Mathematics Grade 11 P1 National Exemplar November 2013 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 11 MATHEMATICS P1 EXEMPLAR 2013 MEMORANDUM MARKS: 150 This memorandum consists of 13 pages. Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/PI 2 DBE/2013 NSC – Grade 11 Exemplar – Memorandum QUESTION 1 1.1.1 (2 x − 1)( x + 5) = 0 1 x= OR x = −5  answers 2 (2) 1.1.2 2x − 4x + 1 = 0 2 − (− 4 ) ± (− 4)2 − 4(2)(1)  substitution into correct x= 2(2 ) formula 4± 8 x= 4 4± 4 . 2 x= 4 4±2 2 x= 4 x= ( 2 2± 2 ) 4 2± 2  answers x= 2 (3) 1.2.1 2 125 3 ( )  53 2 = 53 3 = 52 = 25 answer (accept 25 or 52) (2) 1.2.2 (3 2 − 12)(2 2 + 1)  6.2 + 3 2 = 6 . 2 + 3 2 − 24 2 − 12  − 24 2 − 12 answer = −21 2 (3) 1.3.1 3x − 9 = 0  3x − 9 = 0 3x = 9  answer (2) x=3 1.3.2 x − x − 6 ( x − 3)( x − 2 ) 2  (x − 3)( x − 2 ) = 3x − 9 3( x − 3)  3( x − 3) x−2 answer = (3) 3 [15] Copyright reserved

Downloaded from hlayiso.com Mathematics/PI 3 DBE/2013 NSC – Grade 11 Exemplar – Memorandum QUESTION 2 2.1.1 (x + 2)(x − 3) < −3x + 2 x 2 − x − 6 + 3x − 2 < 0 x 2 + 2x − 8 < 0  standard form (x + 4)(x − 2) < 0  factors + 0 − 0 + or –4 2 –4 2 −4< x − 4 <x < 2 x<2 (4) 2.1.2 x + 2x − 8 < 0 2  − 4 <x < 2 − 4 <x < 2 –3,–2,–1,0,1 Sum of integers = (− 3) + (− 2 ) + (− 1) + (0 ) + (1) answer = −5 (3) 2.2.1 4 x −1 + 4 x +1 4 x . 4 −1 + 4 x.41 = 17 . 12 x 17 . 3 x . 4 x = ( 4 x 4 −1 + 4 ) 17 . 3 x . 4 x 1   factorise numerator 4x  + 4 =  x x 4 17 . 3 . 4  3x . 4 x  simplification of numerator  17  17   to =  x 4 4 17 . 3  answer 1 1 = .3− x or (4) 4 4 .3 x 2.2.2 4 x −1 + 4 x +1 1 − x = .3 17 . 12 x 4 1 = .4t 4 answer =t (1) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/PI 4 DBE/2013 NSC – Grade 11 Exemplar – Memorandum 2.3 3 y = 81x and y = x 2 − 6 x + 9 3 y = 34 x  3 y = 34 x  y = 4x y = 4x  4x = x2 − 6x + 9 4x = x2 − 6x + 9 standard form 0 = x 2 − 10 x + 9 0 = ( x − 9 )( x − 1) factors x=9 or 1 x-values y = 4(9 ) or 4(1) = 36 or 4 y-values (x; y ) = (9 ; 36) or (1 ; 4 ) (7) [19] QUESTION 3 3.1.1 4 −8p = 0  4 −8p = 0 − 8 p = −4  answer 1 p= (2) 2 3.1.2 4 −8p < 0  4 −8p < 0 1 p>  answer 2 (2) 3.2.1 5 − x = x +1 5− x ≥ 0 and x + 1 ≥ 0  5− x ≥ 0  x +1 ≥ 0 x≤5 and x ≥ −1 and Hence − 1 ≤ x ≤ 5 (3) 3.2.2 5 − x = x 2 + 2x + 1  square both sides x 2 + 3x − 4 = 0  standard form (x + 4)(x − 1) = 0  factors x = −4 or x = 1  answers Since − 1 ≤ x ≤ 5 , x = 1 only  selection of 1 (5) 3.2.3 x = −4  answer (1) [13] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/PI 5 DBE/2013 NSC – Grade 11 Exemplar – Memorandum QUESTION 4 4.1 A = P(1 − in ) = 145000[1 − (0,17 )(5)]  substitution  answer = R 21 750 (2) 4.2.1 8% = 2% per quarter  answer 4 (1) 4.2.2 A = P(1 + i ) n After 1 year, A = P (1 + ieff ) and A = P(1+ 0.02 ) 1 4 Hence 1 + ieff = (1 + 0.02 )  1 + ieff = (1 + 0,02 ) 4 4 ieff = (1 + 0.02 ) − 1 4  answer = 0,0824 (2) The effective interest rate is 8,24% p.a. 4.3 3  0,09   0,075  42 0,09 A = 14 0001 +  1 +    2   12  2 3 = R 20 755,08  0,09   14 0001 +   2  0,07  12 42 answer (5) [10] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/PI 6 DBE/2013 NSC – Grade 11 Exemplar – Memorandum QUESTION 5 5.1 R 15 000  answer (1) 5.2 Simple interest  answer (1) 5.3 A = P(1 + in ) 31 = 15(1 + 6i )  substitution of (6 ; 31) 31 = 1 + 6i into correct formula 15  31  i =  − 1 ÷ 6  15  8 = 45 = 0,1778 answer = 17,78% (2) 5.4 A = P(1 + in ) w = 15(1 + 0,1778 × 12) w = 47 = 47 A = P(1 + i ) n 47 = 15(1 + i ) 12  substitutes (12 ; w) 47 12 = 1+ i 47  12 15 15 47 answer i = 12 − 1 = 0,09985 = 9,99% 15 (4) [8] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/PI 7 DBE/2013 NSC – Grade 11 Exemplar – Memorandum QUESTION 6 6.1.1 1 1 1 Multiply by  multiply 8 2 8 1  2 (2) 6.1.2 n −1 1 11  a= Tn =   2 22 n −1 1   OR 2 (2) 1 n  answer Tn =   (2) 2 OR  answer Tn = 2 − n (2) 6.1.3 Continuing the pattern: 1 1 1 1 1 1 1 1 1 1  expand sequence ; ; ; ; ; ; ; ; ; n = 10 2 4 8 16 32 64 128 256 512 1024 Hence n = 10 (2) OR 1 1 = 1 1 2 n 1024  n = 2 1024 2 = 2 −10 −n n = 10 n = 10 (2) 6.2.1 124 answer (1) 6.2.2 Tn = −8n + 164  – 8n +164 (2) 6.2.3 − 8n + 164 < 0  − 8n + 164 < 0 164 < 8n  20,5 < n 20,5 < n Hence T21 is the first term to be negative. answer (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/PI 8 DBE/2013 NSC – Grade 11 Exemplar – Memorandum 6.2.4 2a = −8 a = −4  a = −4 3a + b = 156 3(− 4 ) + b = 156 b = 168  substitutions  b = 168 T5 = −24 − 4(5) + 168(5) + c = −24  substitution 2 c = −764  c = −764 Tn = −4n 2 + 168n − 764 (5) OR T5 = −24 (given) T6 = −24 + 124  T6 = 100 = 100 Tn = −4n 2 + bn + c  a = −4 − 24 = −4(5) + b(5) + c 2 76 = 5b + c ... (1)  substitutions 100 = −4(6) + b(6 ) + c 2 244 = 6b + c ... (2) b = 168 168 = b ... (2) − (1)  c = – 764 c = −764 (5) OR T5 = −24 (given)  T4 = −156 T4 = −24 − 132 = −156 Tn = −4n 2 + bn + c  a = −4 − 24 = −4(5) + b(5) + c 2 76 = 5b + c ... (1)  substitutions − 156 = −4(4 ) + b(4 ) + c 2 − 92 = 4b + c ... (2) b = 168 168 = b ... (1) − (2)  c = – 764 c = −764 (5) [17] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/PI 9 DBE/2013 NSC – Grade 11 Exemplar – Memorandum QUESTION 7 T1 0 T3 0 x y introduce variables 12 12 y = x + 12  T3 = x T3 = 0 + x = x AND T3 + y = 0 y = −x  y = −x AND y = x + 12  y = x + 12 Hence − x = x + 12  − x = x + 12 − 2 x = 12 answer x = −6 [6] OR 2a = 12 a = 6 a=6 Tn = 6n 2 + bn + c n = 2 and n = 4 :  24 + 2b + c = 0 24 + 2b + c = 0  96 + 4b + c = 0 96 + 4b + c = 0 72 + 2b = 0 2b = −72  b = – 36 b = −36 24 − 72 + c = 0  c = 48 c = 48 Tn = 6n 2 − 36n + 48 T3 = 6(3) 2 − 36(3) + 48 = 102 − 108 = −6 answer [6] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/PI 10 DBE/2013 NSC – Grade 11 Exemplar – Memorandum QUESTION 8 8.1 x = 3  answer y=–1  answer (2) 8.2 R;x≠3  R  x≠3 OR (2) (− ∞ ; 3) ∪ (3 ; ∞ )  (− ∞ ; 3)  (3 ; ∞ ) (2) 8.3 d = tan 76°  d = tan 76° d =4  d =4 6 = 4(3) + e e = −6  e = −6 g ( x) = 4 x − 6 (3) 8.4 2 −1 = 4x − 6 x−3  equating 2 = 4x − 5 x−3  simplification 2 = 4 x( x − 3) − 5( x − 3) 2 = 4 x 2 − 12 x − 5 x + 15 0 = 4 x 2 − 17 x + 13  standard form 0 = (4 x − 13)( x − 1)  factors 13  x-values x= or x = 1 4  13  y = 4  − 6 or y = 4(1) − 6 4  y-values y=7 or y = −2  13  Points of intersection are A (1 ; − 2 ) and C  ; 7  4  (6) 8.5 13 13   1≤ x 1 ≤ x < 3 or x ≥ OR x ∈ [1 ; 3) ∪  ; ∞  x<3 4 4  13  x≥ 4 (3) 8.6 y = ( x − 3) − 1  x −3 y = x−4  −1 answer OR (3) y = x+c  substitute (3 ; − 1) Substitute (3; − 1) answer −1 = 3 + c (3) c = −4 y = x−4 [19] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/PI 11 DBE/2013 NSC – Grade 11 Exemplar – Memorandum QUESTION 9 9.1 (1; 4) f:  shape  x-int  y-int  turning point g:  shape  x-int  y-int  asymptote x 2 − 2x − 3 = 0 0 = 1− 2x y = 1 − 20 (x − 3)(x + 1) = 0 2 x = 20 y=0 x = 3 or x = −1 x=0 −2 x= =1 2(−1) y = −(1) + 2(1) + 3 = 4 2 (9) 9.2 f (0) − f (−3)  correct formula Average gradient = 0 − (−3)  f (−3) = −12 3 − (−12) = 3  answer =5 (3) 9.3 − 1 ≤ x ≤ 0 or x ≥ 3  −1 < x  x<0  x>3 (3) 9.4 Given: f ( x ) + c = 0 has one solution/equal roots i.e. f ( x ) = −c has one solution  − c = f (1) ⇒ − c = f (1) = 4 answer (2) ⇒ c = −4 OR h is f translated 4 units down  3 + c = −1 y-intercept of h will then be at – 1 answer ∴ 3 + c = −1 (2) c = −4 9.5 (0 ; 1)  (0 ; 1) (2) 9.6 k ( x) = 1 − 2 −x  answer (1) [20] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/PI 12 DBE/2013 NSC – Grade 11 Exemplar – Memorandum QUESTION 10 Range of f (− ∞;7] ⇒ y-part of turning point [Max value of f ( x ) ] is 7 a <0 and shape b < 0 ⇒ b negative ⇒ axis of symmetry on left of y-axis roots real, unequal & opposite signs ⇒ x-ints on opposite sides of y-axis y  shape 7 turning point at y=7  axis of symmetry on x left of y-axis  roots are on opposite sides [4] QUESTION 11 11.1 No, W and T are not mutually exclusive not mutually exclusive Because P(W and T) ≠ 0  P(W and T) ≠ 0 (2) OR No, W and T are not mutually exclusive not mutually exclusive Because P(W or T) = 0,61 ≠ 0,75 = P(W) + P(T)  P(W or T) ≠ P(W) + P(T) (2) 11.2 P(W and T)= 0,14 (given) and P(W)×P(T)= 0,4 × 0,35 =0,14  P(W)×P(T) = 0,14 ⇒ P(W and T) = P(W)×P(T)  P(W and T) = P(W)×P(T) Therefore yes, W and T are independent events  conclusion (yes) (3) [5] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/PI 13 DBE/2013 NSC – Grade 11 Exemplar – Memorandum QUESTION 12 12.1.1 a=5 a=5 b=4 b=4 c=8 c=8 d=1 d=1 e=6 e=6 (5) 12.1.2 6 answer (1) 12.1.3 4  answer 33 (1) 12.1.4 4 + 3 + 2 + a + b + c 26  answer = (1) 33 33 12.2 0,45 Mathematics P(G and M) = 0,27 girl 0,6 0,55 Mathematical Literacy P(G and ML) = 0,33 Mathematics P(B and M) = 0,14 0,35 0,4 0,4 0,45 boy 0,35 0,65 Mathematical Literacy P(B and ML) = 0,26 P(Mathematics) = P(G and M) + P(B and M) = (0,6)(0,45) + (0,4)(0,35) P(G and M) = 0,27 P(B and M) = 0,14 = 0,27 + 0,14 = 0,41 answer (6) [14] TOTAL: 150 Copyright reserved