Mathematics Grade 11 P1 National Exemplar November 2014 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 11 MATHEMATICS P1 NOVEMBER 2014 MEMORANDUM MARKS: 150 This memorandum consists of 14 pages. Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 2 DBE/November 2014 CAPS – Grade 11 – Memorandum QUESTION 1 1.1.1 7 x=–2 x = – 2 or x = 7 3  x = (2) 3 1.1.2 x 2 − 5x − 2 = 0 standard form − b ± b 2 − 4ac x= 2a correct substitution 5 ± 25 − 4(1)(−2) into correct formula x= 2 5 ± 33 x= 2  x = 5,37 x = 5,37 or x = −0,37 x = − 0,37 (4) OR  25   25   completing the x 2 − 5x +   = 2 +    4   4  square 2  5 33 x −  =  2 4 5 33  33 x− =± 2 2 5 + 33 5 − 33 x= or x = 2 2  x = 5,37 x = 5,37 x = −0,37 x = − 0,37 (4) 1.1.3 x−3 =5+4 isolating ( x − 3 ) = (9) 2 2 squaring both sides x − 3 = 81  simplify x = 84  answer (4) 1.1.4 2x2 − 7x − 4 ≥ 0 (2 x + 1)( x − 4) ≥ 0  factors 1 CV' s : − ; 4 2 + – + + – + method 1 1 − 4 − 4 2 2  notation 1  critical values x≤− or x≥4 (4) 2 OR  notation 1  critical values x ∈ (−∞ ; − ] ∪ [4 ; ∞) 2 Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 3 DBE/November 2014 CAPS – Grade 11 – Memorandum 1.2 x = 2 y +1 ......(1) x 2 − 2 y + 3 xy = 6 ........(2)  substitution of (2 y + 1)2 − 2 y + 3 y(2 y + 1) = 6 x = 2 y +1 4 y2 + 4 y +1− 2 y + 6 y2 + 3y − 6 = 0  simplification 10 y 2 + 5 y − 5 = 0  standard form 2 2 y + y −1 = 0  factors (2 y − 1)( y + 1) = 0  both y values 1  both x values (6) y= or y = −1 2 x=2 x = −1 OR x −1 y= 2  substitution of  x −1   x −1  x 2 − 2  + 3 x =6 x −1  2   2  y= 2 2 x 2 − 2 x + 2 + 3 x 2 − 3 x − 12 = 0  simplification 5 x 2 − 5 x − 10 = 0  standard form x2 − x − 2 = 0 ( x + 1)( x − 2) = 0  factors x = −1 or x=2  both x values  both y values (6) 1 y = −1 y= 2 [20] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 4 DBE/November 2014 CAPS – Grade 11 – Memorandum QUESTION 2 2.1 3 x (3 − 3−1 )  common factor 3 x 2.3 x  3 − 3−1 1 3− = 3  answer (3) 2 8 1 = ×  common factor 3 x −1 3 2 4 simplification = 3 OR  answer (3) 3 x −1 (32 − 1) 2.3 x 3 x.3−1 (8) = 2.3 x 1 = ×4 3 4 = 3 2.2 3 (x − 2)− 2 = 64 [( )] −2 applying exp. law x − 2 = 43 3  43 x − 2 = 4 −2 simplifying 1 x=2+ 16 1 answer (4) ∴x = 2 16 OR ( x − 3) −3 = 64  squaring ( x − 3) −3 = 4096 1  applying exp. law ( x − 2) 3 = 4096 1  simplification x−2= 16  answer (4) 1 x=2 16 Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 5 DBE/November 2014 CAPS – Grade 11 – Memorandum 2.3 1 1 `1 applying surd law x. x .x .x 2 4 8 applying surd law 8 x7 7 simplifying x 8 = 7 8 x answer (4) =x [11] QUESTION 3 3 AC.