Mathematics P1 Feb March 2013 Memo Eng hlayiso.com

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICS P1 FEBRUARY/MARCH 2013 MEMORANDUM MARKS: 150 This memorandum consists of 19 pages. Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 2 DBE/Feb.–Mar. 2013 NSC – Memorandum QUESTION 1 1.1.1 (x − 9)(2 x + 1) = 0 2  (x − 3)(x + 3) (x − 3)(x + 3)(2 x + 1) = 0 ± 3 1 x = ±3 or x=− 1 2  − 2 OR (3) (x − 9)(2 x + 1) = 0 2 –3 1 3 x = ±3 or x=− 1 2 −  2 (3) 1.1.2 x + x − 13 = 0 2 − b ± b 2 − 4ac x= 2a − 1 ± 1 − 4(1)(− 13)  subs into formula = 2 − 1 ± 53  53 = 2 x = 3,14 or x = − 4,14 answer answer (4) 1.1.3 2 ⋅ 3 = 81 − 3 x x  2 ⋅ 3 x + 3 x = 81 2 ⋅ 3 + 3x = 81 x x  3 as common factor 3x (2 + 1) = 81 3x = 27  simplification 3x = 33  answer x=3 (4) OR  2 ⋅ 3 x + 3 x = 81 2.3 = 81 − 3 x x x  3 as common factor 2.3 x + 3 x = 81 3 x (2 + 1) = 81  3 x +1 = 3 4 3 x +1 = 3 4  answer x +1 = 4 (4) x=3 Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 3 DBE/Feb.–Mar. 2013 NSC – Memorandum 1.1.4 (x + 1)(4 − x ) > 0  change of sign ( x + 1)( x − 4) < 0  both critical values  correct inequality sign + 0 − 0 + (3) or –1 4 –1 4 −1 < x < 4 OR  method (x + 1)(4 − x ) > 0  both critical values  correct inequality sign − 0 + 0 − (3) –1 4 –1 4 −1 < x < 4 1.2.1 2 x + 2 x + 2 = −5 y + 20 2 x (1 + 2 2 ) = −5 y + 20  2 x common factor −5 y + 20 2x = 5  answer OR (2) 2 = −y + 4 x 1.2.2 If y = – 4 , 2 x + 2 x + 2 = −5 y + 20 2 x + 2 x + 2 = 40 ( 2 x 1 + 2 2 = 40 ) 2 =8 x  substitution 2 x = 23  answer x=3 (2) 1.2.3 –y+4>0 –y+4>0 y< 4 Largest integer value of y is 3  y=3 2 x = −3 + 4 2x = 1  x=0 x=0 (3) [21] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 4 DBE/Feb.–Mar. 2013 NSC – Memorandum QUESTION 2 2.1.1 32 1 1 r=− =−  − 64 2 2  1  substitution p = 256 −   2  answer p = −128 (3) OR p 64 = 256 p p 64  = p = 16384 2 256 p  p = ±128 p = ±128  answer p = −128 (3) OR p − 32  = 256 64 p − 32  simplification = 256 64  answer 64 p = 8192 (3) p = −128 OR 1 64 1 = 64 = −2  = = −2 r − 32 r − 32  simplification p = −2 × 64  answer p = −128 (3) 2.1.2 Sn = [ a 1− r n ] 1− r  formula  substitution   1 8  256 1 −  −     2   S8 = 1 1+  answer 2 512  255  (3) =   3  256  = 170 OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 5 DBE/Feb.–Mar. 2013 NSC – Memorandum Sn = [ a1− rn ]  formula 1− r   1 8  2 1 −  −   8   2    substitution S8 =  1 1+ 2 2  255  9 =   3  28  = 170  answer (3) 2.1.3 −1 < r < 1  answer (1) OR 1 The common ratio is − which is between – 1 and 1.  