1.1 16 1 ; 13 answers (2) 2.1.2 ⎟⎠ ⎞ ⎜⎝ ⎛ + + + terms to 25 ... 81 41 21 ( ) terms to 25 ... 13 10 7 4 + + + + ( ) 11 −− rr a n = [ ] d n a n ) 1 ( 2 2 − + = 1 21 1 21 21 25 − ⎟ ⎟⎠⎞ ⎜ ⎜⎝⎛ − ⎟⎠ ⎞ ⎜⎝ ⎛ = [ ] ) 3 ( 24 ) 4 ( 2 2 25 + = 0,9999999 = 1 000 00 , 1001 50 = S OR = 50 S 25 terms of 1st sequence + 25 terms of 2nd sequence ( ) terms to terms to S 25 ... 13 10 7 4 25 ... 81 41 21 50 + + + + + ⎟⎠ ⎞ ⎜⎝ ⎛ + + + = [ ] 00 , 1001 1000 ... 999999 , 0 ) 3 ( 24 ) 4 ( 2 2 25 1 21 1 21 21 50 50 25 50 = + = + + − ⎟ ⎟⎠⎞ ⎜ ⎜⎝⎛ − ⎟⎠ ⎞ ⎜⎝ ⎛ = SSS formula for geometric series 1 21 1 21 21 25 − ⎟ ⎟⎠⎞ ⎜ ⎜⎝⎛ − ⎟⎠ ⎞ ⎜⎝ ⎛ answer for geometric series formula for linear series [ ] ) 3 ( 24 ) 4 ( 2 2 25 + 1000 answer (7) Note: If used 50 terms in each series: max 5/7 (answer then is 3876) Answer only: 6 /7 Write out series and then correct answer: full marks Write out both series and not add them: 6 /7 Mathematics/P1 7 DoE/November 2008 NSC – Memorandum Copyright reserved Please turn over 2.2.1 60 ; 78 answers (2) 2.2.2 8 18 30 44 10 12 14 2 2 12 2 == aa c bn n Tn + + = 2 c b + + = 1 8 7 = b + c …(i) 18 = 4 + 2b + c 14 = 2b + c …(ii) (ii) – (i): 14 = 2b + c 7 = b + c ∴ 7 = b c = 0 n n Tn 7 2 + = OR a = 1 substitution solving simultaneously b = 7 c = 0 general term (6) 8 1 = T 10 1 2 = − T T 12 2 3 = −T T = − −1 n n T T nth term of sequence with a = 8 and d = 2 Add both sides ) 7 ( )] 1 ( 2 16 [ 2 terms 25 to ... 12 10 8 + = − + = + + + + = n n T n n TTnnn OR 8 1 = T 10 1 2 = − T T 12 2 3 = −T T Add both sides sequence substitution (6) Mathematics/P1 8 DoE/November 2008 NSC – Memorandum Copyright reserved Please turn over 0 T 1 T 2 T 4 T 5 T 0 8 18 30 44 8 10 12 14 2 2 2 0 T0 = 00 ) 0 ( ) 0 ( 2 == + + cc b a constant second difference = 2 a = 1 7 8 1 T1 = = + = b b ) 7 ( T 7 T 2 + = + = n n n n nn OR [ ] n n T n n n T n n n T n n T T difference ond n difference first first n Tnnnnn 7 8 2 3 10 10 8 ) 1 )( 2 ( ) 1 ( 10 8 )] 2 )( 2 ( ) 10 ( 2 [ 2 1 ) sec )( 2 ( ) ( 2 2 1 2 2 1 + = + + − + − = + − − + − = + − + − = + − + − = OR n n T n n n n T n n n n T n n difference nd T n T n Tnnnn 7 2 3 16 8 18 18 2 ) 2 )( 1 ( 2 ) 8 )( 2 ( ) 18 )( 1 ( 2 ) 2 )( 1 ( 2 ) 2 ( ) 1 ( 2 2 1 2 + = + − + + − − = − − + − − − = − − + − − − = OR finding 0 T c = 0 second difference = 2 a = 1 substitution b = 7 (6) formula substitution simplification answer (6) formula substitution simplification answer (6)Mathematics/P1 9 DoE/November 2008 NSC – Memorandum Copyright reserved Please turn over n n T n n n n n n T n n n n n n T T n n T n n T n n Tnnnn 7 30 45 15 54 72 18 24 20 4 2 ) 30 )( 2 3 ( ) 18 )( 3 4 ( 2 ) 8 )( 6 5 ( 2 ) 1 )( 2 ( ) 3 )( 1 ( 2 ) 3 )( 2 ( 2 2 2 2 2 2 2 3 2 1 + = + − + − + − + − = + − + + − − + − = − − + − − − − − = OR n n TTTTTn 7 11 . 4 44 10 . 3 30 9 . 