Mathematics Grade 12 IEB Paper 1 2016 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE EXAMINATION NOVEMBER 2016 MATHEMATICS: PAPER I MARKING GUIDELINES Time: 3 hours 150 marks These marking guidelines are prepared for use by examiners and sub-examiners, all of whom are required to attend a standardisation meeting to ensure that the guidelines are consistently interpreted and applied in the marking of candidates' scripts. The IEB will not enter into any discussions or correspondence about any marking guidelines. It is acknowledged that there may be different views about some matters of emphasis or detail in the guidelines. It is also recognised that, without the benefit of attendance at a standardisation meeting, there may be different interpretations of the application of the marking guidelines. IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 2 of 12 SECTION A QUESTION 1 4x 2x 1 (a) (1)  =5 2 3 12 x  4 x  2 30 = OR 12x – 2 (2x + 1) = 30 6 6 8x = 32 x=4 (2) ( x  5)( x  6)  56 x 2  11x  30  56 x 2  11x  26  0 ( x  13)( x  2)  0 Critical Values: 13 ;  2 2  x  13 (b) TP (–2 ;–8) Shape Y – Int: (0 ; 0) X – Int. Let y = 0 (x + 2) = ±2 OR x (x + 4) = 0  x = 0 OR x = –4  x-int: (0;0) and (–4;0) (–2; –8) (c) (1) x = –1 and y=2 4 (2) + 2= x x 1  4 + 2 (x + 1) = x (x + 1)  4 + 2x + 2 = x2 + x  x2 – x – 6 = 0  (x – 3)(x + 2) = 0  (3; 3) (2; 2) NB: (For x = 3 OR x = 2 award 3 out of 4) 1 (d) c  1 or c   (other answers possible) 4 (e) 3  k  0 k  3 IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 3 of 12 QUESTION 2 1 (a) (1) LHS = 3    1 3 1 RHS = 6    1  1  3 1 LHS  RHS  x = is incorrect 3 (2) 3x = – 6 x  1  3x  = 6x – 1 2 9x2 – 6x +1 = 0 1 x= 3 from (1), no solution Alternate: Let x be a solution Then 3x < 0 so x < 0 1 But x  6  No solution (b) 7x + a (1 + 3) = 28 7 a   2 28  7  a2 7x + a = 4   7x + a = 7 7a 2 7x + a  71+a 2 x  a2  a  1 QUESTION 3  13,5  (a) 4 800   4 800    100   R4 152   7  ( 512)  1   1    (b)  415 200  x   1 200   Use of correct formula 7     1 200  x  R8 221, 46 (4) [6] IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 4 of 12 QUESTION 4 (a) Amount paid for all 110 laptops: 6 000 110  660 000 5  15  Depreciation over 5 years: A  660000 1    100   292 845,51 Inflation: A  P(1  i)n 5  6  A  660 000 1    100  A  883 228,881 Amount required in 5 years less "buy-back" = 883 228,88  292 845,51 = R590 383,37  (1  i)n  1 (b) Sinking Fund: F  x    i   12  (512)   1    1 590 383,37  x   1200   Use of correct formula  12   1200    x  R7 228,92 QUESTION 5 OR Alternate: (a) T1 = 5(1) + 2 = 7 7  12  17  ...  (5 y  2) T2 = 5(2) + 2 = 12  sequence is arithmetic T3 = 5(3) + 2 = 17 with a  7 and d  5 y  Since T1 + T2 + T3 = 36  [7  5 y  2]  36 2 Then y = 3  9 y  5 y 2  72 5 y 2  9 y  72  0  (5 y  24)( y  3)  0 y 3 (b) (1) 3 p  (2 p  14)  ( p  7)  3 p 3 p  2 p  14  p  7  3 p 3 p  21 p7 (2) a  28 and d  7 38 S38   2(28)  (38  1)(7) 2 S38  3 857 IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 5 of 12 (c) Tn = an2 + bn + c a + b + c = 7 … eq  4a + 2b + c = 13 … eq  9a + 3b + c = 21 … eq   – : 3a + b = 6  – : 5a + b = 8 Sub. b = 6 – 3a Into: 5a + b = 8 5a + 6 – 3a = 8  2a = 2 a = 1, b = 3, c = 3  Tn = n2 + 3n + 3 Alternate 1 Alternate 2  n  1  1 3 7 13 31 T = n n n  3n2  3n  1  1 3 = n = n2 + 3n + 3  Quadratic sequence Tn = T1 +  n  1  f +  n  1 n  2   s 2 f = first term of the first difference = 6 s = second difference = 2 Tn  n 2  3n  3 2 (d) r Alternate 3 The sequence of sums is: 2 a9 and r 2 4 665 2 059 3 9, 15; 19; 21 ; 23 ; ; a(r n  1) 3 9 27 81 Sn  r 1 T6  24,6 and T7 = 25,4  2  n   n = 7 is the smallest. 