Mathematics Grade 12 Paper 1 Limpopo Mock 2018 Memo

Downloaded from hlayiso.com Mathematics P1 1 NSC-Memorandu NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICS P1 MEMORANDUM SEPTEMBER 2018 MARKS: 150 TIME: 3 hours This memorandum consists of 16 pages. Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 2 LimpopoDoE/September 2018 NSC-Memorandum NOTE:  If a candidate answered a question TWICE, mark only the FIRST attempt.  If a candidate crossed out an answer and did not redo it, mark the crossed-out answer.  Consistent accuracy applies to ALL aspects of the marking memorandum.  Assuming values/answers in order to solve a problem is unacceptable. QUESTION 1 1.1.1 ( x  3)( x  1)  0 Values/waardes of/van x x  3  0 or x  1  0  x3 x3 or x  1  x  1 (2) 1.1.2 x2  x  4  0  standard form 2  b  b  4 ac /standaardvorm x 2a 1 (  1) 2  4 (1)(  4 )  correct substitution into  2 (1) correct formula /korrekte vervanging in 1  17 korrekte formule x 2 x  2,56  x  2,56  x  1,56 (5) or x  1,56 ALTERNATIVE /ALTERNATIEF OR x2  x  4  0  standard form/std vorm 1 1 completing the square ( x  )2  4 /voltooiing van vierkant 2 4 1 17 x   x  2,56 2 2 x  2,56 or/of x  1,56  x  1,56 (4) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 3 LimpopoDoE/September 2018 NSC-Memorandum 1.1.3 x x2 4  squaring both sides  x  2 2  4  x 2 /kwadrering beide kante 2 x2 = 16  8 x  x x 2  9 x  18  0  standard form/std vorm  factors/faktore x  3x  6  0 x  3 or/of x  6  both values/beide waardes  select/selekteer x  3 x  3 (5) OR ALTERNATIVE/ALTERNATIEF squaring both sides  x  2 2  4  x 2 /kwadrering beide kante 2 x  2 = 16  8 x  x  standard form/std vorm x  9 x  18  0 2  b  b 2  4 ac x 2a formula/formule 1  (  9) 2  4(1)(18)  2(1) 1 9 both values/beide waardes x 2 x  3 or x  6 x 3  select/selekteer x  3 (5) 1.1.4 1 critical values x  0 or x  2 /kritieke waardes both answers and inequality signs correct (3) /beide antwoorde en ongelykheids tekens moet reg wees Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 4 LimpopoDoE/September 2018 NSC-Memorandum 1.2.1 4 x3  4 x P 8x2  8x 22x6  22x prime numbers with  exponents/priemgetalle met 23x  6  23x eksponente 2 2 x ( 2 6  1)   23x 26  1   factorisation/faktorisering  simplification x /vereenvoudiging x 1 2 or/of    answer/antwoord (4) 2 OR ALTERNATIVE/ALTERNATIEF P   4 x 43  1  factorisation/faktorisering 8 x 8 2  1  factorisation/faktorisering 4 65  x  8 x 65  simplification/vereenvoudiging x 1   answer/antwoord (4) 2 1.2.2 P = 8 x 8  equating/stel gelyk 2 x 2  2 3  x  3  x  3 answer/antwoord (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 5 LimpopoDoE/September 2018 NSC-Memorandum 1.3 3 y  3  x  x 2 4  x ( k  2)  5 y 2 k 3 k  3 : x (3  2) 2  x  k  4 : x (4-2)  4 x both values in terms of x  x  4x  5 y /beide waardes in terme van 5x  5 y x  y equating/stel gelyk (1)...3 x   3  x  x 2 0  x2  2x  3 0  ( x  1)  x  3   x  1 or x3  standard y   1 or y  3 form/standaardvorm  x  1 or x  3  y  1 or y  3 (5) [25] QUESTION 2/VRAAG 2 2.1 10  3 y ; 7 ; 15 ; 8 y  1. 