Mathematics Grade 12 Paper 1 Mpumalanga Mock 2018 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICS PAPER 1 SEPTEMBER 2018 MARKING GUIDELINES MARKS: 150 TIME: 3 hours These marking guidelines consist of 12 pages

Downloaded from hlayiso.com Mathematics P1 2 September 2018 NSC QUESTION 1 [26] 1.1.1 x 2  5x  6  0  Factors x  2x  3  0 x   2 or x   3  Answers (2) 1.1.2 x(2 x  5)  1  0 2 x 2  5x  1  0 Standard form  (5)  52  4(2)(1) x  Substitution into correct formula 2(2) 5  17 x 6  17 x  1,52 or x  0,15  Answers (5) 1.1.3 3x 2  8 x  4  0 3x  2x  2  0  Critical Values  Method OR 2  Notation (3) x  or x  2 3 OR  2 x ϵ    ; 3   2 ;   1.1.4 3 x  3 x  2  24  3 x 1  3 2  24 Common factor 8  Simplification 3 x    24 9  Exponential Law 3  27 x 3 x  33  Answer x3 (4) 1.2 y  12 x  2 x 2  y 2  6x  4 y  4 Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 3 September 2018 NSC y   12 x  2  y the subject Substitute y   12 x  2 in x 2  y 2  6 x  4 y  4 x 2   12 x  2  6 x  4( 12 x  2)  4  Substitution 2 x 2  4  2 x  14 x 2  6 x  8  2 x  4 5 2 Standard form x  6x  0 4 Factors 5  x x  6   0 4  5 x0 or x  6 4 24  x-values x  5 1  Substitution Substitute x  0 in y   x  2 2 1 y   ( 0)  2  2 2 24 1 Substitute x   in y   x  2 5 2 1  24 22 y   ( )  2  y-values (7) 2 5 5 OR y  12 x  2 x 2  y 2  6x  4 y  4 2 y  x  4 x  2 y  4  x the subject Substitute x  2 y  4 in x 2  y 2  6 x  4 y  4 (4  2 y) 2  y 2  6(4  2 y)  4 y  4  Substitution 16  16 y  4 y 2  y 2  24  12 y  4 y  4 Standard form 5 y 2  32 y  44  0 Factors 5 y  22  y  2  0 22  y - values y or y  2 5  22  24 x   2   4    5  5  Substitution x  2(2)  4  0  y-values Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 4 September 2018 NSC 1.3 px 2  x( p  2)  p  0 ∆=0 0   p  2  4 p  p  2  substitution in ∆ 0  p2  4 p  4  4 p2  Simplification  3p  4 p  4  0 2  Factors 3p2  4 p  4  0  answer (5) 3 p  2 p  2  0 2 p or p  2 3 QUESTION 2 [26] 2.1 .... ; .... ; 1 ; 7 ; 15 ; 25 ; ... ... 2.1.1 7 0 1 14  Differences 2 4  Answer (2)  T1  5 2.1.2 2a  2 a 1  Value of a 3a  b  2 31  b  2  Value of b b 1 1 1  c   5  Value of c c  5  Tn  n 2  n  5  General term (4) 2.1.3 Tn  n 2  n  5 415  n 2  n  5  Standard Form n 2  n  420  0  Factors n  21n  20   0  Correct n value (3)  n  21 Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 5 September 2018 NSC 2.2 x ; x  3 ; 3x  1 2.2.1 x  3 3x  1  Ratio  x x3 3x 2  x  x 2  6 x  9 2x 2  7 x  9  0  Standard Form 2 x  9x  1  0  Factors 9 x or x  1 2  Correct x values (4) 2.2.2 1 ; 2 ;  4  Sequence   a r n 1 Sn  r n 1  Substitution into the S 20    1  2 1 20  correct formula  2 1 Answer (3) S 20  349525 2.2.3 9 15 25 ; ; 2 2 2  value of r 5 r 3 Will not converge r is not between -1 and 1  Reason (2) 2.3 Tn  a  n  1d  Substitution into correct Tk  a  k  1d  m formula a  m  k  1d  Substitution into correct Tm  a  m  1d  k formula a  k  m  1d m  d k  1  k  d m  1  Equating  d k  1  d m  1  k  m  Factors d m  1  k  1  k  m k m  Answer (5) d mk 2.4 1  q  2  q  3  q  4  q  5  q  15k  Equation 5q  15  15k  Simplification 5q  15k  15  Answer (3) q  3k  3 Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 6 September 2018 NSC QUESTION 3 [21] 3.1 y  x 2  2x  3  Completing of the square y  ( x 2  2 x  1  1)  3  Value of x y  ( x  1) 2  4  Value of y TP (1 ; 4)  Turning point (4) OR b y  (1) 2  2(1)  3 x 2a 4  Formula for x 2 x  Value of x 2( 1)  Value of y x 1  Turning point TP (1 ; 4) 3.2  x 2  2x  3  0 x 2  2x  3  0  Factors x  3x  1  0  Correct x values x  3 or x  1  Answer (3) AB = 4units 3.3 EF   x 2  2 x  3   x  3  f x   g x    x 2  3x b Max : 2 x  3  0 OR x  Derivative OR formula 2a 3 3 3 x x  2 2(1) 2  Value of x (3) 3.4 p( x)  ( x  1) 2  4  2  Answer (2) p( x)  ( x  1) 2  2 y  2; y  R OR ( ; 2] 3.5.1 Four units down  Translation One unit to the left  Translation  Reflection (3) Reflection over the x-axis Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 7 September 2018 NSC 