9    1  3   25   2 OR 1 3 n  2 2 n Sn = 27  1   2  3     3  27 n 2 2 2 n= log 2 Sn > 25 leads to    3 27 3 27 n  6, 41 ... Try n = 6, it does not work,  Smallest value: n  7 But n = 7 works. IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 6 of 12 (e) V1 = 729 cm3 1 Vn+1 = An+1hn+1 3 1 1 1  =   An  hn  3  3 3  1 = Vn 9  Sequence is geometric 1 with r = 9 729  S  = 820,1 cm3 1 9 1 Alternate 1 Volume of pyramid  = × (9 × 9) × 27 = 729 cm3 3 1  81  27 Volume of pyramid  = ×   × = 81 cm3 3  3 3 1  27  9 Volume of pyramid  = ×   × = 9 cm3 3  3  3 The sequence is geometric 1 a = 729 ; common ratio is 9 a S = 1 r 1 = 820 cm3 8 IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 7 of 12 QUESTION 6 f ( x  h)  f ( x ) (a) f '( x)  lim h 0 h Working: f ( x)  3 x 2  2 x f ( x  h)  3( x  h)2  2( x  h) f ( x  h)  3x 2  6 xh  3h2  2 x  2h 3x 2  6 xh  3h 2  2 x  2h  (3x 2  2 x) f '( x)  lim h 0 h 6 xh  3h  2h 2 f '( x)  lim h 0 h f '( x)  lim(6 x  3h  2) h 0 f '( x)  6 x  2 1 (b) y   x 1  x 2 dy 1 1   x 2  x 2 dx 2 IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 8 of 12 SECTION B QUESTION 7 Shape y-int and Pt of Inflection (0;1) Concave down for x > 0 OR Shape y-int and Pt of Inflection (0;1) Concave down for x > 0 QUESTION 8 3  1 (a) Axis of symm: x =  1 2  f ' (x) > 0 and g(x) < 0 OR f ' (x) < 0 and g(x) > 0  x < – 1 OR x > 0 (4) (b) g ( x)  d x  q sub. (0;0) 0  d0  q q  1 Sub. (1; 2) 2  d1 1 d 3  g ( x )  3x  1 IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 9 of 12 (c) Inverse of g : x  3y  1 3y  x  1 y = log3 ( x  1) Alternative (d) Domain: x > –1 Domain of g–1 = Range of g = (–1; ) (e) f(x) = a(x + 3)(x – 1) Alternate: Given y-int (0;6) 6 = a(3)(–1) 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 6 sub (-3;0) a = –2 0 = 𝑎(−3)2 + 𝑏(−3) + 6 9𝑎+6  f(x) = –2(x + 3)(x – 1) 𝑏= eq. 1 3 = –2x2 – 4x + 6 sub (1;0) 0 = 𝑎(1)2 + 𝑏(1) + 6  a = –2, b = –4, c = 6 𝑎+𝑏+6=0 eq. 2 Sub. Eq. 1 in Eq. 2 −24 = 12𝑎 ∴ 𝑎 = −2 𝑏 = −4 (f) Turning point of f: f(–1) = –2(–1 + 3)(–1 –1) =8  (8; –1) is the T.P of g Shape y-int: –3 and 1 x-int: 6 TP (8;–1) (g) k   6 OR k  (6 ; ) IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 10 of 12 QUESTION 9 (a) f(1) = a(1)3 + b(1)2  a + b f(2) = a(2)3 + b(2)2  f(2) = 8a + 4b 8a  4b  (a  b)  5,5 = 2 1 7a + 3b = 5,5 …… (1) f '(x) = 3ax2 + 2bx –18 = 3a(6)2 + 2b(6) –18 = 108a + 12b …… (2) 4(1)–(2): 28a + 12b = 22 108a + 12b = –18  –80a = 40 1 a=– 2 b=3 Note: No marks for answer only. (b) f ( x ) is increasing when f '(x)  0 Alternate 3 3 – x2  6 x  0 f '(x) = – x 2  6 x 2 2 3x –3x2 + 12x  0 =  x  4 2 –3x(x –4)  0 xc = 4 0x4 f is increasing on 0  x  xc  4 (c) f is concave down when f "(x) < 0 –6x + 12 < 0 x>2 Alternate 04 Point of inflection is: x = 2 2 From graph, f is concave down when x > 2 IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 11 of 12 QUESTION 10 hr 9 h  9  r V   r 2h V   r 2 (9  r ) V  9 r 2   r 3 V  9 r 2   r 3 V '  18 r  3 r 2 0  18 r  3 r 2 3 r (6  r )  0 r 0  r  6 units QUESTION 11 46 45 (a) (1)   0,3 80 79  9 25   25 9  (2)       80 79   80 79  45 45   1264 1264 45   0, 07 (4) 632 8! (b) 2!2!  10 080 (c) P(Khanya will win)  P( RB)  P( RRRB)  P( RRRRRB)  ....  6 1   6  1   6  1  3 5 P(Khanya will win)               + …  7 7   7  7   7  7  This is an infinite geometric series since –1 < r < 1 1 1 2 For and 6 6 6 7 a  2 and r    7 Tree diagram a P (Khanya will win)  1 r 6 2 6 P (Khanya will win)  7 2   0, 46  6  13 1   7 IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 12 of 12 QUESTION 12 1 1  (4 x 2 )  ( x 2   x 2 )  3  8  30 x 2  28 8 4   24 x  2 x 2  30 x 2  28 2 x 2 (54  2 )  28 28  x2  54  2  x  0, 766  x  0,8 Marks allocated as follows: 54𝑥 2 30𝑥 2 4𝜋𝑥 2 6(24𝑥 2 − 6𝜋𝑥 2 ) sum of the three parts = 28 x  0,8 Total: 150 marks IEB Copyright © 2016