Thursday, 10 June 2021 08:57

## GRADE 12 MATHEMATICS PAPER 1 MEMORANDUM - NSC PAST PAPERS AND MEMOS NOVEMBER 2016

MATHEMATICS PAPER 1
GRADE 12
MEMORANDUM
NOVEMBER 2016
NATIONAL SENIOR CERTIFICATE

NOTE:
• If a candidate answers a question TWICE, only mark the FIRST attempt.
• Consistent accuracy applies in all aspects of the marking memorandum.

QUESTION 1

 1.1.1 x(x-7)=0x=0 or x=7 ✓x = 0✓x= 7(2) 1.1.2 x2 -6x + 2 = 0 x = 0,35 or x= 5,65ORx2 - 6x + 2 = 0x2 - 6x + 9 = 2+9(x-3)2 = 7x-3 = ± √7x= 0,35 or x=5,65 ✓correct substitution into✓correct formula✓x = 0,35✓x = 5,65(3)✓(x-3)2 = 7✓x= 0,35✓x=5,65 (3) 1.1.3 √x-1 + 1 = x√x-1 = x-1x-1 = x2 - 2x + 1x2 - 3x + 2 = 0 (x+2)(x-1) = 0 x=2 or x=1Both  answers are validOR OR By inspection:x-1=0 or x-1=1x=2 or x=1 ✓ isolate √x-1✓ x2 - 2x+1✓ standard form✓ factors✓ both answers(5)✓ isolate √x-1✓ k2✓ standard form✓ factors✓ both answers(5)✓ isolate √x-1✓ x-1 = 0✓ x-1 = 1✓ x = 2 ✓ x=1(5) 1.1.4 3x+3 - 3x+2 = 4863x33-3x32=4863x(33-32)=4863x=273x=33x=3OR3x+3 - 3x+2=4863x+2(31-1) = 4863x+2 = 35x+2 = 5x+2 = 5x= 3 ✓expansion✓common factor✓3x=27✓x=3 (4)✓common factor✓(31-1)✓3x+2=243✓x+3(4) 1.2.1 ƒ(x) = x2 + 3x - 40=(x+4)(x+1)x=-4 or x=1 ✓factors✓both answers(2) 1.2.2 x2 + 3x - 4 < 0(x+4)(x-1) < 0 -4 < x< 1 OR X∈(-4;1) ✓✓-4 < x< 1 (2) 1.2.3 2x+3≥0x≥-3/2ƒ(x)≥0 when ƒ is increasingThe turning point occurs at x = (-4+1)/2x≥-3/2 ✓2x+3✓x≥-3/2 (2)✓x = (-4+1)/2✓x≥-3/2 (2) 1.3 x=2y and x2 - 5xy = -24(2y)2 - 5(2y)(y) = -244y2-10y2=-24-6y2 = -24y2=4y=-2 or y=2x=-4 or x=4OR OR ✓substitution of 2y✓-6y2=-24✓ both y – values✓ both x– values(4)✓ substitution of x/2✓ (-3/2)x2=-24✓ both x – values✓ both y – values (4)✓ equating x=x2+24                   2      5x✓ 3x2=48✓ both x – values✓ both y – values (4) 

