Copyright reserved Please turn over MARKS: 150 This memorandum consists of 27 pages. MATHEMATICS P1 NOVEMBER 2010 MEMORANDUM NATIONAL SENIOR CERTIFICATE GRADE 12Mathematics/P1 2 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over NOTE: • If a candidate answers a question TWICE, only mark the FIRST attempt. • If a candidate has crossed out an attempt of a question and not redone the question, mark the crossed out version. • Consistent Accuracy applies in all aspects of the marking memorandum. QUESTION 1 1.1.1 3 ) 5 )( 3 ( = − − x x 0 ) 2 )( 6 ( 0 12 8 3 8 152 2 = − − = + − = + − x x x x x x x = 6 or x = 2 OR 3 ) 5 )( 3 ( = − − x x 4 ) 4 ( 0 12 8 3 8 15 2 2 2 = − = + − = + − x x x x x x – 4 = 2 or x – 4 = – 2 x = 6 or x = 2 expansion factors answers (3) expansion completed square form answers (3) 1.1.2 ) 2 ( 2 3 2 + = x x 0 4 2 3 2 = − − x x 6 52 2 ) 3 ( 2 ) 4 )( 3 ( 4 ) 2 ( ) 2 ( 2 ± = − − − ± − − = x x = 1,54 or – 0,87 OR ) 2 ( 2 3 2 + = x x 3 13 1 313 31 9 13 31 91 34 31 34 32 0 4 2 3 22 2 2 ± = ± = − = ⎟⎠ ⎞ ⎜⎝ ⎛ − + = ⎟⎠ ⎞ ⎜⎝ ⎛ − = − = − − x x xx x x x x x = 1,54 or – 0,87 standard form substitution answers (4) expansion completed square 313 ± answers (4) Note: No penalty for incorrect rounding off of answers. Substitution into incorrect formula, no marks Note: If answer only : Full Marks If the candidate makes it a linear equation, no marks For only 1 answer: 1 /3 Mathematics/P1 3 DBE/November 2010 NSC – Memorandum Copyright Reserved Please turn over 1.1.3 ) 1 2 )( 4 3 ( 0 4 5 6 0 6 5 4 2 2 + − > − − > > + x x x x x x OR 0 ) 1 2 )( 4 3 ( 0 4 5 6 0 4 5 62 2 < + − < − − > + + − x x x x x x critical values: 12121 5 ± = x x = 21 − or 34 OR 34 21 < < − x OR ⎟⎠ ⎞ ⎜⎝ ⎛ − ∈ 34 ; 21 x OR x < − 21 and 34 < x OR 0 ) 1 2 )( 4 3 ( 0 4 5 6 2 > + + − > + + − x x x x critical values: 21 − and 34 OR 34 21 < < − x OR ⎟⎠ ⎞ ⎜⎝ ⎛ − ∈ 34 ; 21 x OR x < − 21 and 34 < x correct inequality factors critical values 21 − and 34 answer (4) correct inequality factors critical values 21 − and 34 answer (4) 1.2 3 22 3 x y x y== 1 3 2 2 9 4 1 3 2 2 3 22 2 2 2 = − + − = ⎟⎠ ⎞ ⎜⎝ ⎛ − + ⎟⎠ ⎞ ⎜⎝ ⎛ − x x x x x x x x OR 0 1 3 4 9 5 1 3 2 2 3 2 2 2 2 = − + = ⎟⎠ ⎞ ⎜⎝ ⎛ − + ⎟⎠ ⎞ ⎜⎝ ⎛ − x x x x x x 0 ) 3 )( 3 5 ( 0 9 12 5 9 6 18 4 9 2 2 2 = + − = − + = − + − x x x x x x x x ( ) ( ) ( ) 9 10 9 20 9 16 3 4 9 5 9 52 3 43 4 2 ) 1 ( 4 + ± − = − − ± − = x 53 = x or 3 − = x x = 0,6 or 3 − = x 3 2x y = substitution simplification standard forms factors or substitution into correct formula x-answers y-answers – 21 34 + 0 + 0 − 34 – 21 – 21 34 − 0 − 0 + 34 – 21 Mathematics/P1 4 DBE/November 2010 NSC – Memorandum Copyright Reserved Please turn over 52 = y or y = – 2 4 , 0 = y or y = – 2 (x ; y) = ⎟⎠ ⎞ ⎜⎝ ⎛ 52 ; 53 or (−3 ; −2) OR ( ) ( ) 4 4 3 4 4 3 4 4 8 4 ) 2 ( 4 4 8 4 4 1 2 2 3 2 2 2 2 2 2 2 2 = − + − = − + − = − + − = − + −= y y y y y x y x y x