answer 2 (1) OR 1 −1< − <1  answer 2 (1) 2.1.4 a S∞ =  formula 1− r 256 =  substitution  1 1− −   2 512 =  answer 3 (3) = 170,67 Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 6 DBE/Feb.–Mar. 2013 NSC – Memorandum 2.2.1 16  answer (1) 2.2.2 Tn = −8 + 6(n − 1)  substitution into equation 148 = 6n − 14 T n = 148 answer 6n = 162 (3) n = 27 2.2.3 Sn = n [2a + (n − 1)d ] 2  n [2(− 8) + (n − 1)(6)] 2 n [2(− 8) + (n − 1)(6)] > 10 140  3n 2 − 11n > 10 140 2 3n 2 − 11n > 10 140 3n 2 − 11n − 10 140 > 0 (3n + 169)(n − 60) > 0  factors When n = 60, S n = 10 140  n = 60 Smallest n = 61  answer (5) 2.3 30 ∑ (3k + 5) k =1  n = 30 a = 8 n = 30 d = 3 30  substitution into correct ∑ (3k + 5) = 2 [2(8) + 29(3)] 30 k =1 formula  answer = 15(103) (3) = 1545 [22] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 7 DBE/Feb.–Mar. 2013 NSC – Memorandum QUESTION 3 3.1 Jacob calculated that the sequence is geometric or Jacob (geometric/exponential) exponential.  Vusi (quadratic) Vusi calculated that the sequence is quadratic. (2) OR Jacob has multiplied each term by 3 to get the next term.  Jacob (multiplied each term by 3) Vusi sees it as a sequence with a constant second difference.  Vusi (constant second difference) (2) OR Jacob calculated that the sequence is geometric or Jacob (geometric/exponential) exponential. Vusi calculated that the sequence can be seen as a  Vusi (exponential and cubic combination of exponential and cubic sequences. combined) (2) 3.2.1 Tn = 3n answer (1) OR Tn = 3.3n −1 answer (1) 3.2.2 3 9 27 57 6 18 30 12 12 2a = 12 3a + b = 6 a+b+c =3 a=6 a=6 18 + b = 6 6 − 12 + c = 3  method b = −12 c=9  b = – 12 c=9 Tn = 6n 2 − 12n + 9 (4) OR 2a = 12 a=6 a=6 T0 = c = 9 c=9 Tn = an 2 + bn + 9 3 = 6(1) 2 + b(1) + 9  method b = −12 Tn = 6n 2 − 12n + 9  b = – 12 (4) OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 8 DBE/Feb.–Mar. 2013 NSC – Memorandum 2a = 12 a=6 Tn = 6n 2 + bn + c a=6 T1 = 3 = 6(1) + b(1) + c 3= 6+b+c  method 2 i.e. T2 = 9 = 6(2 ) + b(2 ) + c 9 = 24 + 2b + c 2 i.e. 6 = 18 + b b = −12 c=9 Tn = 6n 2 − 12n + 9  b = – 12 c=9 (4) OR Tn = 3 n + k (n − 1)(n − 2 )(n − 3)  57 = 3 4 + k (3)(2)(1) Tn = 3n + k (n − 1)(n − 2 )(n − 3) 6k = −24 k = −4  substitution  answer Tn = 3 n − 4(n − 1)(n − 2 )(n − 3) (4) [7] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 9 DBE/Feb.