MATHEMATICS: PAPER I MARKING GUIDELINES

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NATIONAL SENIOR CERTIFICATE EXAMINATION NOVEMBER 016 MATHEMATICS: PAPER I MARKING GUIDELINES Time: 3 hours 150 marks These markig guidelies are prepared for use by examiers ad sub-examiers, all of

1 NATIONAL SENIOR CERTIFICATE EXAMINATION NOVEMBER 016 MATHEMATICS: PAPER I MARKING GUIDELINES Time: 3 hours 150 marks These markig guidelies are prepared for use by examiers ad sub-examiers, all of whom are required to atted a stadardisatio meetig to esure that the guidelies are cosistetly iterpreted ad applied i the markig of cadidates' scripts. The IEB will ot eter ito ay discussios or correspodece about ay markig guidelies. It is ackowledged that there may be differet views about some matters of emphasis or detail i the guidelies. It is also recogised that, without the beefit of attedace at a stadardisatio meetig, there may be differet iterpretatios of the applicatio of the markig guidelies. IEB Copyright 016

NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I MARKING GUIDELINES Page of 1 SECTION A QUESTION 1 (a) (1) 4x x1 = 5 3 1x4x 30 = 6 6 OR 1x (x + 1) = 30 8x = 3 x = 4 () () ( x 5)( x 6) 56 x x 11 30

2 NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I MARKING GUIDELINES Page of 1 SECTION A QUESTION 1 (a) (1) 4x x1 = 5 3 1x4x 30 = 6 6 OR 1x (x + 1) = 30 8x = 3 x = 4 () () ( x 5)( x 6) 56 x x x 11x 6 0 ( x13)( x ) 0 Critical Values: 13 ; x 13 (5) (b) TP ( ; 8) Shape Y It: (0 ; 0) X It. Let y = 0 (x + ) = ± OR x (x + 4) = 0 x = 0 OR x = 4 x-it: (0;0) ad ( 4;0) ( ; 8) (c) (1) x = 1 (5) ad y = () () 4 x 1 + = x 4 + (x + 1) = x (x + 1) 4 + x + = x + x x x 6 = 0 (x 3)(x + ) = 0 (3; 3) (; ) (d) c 1 or NB: (For x = 3 OR x = award 3 out of 4) (4) 1 c (other aswers possible) () 4 (e) 3k 0 k 3 () [] IEB Copyright 016

3 NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I MARKING GUIDELINES Page 3 of 1 QUESTION (a) (1) LHS = RHS = LHS RHS x = 1 is icorrect () 3 () 3x = 6x 1 3x = 6x 1 9x 6x +1 = 0 x = 1 3 from (1), o solutio Alterate: Let x be a solutio (b) 7 x + a (1 + 3) = 8 7 a a x + a = 4 7 x + a = 77 a 7 x + a 1+ 7 a x a a The 3x < 0 so x < 0 1 But x 6 No solutio (4) 1 (3) [9] QUESTION 3 (a) 13, R415 () (b) ( 5 1) x 7 Use of correct formula 1 00 x R8 1,46 IEB Copyright 016 (4) [6]

4 NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I MARKING GUIDELINES Page 4 of 1 QUESTION 4 (a) Amout paid for all 110 laptops: Depreciatio over 5 years: A ,51 Iflatio: A P(1 i) 5 6 A A 883 8,881 Amout required i 5 years less "buy-back" = 883 8, ,51 = R ,37 (4) 5 (b) (1 i) 1 Sikig Fud: F x i (51) ,37 x Use of correct formula x R7 8,9 (4) [8] QUESTION 5 OR Alterate: (a) T 1 = 5(1) + = (5y ) T = 5() + = 1 sequece is arithmetic T 3 = 5(3) + = 17 with a7 ad d 5 Sice T 1 + T + T 3 = 36 y [7 5y ] 36 The y = 3 9y 5y 7 5y 9y 7 0 (5y 4)( y 3) 0 y 3 (3) (b) (1) 3 p ( p 14) ( p 7) 3p 3p p 14 p 7 3p 3p 1 p 7 () () a 8 ad d 7 38 S 38 (8) (38 1)( 7) S (3) IEB Copyright 016

5 NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I MARKING GUIDELINES Page 5 of 1 (c) T = a + b + c a + b + c = 7 eq 4a + b + c = 13 eq 9a + 3b + c = 1 eq : 3a + b = 6 : 5a + b = 8 Sub. b = 6 3a Ito: 5a + b = 8 5a + 6 3a = 8 a = a = 1, b = 3, c = 3 T = Alterate 1 T = = = Alterate Quadratic sequece T = T f + 1 s (d) r 3 The sequece of sums is: 9, 15; 19; 1 3 ; T 6 4,6 ad T 7 = 5,4 = 7 is the smallest. OR S = ; ; S > 5 leads to 3 7 Try = 6, it does ot work, But = 7 works. f = first term of the first differece = 6 s = secod differece = T 3 3 Alterate a9 ad r 3 ar ( 1) S r = log 3 7 6,41... Smallest value: 7 (4) (6) IEB Copyright 016

