Thursday, 25 November 2021 13:07

MATHEMATICS PAPER 1 GRADE 12 MEMORANDUM - NSC PAST PAPERS AND MEMOS NOVEMBER 2019

MATHEMATICS PAPER 1
NOVEMBER 2019
MARKING GUIDELINES
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 x2 + 5x - 6 = 0(x + 6)(x -1) = 0x = -6 or x = 1 factorsx = -6    x = 1  (3) 1.1.2 4x 2 + 3x - 5 = 0x = b ± √b 2 - 4ac             2ax = - 3 ± √(3) 2 - 4(4)(-5)                 2(4)x = -3 ± √89          8x = -1,55    or x = 0,8 substitution into the correct formula x = -1,55    x = 0,8     (3) 1.1.3 4x 2 -1 < 0(2x +1)(2x -1) < 0- 1 < x < 1  2           2 factorsmethodanswer (3) 1.1.4 (√√32 + x)( √√32 - x) = x√32 - x2 = x32 - x2 = x2- 2x 2 = -32 x 2 = 16 x = ±4x = 4 √32 - x2squaring both sidesx2 = 16x = 4 (selection)   (4) 1.2 y + x = 12y = -x +12.......... (1)xy = 14 - 3x........ (2)Sub (1) into (2)x(-x +12) = 14 - 3x- x2 +12x -14 + 3x = 0- x 2 +15x -14 = 0x2 -15x +14 = 0(x -14)(x -1) = 0x = 14    or    x = 1y = -2  or     y = 11ORy + x = 12x = - y +12.......... (1)xy =14 - 3x......... (2)Sub (1) into (2)y(- y +12) = 14 - 3(- y +12)12y - y 2 -14 + 36 - 3y = 0- y 2 + 9y + 22 = 0y 2 - 9y - 22 = 0( y + 2)( y -11) = 0y = -2  or     y = 11x = 14    or    x = 1 y subject of the formula substitutionsimplificationboth values of xboth values of y  (5)ORx subject of the formulasubstitutionsimplification both values of yboth values of x (5) 1.3 3  6  9   12  15   18   21  24   27  30 3  3  32  3     3    32   3     3    33   3 k = 14 identifying multiples of 3ten multiples of 3powers of 3answer(4) 

QUESTION 2

 2.1.1 209 ;   186 209   186  (2) 2.1.2 2a = 2         3a + b = -31      a + b + c = 321a = 1          3(1) + b = -31   1 + (-34) + c = 321                       b = -34                   c = 354Tn = n2 - 34n + 354 2nd diff = 2a = 1b = –34c = 354(4) 2.1.3 n2 - 34n + 354 = 74n2 - 34n + 280 = 0(n -14)(n - 20) = 0n = 14     or      n = 20 equating Tn to 74standard form14 20  (4) 2.1.4 f /(n) = 0 2n - 34 = 02n = 34n = 17Term 17 will have the smallest valueORn = - b      2an = 34        2n = 17Term 17 will have the smallest valueORn = 14 + 20 = 17           2Term 17 will have the smallest value 2n - 34 = 0answer         (2)ORsubstitutionanswer     (2)ORsubstitutionanswer              (2) 2.2.1 a = 5/8 ; r = ½;  n = 21Sn = a(1 - r n )            1 - r = 1,2499...= 1,25 rsubstitution into the correct formulaanswer (3) 2.2.2 Tn >   5          8192arn-1 >   5              8192 n -1 < 10    or   - n + 1 > -10n < 11                  n < 11n = 10                  n = 10OR8 ; 16 ; 32 ; … ; 81928.2n-1 < 81922n-1 < 10242n -1 < 210 n -1 < 10 n < 11n = 10 substitution into the correct formulamethod /same base or logcalculating nanswer(4)ORsubstitution into the correct formulamethodcalculating nanswer    (4) 

QUESTION 3

 3.1 = 1 - 1/9 = 8/9 answer                       (3) 3.2 = 52 m2                        = 52 m2     3                                  3for both sides = 2 × 52 = 104 = 34,67m2                                3       3OR 2  x (1+ 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 +10 +11+12) x 2 9= 34,67 m2ORT1 = 2 x 12  = 8                   l = 2 x 1 = 2                9      3                        9          92S12 = = 34,67 m2 adsubstitution into the correct formula answeranswer for both sides(6)ORa(1 + …. + 12)x 2answer                     (6)ORaT1 = 8/3  l = 2/9substitution into correct formulaanswer                    (6) 