3.5.2 x  0 or x  0  Answers (2) 3.6.1 B3 ; 0   Critical values 0  x 3  Notation (2) 3.6.2 x 1  Answer (2) QUESTION 4 [12] 4.1.1 y  ax OR 1  a1 9  a 2  Substitution 3 1 1 a  a   Answer (2) 3 3 4.1.2 k g ( x)  q x p  Substitution of p and q k g ( x)  1 x2  1  Substitution of 1;  1  k 1  3 3 1 2 2 k  3 3  Value of k k  2 2 g ( x)  1  Equation (4) x2 4.2 y  log 1 x  Answer (1) 3 4.3 h( x)  x  c 9 2c  Substitution of (–2; 9) c  11  Answer (2) 4.4 0 x 1  Answer (2) OR x  ( 0 ;1] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 8 September 2018 NSC QUESTION 5 [15] 5.1 A  P1 i  n  Substitution into correct x  2 x1 0,075  n formula  1  0,075  1 n 2  Substitute A & P 1 log  n log 0,925 2 Use of logs 1 log n 2 log 0,925 n  8,89  Answer n = 8 years 11 months  Rounding to months (5) 5.2.1  Answer (1) 30 18 480  100 = R5 544 5.2.2 P   x 1 1  i  n   Substitution in correct i Formula   0,08   300 5544 1 1      12   value of n P  0,08 12  value of i P  R 718 305,71  Answer (4) 5.3 A  P1  i  n 18  0,15   Substitution in correct A  1000 1    12  Formula A  R 1 250,58  Value of A F   x 1  i   1 n  i  18   Substitution into correct 0,15  700 1    1 formula  12    0,15 12  Value of F x = R 14 032,33  answer (5) Total: R 15 282,91 Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 9 September 2018 NSC QUESTION 6 Penalty of –1 for notation in Question 6 [13] 6.1 f x    2 x 2  5 f x  h   2x  h2  5  Finding f(x + h) f x  h   2 x 2  4 xh  2h 2  5  Simplification f ( x  h) 2  f ( x ) f  x   lim h 0 h f x   lim   2 x 2  4 xh  2h 2  5   2 x 2  5   Correct substitution into Formula and notation h 0 h  2 x 2  4 xh  2h 2  5  2 x 2  5 f x   lim h 0 h  4 xh  2h 2 f x   lim h 0 h  Common factor h 4 x  2h  f x   lim h 0 h  Answer (5)   4x 6.2 6 y  5x   x½ x 6 y  5x  1 x2  Exponent Law  12 y  5x  6 x dy 3   Derivative  5  3x 2 dx (4) 6.3.1 2 x  x  3x 3 2 g ( x)  x 1 x2 x  3 x  1  Factorise  x 1  Simplify  x(2 x  3)  2 x 2  3x  Derivative g ' ( x)  4 x  3 g ' (2)  11  Answer (4) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 10 September 2018 NSC QUESTION 7 [16] 7.1 y  ax  x1 x  x2 x  x3   Substitution of x values  1 y  2x  1 x  x  3  2 into correct formula    1 y  2 x 2  2x  3  x    Simplification  2  5 3 y  2 x 3  x 2  2 x    2 2  Simplification y  2 x 3  5x 2  4 x  3  Answer (4) 7.2 y  2 x 3  5x 2  4 x  3  Derivative 6 x 2  10 x  4  0 =0 3x 2  5 x  2  0 x  23x  1  0  Factors 1 x  2 or x   3  x - values 3 2  1  1  1 y  2    5    4    3  3  3  3  select one value 100 y 27  1 100  A  ;   y – value (6)  3 27  7.3 f ( x)  2 x 3  5 x 2  4 x  3  Value of f(-2) f ( x)  2(2) 3  5(2) 2  4(2)  3  25  D(2;25) f ' ( x)  6 x 2  10 x  4  Value of f’’(-2) f ' ( x)  6(2) 2  10(2)  4  40 y  y1  m( x  x1 )  Substitution into correct y  25  40( x  2) formula y  40 x  55  Answer (4) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 11 September 2018 NSC 7.4 12 x  10  0 5  Inequality x  6  Answer (2) QUESTION 8 [7] 8.1 Area = 2(½b x H) + Lx B + Lx B+ Lx B  area 600 = 5x.12x +5xy + 12xy + 13xy  = 60 x2 + 30xy method 600  60 x 2  y 30 x Answer 20  2 x 2 (3) = x 8.2 Volume = area of base x h  Volume 20  2 x 2  = ½ 5x. 12x . x  simplification = 600x – 60x3  Max : 600 – 180x = 0 2  f ' ( x)  0  10 answer x2  3 (4) 10  x 3 Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P1 12 September 2018 NSC QUESTION 9 [15] 9.1.1 Event A : 1,2 B: 2,4,6 C:6  P(A) 2 1 P(A) =  6 3 P(B) 3 1 P(B) =  6 2 P(A or B) = P(A) + P(B) – P(A and B)  Substitution into correct 2 1 1 formula =   6 2 6 2  Answer = 3 (4) 9.1.2 A and C are mutually exclusive  Events P(A and C) = 0  Reason (2) 9.1.3 1 1 1  P(B) x P(C) P(B) x P(C)=   2 6 12 1 P( B  C )   P( B  C ) 6 P( B)  P(C )  P( B  C ) Conclusion (3) B and C are not independent events 9.2.1 6!  720  6!  Answer (2) 9.2.2 1! 4!1!  24  1! 4!1!  Answer (2) 9.2.3 2!  4!  5  240  2!  4!  5 240 1  Answer P(Visits )   (2) 720 3 TOTAL: 150 Copyright reserved Please turn over