y x y x y x x y 0 ) 2 )( 2 5 ( 0 4 8 5 4 8 4 9 2 2 2 = + − = − + = + − y y y y y y y 52 = y or y = – 2 53 = x or 3 − = x (x ; y) = ⎟⎠ ⎞ ⎜⎝ ⎛ 52 ; 53 or (−3 ; −2) OR 1 3 4 9 1 2 3 2 2 3 2 32 3 2 2 2 2 = − + − = − ⎟⎠ ⎞ ⎜⎝ ⎛ + − ⎟⎠ ⎞ ⎜⎝ ⎛ == y y y y y y y y y x x y 0 ) 2 )( 2 5 ( 0 4 8 5 4 8 4 9 2 2 2 = + − = − + = + − y y y y y y y 52 = y or y = – 2 53 = x or 3 − = x (x ; y) = ⎟⎠ ⎞ ⎜⎝ ⎛ 52 ; 53 or (−3 ; −2) (7) simplification of original quadratic substitution 2x = 3y simplification standard form factors or substitution into correct formula y-answers x-answers (7) 2 3y x = substitution simplification standard forms factors or substitution into correct formula y-answers x-answers (7) Note: If mathematical breakdown eg. if y = 2x – 3 is used, max 3 /7 Mathematics/P1 5 DBE/November 2010 NSC – Memorandum Copyright Reserved Please turn over 1.3 2009 2008 2010 2007 5 5 5 5 ++ 4 51 45 21 30 1266 5126 ) 5 1 ( 5 ) 5 1 ( 5 5 . 5 5 5 . 5 5 2008 3 2007 2008 2008 3 2007 2007 ≈=== × = + + = + + = OR 430 1265 5 5 1 5 . 5 5 . 5 5 . 5 5 ) 5 by each term (divide 5 5 5 5 23 2 2007 2007 3 2007 2007 2007 2009 2008 2010 2007 ≈= ++ = + + = ++ OR 4 51 45 21 1 51 5 25 1 ) 1 5 ( 5 ) 5 5 ( 5 5 5 5 5 2009 let 12 1 1 2 ≈== ++ = ++ = + += − − − + − x x x x x x x OR 4 51 45 21 30 126 25 5 125 1 ) 5 5 ( 5 ) 5 1 ( 5 5 5 5 5 2007 let 23 2 1 3 ≈=== + + = ++ = + += + + + xx x x x xx OR 4 51 45 21 30 126 1 ) 5 5 ( 5 ) 1 5 ( 5 5 5 5 5 2010 let 51 25 1 125 1 1 2 3 1 23 ≈=== ++ = ++ = ++ = − − − − −−x x x x x xx 2 2008 2008 3 2007 2007 5 . 5 5 . 5 5 . 5 5 + + simplification to 23 5 5 5 1 ++ or 30 126 or 5 21 answer = 4 (3) 2 2007 2007 3 2007 2007 5 . 5 5 . 5 5 . 5 5 + + simplification to 23 5 5 5 1 ++ or 30 126 or 5 21 answer = 4 x x x x 5 5 5 5 1 1 2 + + − + − or 2 1 3 5 5 5 5 + + + + + x x x x or 1 23 5 5 5 5 − −− ++ x x x x simplification to 1 51 5 25 1 ++ or 25 5 125 1 + + or 51 25 1 125 1 1 ++ answer = 4 (3) [21] Note: If the candidate leaves the answer as 4,2 max 2 /3 marks Answer only of 4,2 0 /3 marks Mathematics/P1 6 DBE/November 2010 NSC – Memorandum Copyright Reserved Please turn over QUESTION 2 2.1 ( ) 3 581130733, OR 31 581130733 OR 33 , 581130733 6 1 3 20 ; 3 ; 1 3 1 3 31 terms 20 to ... 3 1 31320 20 201 2 = − = = = − − = + + + =∑= − n r n n a = 31 r = 3 n = 20 answer (4) 2.2.1 x 5 ; 2 x ; 53 x ; … 5x r = 5 5 1 5 1 < < − < < − xx Answer can be written as ) 5 ; 5 (− ∈ x 5x r = or x x52 1 1 < < − r answer (3) 2.2.2 52 = r and a = 10 16,67 or 3 50 52 110 = − = ∞ S a = 10 answer (2) Note: If 1 1 < < − x 1 mark Note: If answer is 5 5 ≤ ≤ − x then 2 /3 Note: If leave only as 387420489 129140163 43046721 14348907 4782969 1594323 531441 177147 59049 19683 6561 2187 729 243 81 27 9 3 1 31 + + + + + + + + + + + + + + + + + + + only, then 2 /4 Note: The 20th term is 387 420 489 Answer only: 3 /4 marks Note: The mark for n = 20 can be implied in the substitution to the formula Mathematics/P1 7 DBE/November 2010 NSC – Memorandum Copyright Reserved Please turn over 2.3.1 28 3 84 3 ) 1 ( 20 101 ) 1 ( 3 20 == − + = − + = n n nn Tn OR 28 3 84 17 3 101 17 3 == + = + = n nnn Tn ) 1 ( 3 20 101 − + = n or 17 3 101 + = n answer (2) substitution answer (2) ... 