QUESTION 2

 2.1 T4=-7 ✓-7 (1) 2.2 Tn = a+(n-1)d-87=5+(n-1)(-4)-87=5-4n+44n=96n=24OR-4n+9=-87-4n=-96n=24 ✓a=5 and d=-4✓-87 = 5+(n-1)(-4)✓n=24 (3)✓-4n+9✓-4n+9=-87✓n=24 (3) 2.3 -3;-7......;-87 OR ORAll negative terms can be written down and added to get the answer of –990OR ✓n=22✓a=-3✓answer (3)✓n=22✓a=-3✓answer(3)✓a=-3answer(3)✓ ✓-6✓answer(3) 2.4 5;-15;-35......d=-20tn=-20n +25Last term in the sequence divisible by 5 is:-4187+4(3)=-4175Tn =-20n+25-4175 =-20n+2520n=4200n=210There will be 210 terms in the sequence that is divisible by 5.OR5 ; 1 ; –3 ; ... ; –83 ; –87; ……; – 4187Tn=-4n+9-4187 =-4n+94n=4196n=1049There are 1049 terms in the sequenceT1;T6;T11;T16 are divisible by 5The largest integer value of k such that5k-4≤10495k ≤ 1053k≤210.6k=210OR5 ; 1 –3 ; –7 ; … ; –4175; –4179 ;–4183 ; –4187Tn=a+(n-1)d-4175 = 5+(n-1)(-4)-4180 = -4(n-1)n=1046Number of terms divisible by 5=1046 - 1 +1        5=210 ✓d=-20✓Tn=-20n+25✓-4175 =-20n+25✓n=210 (4)✓-4n+9=-4187✓n=1049✓5k-4 ≤1049✓k=210 (4)✓d=-4✓-4175=-4n+9✓1046✓n=210 (4)

QUESTION3

 3.1.1 – 1 ;   x ;   3   ;   x + 8 ; ... - 2x + 2 = 2x + 24x = 0x = 0 ✓x + 1; 3 - x and x✓calculating second differences✓- 2x + 2 = 2x + 2 ✓x = 0(4) 3.1.2 First differences 1 ; 3 ; 5 ; ……Sn = n [2(1) + (n - 1)(2)]         2= n2250 < n 2n > √ 250∴ n > 15,8The sum of the 16 first differences will be greater than 250. Therefore the 17th term of the quadratic number pattern is the first satisfying this condition. ✓S = n 2✓Sn > 250✓n > 15,8✓n = 17(4) 3.2.1 21 + 21(0,85) + 21(0,85)2  + .........T = ar n-1 nT   = (21)(0,85)9= 4,86 cm ✓n = 10 ; r = 0,85 or 17                                  20✓substitution into correct formula✓answer(3)

 3.2.2 Sn a(1 - r n )         1 - rS15= 21(1 - (0,85)15 )              1 - 0,85= 127,77 Area of the page = 30 x 21 = 630 Percentage of paper covered in grey ink:= 127,77 x 100%      630= 20,28% ✓n = 15✓127,77 ✓630✓20,28(4)

QUESTION 4

 4.1 y = 0 ✓y = 0 (1) 4.2 R(0 ; 1) ✓answer (1) 4.3 y = ax9 = a2∴ a = 3 ✓substitution✓a = 3(2) 4.4 DP = 2 – by = 3x1 = 3b813-4 = 3bb = - 4DP = 2 - (-4)= 6 units ✓1 = 3b   81✓3-4 or use of logs✓b = - 4✓DP = 6 units(4) 4.5 h(x + 2) + k = 0h(x + 2) = -k0 < -k < 1             81- 1 < k < 0 81 ✓✓- k < 1    or k > - 1        81              81✓- 1 < k < 0    81(3)

QUESTION 5

 5.1 f (x) = -x 2 + 4x - 3f '(x) = 0      or      x = -    4 2(- 1)- 2x + 4 = 0                     x = 2x = 2y = -(2)2  + 4(2) - 3= 1B(2 ; 1)OR - x 2 + 4x - 3 = 0x 2 - 4x + 3 = 0(x - 3)(x - 1) = 0x = 3 or x = 1x = 3 + 1        2x = 2y = -(2)2  + 4(2) - 3= 1B(2 ; 1) ✓- 2x + 4 = 0 orx = -    4           2(-1)✓y = -(2)2  + 4(2)-3(2)✓x = 3 +1         2 ✓y =-(2)2  +4(2)-3(2) 5.2 Range: y ≤ 1ORRange: y ∈ (-∞ ; 1] ✓y ≤ 1 (1)✓(-∞ ; 1] (1) 5.3 x ≤ -1    or     x > 2OR( -∞; -1]∪(2 ;-∞) ✓critical values✓x ≤-1  or x>2(2)✓critical values✓x ≤-1  or x>2(2) 5.4 (x - p) (y + t ) = 3Vertical asymptote of h(x)  at x = 2Translation 4 units to the left x = 2 – 4 = –2 is the equation of the vertical asymptote of h(x + 4)OR ✓x = -2 (1)

