Mathematics Grade 12 P1 Gauteng Mock 2022 Memo (English)

Downloaded from hlayiso.com PREPARATORY EXAMINATION 2022 MARKING GUIDELINES MATHEMATICS (PAPER 1) (10611) 23 pages

Downloaded from hlayiso.com 10611/22 INSTRUCTIONS AND INFORMATION A – Accuracy C.A. – Consistent Accuracy 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 guidelines.  It is unacceptable for candidates to use adopted values/answers in solving questions. 2

Downloaded from hlayiso.com 10611/22 QUESTION 1 1.1 1.1.1 2 x( x 2  1)  0  factors 2 x( x  1)( x  1)  0 x  0 or x  1 or x  1  answers NOTE: Any other valid method (2) 1.1.2 2 x6 0 x x2  6x  2  0  standard form  b  b  4ac 2 x 2a  (6)  (6) 2  4(1)(2) x 2(1)  substitution into correct formula 6  28 x 2  x  5,65  or  x  0,35  answers NOTE: Penalise for rounding-off in this question ONLY. Any other valid method (4) 1.1.3 (x – 1)(x + 4)  6  x2 + 3x – 4  6  x2 + 3x – 10  0  standard form –5 2  (x + 5)( x– 2)  0  factors  x  –5 OR x  2  answers (3) 3

Downloaded from hlayiso.com 10611/22 1.1.4 10 x2 3 x2 Let k  x  2 10 k 3 k k 2  3k  10  0  standard form ( k  5)( k  2)  0  factors k  5 or k  2  both answers for k  selection x  2  5 or x2  2  x  2    2 2 2 x  6  answer OR 10 x2 3 x2  x  2  x  2   3 x  2  10 x  2  3 x  2  10 3 x  2  12  x  simplified both sides  3 x  2   12  x   squaring both sides 2 2 9 x  18  144  24 x  x 2 x 2  33 x  162  0  standard form ( x  6)( x  27)  0  factors x  6 or x  27  selection (5) 4

Downloaded from hlayiso.com 10611/22 1.2 x  2 y  1 and 2 x2  xy  5 y  3 y 2  2  0  equation for x x  2 y  1 …(1) 2 x2  xy  5 y  3 y 2  2  0 ….(2) Substitute (1) into (2): 2  2 y  1   2 y  1 y  5 y  3 y 2  2  0 2 3 y2  2 y  0  standard form y (3 y  2)  0 2  y-values y  0 or y   3 1 x  1 or x    x-values 3 NOTE: Any other valid method (4) 1.3 2 x 1  2 x  3 y  2  3 y 2 x (21  1)  3 y (32  1)  factorise 2 x (3)  3 y (8) 2x 3y   8 3 x 2 3y  3  2 3  2  3 y 1 x 3  simplified equated bases x3  0 and y 1  0  x  3 and y  1  answers OR  x 1 y2 2 2 3 x 3 y  2 (2  1)  3 (3  1) x 1 y 2  factorise 2 (3)  3 (8) x y  2 x (3)  3 y (23 )  simplified equated bases  x  3 and y  1  answers (3) 5

Downloaded from hlayiso.com 10611/22 1.4 x 2  rx  m  0 and x 2  mx  r  0 x 2  rx  m  0 For real and equal roots,   0 b 2  4ac  0  substitute into   0  r   4(1)(m)  0 2 r 2  4m  0 r 2  4m r2 m ... (1)  equation for m 4 x 2  mx  r  0 b 2  4ac  0 m2  4(1)(r )  0 m 2  4r  0 ... (2)  equation 2 Substitute (1) in (2) 2  r2     4r  0  substitute for m 4 r4  4r  0 16 r 4  64r  0 r (r 3  64)  0 r (r  4)(r 2  4r  16)  0 r  4  value of r r2 m 4 42 m 4 m  4  value of m OR OR 6

Downloaded from hlayiso.com 10611/22 x 2  rx  m  0 and x 2  mx  r  0 For real and equal roots quadratic must be a perfect square. x 2  rx  m  0 x  m  0 2 x2  2 mx  m  0 x  m  0 2  r2 m r 2  4m r2  m ... (1) equation for m 4 x 2  mx  r  0 x  r   0 2 x2  2 r  r  0 m2 r  equation 2 m  4r 2 ….(2) Substitute (1) in (2) 2  r2   substitute for m    4r  0 4 r4  4r  0 16 r 4  64r  0 r (r 3  64)  0 r (r  4)(r 2  4r  16)  0 r  4  value of r r2 m 4 42 m 4 m  4  value of m (6) [27] 7