29 23 + + to 14 terms 868 ] 101 23 [ 2 14 OR ] 6 ) 1 14 ( ) 23 ( 2 [ 2 14 = + − + = OR Even numbers = 20 ; 26 ;…; 98 2.3.2 nnnn Tn == + = + = 14 6 84 14 6 98 14 6 OR nn nn Tn == − + = − + = 14 6 84 6 ) 1 ( 20 98 6 ) 1 ( 20 S remaining [ ] [ ]) 6 )( 13 ( ) 20 ( 2 2 14 ) 3 )( 27 ( ) 20 ( 2 2 28 + − + = 868 826 1694 ) 118 ( 7 ) 121 ( 14 = − = − = OR Sequence is 20; 23; 26; 29; 32; 35; 38; 41; 44; 47; 50; 53; 56; 59; 62; 65; 68; 71; 74; 77; 80; 83; 86; 89; 92; 95; 98; 101 Sum of odd numbers = 23 + 29 + 35 + 41 + 47 + 53 + 59 + 65 + 71 + 77 + 83 + 89 + 95 + 101 = 868 23 + 29 + ... a = 23 n = 14 d = 6 or l = 101 substitution into correct formula answer (6) OR 14 6 98 + = n or ) 1 ( 20 98 − + = n n = 14 substitution into correct formula 1694 826 answer (6) Full marks (6) [17] Note: If 3 17 − = n Then 1 /2 marks Answer only: Full marks Note: If “to 14 terms” is left out, do not penalise Note: If the candidate only works out the even numbers i.e. 826, then 3 /6 marks If only 1694 max 1 /6 marks Note: If incorrect value for n, max 4 /6 Note: If incorrect formula, max 2 /6 Mathematics/P1 8 DBE/November 2010 NSC – Memorandum Copyright Reserved Please turn over QUESTION 3 3.1 First difference : x x − − 37 ; 9 ;5 Second difference : x – 14; − 2x + 46 20 60 3 2 46 14 == − = − xx x x OR 20 60 3 37 23 2 37 ) 14 ( ) 9 ( == − = − − = − + − xx x x x x x OR 20 60 3 37 23 3 37 ) 14 ( ) 9 ( === − = − + − + xx x x x x OR 20 60 3 46 2 14 ) 9 ( ) 37 ( 5 ) 9 ( == + − = − − − − = − − xx x x x x x 4 9 x 37 5 x − 9 37 − x x − 14 37 − x − (x − 9) first differences 5; x − 9; 37 − x seconds difference answer (3) equating manipulation answer (3) first differences 5; x − 9; 37 − x equating answer (3) 3.2 5 4 3 5 ) 1 ( 4 and 4 5 9 3 2 9 2 12 1 4 3 3 36 2 2 21 2 + − = ∴ = − − = − = = + ∴ − = + = + + = + = + + + + = == n n T cb b c b T c b c b T c b c bn n Taa n n KK OR 4 9 20 37 5 11 17 6 6 a = 3 c bn n Tn + + = 2 3 b = – 4 c = 5 (4) Note: Answer only: Full Marks Note: If x is incorrect in 3.1 then max 2 /4 marks Mathematics/P1 9 DBE/November 2010 NSC – Memorandum Copyright Reserved Please turn over 5 4 34 5 ) 1 ( 3 4 5 35536 2 2 2 2 0 + − = − = + + = + + ===== n n Tb b bn n Tc Taann OR 5 4 354 4 3 44 5 3 36 2 2 + − === + − = + + − == + == n n Tccc b a bb a aan OR 5 4 354 39 3 14 6 28 2 12 5 3 iii 37 4 16 ii 9 2 4 i 4 2 + − == − ==== + = + = + = + + = + + = + + n n Tcbaab a b a b a c b a c b a c b a n KKK OR 5 4 3 6 9 3 5 5 4 2 ) 2 )( 1 ( 6 5 ) 1 ( 42 2 + − = + − + − + = − − + − + = n n n n n n n n Tn OR a = 3 c = 5 method b = – 4 (4) a = 3 c = 5 method b = – 4 (4) [7]Mathematics/P1 10 DBE/November 2010 NSC – Memorandum Copyright Reserved Please turn over QUESTION 4 x y A(2 ; 3) (1 ; 0) D(0 ; 1,5) f f 0 4.1 x = 2 OR x-asymptote = 2 y = 3 y-asymptote = 3 If x = p ; y = q then 1 mark Note: If the candidate just writes down the number 2 or 3 or just coordinates (2 ; 3), then no marks answer answer (2) 4.2 3 2 ) ( + − = x a x f 3 3 0 3 2 1 0 = + − = + − = a aa 3 2 3 ) ( + − = x x f OR 2 3 3 2 − = − + − = x a y x a y a y x = − − ) 3 )( 2 ( But (1;0) lies on the graph 3 ) 3 )( 1 ( = = − − ∴ a 3 ) 3 )( 2 ( = − − ∴ y x subs in of asymptotes subs in (1 ; 0) answer (3) equation subs in (1 ; 0) answer (3) 4.3 When x = 0, y = 3 2 0 3 + − = 23 D ⎟⎠ ⎞ ⎜⎝ ⎛ 23 ; 0 x = 0 y = 23 (2) If the asymptotes are swopped in 4.1, then 2 3 4 ) ( 4 2 3 1 0 2 3 ) ( + − = = + − = + − = x x f a a x a x f If asymptotes swopped: 32 2 3 0 4 0 = + − == yyx D ⎟⎠ ⎞ ⎜⎝⎛ 32 ; 0 Mathematics/P1 11 DBE/November 2010 NSC – Memorandum Copyright Reserved Please turn over 4.4 0 2 5 , 1 3 − − = AD m = 43 23 43 + = x y OR 6 3 4 + = x y OR 23 4343 23 ) 2 ( 3 23 + == + = + = x ym mmx y substitution into gradient 43 answer (3) substitution of point (2 ; 3) and c = 23 43 answer (3) 4.5 21 2 3 43 2 42 2 0 == + == + q q p p Other point of intersection is ⎟⎠ ⎞ ⎜⎝ ⎛ 21 4 ;4 OR By symmetry the rule to calculate the point of intersection is (x ; y) → ⎟⎠ ⎞ ⎜⎝ ⎛ + + 23 ; 2 y x Other point of intersection is ⎟⎠ ⎞ ⎜⎝ ⎛ = ⎟⎠ ⎞ ⎜⎝ ⎛ + + 21 4 ; 4 23 3 ; 2 2 2 2 0 = + p 3 2 2 3= + q x = 4 y = 21 4 (4) x-answer y-answer (4) If asymptotes swopped: 32 6767 21 37 0 23 3 2 + == × = −− = x y mAD Answer only: Full Marks To help with applying CA the y-coordinate will be 3 + (3 – y) Mathematics/P1 12 DBE/November 2010 NSC – Memorandum Copyright Reserved Please turn over OR 3 2 3 23 43 + − = + x x 0 ) 4 ( 3 0 12 3 24 12 12 12 6 6 3 ) 2 ( 12 12 ) 2 ( 6 ) 2 ( 3 2 2 = − = − − + = − + − − + = − + − x x x x x x x x x x x x x = 0 and x = 4 Other point of intersection is ⎟⎠ ⎞ ⎜⎝ ⎛ 21 4 ;4 equating standard form x-values y-value (4) [14] QUESTION 5 5.1 2 4 ) ( − = − x x f y-intercept: x = 0; 1 2 40 − = − = y ; (0 ; – 1) x-intercept: 21 1 2 2 2 2 4 0 2 4 2 − == − === −−− − xxxx x OR 21 2 log 2 2 log4 log 2 log2 log 4 log 2 4 0 2 4 − == − = − === −−− − xxxxx x OR 21 214 4 2 4 0 2 4 21 − == − === −−− − xxxx x x-intercept is ⎟⎠ ⎞ ⎜⎝ ⎛ − 0 ; 21 y-intercept x-intercept (4) 5.2 y = – 2 equation (1) If asymptotes swopped: 7 29 or 0 0 ) 29 7 ( 0 29 7 ) 3 )( 6 ( 2 ) 6 ( 4 ) 3 ( 4 ) 3 ( 7 2 3 4 32 67 2 = = = − = − − + = − + − + − = + x x x x x x x x x x x x Other point of intersection is ⎟⎠ ⎞ ⎜⎝⎛ 2 11 ; 7 29 Note: If the candidate does not select the x-value greater than 2 i.e. a realistic answer, max 3 /4 marks Note: No penalty if the answer is not left as a coordinate.Mathematics/P1 13 DBE/November 2010 NSC – Memorandum