 h(x) =         3       + 1             x - 2 + 4=     3    + 1   x + 2x = -2is the equation of the vertical asymptote ✓x = -2 (1) 5.5 (x - p)(y + t ) = 3(y + t ) =    3   - t               x - py =    3    - t       x - pB(2 ;1)Point of intersection of the asymptotesp = 2- t = 1t = -1 ✓  3        x - p✓- t✓p = 2✓t = -1(4) 5.6 x-intercepts of f :x 2 - 4x + 3 = 0(x - 3)(x -1) = 0x = 1 or     x = 3g / (x) < 0 for x ∈ R ; x ≠ 2Hence f (x) < 0x ≤ 1 or x ≥ 3       OR    (- ∞ ; 1]∪[3 ; ∞) ✓both critical values✓x ≤ 1✓or✓x ≥ 3(4)

QUESTION6

 6.1 g✓shape: increasing curve✓(1 ; 0):only on log graphf:✓(3 ; 0)✓(0 ; 3)(4) 6.2 y = log2 xg -1 : x = log2 yy = 2x ✓interchange x and y✓y = 2 x(2) 6.3 log2 (3 - x) = x2 x = 3 - x2 x = -x + 3Reflect the graph of g about the line y = x to obtain g -1 and determine the point of intersection of f and g -1 ✓✓2x = -x + 3✓point of intersection off and g -1(3) 6.4 x = 1 ✓answer(1)

QUESTION 7

 7.1 A = P(1 + i )n 🗸substituting i and values in correct formula 🗸 answer(2) 7.2 ✓i = 0,15        12✓n = 46✓substitution into correct formula✓answer(4)✓i = 0,15      12✓n = 46✓substitution into correct formula✓answer(4) 7.3 n = 35,41872568 months∴ 36 payments are required∴Thabiso will pay his loan off 10 months soonerOR ✓x = 9 000✓substitute into correct formula✓use of logs✓n =35,42✓10 months(5) n = 35,41872568∴ 36 payments are required∴Thabiso will pay his loan off 10 months sooner. ✓9 000✓substitute into correct formula✓use of logs✓n =35,42✓10 months(5) 7.4 The balance of his loan after the 35th payment was made: OR OR (4)  ✓✓✓✓ (4)

QUESTION 8

 8.1 f (x + h) = 3(x + h) 2= 3(x 2 + 2xh + h 2 )= 3 x 2 + 6xh + 3h 2f (x + h) - f (x) = 3 x 2 + 6xh + 3h 2 - 3x 2= 6xh + 3h 2OR  (5)

 8.2 ✓answer✓answer(2) 8.3  (4) 8.4 f (x) = x3 + ax2 + bx +18f / (x) = 3x2 + 2ax + bAt x = 1, mtan = -8f / (1) = -83(1)2  + 2a(1) + b = -83 + 2a + b = -82a + b = -11........... (1)y = f (1)= g(1)= -8(1) + 20= 121 + a + b + 18 = 12a + b = -7............ (2)a = -4b = -3 ✓3x2 + 2ax + b✓f / (1) = -8 or3(1)2  + 2a(1)+ b = -8✓1+ a + b +18 = 12✓a = -4✓b = -3(5)