Downloaded from hlayiso.com 10611/22 QUESTION 2 2.1 2.1.1 20 12 10 14 –8 –2 4 6 6  2nd diff. 2a = 6 3a + b = –8 a + b + c = 20  a=3 a=3 3(3) + b = –8 3 – 17 + c = 20  b = –17 b = –17 c = 34  c = 34 Tn  3n 2  17 n  34 (4) 2.1.2 Tn  a  (n  1)d 148  8  (n  1)(6)  substitution into correct formula 148  6n  14 n  27  value for n Between 27 th and 28 th terms.  conclusion (3) 2.1.3 Sn = n 2a  (n  1)d  2 n 2(8)  (n  1)6  > 10140  substitution 2 3n2 – 11n > 10140 3n2 – 11n – 10140 > 0  standard form (3n + 169) (n – 60) > 0  factors  n > 60  selection, n > 60 n = 61  answer (5) 5 2.2  (r  b)  (1  b)  (2  b)  (3  b)  (4  b)  (5  b) r 1  expansion 5  (r  b)  15  5b r 1 5   (r  b)  10 a r 1  equating 15  5b  10 a  5b  10 a  15  b  2a  3  answer (3) [15] 8

Downloaded from hlayiso.com 10611/22 QUESTION 3 3.1 24 a= x 6x 3x 2 x r= or or 12 6x 2  value of r a S = 1 r 24  substitution in correct S = x formula x 1 2 48  correct numerator S = 2x  x 2  correct denominator (4) 3.2 It exists when: x  –1 < r < 1 –1 < < 1 i.e.: –1 < r < 1 2  –2 < x < 2  answer ANSWER ONLY: Award full marks. (2) 3.3 r>1  r >1 x  >1 2  x >2  answer ANSWER ONLY: Award full marks. (2) 3.4 24 a= 4 a=6  value of a x r= =2  value of r 2 n = 15 a( r n  1) Sn = r 1 6( 215  1)  substitution into correct  S15  formula 2 1 S15  196 602  answer (4) [12] 9

Downloaded from hlayiso.com 10611/22 QUESTION 4 4.1 a y 2  x 1 substitute b and c a  substitute point (0 ; 0) 0 2 0 1 0  a  2 a2  value for a 2 y 2 x 1 (3) 4.2 f ( x)  ( x  p ) 2  q 5 0  (  1) 2  q  substitute p and point A 2 9 9 q  q 4 4  9 Turning Point :  1;    4  correct x-value of turning point OR 1 5 y  a( x  )( x  ) 2 2 1 5 y  1( x  )( x  ) 2 2  substitute a and both roots 5 y  x2  2x  4 (2) x 2(1) x 1  value for x 9 y  value for y 4 NOTE: Answer does not have to be in coordinate form. (3) 4.3 x2  answer NOTE: Answer only, award FULL marks (1) 4.4 y  y1  m( x  x1 ) y  2  1( x  1)  substitute m = –1 and pt. (1;2) y  x  3  answer NOTE: Answer only, award FULL marks (2) 10

Downloaded from hlayiso.com 10611/22 4.5 k ( x)   g ( x)  k(x) = –g(x) 2  k ( x )  (  2) x 1 2  k ( x)  2  answer x 1 (2) 4.6 x  1 or 0  x  1  answer  answer OR  answer x  (1; ) or (0<x  1)  answer OR x  0 ; x 1  answer  answer OR x  (0; ) ; x  1  answer  answer (2) 4.7 k > 0  answer (1) [14] 11

Downloaded from hlayiso.com 10611/22 QUESTION 5 5.1 (0;1)  answer NOTE: Answer MUST be in coordinate form. (1) 5.2 1 a 1   substitute 2 1 1 point  a 2  answer a2 (2) 5.3 2 x y  interchange x and y y  log 2 x  answer NOTE: Answer only, award FULL marks (2) 5.4  shape  x-intercept (2) 5.5 x > 0  answer (1) 5.6 x2  answer (1) 12