Copyright Reserved Please turn over 5.3 -2 -1.5 -1 -0.5 0.5 1 1.5 2 2.5 -2.5 -2 -1.5 -1 -0.5 0.51 x y asymptote y-intercept or x-intercept shape (decreasing) (3) 5.4 xx x g x g −− = + − = 4 ) ( 2 2 4 ) ( OR x x g ⎟⎠ ⎞ ⎜⎝ ⎛ = 41 ) ( OR x x g 2 2 ) ( − = OR x x g 2 21 ) ( ⎟⎠ ⎞ ⎜⎝ ⎛ = equation (1) 5.5 5 log 4 log5 4 3 2 4 = − = = − −−xxx 4 log 5 log − = x OR 5 log4 − = x OR 5 log 41 = x OR 51 4 log = x OR x = – 1,16 OR 41 log5 log = x OR 4 log log 51 = x 5 4 = − x 5 log 4 log = − x answer (3) [12] Mathematics/P1 14 DBE/November 2010 NSC – Memorandum Copyright Reserved Please turn over QUESTION 6 6.1 36 8 36 8 ) 6 ( 8) ( 2 2 − == − − = − = a a aax x f OR 92 − = a substitution answer (2) 6.2 2 92 : ) ( x y x f − = 2 92 y x − = 2 3 2 9 0 since , 29 2 9 2 9 22 x y x y y x y y x y x − − = − − = ≤ − ± = = − − = OR swop x and y 2 9 2 x y − = or 29x y − ± = 2 9x y − − = (3) 6.3 0 ≤ y OR ] 0 ; (−∞ ∈ y answer (1) 6.4 shape (third quadrant) (concave upward) Any point other than (0 ; 0) that lies on the graph Point corresponding from original graph will be (– 8 ; – 6) (2) -10 -8 -6 -4 -2 2 4 -6 -4 -22468 y O (-8;-6) ● Note: If candidate does not substitute the value of a the answer is ax y − = then 2 /3 marks Mathematics/P1 15 DBE/November 2010 NSC – Memorandum Copyright Reserved Please turn over 6.5 29 ) ( 1xx f y − = − = − OR 2 3 ) ( 1xx f y − = − = − OR 2 92 x y − = Reflection in y = x: 2 92 y x − = 2 9 29 2 x y y x − − = = − Reflection about y-axis: 2 9x y − = )( 1 x f − − answer (3) 2 92 y x − = 2 9x y − − = 2 9x y − = (3) [11] Note: If candidate does not substitute the value of a the answer is ax y = then full marks Note: If candidate has (x ; y) → (y ; – x) then 2 /3 marks Mathematics/P1 16 DBE/November 2010 NSC – Memorandum Copyright Reserved Please turn over QUESTION 7 7.1 ( ) 4 2 4 1 2 42 4 1 4 1 2 4 1 2 1 24 1 24 1 24 1 24 4 6 − ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ = ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ − == + ⎟⎠ ⎞ ⎜⎝ ⎛ + = ⎟⎠ ⎞ ⎜⎝ ⎛ + = + = × rrr r r P P i P A n r = 0,1172 … rate = 11,72% p.a. compounded quarterly OR ( ) 400 2 400 1 2 400 2 400 1 400 1 2 400 1 2 1 24 1 24 1 24 1 24 4 6 − ⎟ ⎟⎠⎞ ⎜ ⎜⎝⎛ = ⎟ ⎟⎠⎞ ⎜ ⎜⎝⎛ − == + ⎟⎠ ⎞ ⎜⎝ ⎛ + = ⎟⎠ ⎞ ⎜⎝ ⎛ + = + = × rr r r r P P i P A n r = 11,72% p.a. 2P 4r and 24 24 1 2 4 1 = + r 4 2 4 24 1 − ⎟ ⎟⎠⎞ ⎜ ⎜⎝⎛ = r answer (5) 2P 400 r and 24 24 1 2 400 1 = + r 400 2 400 24 1 − ⎟ ⎟⎠⎞ ⎜ ⎜⎝⎛ = r answer (5) 7.2.1 5 12 095 , 0 1 10000 ⎟⎠ ⎞ ⎜⎝ ⎛ + = A = R 10 402,15 substitution in correct formula answer (2) Note: Penalty 1 for incorrect rounding off. Mathematics/P1 17 DBE/November 2010 NSC – Memorandum Copyright Reserved Please turn over 7.2.2 months nn n nn n n 63 , 25 .... 63151282 , 25 ... 