QUESTION 9

 9.1 f'(x) = 3x2 + 8x - 3 = 0(3x - 1)(x + 3) = 0x = 1     or    x = -3      3 ✓equating derivative to zero✓factors✓x – values (3) 9.2 f // (x) = 6x + 8 6x + 8 < 0x < - 4        3OR ✓6x + 8✓✓x < - 4        3 (3) (3) 9.3 x ≤ -3   or     x ≥ 1                          3OR ✓x ≤ -3✓x ≥ 1        3 (2)✓[-∞;-3]✓ (2) 9.4 f (0) = -18d = -18f (x) = ax3 + bx 2 + cx -18f'(x) = 3ax 2 + 2bx + c f'(x) = 3x 2 + 8x - 33a = 3              2b = 8a = 1                  b = 4           c = -3f (x) = x3 + 4x 2 - 3x -18ORf'(x) = 3x2 + 8x - 3By integrationf (x) = x3 + 4x 2 - 3x + d f (0) = d = -18a = 1b = 4c = -3 ✓d = -18✓f '(x) = 3ax2 + 2bx + c✓a = 1✓b = 4✓c = –3(5) ✓f (x) = x3 + 4x2 - 3x + d✓d = -18✓a = 1✓b = 4✓c = –3(5) 

QUESTION 10

 10.1 M(t ) = -t 3 + 3t 2 + 72tM(3) = -(3)3  +  3(3)2  + 72(3)= 216216 molecules M(3) =✓-(3)3+3(3)2+72(3)✓216(2) 10.2 M (t ) = -t 3 + 3t 2 + 72tM / (t ) = -3t 2 + 6t + 72M / (2) = -3(2)2  + 6(2) + 72= 7272 molecules per hour ✓M/ (t ) = -3t 2 + 6t + 72✓M/ (2)✓72(3) 10.3 M (t ) = -t3 + 3t 2 + 72tM /(t) = -3t 2 + 6t + 72M // (t) = 0- 6t + 6 = 0t = 1Maximum rate of change of the number of molecules of the drug in the bloodstream is after 1 hour ✓M// (t )✓M // (t ) = 0✓answer(3)

QUESTION 11

 11.1 ✓a = 20✓b = 128(2) Watches TV during exams Do not watch TV during exams Male 80 a Female 48 12 Total b 32 a +12 = 32a = 20b = 80 + 48= 128 11.2 NoP(M and not watching TV) = 20  ≠0                                             160 ✓No✓reason(2) 11.3.1 P(watching TV) = 128 = 4 = 0,8 = 80%                             160    5 ✓128✓160(2) 11.3.2 P(female and not watching TV) = 12 = 3 = 0,075 = 7,5%                                                    160   40 ✓12✓160(2) 

QUESTION 12

 12. We want to create codes that are even numbers greater than 5000. The digit 6 can be used in one of two places in these codes and therefore this presents two scenarios.CASE 1: The first digit is a 6.   6                                                                2                                                                     4   1         ×         5         ×         4         ×         2    Number of codes starting with 6.= 1 × 5 × 4 ×2 = 40CASE 2: The first digit is a 5 or 7   5                                                                2   7                                                                4                                                                     6   2         ×         5         ×         4         ×         3                                                                                                                               Number of codes not starting with 6.begin = 2 × 5 × 4 ×3 = 120Therefore total number of possible codes.= 40 + 120 = 160.OR(3 ´ 5 ´ 4 ´1) + (3 ´ 5 ´ 4 ´1) + (2 ´ 5 ´ 4 ´1)=      60         +        60            +         40= 160OR(3 ´ 5 ´ 4 ´ 3) - (1´ 5 ´ 4 ´1)= 180 - 20= 160 ✓1 × 5 × 4 ×2 ✓40✓2 × 5 × 4 ×3✓120✓160✓(3×5×4×1)✓(3×5×4×1)✓(2×5×4×1)✓160✓(3×5×4×3)✓(1×5×4×1)✓160

 TOTAL 150
Last modified on Tuesday, 15 June 2021 07:13