Downloaded from hlayiso.com 10611/22 5.7 g ( x)  100.3x h( x )  2 x if g ( x )  h( x ) 100.3x  2 x 2x 100  3x 2  simplification 100  ( ) x 3  x  log 2 100  x as subject 3  x  11,36  answer OR OR 2  x log 3  x.log 2  2  x.log 2  x log 3 2  2  x.log 3  equating simplified 2 sides x   x as subject 2 log 3  x  11,36  answer (3) [12] 13

Downloaded from hlayiso.com 10611/22 QUESTION 6 6.1 0 x3  critical values NOTE: The answer may be written as separate inequalities.  notation OR  critical values x  [0;3]  notation (2) 6.2 Equation of f : y   27 x for x  0  interchange x and y Equation of f 1 : x   27 y for y0  constraint of f 1 x2  equation of y y  for x  0 27  constraint of y (4) 6.3  shape  x-intercept  point (–9 ; 3) (3) 6.4 Reflection in the x-axis  answer (1) [10] 14

Downloaded from hlayiso.com 10611/22 QUESTION 7 NOTE: Incorrect formula, STOP marking, unless independent marks are being awarded. 7.1 A  P(1  i ) n i 2 x  x(1  ) 24  n = 24 4 i  substitution  (1  ) 24  2 4 i 1   24 2 4 i   0,0293  simplified sides 4  i  0,1172  value of i The annual interest rate  11,72%  answer NOTE: If a candidate substitutes in their own values for A and P, award 4/5 marks. (5) 7.2.1 Fv = Pv (1 + i)n 9 ,5 Fv =10 000 (1  12 )  substitution 5 100 Fv =10 402,15  answer NOTE: Any other valid method (2) x[1  (1  i )  n ] 7.2.2 Pv = i 450[1  (1,0079)  n ] 10 402,15 =  substitution into the 0.0079 correct formula [1 – (1,0079)–n] =0,183 1 – 0,183 = (1,0079) –n  0,8169 = (1,0079) –n  simplified sides –n = log1,0079 0,8169  correct use of logs –n = –25,7008  n = 26 months  answer NOTE: Answer mark is for 26 months. (4) 15

Downloaded from hlayiso.com 10611/22 7.2.3 Balance of loan after 25 th payment = value of loan – value of annuity at that time 450 [(1,0079 ...) 25  1] = 10 402,15(1,0079 …)25 –  substitution 0,0079 = 12 668,89 – 12 386,53 = R282,36  answer OR After the 25 th payment, the remaining number of payments = 25,63 – 25 = 0,63 0,095 0,70 450 [1  (1  ) ] balance = 12 0,095  substitution 12 balance = R282,36  answer (2) [13] QUESTION 8 8.1 f ( x)  3 x 2  2 x f ( x  h)  3( x  h)2  2( x  h)  f (x + h) f ( x  h)  3x  6 xh  3h + 2 x  2h 2 2 f ( x  h)  f ( x)  3x 2  6 xh  3h 2  2 x  2h  3x 2  2 x  f(x + h) - f(x)  6 xh  3h 2  2h f x  h   f x  f / x   lim h0 h ℎ (6𝑥 +3ℎ + 2)  factorise =lim ℎ→0 ℎ =lim (6𝑥 + 3ℎ + 2) ℎ→0  6x  2  answer NOTE: Penalise 1 mark for notation error in this question. (4) 16

Downloaded from hlayiso.com 10611/22 8.2 8 y4 x   x3 x 1 1  y  4 x 2  8 x 2   x3  change surds 1 3 to rational dy    2 x 2  4 x 2  3 x 2 exponents dx  answers 2 4 NOTE: Accept 1  3  3x 2 2 2 x x OR 2 4   3x 2 3 x x (4) 8.3 This question has been removed from the question paper. Do not mark this question. (0) [8] 17

Downloaded from hlayiso.com 10611/22 QUESTION 9 9.1 h(x) = – [x(x – 6)2]  substitute roots h(x) = – [x3 – 12x2 + 36x]  simplification h(x) = – x3 + 12x2 – 36x  simplification  a = 12  b = –36 (3) 9.2 h( x)   x3  12 x 2  36 x h( x)  3x2  24 x  36 3x 2  24 x  36  0  derivative = 0 3( x  6)( x  2)  0  x  6 or x  2 x  2 at point A  choice of x-value  h(2)  32  y-value A(2; – 32) NOTE: Answer does NOT have to be in coordinate form. Equating to zero must NOT be implied. (3) 9.3 h(3) = –(3)3 + 12(3)2 – 36(3) y = –27  answer Answer only, FULL marks (1) 18