QUESTION 4

 4.1 p = -1 p = –1                     (1) 4.2 y =   a         x  - 1- 3 =   a           0 -1a = 3y = x 2 + bx - 3 0 = (1)2  + (1)b - 3b = 2 coordinates D(0 ; –3)substitute (0 ; –3)substitute (1 ; 0)(3) 4.3 y = x 2 + 2x - 3axis of sym: x = - b                          2ax = - 2      2(1)x = -1y = (-1)2 + 2(-1) - 3 = -4C(-1; - 4)ORdy = 0dx2x + 2 = 0x = -1y = (-1)2 + 2(-1) - 3 = -4C(-1; - 4) substitutionx = -1substitutiony = –4   (4) ORderivativex = -1substitutiony = –4   (4) 4.4 y ∈ [-4;∞)  or  y ≥ -4 - 4answer                      (2) 4.5 m = tan45° = 1y = mx + c- 4 = (1)(-1) + c c = -3y = x - 3 gradientsubs m and (-1 ; - 4)equation         (3) 4.6 No, the line passes through C and DORNo, a tangent through turning point C will have a gradient of 0 Noreason                      (2)ORNoreason                      (2) 4.7 f (m - x) = f [-(x - m)]f is reflected in the y-axis and translated 1 unit to the left and 4 units upwards.Therefore: m = -1q = 4ORSubstitute x = 0  and     q = 4 for one x- intercepth(x) = (m - x)2 + 2(m - x) - 3 + q h(0) = (m - 0)2 + 2(m - 0) - 3 + 40 = m2 + 2m +10 = (m +1)2m = -1q = 4 value of mvalue of qORvalue of mvalue of q(4)

QUESTION 5

 5.1 f (x) = k x16 = k 4k = 2 substitution (4 ; 16)answer       (2) 5.2 f : y = 2 xf -1 :       x = 2 yy = log 2 x x = 2yy = log2 x (2) 5.3 asymptoteshapefor any two valid pointseg.(16 ; 4) or(2 ; 1) or (4 ; 2)or (1 ; 0)(4) 5.4.1 x ∈ (1 ; ∞) or   x > 1 1answer(2) 5.4.2 0 < x ≤ ½   or   x ∈ [0; ½] ½ answer(2) 5.5 2x - 2-x = 15                 42x - 1  = 15       2x     422x -1 = 15  x 2x              44.22x - 4 = 15´ 2x4.22x -15.2x - 4 = 0(4.2x +1)(2x - 4) = 04.2x +1 = 0 or 2x - 4 = 02x = -1 or 2x = 22        4N/A       x = 2OR2 x - 2- x = 15                   42x - 1  = 15       2x     4Let k = 2xk 2 - 1 = 15 x k             44.k 2 - 4 = 15 x k4.k 2 -15.k - 4 = 0 (4.k +1)(k - 4) = 0k = -1 or k = 4      42 x = -1 or 2 x = 22         4N/A     x = 2 2x - 2-x = 15                 4standard formfactorsanswer(4)OR2x - 2-x = 15                 4standard formfactorsanswer(4)

QUESTION 6

 6.1 Kuda :     A = P(1+ in)= 5 000(1+ 0,083´ 4)= R6 660,00Final Answer: R6 660,00 + R266,40= R6 926,40ORKuda :    A = P(1+ in) x 1,04= 5 000(1+ 0,083´ 4) x 1,04= R6 926,40Thabo :     A = P(1 + i)n= 5 000 (1 + 0.081 )12 x 4                       12= R6 905,71Kuda will have a better investment substitution into the correct formulafinal answerORsubstitution into the correct formulafinal answersubstitution into the correct formulaanswerconclusion   (5) 6.2.1 n = 157,40n = 158 paymentsOR Number of payments = 13,11686841 x 12=157,40n = 158 payments 0,112substitution into the correct formulasimplificationuse of logsanswer   (5) ORsubstitution into the correct formulasimplification0,112substitution into the correct formulasimplificationuse of logs(5) 6.2.2 Difference: R6 000 – R5 066,36 = R933,64 F = x[(1 + i)n - 1]              i = R162 503,51OR F = R881 818,77.....Amount available for withdrawal= R1 044 322,28 – R 881 818,77= R162 503,51 OROutstanding balance with monthly repayment of R5 066,35 Outstanding balance with monthly repayment of R6 000 Amount available for withdrawalR404 666,23 – R242 162,72 = R162 512,18 R933,64n = 108substitution into the correct formulaanswer      (4)ORn = 108substitutioninto correct formulasubstitution into correct formulafinal answerORn = 108substitution into the correct formulasubstitution into the correct formulafinal answer(4)