10 3y ; 7 ; 15 ; 8y +1 3 + 3y 8 8y 14  First differences /eerste verskil  Second differences 11 3y 8y 22 /tweede verskil  equating 11 3y = 8y 22 /gelykstelling y = 3  value of y /waarde van (4) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 6 LimpopoDoE/September 2018 NSC-Memorandum 2.2 1 ; 7 ; 15; 25 The quadratic row /kwadratiese ry 2a  2 a 1  value of a /waarde van 6 = 3(1) + b 3= b value of b /waarde van 1+ 3 + c = 1 c  3 value of c /waarde van (4) Tn  n 2  3n  3 2.3 n 2  3n  3 > 2157 inequality /ongelykheid n 2  3n  2160  0  factorization  n  48  n  45   0 faktorisering n  48 or n  45  select correct value of n /selekteer korrekte waarde  T46 (3) ALTERNATIVE: n 2  3n  3  2157 OR  inequality n 2  3n  2160  0 /ongelykheid  b  b 2  4 ac n 2a  3  8649  formula n 2 /formule n   48 or/of n  45  select correct value of n /selekteer regte waarde (3)  T46 [11] QUESTION 3/VRAAG 3 3.1.1 The first term a  p and  p and ( q  p ) the constant difference d  q  p  9q  8 p T10  p  9(q  p )  9q  8 p (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 7 LimpopoDoE/September 2018 NSC-Memorandum 3.1.2 Sn  n 2 a  ( n  1) d  correct substitution into 2  S 20  10 ( 2 a  19 d ) correct formula /korrekte vervanging in S 20  20 p  190 ( q  p ) korrekte waarde = 190 q  170 p answer /antwoord (2) 3.2.1 T3  T4  167 a  2 d  a  3d  167 equation/vergelyking 1 2 a  5 d  167 .................(1) T21   4 a  20 d   4  equation/vergelyking 2 a   20 d  4 .....................(2) equating : Substitute (1) into (2) : 2( 20 d  4)  5 d  167 Equation 1 = equation 2  40 d  8  5 d  167  value of d (4)  35 d  175 /waarde van d  5 3.2.2 a  20 d   4  value of a a   20 (  5)  4  96 /waarde van correct substitution in S 21  21 2 a  20 d  correct formula 2 /korrekte vervanging in  21 2(96 )  20 ( 5)  korrekte waarde 2  966  answer/antwoord OR OR a  20 d   4  value of a a  20( 5)  4  96 /waarde van 21 correct substitution in S 21  a  l  correct formula 2 /korrekte vervanging in 21   96  4  korrekte waarde 2  966  answer/antwoord (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 8 LimpopoDoE/September 2018 NSC-Memorandum 3.3 1 a  3; r   3 1 1 1 Sequence is:  3 ; 1 ;  ; ;  geometric sequence 3 9 27 /meetkundige ry Sequence of negative numbers: 1 1 3;  ;  ;…… 3 27 sequence of negative numbers 1 a  3 ; r  /ry met negatiewe getalle 9 a S  1 r 3  value of r and a 1 /waarde van 1 9 correct substitution in 39    correct formula 1 8 /korrekte vervanging in  27 korrekte formule  8 3  3 8 correct answer /korrekte antwoord (5) [16] QUESTION 4/VRAAG4 4.1 y  2x x0  f (0)  2 0  1  C(0;1) g ( x )    x  1  q 2 y - intercept: C(0;1)  substitute C(0;1) 1   (0  1) 2  q /vervang 1  1  q q  2  q2 (2) 4.2 g ( x )   ( x  1) 2  2 answer /antwoord (2) D(1; 2) 4.3 t2  answer/antwoord (1) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 9 LimpopoDoE/September 2018 NSC-Memorandum 4.4 f 1 : x  2 y  answer/antwoord (2)  y  log 2 x 4.5  form of the graph /vorm van die grafiek  x  intercept/afsnit  any other coordinate on the graph /enige ander koordinaat op die grafiek (3) 4.6 g ( x )    x  12  2 g ( x  1)  2   ( x  1  1) 2  2  2  +1 and  2  h( x)   x 2  answer/antwoord (2) 4.7 Domain: x  0 or x  0  answer /antwoord (1) [13] QUESTION 5/VRAAG 5 x  2  