2 18 8 . 1 8 2 4321 + = = = = = = = = = formula substitution simplification answer (6) observation answer (6) Note: By trial and error: 6 /6 Answer only: 6 /6 2.2.3 0 ) 15 )( 22 ( 0 330 7 330 ) 7 (2 = − + = − + = + n n n n n n n = – 22 or n = 15 n = 15 ∴15th term is 330. substitution standard form factorisation answer (4) [21] Note: 3 /4 if did not reject n = – 22 Answer only: 4 /4 By trial and error and then write n = 15: 4 /4 1 /4 if just equate Tn that they found If linear Tn and valid answer : 2 /4 Mathematics/P1 10 DoE/November 2008 NSC – Memorandum Copyright reserved Please turn over QUESTION 3 3.1 ( ) 1 2 2 8 − ⎟⎠ ⎞ ⎜⎝ ⎛ = n n x x T OR 1 1. 21 8 + − ⎟⎠ ⎞ ⎜⎝ ⎛ = n n n x T OR n n x x T ⎟⎠ ⎞ ⎜⎝ ⎛ = 2 16 OR 1 4 2 + − = n n n x T answer (1) 3.2 2x ratio = 2 2 1 2 1 < < − < < − xx ratio inequality answer (3) 3.3 72 23 21 1 23 8 2 18 1 2 2 = ⎟⎠ ⎞ ⎜⎝ ⎛ − ⎟⎠ ⎞ ⎜⎝ ⎛ = − = − = ∞∞∞∞ SS x x S r a S OR ... 8 81 2 27 18 + + + 72 41 18 43 118 == − = ∞∞∞ SSS substitution into formula for ∞ S substitution of 23 = x answer (3) series substitution answer (3) Formula Incorrect: 0 /3 [7] Mathematics/P1 11 DoE/November 2008 NSC – Memorandum Copyright reserved Please turn over QUESTION 4 4.1 p = 4 q = 2 11 1 2 4 5 3 = = + − = a a a answer p answer q substitution of (5; 3) answer (4) Answer for p 1 mark Answer for q 1 mark Answer for a 2 marks 4.2 y = – x + c substitute (4 ; 2) 2 = – 4 + c c = 6 OR Translation of the line y = – x 2 units up and 4 units right 6 2 ) 4 ( + − = + − − = x y x y correct point (4 ; 2) substitution answer (3) substitution of x – 4 adding 2 answer (3) Substitution of T(3 ; 5): 0 /3 Answer only: 3 /3 [7] Mathematics/P1 12 DoE/November 2008 NSC – Memorandum Copyright reserved Please turn over QUESTION 5 5.1 & 5.2 -9 -8 -7 -6 -5 -4 -3 -2 -11234 x y EXPONENTIAL shape (must be increasing above x-axis) y-int PARABOLA shape turning point y-intercept x-intercepts (8) INVERSE/LOG x-int shape (must be increasing on the right of the yaxxis (2) Note: If x-intercepts not shown but correct on graph 2/2 for xinterrcepts Calculation of x-intercepts of parabola 1 3 1 2 1 2 ) 1 ( 4 ) 1 ( 2 8 8 ) 1 ( 2 0 2 22 − = = − = − − = − = − = − − = x or x x or xxxx OR 1 3 ) 1 )( 3 ( 0 3 2 0 6 4 2 0 8 ) 1 2 ( 2 0 8 ) 1 ( 2 0 2 22 2 − = = + − = − − = − − = − + − = − − = x or x x x x x x x x xx 5.3 8 ) 1 ( 2 2 − + = x y OR 6 4 2 2 − + = x x y 8 − 1 + (2) 6 − 4 + (2) g f h (1 ; – 8) 3 1 – 1 Mathematics/P1 13 DoE/November 2008 NSC – Memorandum Copyright reserved Please turn over 5.4 ( )) ( 2 4 2 4 . 44 21 21 21x h x h x xx ==== ⎟⎠ ⎞ ⎜⎝ ⎛ + + OR ( )) ( 2 4 . 2 2 . 22 ) 2 (4 21 2 1 2 21 2 21x h x h x xx x x ====== ⎟⎠ ⎞ ⎜⎝ ⎛ + + + + substitution 21 4 . 