QUESTION 7

 7.1 f (x) = 4 - 7xf '(x) = lim    f (x + h) - f (x)           h→0         h= lim   4 - 7(x + h) - (4 - 7x)  h→0                h= lim    h(-7)   h→0     h= –7 4 - 7(x + h)substitutionsimplificationanswer(4) 7.2 y = 4x8 +  √x3= 4x8 + x3/2 dy = 32x7 +  3 x½dx                 2 x3/232x7 3 x½ 2 7.3.1 y = ax2 + ady = 2ax + 0dxdy = 2ax dx 2ax (1) 7.3.2 y = ax2 + ady = x2 + 1da answer    (2) 7.4 Substitute (2 ; b) in y = x + 12                                            xb = 2 + 12             2b = 8mtangent =dy                  dxmtangent =1 - 12  = -2                      x2mperp = ½Equation of perpendicular line:y - y1 = m(x - x1)   OR     y = mx + cy - 8 = ½ (x - 2)          8 = ½ (2)+ cy = ½ x + 7                         c = 7y = ½ x + 7 value of bdy = 1 - 12 dx         x2 gradient of perpendicular lineequation                    (4)

QUESTION 8

 8.1 36cm answer     (1) 8.2 \t = 6        (-2t2 + 3t - 6) have no real roots Insect reaches the floor only once. only once      (3) 8.3 h(t) = -2t3 +15t 2 - 24t + 36h¢(t) = -6t 2 + 30t - 24- 6t2 + 30t - 24 = 0t 2 - 5t + 4 = 0 (t - 4)(t -1) = 0t = 4     or     t = 1Only   t = 4 because maximum value requiredh = -2(4)3 +15(4)2 - 24(4) + 36 = 52 cm expansion- 6t 2 + 30t - 24 = 0 both valuesanswer(4) 

QUESTION 9

 9.1 f /(x) = 9x23x3 = 9x23x3 - 9x2 = 03x2(x - 3) = 0x = 0     or    x = 3 f /(x) = 9x2x = 0x = 3  (3) 9.2.1 For f and f / answer     (1) 9.2.2 The point (0 ; 0) is :A point of inflection of  fA turning point of f / f : inflection pointf / : turning point (2) 9.3 f // (x) = 18xDistance = f //(1) - f / (1)= 18(1) - 9(1)2= 9 f // (x) = 18xsubstitutionanswer     (3) 9.4 3x3 - 9x2 < 03x2 (x - 3) < 0but 3x2 > 0 x - 3 < 0x < 3 , x ≠ 0 3x3 - 9x2 < 0factors   x < 3x ≠  0                     (4) 

QUESTION 10

 10.1 P(same day) = 4 or 1                         16    4or 0,25 or 25% 4 numerator16 denominator (2) 10.2 P(2 consecutive days) = 3 x 2 = 3                                          16      8 3 x 2answer (3) 

QUESTION 11

 11.1.1 P( A) x P(B)       independent events= 0,40 x 0,25 = 0,1 0,10,15 and 0,30,45     (3) 11.1.2 P(A or not B) = P(A) + P(not B)- P(A and not B)= 0,4+ 0,75 - 0,3= 0,85ORP(A or not B) = 1 - P(only B)= 1 - 0,15= 0,85ORFrom Venn diagram: 0,3 + 0,1 + 0,45 = 0,85 substitutionanswer(2)OR1 – 0,15answer(2)ORsubstitutionanswer                 (2) 11.2 (5 × 1 × 5) + (5 × 1 × 6 )+ (5 × 1 × 6) + (5 × 1 × 5) =110110 × 5 = 550 > 500Not possible, because not enough spaceOR(5 × 2 × 5) + (5 × 2 × 6) = 110110 × 5 = 550 > 500Not possible because not enough spaceOR5 x 4 x 6 = 1205 x 2 = 10120 -10 = 110 110 × 5 = 550 > 500Not possible because not enough space 5 × 1 × 55 × 1 × 65 × 1 × 65 × 1 × 5110conclusion         (6)OR5 × 2 × 55 × 2 × 6110conclusion          (6)OR5 x 4 x 6 = 1205 x 2 = 10120 -10 = 110conclusion  (6) 

TOTAL : 150