–Mar. 2013 NSC – Memorandum QUESTION 4 4.1 R OR (− ∞; ∞ ) answer (1) 4.2 y=0 y=0 (1) 4.3 1 y 1 y x=   x=  3 3 y = log 1 x  y = log 1 x 3 3 OR (2) y 1 y x=  1  x=  3  3 x = 3− y − y = log3 x  y = − log 3 x y = − log3 x (2) 4.4 y  shape intercept at (1 ; 0) any other correct point (3) 1 x (3 ; -1) 4.5 x = −2  x = −2 (2) 4.6 LHS = [ f ( x)] − [ f (− x)] 2 2 2 2  1  x   1  − x  2 =    −     1  x   1  − x   3    3       −     3    3   = 3− 2 x − 32 x RHS = f (2 x) − f (−2 x)  3−2 x − 32 x 2x −2x 1 1 =   −  2x −2 x  3  3 1 1    −  −2x = 3 −3 2x 3 3 ∴ LHS = RHS [ f ( x)]2 − [ f (− x)]2 = f (2 x) − f (−2 x) (3) [12] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 10 DBE/Feb.–Mar. 2013 NSC – Memorandum QUESTION 5 y C(2 ; 6) 0 B(2,5 ; 0) x 5.1 g ( x) = a +6  p=2 x−2  q=6 a 0= +6 2,5 − 2  substitute B(2,5 ; 0) 0 = 2a + 6  a = −3 a = −3 (4) −3 g ( x) = +6 x−2 5.2 1 xf = 2− 2 3 xf = 2 yf = 6+6 y f = 12 3  x-coordinate F  ; 12  2  y-coordinate (2) [6] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 11 DBE/Feb.–Mar. 2013 NSC – Memorandum y QUESTION 6 S(1 ; 18) f ( x) = ax 2 + bx + c g ( x) = −2 x + 8 f R g P T x 0 6.1 0 = −2 x + 8 y=0 2x = 8 x=4 x = 4 T (4 ; 0) (2) 6.2 By symmetry, P(– 2 ; 0)  f ( x ) = a(x + 2 )( x − 4 ) f (x ) = a (x + 2 )(x − 4 ) 18 = a (1 + 2 )(1 − 4 )  substitutes S(1 ; 18) a = −2 f (x ) = −2(x + 2 )(x − 4 )  a = −2 ( = −2 x 2 − 2 x − 8 )  multiplies out correctly to get = −2 x + 4 x + 16 2 − 2 x 2 + 4 x + 16 (4) OR f ( x ) = a( x − 1) + 18  f ( x ) = a(x − 1) + 18 2 2 0 = a(4 − 1) + 18  substitutes T(4 ; 0) 2 a = −2  a = −2 f ( x ) = −2(x − 1) + 18 2 ( ) = −2 x 2 − 2 x + 1 + 18  multiplies out correctly to get = −2 x + 4 x + 16 2 − 2 x 2 + 4 x + 16 (4) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 12 DBE/Feb.–Mar. 2013 NSC – Memorandum 6.3 − 2 x + 8 = −2 x 2 + 4 x + 16  − 2 x + 8 = −2 x + 4 x + 16 2  2x − 6x − 8 = 0 2 2x2 − 6x − 8 = 0 x 2 − 3x − 4 = 0 (x − 4)(x + 1) = 0  x = −1 x = 4 or x = −1 at R, y = −2(− 1) + 8 = 10  y = 10 i.e. R(– 1; 10) (4) 6.4.1 −1 ≤ x ≤ 4  −1 ≤ x x≤4 (2) 6.4.2 − 2x + 4x − 2 < 0 2  − 2 x 2 + 4 x − 2 + 18 < 18 − 2 x 2 + 4 x − 2 + 18 < 18 − 2 x 2 + 4 x + 16 < 18  − 2 x 2 + 4 x + 16 < 18 f ( x) < 18  f ( x) < 18 (−∞ ; 1) ∪ (1 ; ∞)  (−∞ ; 1) ∪ (1 ; ∞) (4) OR − 2x 2 + 4x − 2 < 0  − 2 x 2 + 4 x − 2 + 18 < 18 − 2 x 2 + 4 x − 2 + 18 < 18  − 2 x 2 + 4 x + 16 < 18 − 2 x + 4 x + 16 < 18 2  f ( x) < 18 f ( x) < 18  x ∈R ; x ≠ 1 x ∈R ; x ≠ 1 (4) [16] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 13 DBE/Feb.