answer/antwoord 5.1 y 1  answer/antwoord (2) 5.2 .1 A: x0 4 y 1 02 y  2 1 y3  answer /antwoord (1) A(0 ; 3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 10 LimpopoDoE/September 2018 NSC-Memorandum 5.2.2 y0 4 equating to 0 0 1 /gelykstel aan x2  1( x  2)  4 x  2  4  simplification x6 /vereenvoudiging D(  6 ; 0)  answer/antwoord (3) 5.2.3 4  equating /gelykstelling  1  2x  3 x2 4  1( x  2 )  ( 2 x  3)( x  2 ) 4  2x2  6x  4  standard form /std vorm B: 0  2x2  6x 0  2 x ( x  3)  x  0 or x   3  value of x /waarde van g (  3)  2 (  3)  3   3  value of y  B (  3; 3) /waarde van (4) 5.3 y  y1 y 2  y1 m 2 applying x 2  x1 x 2  x1 Average gradient: 03 1 /toepassing   60 2 answer /antwoord (2) 5.4 g ( x)  2 x  3 / g / ( x)  2  g ( x)  f ( x ).g ( x )  0  explaining g  x   0 / g / ( x )  0,  x  R /verduidelik f ( x)  0 for   answer of f x  0 /antwoord (3) [15] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 11 LimpopoDoE/September 2018 NSC-Memorandum QUESTION 6/VRAAG 6 6.1 12 500 1  i 5 =5 546,32 correct substitution in correct formula /korrekte vervanging in korrekte formule 1  i   0, 4437056 5 5 (1  i ) 5  5 0, 4437056 applying 5 /toepassing 1  i  0,8500001102 i  0,1499998898  value of i (3) i  15% /waarde van 6.2.1 i ieff  (1  m ) m m  m  12 0,08 12 1  i eff  (1  ) correct substitution in 12 correct formula ieff  0,082999  100 /korrekte vervanging in korrekte formule ieff  8, 299  8,3 simplification/vereenvoudig answer/antwoord (4) 6.2.2 60 Deposit paid = 1 800 000  R1 080 000 100 Deposito betaal 40  deposit paid/deposito Balance owed = OR 180 000   R 720 000 betaal 100 Therefore/dus : R720 000 was financed by the bank was paid by the bank.  balance owed Was gefinansier deur die bank /betaal deur die /balans verskuldig bank.   119    correct substitution in 8  1   1    correct formula  0,08    1 200   /korrekte vervanging in 720 000  1    x  12  8  korrekte formule    1 200   n  119 4832  ( 0,5464730496 ) x 8842 ,16  x  answer of R8 842,16 (5) /antwoord Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 12 LimpopoDoE/September 2018 NSC-Memorandum 6.2.3  312   8  A  724800  1    correct substitution in  1 200  correct formula A  R920 667,815 /korrekte vervanging in korrekte formule  8  36   1   1  A = R920 667,815 F  8 842,16   1 200    8   correct substitution in  1 200  correct formula   /korrekte vervanging in F= R358 421,8873 korrekte formule Balance of the loan = R 920 667,815 - R35 8421,88734 119  difference/verskil =R 562 246  balance/balans (5) ALTERNATIVE/ALTERNATIEF x[1  (1  i )  n ] OR P i   83    correct formula  8  1   1    /korrekte formule   1 200    n = – 83 P = 8 842,16   8    8 842,16  1 200  Balance of the loan: P = R562 246  correct substitution  balance/balans (5) [17] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 13 LimpopoDoE/September 2018 NSC-Memorandum QUESTION 7/VRAAG7 Penalty for notation in 7.1 or 7.2- only once. 7.1 f ( x  h)  f ( x) lim  correct formula h0 h 3( x  h ) 2  2  3 x 2  2  lim  subst. in formula h0 h 3( x 2  2 hx  h 2 )  2  3 x 2  2  lim h0 h 3 x  6 hx  3h 2   3 x 2  2 2  lim h0 h 6 hx  3h 2  simplifying  lim h0 h h (6 x  3h )  factorising  