4x ( ) x 42 (3) substitution 21 2 ) 2 ( + x ( ) x 42 (3) Note: If numerical examples are used : 1 /3 [15] Mathematics/P1 14 DoE/November 2008 NSC – Memorandum Copyright reserved Please turn over QUESTION 6 6.1 ° − = 45 x ° = 135 x answer answer (2) Note: If correct numbers but not writing as an equation 1 /2 If units left out: 2 /2 6.2 ) ( ) 45 tan( ) ( ) 45 tan( ) ( x f x x h x x h − = ° − − = − ° = h is the reflection of f about the x-axis OR h is the reflection of f about the line y = 0 reflection about x-axis (2) reflection about y = 0 (2) Note: If calculation only: 1 /2 If answer is: Reflection only: 0 /2 If do calculation and say reflection: 1 /2 Only ) ( ) 45 tan( ) ( ) 45 tan( ) ( x f x x h x x h − = ° − − = − ° = 1 /2 6.3 x y 2 sin 3 = 3 2x (2) [6]Mathematics/P1 15 DoE/November 2008 NSC – Memorandum Copyright reserved Please turn over QUESTION 7 Penalise ONCE in question 7 for early rounding off. 7.1 ( )( ) ..) (23,68701. years 69 , 23 .. 64968153 , 14 log 12 , 1 log .. 64968153 , 14 12 . 1 ) 12 . 1 ( 1570 23000 ) 1 ( = = = =+ = nn i P A n n n or n = 24 years or n = 23 years 8 months or n = 23,7 years OR ( )( ) ..) (23,68701. years 69 , 23 .. 64968153 , 14 log 12 , 1 log .. 64968153 , 14 12 . 1 ) 100 12 1 ( 1570 23000 ) 1 ( = = = + =+ = nn i P A n n n or n = 24 years or n = 23 years 8 months or n = 23,7 years formula substitution apply log function answer (4) formula substitution of 100 12 apply log function answer (4) 7.2.1 46 , 975462 000 200 46 , 1175462 46 , 1175462 ) 08 . 1 ( 800000 ) 1 ( 5 R R RR i P A n = − ∴== + = Some calculators give R 975 462,50 substitution R 1 175 462,46 R 975 462,46 (3) Incorrect Formula: 0/3 7.2.2 00 , 11944 1 ] 01 , 1 [ 01 . 0 46 , 975462 01 , 0 1 ] 01 , 1 [ 46 , 975462 ] 1 ) 1 [( 60 60 R x x x ii x F n = = −× − = − + = F = R975462,46 or answer in 7.2.1 n = 60 i = 1,01 formula simplification answer (6) Note: Accept 24 years : 4 /4 Incorrect Formula: 0/4 Mathematics/P1 16 DoE/November 2008 NSC – Memorandum Copyright reserved Please turn over OR 00 , 11944 66966986 , 81 46 , 975462 01 , 0 1 ] 01 , 1 [ 46 , 975462 60 R x x x = = − = F = R975462,46 n = 60 i = 1,01 formula simplification answer (6) Note: Continued Accuracy applies. 7.2.3 ( ) ( ) ( ) ( ) 24 , 12338 77 , 32197 66966986 , 81 46 , 975462 01 , 0 1 ] 01 , 1 [ 46 , 975462 77 , 32197 ] 5000 01 , 1 5000 01 , 1 5000 01 , 1 5000 01 , 1 5000 [ 60 12 24 36 48 R x x Service x Service = − = − − = = + + + + = OR 24 , 12338 77 , 32197 66966986 , 81 46 , 975462 01 , 0 1 ] 01 , 1 [ 46 , 975462 77 , 32197 1 01 , 1 ] 1 01 , 1 [ 5000 60 12 60 R x x Service x Service = − = − − == − − = OR Present Value payment of R 5000 { } 25 , 723 17 R ) 01 , 1 ( 1 ) 01 , 1 ( 1 ) 01 , 1 ( 5000 ) 01 , 1 ( ) 01 , 1 ( ) 01 , 1 ( ) 01 , 1 ( ) 01 , 1 ( 5000 12 60 12 60 48 36 24 12 = ⎭ ⎬ ⎫ ⎩ ⎨ ⎧−− = + + + + = −− − − − − − − Present Value of the sinking fund 94 , 942 536 