816999213 , 0 log 12 095 , 0 1 log 816999213 , 0 log 12 095 , 0 1 log 816999213 , 0 12 095 , 0 1 12 095 , 0 1 1 183000787 , 0 12 095 , 0 12 095 , 0 1 1 450 15 , 10402 === ⎟⎠ ⎞ ⎜⎝ ⎛ + − = ⎟⎠ ⎞ ⎜⎝ ⎛ + = ⎟⎠ ⎞ ⎜⎝ ⎛ + ⎟⎠ ⎞ ⎜⎝ ⎛ + − = ⎥ ⎥⎦ ⎤ ⎢ ⎢⎣⎡ ⎟⎠ ⎞ ⎜⎝ ⎛ + − = −− − − n = 26 Accept: n = 31 (because of first 5 months) OR months 63 , 25 12 095 , 0 1 log 223991387 , 1 log 12 095 , 0 1 log 223991387 , 1 log 12 095 , 0 1 95526 , 46439 10526 , 56842 1 12 095 , 0 1 10526 , 56842 12 095 , 0 1 15 , 10402 12 095 , 0 1 12 095 , 0 1 450 12 095 , 0 1 15 , 10402 = ⎟⎠ ⎞ ⎜⎝ ⎛ + = ⎟⎠ ⎞ ⎜⎝ ⎛ + = ⎟⎠ ⎞ ⎜⎝ ⎛ + = ⎥ ⎥⎦ ⎤ ⎢⎢⎣ ⎡ − ⎟⎠ ⎞ ⎜⎝ ⎛ + = ⎟⎠ ⎞ ⎜⎝ ⎛ + ⎥ ⎥⎦ ⎤ ⎢ ⎢⎣ ⎡ − ⎟⎠ ⎞ ⎜⎝ ⎛ + = ⎟⎠ ⎞ ⎜⎝ ⎛ + nn n nn n n n n = 26 Accept: n = 31 (because of first 5 months) 10 402,15 substitution into present value formula application of logs answer (4) 10 402,15 substitution into future value formula application of logs answer (4) Note: Incorrect Formula No marks Note: If the Present value of R 10 000 is used, then n = 25,53 months is obtained. Max 3 /4 marks. Mathematics/P1 18 DBE/November 2010 NSC – Memorandum Copyright Reserved Please turn over 7.2.3 Balance outstanding after 25 months 12 095 , 0 1 12 095 , 0 1 450 12 095 , 0 1 15 , 10402 25 25 ⎥ ⎥⎦ ⎤ ⎢ ⎢⎣ ⎡ − ⎟⎠ ⎞ ⎜⎝ ⎛ + − ⎟⎠ ⎞ ⎜⎝ ⎛ + = = R 282,36 OR Balance Outstanding after 25 months 12 095 , 0 1 12 095 , 0 1 450 12 095 , 0 1 10000 25 30 ⎥ ⎥⎦ ⎤ ⎢ ⎢⎣ ⎡ − ⎟⎠ ⎞ ⎜⎝ ⎛ + − ⎟⎠ ⎞ ⎜⎝ ⎛ + = = R 282,36 OR n = .... 6315128204 , 25 – 25 = 0,6315128204 ... Balance Outstanding after 25 months 12 095 , 0 12 095 , 0 1 1 450 631512804 , 0 ⎥ ⎥⎦ ⎤ ⎢ ⎢⎣ ⎡ ⎟⎠ ⎞ ⎜⎝ ⎛ + − = − = R 282,36 OR Present value at beginning of 25 months 12 095 , 0 12 095 , 0 1 1 450 15 , 10402 25 ⎥ ⎥⎦ ⎤ ⎢ ⎢⎣ ⎡ ⎟⎠ ⎞ ⎜⎝ ⎛ + − − = − = R 231,84 Balance Outstanding 25 12 095 , 0 1 84 , 231 ⎟⎠ ⎞ ⎜⎝ ⎛ + = = R 282,36 correct formula substitution into 12 095 , 0 1 12 095 , 0 1 450 25 ⎥ ⎥⎦ ⎤ ⎢ ⎢⎣ ⎡ − ⎟⎠ ⎞ ⎜⎝ ⎛ + answer (3) correct formula 12 095 , 0 1 12 095 , 0 1 450 25 ⎥ ⎥⎦ ⎤ ⎢ ⎢⎣ ⎡ − ⎟⎠ ⎞ ⎜⎝ ⎛ + answer (3) correct formula substitution into 12 095 , 012 095 , 0 1 1 450 631512804 , 0 ⎥ ⎥⎦⎤ ⎢ ⎢⎣⎡ ⎟⎠ ⎞ ⎜⎝ ⎛ + − − answer (3) correct formula substitution into 12 095 , 0 12 095 , 0 1 1 450 25 ⎥ ⎥⎦ ⎤ ⎢ ⎢⎣ ⎡ ⎟⎠ ⎞ ⎜⎝ ⎛ + − − answer (3) [14] Note: Accept If a candidate uses – 0,63, the final answer is R 281,68 Mathematics/P1 19 DBE/November 2010 NSC – Memorandum Copyright Reserved Please turn over QUESTION 8 8.1 5 ) ( 2 − = x x g x h x h h x h h h xh h x h xh x h x h x h x g h x g x g hhhhhh2 ) 2 ( lim ) 2 ( lim 2 lim 5 5 2 lim ) 5 ( 5 ) ( lim ) ( ) ( lim ) ( ' 000000 2 2 2 2 2 2 = + = + = + = + − − + + = − − − + = − + = →→→→→→ OR 5 ) ( 2 − = x x g x h x h h x h h x h x x h x h x h x h x h x h x g h x g x g hhhhhh2 ) 2 ( lim ) 2 ( lim ) )( ( lim ) ( lim ) 5 ( 5 ) ( lim ) ( ) ( lim ) ( ' 000000 2 2 2 2 = + = + = − + + + = − + = − − − + = − + = →→→→→→ formula substitution expansion 2x + h answer (5) formula substitution expansion 2x + h answer (5) 8.2 x x y 4 26 + = 21 5 21 6 2 3 4 21 − + = + = x x dx dy x x y 21 