Mathematics Grade 12 Paper 1 KwaZulu Natal Mock 2016 Memo

Downloaded from hlayiso.com Basic Education KwaZulu-Natal Department of Education REPUBLIC OF SOUTH AFRICA MATHEMATICS P1 PREPARATORY EXAMINATION SEPTEMBER 2016 MEMORANDUM NATIONAL SENIOR CERTIFICATE GRADE 12 MARKS: 150 This memorandum consists of 12 pages.

Downloaded from hlayiso.com Mathematics P1 2 September 2016 Preparatory Examination NSC-MEMORANDUM(KZN) QUESTION 1 1.1.1 2 x(3x  5)  0 5 x  0 or x AA Answers (2) 3 1.1.2 x 2  3x  7 x 2  3x  7  0 A equation in std. form  (3)  (3) 2  4(1)(7) CA substitution x 2(1) Penalise rounding off x  4,54 or x  1,54 CACA answers (4) 1.1.3 2x  5 x  3 2x  3  5 x A isolating 5 x 4 x 2  12 x  9  25 x M squaring both sides 4 x 2  37 x  9  0 CA standard form 4 x  1x  9  0 CA factors 1 CA both answers x or x9 CA rejecting 4 (6) n/a OR 1 2x  5x  3  0 2 A standard form  12  12   2 x  1 x  3   0    CAfactors    1 1 1 1 1 1 x  2 or x 3 2 2 CA x 2  CA x 2  3 2 No solution x9 CAanswer CArejecting (6) 1.1.4 2 x (3x  1)  0 2 x  0 for all x  R A 2 x  0 for all x  R  3x  1  0 A 3x  1  0 1 x A x   1 (3) 3 3 OR [Graphically] A sketch of y  2 x A sketch of y  3x  1 1 1 A x   x 3 3 (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 3 September 2016 Preparatory Examination NSC-MEMORANDUM(KZN) 1.2 299 . 21 – 299 M for taking 299 as a common 299 (2 – 1) factor = 299 A (2 – 1) (3) A answer 1.3 2 x  y  3  (1) x 2  5 xy  y 2  15  (2) From (1) : y  2 x  3 A y as subject Subst . (1) into (2) x 2  5 x(2 x  3)  (2 x  3) 2  15 CA substitution y  2 x  3 x 2  10 x 2  15 x  4 x 2  12 x  9  15  0 15 x 2  27 x  6  0 CA std. form 5 x  13x  6  0 CAfactors 1 CAx values x or x  2 5  17 CAy values (6) y or y  1 5 OR 2 x  y  3  (1) x 2  5 xy  y 2  15  (2) y3 From (1) : x  A x as subject 2 Subst. (1) into (2)  y  3  y  3 y3 2    5 y   y  15 2 CA substitution x   2   2  2  y 2  6 y  9  5 y 2  15 y     y 2  15  4  2 y  6 y  9  10 y 2  30 y  4 y 2  60 2 15 y 2  36 y  51  0 CA std. form 5 y 2  12 y  17  0 5 y  17  y  1  0 CAfactors  17 y or y  1 CAy values 5 1 x or x  2 CAx values 5 (6) [24] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 4 September 2016 Preparatory Examination NSC-MEMORANDUM(KZN) QUESTION 2 2.1 2a  4 a2 A a value 3a  b  0 b  6 (4) CAb value abc  4 c8 CAc value Tn  2n  6n  8 2 CAgeneral term OR 2a  4 a2 A a value 3a  b  0 CAb value b  6 T0  c  8 CAc value Tn  2n  6n  8 2 (4) CAgeneral term OR T  T  n  1d  n  1n  2 d Aformula n 1 1 2 2 Asubstituting first and second  4  n  10   n  1n  2 4 difference values 2  5  0  2n  6n  4 2 CAsimplifying  2n  6n  8 2 CAgeneral term (4) 2.2 Tn  4n  4  28088 A 4n  4 4n  28092 CA 4n  4  28088 n  7023 CA(n ЄN) n  7023 Between the 7023rd and 7024th terms. CA(n ЄN)conclusion OR (4) Tn  Tn 1  28088 A Tn  Tn 1  28088  2n  6n  8  2(n  1)  6(n  1)  8  28088 2 2  2n  6n  8  2n  4n  2  6n  6  8  28088  0 2 2 4n  28096  0 CA 4n  28096  0 4n  28096 n  7024 CA(n ЄN) n  7024 (4) Between the 7023rd and 7024th terms. CA(n ЄN)conclusion [8] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 5 September 2016 Preparatory Examination NSC-MEMORANDUM(KZN) QUESTION 3 3.1.1 6 ; 2 ; 10 ; 2 ; 14 ; 2 ; ... 18 ; 2 AA answers (2) 3.1.2 a  2 and is a constant sequence  S50  50  2 A100 (1)  100 3.1.3 a  6; d  4; n  50 A for a, d and n values S50  26   450  1 50 CA for substitution into formula 2  5200 CA 5200 S100  5200  100 CAanswer (4)  5300 3.2 S n  a  ar  ar 2  ...  