( x − 2) = x 2 + 2 x − 8  statement  factors AC.( x − 2) = ( x + 4)( x − 2)  AC = ( x + 4) cm AC = ( x + 4) cm ∴ FD = ( x + 4) cm ∴ ED = x + 4 − ( x − 2)  method  answer (6) ED = 6 cm [6] QUESTION 4 4.1 –7 0 9 20 7 9 11 2 2  2a = 2 2a = 2 avalue a=1 3(1) + b = 7 bvalue b=4 (1) + (4) + c = – 7 c value (4) c = – 12 ∴Tn = n 2 + 4n − 12 OR  2a = 2 2a = 2 avalue a =1 T2 = 2 2 + b(2) + c = 0 2b + c = −4 (1) 3(1) + b = 7 2 T3 = 3 + b(3) + c = 9 b=4 OR 3b + c = 0 (2) 1 + a + c = −7 bvalue c = −12 c value (4) (2) − (1) b = 4 ∴ c = −4 − 2(4) = −12 Tn = n 2 + 4n − 12 Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 6 DBE/November 2014 CAPS – Grade 11 – Memorandum OR (n − 1)(n − 2) formula Tn = T1 + (n − 1)d 1 + .d 2 substitution 2 (n − 1)(n − 2) = −7 + (n − 1).7 + .2 simplification (4) 2 = −7 + 7n − 7 + n 2 − 3n + 2 = n 2 + 4n − 12 4.2 n 2 + 4n − 12 = 128  equation n 2 + 4n − 140 = 0 (n + 14)(n − 10) = 0 factors n ≠ −14 or n = 10 answers for n n = 10 (choice) (4) invalid ∴ n = 10 4.3 – 7 ; 0 ; 9 ; 20 ; … first difference7 9 11 first differences second difference2 2 Fn = 2n + c Answer only: Full Marks F1 = 2(1) + c = 7  2n ∴c = 5 c = 5 (3) Fn = 2n + 5 4.4 Fn = 2n + 5 = 599  equating 2n = 594 297 298(3) ∴ n = 297 [14] this difference will be between term 297 and term 298 QUESTION 5 5.1 Pattern 1 2 3 White squares 4 12 24 40 answer(2) 5.2 Wn = 2n 2 + 2n Wn substitution  W157 = 2(157) 2 + 2(157) answer (3) = 49612 Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 7 DBE/November 2014 CAPS – Grade 11 – Memorandum 5.3 2n 2 + 2n + 1 < 613 setting up inequality 2n 2 + 2n − 612 < 0 n 2 + n − 306 < 0 (n − 17 )(n + 18) < 0 factors - 18 17 method ∴n = 16 answer (4) 5.4 Pn = 4n 2 + 4n + 1  Pn = 4n 2 + 4n + 1 = (2n) 2 + 2(2n) + 1 rewriting P n 2n is even for all n ∈ Z conclusion ∴ Total squares used in the n th pattern is always odd. (3) OR Pn = 4n 2 + 4n + 1  Pn = 4n 2 + 4n + 1 = 2(2n 2 + 2n) + 1  rewriting P n 2(2n 2 + 2n) is odd for all n ∈ Z 2(2n 2 + 2n) + 1 is odd for all n ∈ Ζ  conclusion (3) ∴ Total squares used in the n th pattern is always odd. [12] QUESTION 6 6.1 x=2 x=2 y=3  y = 3 (2) 6.2 8 8 x. int : +3=0  +3= 0 x−2 x−2 8 + 3( x − 2) = 0 3x + 2 = 0 2 ∴x = − 3  2   2  ∴ x − int  − ; 0    − ;0  3   3  8 y= +3 0−2 y = −1  (0 ;−1) (3) y. int : (0 ;−1) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 8 DBE/November 2014 CAPS – Grade 11 – Memorandum 6.3 y asymptotes 3 intercepts with axes shape (3) − 2/ 3 O 2 x −1 6.4 3=2+ k  substitute k =1  answer (2) OR y = ( x − 2) + 3 y = x +1  y = x +1 ∴ k =1 answer (2) [10] QUESTION 7 7.1 q=–6 answer (1) 7.2 1 