4x + 5 3x 21 2 − x (3) Note: If the notation is incorrect, penalty 1 mark If candidate subtracts and gets 5 5 2 2 2 2 − − − + + x h xh x in the numerator and then candidate corrects themselves, max 2 /5 Answer only: 0 /5 Note: If dx dy or y′ is left out, penalty 1 mark If a candidate shows evidence of how to differentiate from an incorrect function which involves breakdown, then max 1 /3 Mathematics/P1 20 DBE/November 2010 NSC – Memorandum Copyright Reserved Please turn over 8.3 xb ax x g + = 2 ) ( a b b a b a bx ax x g bx ax x g 128 16 8 ) 4 ( ) 4 ( 2 0 2 ) ( ) ( 2 2 1 2 == − = − = ′ + = − − 256 248 96 ) 128 ( 41 16 96 4 ) 4 ( 96 2 === + = + = ba a a a b a OR 256 2 96 48 0 4 32 96 4 16 ) 4 ( 0 16 8 ) 4 ( 2 ) ( 2 == = = − = + = = − = ′ − = ′ ba a b a b a g b a g xb ax x g 2 2 ) ( − − = ′ bx ax x g ) ( 0 x g′ = 2 ) 4 ( ) 4 ( 2 b a − subs (4 ; 96) a = 2 b = 256 (6) 2 2 ) ( xb ax x g − = ′ 16 8 ) 4 ( b a g − = ′ = ′ ) (x g 0 96 4 16 ) 4 ( = + = b a g a = 2 b = 256 (6) [14] Note: In the equation 0 ) ( = ′ x g ; = 0 must be shown in the equation. Mathematics/P1 21 DBE/November 2010 NSC – Memorandum Copyright Reserved Please turn over QUESTION 9 x y 9.1 The y-intercept of g is E(0 ; – 4) OR x = 0 and y = – 4 answer (1) 9.2 4 34 31 ) 6 )( 2 ( 3131 12 4 ) 6 0 )( 2 0 ( 4 ) 6 )( 2 ( 2 − − = − + == − = − − + = − − + = x x y x x ya a a x x a y OR 4 34 31 34 314 2 2 4 0 ) 2 ( 2 0 ) 2 ( 2 20 4 2 4 0 ) 2 ( 4 ) ( 4 ) 0 ( 2 2 − − = − == − = − − == + = ′ − == − − = − ′ − + = ′ = − = ′ x x yba a a a bb a g a b b a g bx ax x g c g setting up of equation subs (0 ; – 4) a = 31 4 34 31 2 − − = x x y (4) substitution x = – 2 and = ′ ) (x g 0 0 ) 2 ( = ′ g a = 31 4 34 31 2 − − = x x y (4) C(– 2 ; 0) D(6 ; 0) A B E O y = ) (x g′ h Mathematics/P1 22 DBE/November 2010 NSC – Memorandum Copyright Reserved Please turn over OR 4 34 3134 31 0 16 48 0 4 6 36 0 4 2 4 4 2 − − = − == = − = − + = − −− = x x yba a b a b a c OR 4 34 31314 12 12 4 ) 12 4 ( ) 6 )( 2 ( 2 2 2 − − == − = − − − = − − = − + = x x yaa a ax ax x x a x x a y OR a b b a b ax dx dy 4 ) 2 ( 2 0 2− = + = + = EITHER OR 4 34 31 34 316 2 ) 4 ( 3 18 2 3 18 2 6 36 4 4 6 36 0 ) 0 ; 6 ( subs2 − − = − === − + = + = + = − + = x x yba a a a b a b a b a 4 34 31 34 314 12 4 ) 4 ( 2 4 0 4 2 4 0 2 − − = − === − − − = − − = x x ybaa a a b a setting up of equation simultaneous equation a = 31 4 34 31 2 − − = x x y (4) setting up of equation a ax ax 12 4 2 − − a = 31 4 34 31 2 − − = x x y (4) a b 4 − = simultaneous equation a = 31 4 34 31 2 − − = x x y (4) Mathematics/P1 23 DBE/November 2010 NSC – Memorandum Copyright Reserved Please turn over 9.3 At turning point 0 ) ( = ′ x g x = – 2 and x = 6 0 ) ( = ′ x g x = 6 and x = – 2 (2) 9.4 2 2 6 2 = + − = xx OR x-value of point of inflection of g is at A. 24 2 0 4 2 0 34 3 2 0 ) ( === − = − = ′ xx x x x g OR ( ) 22 231 3 4 == − = xx a b x OR ( ) 2 3 16 2 31 ) ( 2 = − − = ′ x x x g x = 2 6 2 + − answer (2) 0 4 2 = − x answer (2) ( ) 31 3 4 2 = x answer (2) ( ) 3 16 2 31 ) ( 2 − − = ′ x x g answer (2) 9.5 0 ) ( /> x g for 2 − < x , so g is increasing for 2 − < x . 