6 NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I MARKING GUIDELINES Page 6 of 1 (e) V 1 = 79 cm 3 V +1 = 1 3 A +1 h +1 = 1 3 = 1 V A h 3 3 Sequece is geometric with r = S = 80,1 cm 3 1 Alterate 1 9 Volume of pyramid = 1 3 (9 9) 7 = 79 cm3 Volume of pyramid = = 81 cm3 3 Volume of pyramid = = 9 cm3 The sequece is geometric a = 79 ; commo ratio is 1 9 a S = 1 r = cm3 (5) [3] IEB Copyright 016

7 NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I MARKING GUIDELINES Page 7 of 1 QUESTION 6 (a) f ( x h) f ( x) f '( x) lim h 0 h Workig: f ( x) 3x x f x h x h x h ( ) 3( ) ( ) f ( x h) 3x 6xh 3h x h (3 ) x xh h x h x x f '( x) lim h 0 h 6xh 3h h f '( x) lim h 0 h f '( x) lim(6x 3h ) h0 f '( x) 6x (5) (b) 1 1 y x x 1 dy 1 x x (4) dx [9] 77 marks IEB Copyright 016

8 NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I MARKING GUIDELINES Page 8 of 1 SECTION B QUESTION 7 Shape y-it ad Pt of Iflectio (0;1) Cocave dow for x > 0 OR Shape y-it ad Pt of Iflectio (0;1) Cocave dow for x > 0 [3] QUESTION 8 31 (a) Axis of symm: x = 1 f ' (x) > 0 ad g(x) < 0 OR f ' (x) < 0 ad g(x) > 0 x < 1 OR x > 0 (4) x (b) g( x) d q sub. (0;0) 0 0 d q q 1 Sub. (1;) 1 d 1 d 3 gx ( ) 3 x 1 (4) IEB Copyright 016

9 NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I MARKING GUIDELINES Page 9 of 1 (c) Iverse of g : y x 3 1 y 3 x 1 y = log 3( x 1) (3) (d) Domai: x > 1 (e) f(x) = a(x + 3)(x 1) 6 = a(3)( 1) a = f(x) = (x + 3)(x 1) = x 4x + 6 a =, b = 4, c = 6 Alterative Domai of g 1 = Rage of g = ( 1; ) Alterate: Give y-it (0;6) y = ax + bx + 6 sub (-3;0) 0 = a( 3) + b( 3) + 6 b = 9a+6 3 eq. 1 sub (1;0) 0 = a(1) + b(1) + 6 a + b + 6 = 0 eq. Sub. Eq. 1 i Eq. 4 = 1a a = b = 4 () (f) Turig poit of f: f( 1) = ( 1 + 3)( 1 1) = 8 (8; 1) is the T.P of g Shape y-it: 3 ad 1 x-it: 6 TP (8; 1) (4) (g) k 6 OR k ( 6 ; ) () [4] (5) IEB Copyright 016

10 NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I MARKING GUIDELINES Page 10 of 1 QUESTION 9 (a) f(1) = a(1) 3 + b(1) a + b f() = a() 3 + b() f() = 8a + 4b 5,5 = 8 a 4 b ( a b ) 1 7a + 3b = 5,5 (1) f '(x) = 3ax + bx 18 = 3a(6) + b(6) 18 = 108a + 1b () 4(1) (): 8a + 1b = 108a + 1b = 18 80a = 40 a = 1 b = 3 Note: No marks for aswer oly. (8) (b) f( x ) is icreasig whe f '(x) 0 Alterate x x 3 6 x x 3x + 1x 0 3x = x 4 3x(x 4) 0 x c = 4 0 x 4 f is icreasig o 0 x x c 4 (4) (c) f is cocave dow whe f "(x) < 0 6x + 1 < 0 x > Alterate Poit of iflectio is: x = 0 4 From graph, f is cocave dow whe x > (3) [15] IEB Copyright 016

11 NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I MARKING GUIDELINES Page 11 of 1 QUESTION 10 hr 9 h 9r V r h V r (9 r) V 9r r 3 3 V 9r r V ' 18r 3 r 0 18r3 r 3 r(6 r) 0 r 0 r 6 uits [7] QUESTION 11 (a) (1) () ,3 (3) ,07 (4) 63 (b) 8!!! (3) (c) P(Khaya will wi) P( RB) P( RRRB) P( RRRRRB)... IEB Copyright P(Khaya will wi) This is a ifiite geometric series sice 1 < r < 1 For ad a ad r 7 7 Tree diagram a P (Khaya will wi) 1 r P (Khaya will wi) 0, (7) [17] (7)

12 NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I MARKING GUIDELINES Page 1 of 1 QUESTION (4 x ) ( x x ) x x x 30x 8 x (54 ) 8 8 x 54 x 0,766 x 0,8 Marks allocated as follows: 54x 30x 4πx 6(4x 6πx ) sum of the three parts = 8 x 0,8 [7] 73 marks Total: 150 marks IEB Copyright 016

NATIONAL SENIOR CERTIFICATE GRADE 12

NATIONAL SENIOR CERTIFICATE GRADE 12 NAIONAL SENI CERIFICAE GRADE MAHEMAICS P NOVEMBER 00 MEMANDUM MARKS: 50 his memoradum cosists of 7 pages. Mathematics/P DoE/November 00 QUESION.. x ( x 4) 5.... x 4x 5 0 ( x 5)( x + ) 0 x 5 or x 4x 0x