–Mar. 2013 NSC – Memorandum QUESTION 7 F = P(1 + i )  formula 7.1 n  substitution = 4 000 000(1 + 0,06) 3  answer = R 4 764 064 (3) 7.2.1  formula   0,06  − n  30 0001 − 1 +   0,06   12   i = 4 000 000 = 12 0,06 12 substitution into correct formula  0,06  −n 4000000 ×   −n 1  0,06   12  = 1 − 1 + 0,06   = 1 +    3  12  30000  12  1  0,06  −n correct use of logs = 1 +  3  12  answer of 220 withdrawals 1 (6) log  0, 06  = −n  1+  3  12  n = 220,27 Therefore she will make 220 withdrawals of R30 000. OR  formula   0,06  − n  0,06 30 000 1 − 1 +   i =   12   12 4 000 000 = substitution into correct formula 0,06 12  0,06  4000000 ×   −n 1  0,06  −n  12  = 1 − 1 + 0,06   = 1 +    3  12  30000  12  −n 1  0,06  correct use of logs = 1 +  3  12  answer of 220 withdrawals 1  0,06  log = −n log1 +  (6) 3  12  n = 220,27 Therefore she will make 220 withdrawals of R30 000. Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 14 DBE/Feb.–Mar. 2013 NSC – Memorandum 7.2.2   0,06  − n   20 0001 − 1 +     0,06  −n    12   20 000 1 − 1 +   4 000 000 =   12   0,06 4 000 000 = 12 0,06 −n 12  0,06  0 = 1 +   0,06  −n  12   0 = 1 +   12  She can make as many withdrawals as she pleases.  conclusion (3) [12] QUESTION 8 12 2 12  0,08   r  0,08  1 +  = 1 +  1 +   12   2  12  r 2 = 0,040672622  i 2 1 +  r = 8,13452446%  2  answer r = 8,13% [3] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 15 DBE/Feb.–Mar. 2013 NSC – Memorandum QUESTION 9 9.1 f ( x) = 2 x 3 f ( x + h) = 2( x + h ) 3 ( = 2 x 3 + 3 x 2 h + 3 xh 2 + h 3 )  substitution = 2 x 3 + 6 x 2 h + 6 xh 2 + 2h 3 f ( x + h ) − f ( x) = 2 x 3 + 6 x 2 h + 6 xh 2 + 2h 3 − 2 x 3 = 6 x 2 h + 6 xh 2 + 2h 3  expansion 6 x 2 h + 6 xh 2 + 2h 3 f ′( x ) = lim  formula h →0 h = lim ( h 6 x + 6 xh + 2h 2 2 ) h →0 h ( = lim 6 x + 6 xh + 2h 2 h →0 2 )  6 x 2 + 6 xh + 2h 2 f ′( x) = 6 x 2  answer (5) OR f ( x + h) − f ( x ) f ′( x ) = lim  formula h →0 h 2( x + h ) − 2 x 3 3  substitution = lim h →0 h = lim ( 2 x + 3 x 2 h + 3 xh 2 + h 3 − 2 x 3 3 )  expansion h →0 h 6 x h + 6 xh + 2h 3 2 2 = lim h →0 h = lim ( h 6 x + 6 xh + 2h 2 2 ) h →0 h  6 x 2 + 6 xh + 2h 2 ( = lim 6 x + 6 xh + 2h 2 h →0 2 ) f ′( x) = 6 x 2  answer (5) 9.2 2 x +1 y= − 3 x2  2x 2 3 − −2 = 2x 2 + x −2  x 5 −  − 3x 5 2 dy − = −3 x 2 − 2 x −3 −3 dx  − 2x (4) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 16 DBE/Feb.–Mar. 2013 NSC – Memorandum 9.3 f ′(−1) = −7 f ′( x) = 2ax + b  f ′( x) = 2ax + b − 7 = −2a + b  substitution of x = – 1  − 7 = −2a + b f (−1) = −7(−1) + 3 = 10  f(-1) = 10 ∴ a − b + 5 = 10 a − b = 5............