R ) 01 , 1 ( 46 , 975462 60 == − Total Value of sinking fund = R 17 723,25 + R 536 942,94 = R 554 666,19 ∴ 24 , 338 12 R 01 , 0 ) 01 , 1 ( 1 19 , 554666 60 = ⎭ ⎬ ⎫ ⎩ ⎨ ⎧− = − x x OR 32 197,77 setting up of correct equation answer (4) 32 197,77 setting up of correct equation answer (4) 17723,25 554666,19 setting up of correct equation answer (4)Mathematics/P1 17 DoE/November 2008 NSC – Memorandum Copyright reserved Please turn over ...... 12682503 , 0 ) 01 , 0 1 ( ) 1 ( 12 = + = + eff eff i i 24 , 12338 77 , 32197 66966986 , 81 46 , 975462 77 , 32197 01 , 0 1 ] 01 , 1 [ 46 , 975462 77 , 32197 12682503 , 0 1 ) 12682503 , 1 ( 5000) 1 ( 60 5 R x x x i P n = − = − − = = − = + OR 24 , 394 1 01 , 1 01 , 0 5000 01 , 0 ) 1 ) 01 , 1 [( 5000 12 12 = − × = − = xx x So monthly deposit must be increased by R 394,24 New monthly deposit = R 11 944 + R 394,24 = R 12 338,24 substitution into formula 32 197,77 setting up of correct equation answer R 12 338,24 (4) substitution into formula 394,24 setting up of correct equation answer R 12 338,24 (4) [17] Mathematics/P1 18 DoE/November 2008 NSC – Memorandum Copyright reserved Please turn over QUESTION 8 8.1 x h x h h x h h h xh h x h xh x h x h x h x f h x f x f hhhhhh 6 ) 3 6 ( lim ) 3 6 ( lim 3 6 lim 3 3 6 3 lim ) 3 ( ) ( 3 lim ) ( ) ( lim ) ( 00 2 0 2 2 2 0 2 2 00 − = − − = − − = − − = + − − − = − − + − = − + = ′ →→→→→→ definition h x f h x f h ) ( ) ( lim0 − + → 2 ) ( 3 h x + − substitution of 2 3x − correct answer (5) Note: Penalty 1 for incorrect notation If a candidate has used the rules only: 0/5 8.2 3 61 2 x x y − = 4 4 21 4 21 3 21 21 4 1 21 41 63 41 . 61 . 21 x x dx dy x x dx dy x x dx dy x x y + = + = + = − = − − − − − Simplification 21 41 − x 4 21 − x or 4 63 − x (3) [8] Note: If removed coefficients, or moved the numbers from the denominator to the numerator: Continued accuracy applies for each correct derivative Max 2/3 If leave out dx dy penalise 1 mark. Mathematics/P1 19 DoE/November 2008 NSC – Memorandum Copyright reserved Please turn over QUESTION 9 9.1 2 25 0 ) 2 )( 5 2 ( − = = + − − or x x x AB = 4,5 units OR 2 25 0 ) 2 )( 5 2 ( − = = + − − or x x x AB = 2 2 ) 0 0 ( ) 2 ( 5 , 2 ( − + − − AB = 4,5 units 2 ; 25 − = = x x answer (2) 2 ; 25 − = = x x answer (2) 9.2 0 ) ( = ′ x g 0 ) 1 )( 2 ( 0 2 0 12 6 6 2 2 = − + = − + = + − − x x x x x x x = – 2 or x = 1 at T: x = 1 0 ) ( = ′ x g 12 6 6 ) ( 2 + − − = ′ x x x g factorisation answer (4) 9.3 12 6 6 ) ( 2 + − − = ′ x x x g 24 ) 3 ( 12 18 54 ) 3 ( 12 ) 3 ( 6 ) 3 ( 6 ) 3 ( 2 − = − ′ + + − = − ′ + − − − − = − ′ ggg 61 24 61 ) 3 ( 24 11 − − = − = + − − = + = x yq q q ax y OR 12 6 6 ) ( 2 + − − = ′ x x x g 24 ) 3 ( 12 18 54 ) 3 ( 12 ) 3 ( 6 ) 3 ( 6 ) 3 ( 2 − = − ′ + + − = − ′ + − − − − = − ′ ggg 61 24 72 24 11 ) 3 ( 24 11 − − = − − = − + − = − x y x y x y )3 (− ′ g – 24 method of setting up straight line equation substitution of point (– 3 ; 11) answer in equation form (5) )3 (− ′ g – 24 formula substitution of point (– 3 ; 11) answer in equation form (5)Mathematics/P1 20 DoE/November 2008 NSC – Memorandum Copyright reserved Please turn over 9.4 y-coordinate of T is 27 20 ) 1 ( 12 ) 1 ( 3 ) 1 ( 2 ) 1 ( 2 3 = + + − − = g T(1 ; 27) ∴ 27 0 < < k OR 27 0 0 27 20 20 7 0 20 12 3 2 20 12 3 2 2 3 2 3< < < − < − < − < − = − + + − − = + + − − k k k k x x x k x x x y-coordinate of T (27) answer (3) 20 20 7 < − < − k answer (3) Answer Only: 3/3 27 0 ≤ ≤ k : 2 /3 k > 0: 1 /3 k < 27: 1 /3 9.5 12 6 6 ) ( 2 + − − = ′ x x x g 21 0 6 12 6 12 ) ( − = = + − − = ′ x x x x g ) (x g ′ changes sign at 21 − = x ∴ point of inflection at 21 − = x OR Turning points A(-2;0); T(1;27) Now x co-ordinate of point of inflection is 21 2 1 2 − = + − − = x – 12x – 6 = 0 21 − = x (4) points 21 − =x (4) [18] ( ) 0 < ′ x g ( ) 0 > ′ x g 21 − = x Mathematics/P1 21 DoE/November 2008 NSC – Memorandum Copyright reserved Please turn over QUESTION 10 10.1 2 2 2 200 200 r h h rh r V π π π = = = formula substitution (2) 10.2 r r r S r r r r S r rh 400 ) ( 2 . 200 ) ( 2 Area Surface 2 2 2 2 + = + = + = π π π π π π formula substitution (2) 10.3 0 200 0 400 2 0 : minimum At 400 2 400 ) ( 3 2 2 1 2= − = − = − = + = − − r r r dr dSr r dr dS r r r Sππ ππ π 200 3 = r r = 3,99 cm exponents correct 2 400 2 − − = r r dr dS π 0 = dr dS π 200 3 = r r = 3,99 or 3 200 π = r (5) Note: If did not put = 0, penalise 1 mark If notation is dx dy , ignore notation. [9] Mathematics/P1 22 DoE/November 2008 NSC – Memorandum Copyright reserved Please turn over QUESTION 11 11.1 800 8 10 ≤ + y x 360 4 3 ≤ + y x 60 ≥ y 0 , N y x ∈ answer answer answer (3) 11.2 & 11.3 See attached graph … (5) See attached graph … (1) 11.2 90 43 + − = x y 100 45 + − = x y 60 = y (5) 11.3 feasible region (1) Note: If shading only, and did not state feasible region 1/1 11.4 P = 200x + 250y answer (1) 11.5 250 54 200 250 P x y P x y + − = + − = Maximum at (20 ; 75) gradient search line answer (3) Note: Read correctly from the candidate’s graph for the point for maximum profit. If used vertices method: 1/3 for accurate answer. 11.6 43 − = m Since the gradient of the new profit function is equal to the gradient of the constraint 360 4 3 ≤ + y x , there are points other than (20 ; 75) that give an optimal solution. 43 − = m more points in optimal solution (more than one solution) (3) Note: If just answer Yes 0 /3 If just answer No 0 /3 [16] Mathematics/P1 23 DoE/November 2008 NSC – Memorandum Copyright reserved 10 20 30 40 50 60 70 80 90 100 110 120 130 140 10 20 30 40 50 60 70 80 90 100 x y FEASIBLE REGION Search Lines ACUNA (20 ; 75) MATATA x, y ∈ N0