0 ) ( /< x g for 2 − > x , so g is decreasing for 2 − > x . ∴ g has a local maximum at x = 2 − because the graph is increasing followed by decreasing OR ∴ g has a local maximum at x = 2 − OR 0 ) ( /> x g g is incr for 2 − < x g is decr for 2 − > x (3) 0 ) ( /> x g for 2 − < x 0 ) ( /< x g for 2 − > x max at x = – 2 (3) g' g −2 −2 6 6 Note: Answer only Full marks Answer only: Full marks If only 1 value given, max 1 /2 Mathematics/P1 24 DBE/November 2010 NSC – Memorandum Copyright Reserved Please turn over OR 0 ) 2 ( = − ′ g 0 ) 2 ( < − ′ g so graph is concave down at x = – 2, so g has a local maximum 0 ) ( /> x g for 2 − < x 0 ) ( /< x g for 2 − > x max at x = – 2 (3) 0 ) 2 ( = − ′ g 0 ) 2 ( < − ′ g max at x = – 2 (3) [12] g'(x)=0 x = −2 g'(x) < 0 g'(x) > 0 g Mathematics/P1 25 DBE/November 2010 NSC – Memorandum Copyright Reserved Please turn over QUESTION 10 10.1 3 4 61 34 6 34 6 34 6 34 34 21 2 2 23 2 3 2 3 2 3 2 3 2 r r h rr r h r h r r h r r h r V r h r V − = − = − = + = + = × × + = ππ ππ π π π π π π π π π π volume equation substitution of 6π 23 2 4 6 rr r h ππ ππ − = (3) 10.2 ⎟⎠ ⎞ ⎜⎝ ⎛ − + = + = + × = 3 4 61 2 4 2 4 2 2 2 2 22 2 r r r r S rh r S rh r S π π π π π π r r r r r S 3 34 3 8 3 4 2 2 2 π π π π π + = − + = surface area equation substitution of h simplification (3) 10.3 metres square 14 , 3 metres square ) 2 ( 3 21 34 Then 21 1 8 1 8 0 3 3 8 3 34 2 3 2 2 1 2 == + ⎟⎠ ⎞ ⎜⎝ ⎛ ==== = − = + = − π π π π π π π SSr r r r r r dr dS r r S 1 3 − r π ⎟⎠ ⎞ ⎜⎝ ⎛ − = 2 1 8 3 r r dr dS π or ( ) 2 8 3 − − = r r dr dS π 0 = dr dS 2 1 8 r r = 21 = r S = π (6) [12] Mathematics/P1 26 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over QUESTION 11 11.1 0 N , ∈ y x 28 2 ≤ + y x or 14 2 + − ≤ x y 24 3 ≤ + y x or 24 3 + − ≤ x y First inequality Second inequality (4) 11.2 1 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 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 x y 3x + y = 24 x + 2y = 28 Feasible Region P(6 ; 6) 11.5 y = x 11.5 FR graph of 28 2 ≤ + y x graph of 24 3 ≤ + y x feasible region (quadrilate ral) (3) 11.3.1 8 answer (1) 11.3.2 14 answer (1) 11.4 Maximise x + y Use search line with gradient – 1 4 Type A 12 Type B 4 Type A 12 Type B (2) Note: If inequality signs incorrect or equal signs used: max 3 /4 marks Mathematics/P1 27 DBE/November 2010 NSC – Memorandum Copyright reserved 11.5 x y y x ≤≥ New Feasible region (triangle) in diagram Maximise x + y. Maximum at (6 ; 6) Answer: 6 + 6 = 12 braai stands Machine Time = x + 2y = 6 + 2×6 = 6 + 12 = 18 hours y ≤ x (6 ; 6) 12 18 hours (5) [16] TOTAL: 150 Note: Answer only of machine time 18 hours and braai stands 12 Full marks