lim h0 h  lim(6 x  3h ) h0  6x (5)  6x 7.2 y   ( 4 x 2  12 x  9 )  simplifying y   4 x 2  12 x  9 dy  – 8x   8 x  12 dx  12 (3) 7.3 f ( x)  x  x  2 f ( x)  x x  2 x 3 1  simplifying  x  2x 2 2 1 1 3 f / ( x)  x2  x 2  determining derivative of 2 3 1 1 each fraction f / 4  42  4 2 /bepaal eerste afgeleide 2  3  2 1 van elke term  substitution of 4 1 3  answer (4) 2 [12] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 14 LimpopoDoE/September 2018 NSC-Memorandum QUESTION 8/VRAAG 8 8.1 g ( x )  ( x  1)( x  1)( x  5)  0  A(-1; 0)  x  1 or x  5 A (  1 ; 0) and D(5 ; 0)  D(5 ; 0) (2) 8.2 g / ( x)  3x 2  6 x  9  0  first derivative = 0 3 : x 2  2 x  3  0 /eerste afgeleide ( x  3)( x  1)  0 x3 or x  1 g (3)  32  C(3;-32)  C(3; 32) and A(  1;0)  A(-1;0) (3) 8.3 EC: y   32  equation of tangent /vergelyking van raaklyn   32  x 3  3 x 2  9 x  5 0  x 3  3 x 2  9 x  5  32 equating /gelykstelling 0  x 3  3 x 2  9 x  27 0  ( x  3)( x  3)( x  3)  x  3 or x   3 value of x /waarde van  x  coordinate at E : x   3 (3) 8.4.1 g ( x)  3x  6 x  9 / 2  second derivative=0 /tweede afgeleide g // ( x)  6 x  6  0  value of x x 1 /waarde van g (1)  (1)  3(1) 2  9(1)  5   16 3  value of y  point of inflection: (1;  16) /waarde van OR OR  average x x y  yC x A+ C y A 2 and 2  value of x 1  3 0  32  value of y x  1 y    16 2 2  point of inflection: (1 ; 16) (3) 8.4.2 Gradient of tangent: g / (1)  3 (1) 2  6(1)  9  12  first derivative /eerste afgeleide Point of inflection (1;  16 ) value of gradient y   12 x  c /waarde van gradient  16   12 (1)  c  substitution of Equation of tangent: coordinates of the point of 4c inflection  y   12 x  4  equation of tangent (4) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 15 LimpopoDoE/September 2018 NSC-Memorandum QUESTION 9 /VRAAG9 9.1 2 Area shaded= 8 x  2x 2     2 x  2  formula /formule 3   16  2 = 8 x    4  x  9 factorization (4) 52 /faktorisering = 8x x 2 9 9.2 A=8x 52  x2 9 104  first derivative = 0 Maximum area: D x ( A)  8  x0 /eerste afgeleide 9  104  x  72  simplifying /vereenvoudiging 72 x  0, 22 m  x value /waarde (4) 104  9.3 52 maximum area = 8(0,22)  (0,22)2 = 0,88 9  answer /antwoord (1) 2 m [9] QUESTION 10/VRAAG 10 10.1 For an odd product,the two numbers must be odd. Vir ‘n onewe produk moet albei getalle onewe wees First draw: 4 odd out of 7 cards Eerste trekking: 4 onewe van 7 kaarte 4  Second draw: 3 odd out of 6 7 Tweede trekking: 3 onewe van 6 3 4 3 2  P(odd product) =    0,3 6 7 6 7 2 4 3 2  0r 0,3 P(onewe produk)=    0,3 7 7 6 7 (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 16 LimpopoDoE/September 2018 NSC-Memorandum 10.3 S P(A) P(B)  0,3 and 0,35 0,3 0,25 0,35  0,25  0,1 0,1 (3) 10.4.1 P(A and B/) = 0,55  0,25 = 0,3  0,55-0,25  0,3 (2) 10.4.2 P(A or B/) =1  0,35 = 0,65 answer /antwoord (1) . [9] QUESTION 11/VRAAG 11 11.1 6! = 6  5  4  3  2  1  720  answer /antwoord (2) 11.2 2!  5! 2  120  240  2!  5!  240 (2) 11.3 Left/links : 1 5! Right /regs: 5! 1  1 5! 1  5! 5! 1  1 5! 6!  6! 120  120 720  240  answer /answer 720 1  3 OR OR 1 Left/links : (4) 6 1 Right /regs: 6 1 1 + 6 6 2 1   6 3 [8] TOTAL 150 Copyright reserved Please turn over