ar n 1  (1) A for equation (1) rS n  ar  ar 2  ...  ar n  (2) A for equation (2) (2)  (1) : rS n  S n  ar n  a A subtraction on LHS and RHS  S n r  1  a r  1n  A factorising  Sn   a r n 1    a 1 rn r 1 if S n  penalize 1 mark (4) 1 r [11] QUESTION 4 4.1 2 a 1 r Aa and r value 3 a S  1 r CAsubstitution of r into formula 1  2 1 3 CAanswer 3 (3) 4.2 S n  3n 2  n T4  S 4  S3 M T4  S 4  S3     3(4) 2  (4)  3(3) 2  (3)  AAsubstitution  44  24  20 A answer (4) OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 6 September 2016 Preparatory Examination NSC-MEMORANDUM(KZN) S1  T1  a  2 A a value S 2  T1  T2  3(2) 2  2  10 T2  10  2  8 Asecond term value S3  T1  T2  T3  3(3)  3  24 2 T3  24  8  2  14 CAd value d 6 CAanswer T4  a  3d  2  3(6)  20 (4) [7] QUESTION 5 5.1 A  P(1  i ) n A formula A  255000 1  0,125  7 Acorrect substitution  R 100 137,46 CAanswer (3) 5.2.1 A  P (1  i ) n 5  0,095   10 000 1   Acorrect substitution into  12  correct formula  R10 402 ,15 CAanswer (2) 5.2.2 x[1  (1  i )  n ] P i   0,095   n  CA correct substitution of P into 450 1  1      12   formula 10 402 ,15  0,095 A x  450 12 0,095 10 402 ,15  n 12  1  1  0,095  450  12  n  0,095  1    0,816999213  12  log 0,816999213 CA n as subject n   0,095  log 1    12  Musing logs n  25,6315128  n  26 payments (5) CA(n Є N) answer Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 7 September 2016 Preparatory Examination NSC-MEMORANDUM(KZN) 5.2.3 Balance on Loan after the 25th payment M (A - F v ) formula A n = 25  0,095  25   0,095  25 450 1    1 CA 10 402 ,151    0,095  25  12    12   10 402 ,151     12  0,095  0,095  25  450 1    1 12  12    R12668 ,90  R12386 ,54 CA 0,095  R 282 ,36 12 (5) CA answer OR OR x[1  (1  i )  n ] P Aformula i   0,095  0, 6315128 Asubst. 450 450 1  1    0,095   12   CAsubst (5)  12 0,095 CAsubst 0 , 6315128 12  R 282 ,36 CAanswer [15] QUESTION 6 (For learners who did not receive the errata and used P(-4;4) , should be marked accordingly) 6.1 p  3 A p value (2) q  2 A q value 6.2 a h( x)   q . P(4 ;4) is a point on h x p a 4 2 43 CA subst. p, q and point P  a  2 CA(negative)a value (2) 6.3 2 h( x)  2 x3 2 h ( 0)  2 Asubstituting x = 0 03 1 CA(negative)answer (2)  1 3  1   0 ; 1   3  6.4 x 1 CACA answer (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 8 September 2016 Preparatory Examination NSC-MEMORANDUM(KZN) 6.5 y  ( x  p )  q y  ( x  3)  2 CAsubstitution of p and q values into equation of line of symmetry y  x  1 c  1 CAanswer (2) OR y  x  c Point of intersection of asymptotes 3 ;  2 CAsubstitution of 3 ;  2  2  3  c into equation of line of symmetry CAanswer (2) c 1 [10] QUESTION 7 7.1 D0 ;  10  A(must be in coordinate form) (1) answer 7.2 x 2  3 x  10  0 A x 2  3x  10  0 x  2x  5  0 CAfactors x  2 or x5 CACA(negative and positive) A 2 ; 0  each x – value B(5; 0) A 2 ; 0 (4) B(5 ; 0) 7.3 a  2 and q  10 CA(positive)a – value (2) A q – value 7.4 b (3) 3 25 3 b (3) 3 x   or x   A x     2a 2(1) 2 2 2 2a 2(1) 2 2 or 3 3 49 1 y     3   10   /  12,25 /  12 25 3 2 2 4 4 CA x   2 2  C(1,5 ; 12,25) (3) CAsubstitution CAminimum value 7.5 (  1,5 ;  9,25) CA x – value CA y – value (2) 7.6 3 x CACAanswer (2) 2 OR g / ( x).h / ( x)  0 CAproduct of derivatives 2 x  3 .2  0 3 CA(positive)answer x 2 penalize 1 mark for incorrect notation (2) [14] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 9 September 2016 