Downloaded from hlayiso.com 10611/22 9.4 h( x)   x 3  12 x 2  36 x h( x)  3 x 2  24 x  36 h( x)  6 x  24  h(x ) h(3)  6(3)  24 h(3)  6  value of h(3)  h(3)  0 h is concave UP at point T.  conclusion OR h "( x)  0  h(x ) 6 x  24  0  point of inflection x  4 Point of Inflection is at x  4 at x = 4 at Point T, the graph is CONCAVE UP  conclusion NOTE: Candidate must have a valid CALCULATION to reach a conclusion for full marks. If only the conclusion is given, award 1 mark. (3) 9.5 h( x)  6 x  24  0  6 x  24  6 x  24 x  4  value of x  h(4)  16  value of y Point of inflection is: (4 ; –16) NOTE: Answer does not have to be in coordinate form. (2) [12] 19

Downloaded from hlayiso.com 10611/22 QUESTION 10 10.1 Rent Occupation (Rented) Income R500 72 R36 000 R600 72 – 2 x  500 x  500 Number of increases:  100 100 x  500 x  500 Occupation (Rented): 72  2( )  72  2( 100 ) 100 2 x 1 000 Income(I) = x[72   ]  substitution 100 100 7 200  2 x  1 000 7 200  2 x  1 000 = x[ ]  x[ ] 100 100  8 200  2 x   x  8 200  2 x   100    x  100  x2  82 x  50 (5) 10.2 x2 I  82 x  50 2x I   82  50 2x 82  0  derivative = 0 50 2x 82  50 x = R2 050  answer NOTE: Equating to zero MAY be implied. (2) [7] 20

Downloaded from hlayiso.com 10611/22 QUESTION 11 11.1 P(A or B) = P(A) + P(B) 0,57 = P(A) +2 P(A)  substitution  3P(A) = 0,57  P(A) = 0,19  value of P(A)  P(B) = 2 P(A) = 2(0,19) P(B) = 0,38  answer (3) 11.2 11.2.1 Not Defective (ND) 35 7 P(ND) = or or 0,88  answer 40 8 (1) 11.2.2 Not Defective (ND)/Defective (D) P(ND and D) + P(D and ND)  formula ( may be implied) 35 5 5 35 =( . )+( . ) 40 39 40 39 35  substitution = or 0,22 156 (3)  answer 11.2.3 Defective (D) P(D and D) 5  5 4 40  . 40 39 4  20 1 39 = or or 0,01  answer 1 560 78 (3) 21

Downloaded from hlayiso.com 10611/22 11.3 11.3.1 7x6x5x4x3x2x1 OR 7ǃ OR 5 040  answer (1) 11.3.2 Total number of arrangements = 2ǃ x 6ǃ  answer OR 1 440 (1) 11.3.3 Economics books = 4ǃ Life Sciences books = 3ǃ  combinations for Economics and Life Sciences = 2ǃ all criteria Total number = 4ǃ x 3ǃ x 2ǃ OR 24 x 6 x 2  answer OR 288 (2) [14] TOTAL: 144 Note: The mark out of 144 MUST be converted to 150. 22

Downloaded from hlayiso.com 10611/22 DETAILED EXPLANATION OF QUESTION 10  For every R100 that the price of a room increases, 2 rooms less are let.  To determine the amount of rooms that been let, the number of increases of R100 must be determined because that value must be multiplied by 2 to subtract from 72.  Example: If the rooms are let for R800 per day, then the number of increases of R100 is 3. 800 − 500 Therefore: Number of increases = 100 Therefore, 72 - 2(3) = 66 rooms let. Income will be: R800 x 66 = R52 800. Now, the new price is not R800, but Rx, making the formula: 𝑥 − 500 Number of increases = 100 𝑥 − 500 Therefore, the number of rooms let = 72 − 2 ( 100 ) 𝑥 − 500 Therefore, the income = = 𝑥 [72 − 2 ( 100 )] 23