[1] − 2a + b = −7.........[2] − a = −2.........[1] + [2] a=2  a=2 b = −3  b= –3 (6) [15] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 17 DBE/Feb.–Mar. 2013 NSC – Memorandum QUESTION 10 f ( x) = − x 3 − x 2 + x + 10 10.1 (0 ;10)  (0 ;10 ) (1) 10.2 0 = − x − x + x + 10 3 2  (x − 2) ( 0 = −( x − 2 ) x 2 + 3 x + 5 )  (x 2 + 3x + 5) x−2 = 0 or x 2 + 3x + 5 = 0 x=2 − 3 ± 32 − 4(1)(5) − 3 ± − 11 x= x= 2(1) 2 − 3 ± − 11 = 2  no solution which has no solution (4) Therefore the only x-intercept of f is (2; 0 ) 10.3 f ′(x ) = −3 x 2 − 2 x + 1  0 = −3 x 2 − 2 x + 1 f ′(x ) = −3 x 2 − 2 x + 1  f (x ) = 0 0 = (3x − 1)(x + 1) ′ 1  factors x= or x = −1 3 3 2  x-values 1 1 1 y = −  −   +   + 10 or y = −(− 1) − (− 1) + (− 1) + 10 3 2 3 3 3 275 = =9   ; 10  1 5 27 3 27  1 5   ; 10  (− 1 ; 9)  (– 1 ; 9) 3 27  (6) y 10.4 (0,33 ; 10,19) Turning point 10 (-1 ; 9) Turning Point  shape  intercepts  turning points 2 x (3) 0 [14] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 18 DBE/Feb.–Mar. 2013 NSC – Memorandum QUESTION 11 11.1 Length of box = 3 x  length of box = 3x Volume = l × b × h  9 = 3x ⋅ x ⋅ h 9 = 3x ⋅ x ⋅ h 9 = 3x 2 h 3  h= 3 x2 h= 2 x (3) 11.2 ( ) C = (2(3xh ) + 2 xh ) × 50 + 2 × 3x 2 × 100  (2(3 xh ) + 2 xh ) × 50  3 = 8 x 2  × 50 + 600 x 2 ( )  2 × 3 x 2 × 100 x   substitution of h = 3 1200 x2 = + 600 x 2 (3) x OR  (h × 8 x )× 50 ( ) C = (h × 8 x ) × 50 + 2 × 3x 2 × 100  3 ( )  2 × 3 x 2 × 100 = 8 x 2  × 50 + 600 x 2 3 x   substitution of h = x2 1200 = + 600 x 2 (3) x 11.3 C = 1200 x −1 + 600 x 2 dC  = −1200 x − 2 + 1200 x dC dx = −1200 x − 2 + 1200 x dC dx =0 0 = −1200 x − 2 + 1200 x  dx 1200 x 3 = 1200  x3 = 1 x =1 3 x =1 x =1 Therefore the width of the box is 1 metre. (4) [10] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 19 DBE/Feb.–Mar. 2013 NSC – Memorandum QUESTION 12  region ABIJ shaded y 12.1 60 55 (2) 50 A B C D 45 NOTE: If region BCEFGI is shaded: 40 award ONE mark 35 J 30 If any other region is shaded: award 0 marks 25 20 E 15 10 I 5 H G F x 5 10 15 20 25 30 35 40 45 50 55 60 65 12.2 x ≤ 40  x ≤ 40 x + y ≤ 60  x + y ≤ 60 y≥0 y≥0 (3) 12.3 x = 25 answer (1) 12.4 At I(25 ; 10), P = 4(25) + 10 = 110  x = 25 Maximum value of P is 110 when x = 25 and y = 10  y = 10 substitution  maximum value of P is 110 (4) 12.5 C = kx + y y = −kx + C  y = −kx + C − k < −1 k >1 (2) k >1 NOTE: Answer only: award TWO marks [12] TOTAL: 150 Copyright reserved