substitute x −5 = a.2 −1−1 − 6 substitute y 4 3 1 = a simplifying 4 4 answer a=3 (4) 7.3 x int : 2 x −1 = 2 ∴ x = 2 ∴ (2 ; 0)  2 x −1 = 2 x=2 1  1 y int : y = 3.2 −1 − 6 = −4 ∴ 0; − 4  1 2  2 y= − 4 2 Average Gradient 1 subst. into gradient 0+4 2 formula = 2−0 answer 9 1 (5) = or 2 4 4 7.4 y = 3.2 x −3 − 6 answer (2) [12] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 9 DBE/November 2014 CAPS – Grade 11 – Memorandum QUESTION 8 8.1 C(– 1 ; 0)  C(– 1 ; 0) (1) 8.2 y = ( x − 3)( x + 1)  ( x − 3) y = x 2 − 2x − 3  (x + 1)  y = x 2 − 2 x − 3 (3) 8.3 TP : y = (1) 2 − 2(1) − 3  y = −4 y = −4  [−4 ; ∞) (2) R: y ∈ [−4 ; ∞) OR  y ≥ −4 y ≥ −4 8.4 0+4 substituting into gradient m= =2 formula 3 −1 y − 0 = 2( x − 3) m = 2  equation (3) y = 2x − 6 8.5.1 x ≤ −1 or x≥3  x ≤ −1 OR  x ≥ 3 (2) x ∈ (−∞ ; − 1] ∪ [3; ∞)  (−∞ ; − 1]  [3 ; ∞) (2) 8.5.2 − 1 < x < 3 or x > 3  critical values OR  notation (2) x > −1 ; x ≠ 3 OR  x > −1  x ≠ 3 (2) (−1; 3) ∪ (3; ∞)  (−1; 3)  (3 ; ∞) (2) 8.5.3 − 1 < x < 0 or x > 3  critical values OR  notation (2) (−1; 0) ∪ (3; ∞)  (−1; 0)  (3 ; ∞) (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 10 DBE/November 2014 CAPS – Grade 11 – Memorandum 8.6 x2 − 2x − p = 0 ∆ = (−2) 2 − 4(1)(− p) = 4+4p for non - real roots ∆ < 0 4+ 4p < 0 4+ 4p < 0 4 p < −4  p < −1 (2) ∴ p < −1 OR A(1; − 4) x2 − 2x − 3 = 0 x2 − 2x − p = 0 − p >1 − p >1  p < −1 (2) ∴ p < −1 8.7 ( PM = (2 x − 6 ) − x 2 − 2 x − 3 ) subtraction quadratic expression = −x2 + 4x − 3 b  method x=− 2a 4 =− =2 2(− 1) maximum value (4) Max. PM = −(2 ) + 4(2 ) − 3 = 1 unit 2 OR ( PM = (2 x − 6 ) − x 2 − 2 x − 3 ) subtraction quadratic expression = −x 2 + 4x − 3 ( = − x 2 − 4x + 4 − 4 + 3 ) method [ = − (x − 2) − 1 2 ] = −( x − 2 ) + 1 2 Max. PM = 1 unit maximum value (4) [21] QUESTION 9 9.1 A = P(1 − i ) n substitution 11090,41 = 120000(1 − i ) 12 11090,41 making i subject ∴ i = 1 − 12 120000 i value as decimal Thus i = 0,179999... Rate of Depreciation = 18 % answer (4) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 11 DBE/November 2014 CAPS – Grade 11 – Memorandum 9.2  i  m formula ieff = 1 +  − 1  m 12  0,098  substitution into = 1 +  −1 formula  12  = 0,10252..... 10,25% (3) rate = 10,25% 9.3 A = P(1 + i1 ) 1 (1 + i2 ) 2 n n 16  0.075   1 +   0.075   0,092  16 36  4  = 80 0001 +  1 +  36  4   12   0,092   1 +  = R141768,60  12  multiplication OR answer (4) 16  0,075  A1 = 800001 +  16  4   0,075   1 +  = 107689,1465..  