Preparatory Examination NSC-MEMORANDUM(KZN) QUESTION 8 8.1 y  log 3 x AA answer (2) 8.2 y p (1 ; 3) AShape of p and p-1 1 Ay – intercept of p (3 ; 1) p-1 A x – intercept of p-1 A point on each graph O 1 x (4) 8.3 log 3 x  3 Msetting up equation x  27 CAx = 27 0  x  27 CACAfor end points and (4) inequality ANSWER ONLY full marks [10] QUESTION 9 [Penalise once for notational error in questions 9, 10 and 11] 9.1 f ( x)  x 3 f x  h   f x  f ( x)  lim A formula h 0 h  lim x  h 3  x3 h 0 h A substitution into correct formula x 3  3 x 2 h  3 xh2  h3  x 3  lim CA simplifying h 0 h  lim  h 3 x 2  3 xh  h 2  CAfactors h 0 h   lim 3x 2  3xh  h 2  h 0 CA answer (5)  3x 2 9.2.1 1 AA exponential form f ( x)  x  4 x  2 2 1 2 1 CACA answer f ( x)  x  8x 3 (4) 2 Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 10 September 2016 Preparatory Examination NSC-MEMORANDUM(KZN) 9.2.2 1 2  2  x  3  y Asquaring both sides 2  1 Asimplifying y  x 4  3x 2  9 4 dy  x3  6x CACAanswers dx (4) OR OR dy 1   2 x 2  3  . x dx 2  AAderivative using chain rule  x3  6x CACAanswers (4) [13] QUESTION 10 10.1.1 f (3)  (3) 2  8  1 A y = 1 ( y – value) (1) 10.1.2 f ( x)  2 x A value of gradient of tangent (1) m  f (3)  2(3)  6 10.1.3 Equation of tangent : - y  mx  c 1  6(3)  c c  17 CA substitution y  6 x  17 CAanswer (2) OR y  1  6( x  3) CA substitution y  6 x  18  1 CAanswer y  6 x  17 (2) 10.2.1 f ( x)  x 3  3x  2 f ( x)  3x 2  3  0 Aderivative and equating to 0 3( x  1)( x  1)  0 x  1 or x  1 CAfactors y0 or y  4 CAx - values CA y - values (4) A(1 ; 0) B(1 ;  4) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 11 September 2016 Preparatory Examination NSC-MEMORANDUM(KZN) 10.2.2 PQ  g ( x)  f ( x)  (2 x  2)  ( x3  3x  2) Aexpression for PQ   x3  5 x PQ   3x 2  5  0 A PQ = 0 5 x2  3 5 CA x – values x  1,29 3 For x  1,29 y  1,29   5(1,29)  4,3 3 CAy value for x = 1,29 Maximum Length of PQ  4,3units (4) 10.2.3 k  4 or k  0 k = 0 A k = -4 CAanswers (2) 10.2.4 f ( x)  6 x  0 A 6x x0 A 6 x  0 OR Aanswer (3) x1  x2 (1)  (1) x  0 Amidpoint formula 2 2 CA x  0 x0 CAanswer (3) OR b 0 x  0 3a 3(1) Aformula x0 A x  0 Aanswer (3) [17] QUESTION 11 (Ignore units in the question) 11.1 V  s(t )  2t  15 m / s Aanswer (1) 11.2 V  s(25)  2(25)  15  65 m / s CAanswer (1) 11.3 V  s(t )  2t  15 m / s (2) A  s(t )  2 m / s 2 CACA 2 11.4 A  s / (5)  2 m / s 2 / CAanswer (answer only 1 mark) (1) 11.5 s(t )  t 2  15t  250 Aequating s to 250 t 2  15t  250  0 CAfactors t  25 t  10   0 t  10 s CAt = 10 V  s(10)  2(10)  15  35 m / s CAanswer (4) [9] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 12 September 2016 Preparatory Examination NSC-MEMORANDUM(KZN) QUESTION 12 12.1.1 150 Aanswer (1) 12.1.2(a) 70 7 Aanswer (1) P(Male )   150 15 12.1.2(b) 80 8 Aanswer (1) P(Eating Chocolate)   150 15 12.1.3 45 AP(male eating chocolate ) P(Male eating chocolate)   0,3 150 P(Male) P(Eating chocolate) CA P(Male) x P(eating chocolate) value 70 80 56     0,249 150 150 225 P(Male and Eating chocolate)  P(Male) P(Eating chocolate) CAconclusion (3) Events are not independent 12.2.1 6 x 7 x 7 x 7 = 2058 A6 x 7 x 7 x 7 (2) A2058 12.2.2 6 x 6 x 5 x 4 = 720 A6 x 6 x 5 x 4 (2) A720 12.2.3 Four digit codes divisible by 5:- A6 x 7 x7 x 1 6 x 7 x 7 x 1 = 252 A252 Probability of a four digit code divisible by 5 252 7 =   0,1428  14,28 % CAanswer (3) 2058 49 [13] TOTAL MARKS: 150 Copyright reserved Please turn over