4  36 A 1  0,092  36 A2 = 107689,14651 +   0,092   12   1 +   12  = R141768,60 answer (4) 9.4.1 Investment : end of third year : A = P(1 + i ) n 0,065  96 12  0,065  subst. into correct = 30 0001 +   12  formula = R50390,07 answer (3) 9.4.2 T0 T3 T5 T8  0,65  96  300001 +  30000 – 10000 +10000  12  60 96 60 36  0,65   0,65   0,65   0,65   − 100001 +  A = 300001 +  − 100001 +  + 100001 +   12   12   12   12  36  0,65  A = R 48708,61  100001 +   12  ∴ difference = 48708,61 − 50390,07  R 48708,61 = −R1681,46 subtracting answer (7) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 12 DBE/November 2014 CAPS – Grade 11 – Memorandum Investment : end of third year : A = P(1 + i ) n 36  0,065  subst. into formula = 30 0001 +   12  answer = R 36 440,14881 Principal(new) : R36440,14881 − R10000,00 = R 26 440,14881 Investment : end of fifth year : A = P(1 + i ) n 24  0,065  subst. into formula = 26440.148811 +   12  answer = R 30100,2304 Principal(new) : R30100,2304 + R10000,00 = R 40100,2304 Investment : end of eighth year : A = P(1 + i ) n 24  0,065  subst. into formula = 40100,23041 +   12  answer conclusion(7) = R 48708,61 [21] Tashil had a deficit of R1681,46 QUESTION 10 10.1 5 customers answer (1) 10.2 P(C and B) ≠ 0 P(C and B) ≠ 0 Thus events B and C are not mutually exclusive  conclusion (2) 10.3.1 58 29  answer (1) P(V only) = = 240 120 10.3.2 29  answer (1) P(C and B) = 240 10.3.3 P( not C) = 1 – P(C)  formula 122 59  substitution =1− = answer (3) 240 120 OR 52 + 3 + 58 + 5 numerator and P(not C) = denominator 240 answer (3) 118 59 = = 240 120 Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 13 DBE/November 2014 CAPS – Grade 11 – Memorandum 10.3.4 P(B or V) = P(B) + P(V) – P(B and V) 84 84 82 15  = + − 240 240 240 240 82  240 151 15 =  240 240 151  (4) OR 240 17 + 52 + 12 + 3 + 9 + 58 P(B or V) = 240 151 numerator and = denominator 240 answer (4) [12] QUESTION 11 P(A or B) = P ( A) + P ( B ) − P ( A ∩ B ) 0,428 = 0,12 + 0,35 − P( A ∩ B) substitution value of P ( A ∩ B ) P( A ∩ B) = 0,042 value of P( A) × P( B) P( A) × P( B) = 0,12 × 0,35 = 0,042 conclusion (4) ∴ P( A ∩ B) = P( A) × P( B) [4] Thus A and B are independent events Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 14 DBE/November 2014 CAPS – Grade 11 – Memorandum QUESTION 12 12.1 There are 100% – 60% – 10% = 30% red marbles 30% 30 ∴ × 80 = 24 red marbles 24 (2) 100 12.2 Outcome R R,R 23 79 8 Y R,Y 79 48 R 24 79 G R,G first branch 80 24 R second branch 79 7 Y,R 8 values on diagram 80 Y 79 Y (3) 48 48 G Y,Y 80 79 24 R G 79 8 Y,G 79 Y G,R 47 G,Y 79 G 12.3 P(G and Y) = P(G, Y) + P(Y, G) 48 8 8 48  multiplication rule = × + ×  addition 80 79 80 79 48 = answer (3) 395 [8] TOTAL: 150 Copyright reserved