Mathematics P LimpopoDoE/September 08 NATIONAL SENI CERTIFICATE GRADE MATHEMATICS P MEMANDUM SEPTEMBER 08 MARKS: 50 TIME: hours This memorandum consists of 6 pages.
Mathematics P LimpopoDoE/September 08 NOTE: If a candidate answered a question TWICE, mark only the FIRST attempt. If a candidate crossed out an answer and did not redo it, mark the crossed-out answer. Consistent accuracy applies to ALL aspects of the marking memorandum. Assuming values/answers in order to solve a problem is unacceptable. QUESTION.. ( x )( x ) 0 x 0 or x 0 x or x Values/waardes of/van x x x ().. x x 4 0 b x 7 x x,56 b 4ac a or x, 56 ( ) 4()( 4) () ALTERNATIVE /ALTERNATIEF x x 4 0 ( x ) x x,56 7 or/of 4 4 x,56 standard form /standaardvorm correct substitution into korrekte formule x, 56, 56 x (5) standard form/std vorm completing the square /voltooiing van vierkant x, 56, 56 x (4)
Mathematics P LimpopoDoE/September 08.. x x 4 x 4 x x = x 9x 8 0 x x 6 0 6 8x x x or/of x 6 x ALTERNATIVE/ALTERNATIEF x 4 x x x = 6 8x x 9x8 0 x b b 4ac a ( 9) 4()(8) () 9 x x or x 6 x squaring both sides /kwadrering beide kante standard form/std vorm factors/faktore both values/beide waardes select/selekteer x (5) squaring both sides /kwadrering beide kante standard form/std vorm formula/formule both values/beide waardes select/selekteer x (5)..4 x 0 or x critical values /kritieke waardes both answers and inequality signs correct () /beide antwoorde en ongelykheids tekens moet reg wees
Mathematics P 4 LimpopoDoE/September 08.. x x 4 4 P x x 8 8 x 6 x x 6 x x 6 ( ) x 6 x x or/of ALTERNATIVE/ALTERNATIEF x 4 4 P x 8 8 x 4 65 x 8 65 x prime numbers with exponents/priemgetalle met eksponente factorisation/faktorisering simplification /vereenvoudiging answer/antwoord (4) factorisation/faktorisering factorisation/faktorisering simplification/vereenvoudiging answer/antwoord (4).. P = 8 x 8 x x x equating/stel gelyk answer/antwoord ()
Mathematics P 5 LimpopoDoE/September 08. y x x 4 k x k ( ) 5 k : x( ) x k 4 : x(4-) 4x x 4x 5y 5x 5y x y ()...x x x 0 x x 0 ( x) x x or x y or y y both values in terms of x /beide waardes in terme van equating/stel gelyk standard form/standaardvorm x or x y or y [5] (5) QUESTION /VRAAG. 0 y ; 7 ; 5 ; 8 y. 0 y ; 7 ; 5 ; 8y + + y 8 8y 4 y 8y y = 8y y = First differences /eerste verskil Second differences /tweede verskil equating /gelykstelling value of y (4)
Mathematics P 6 LimpopoDoE/September 08. ; 7 ; 5; 5. a a 6 = () + b = b + + c = c Tn n n n n n > 57 n 60 0 n n 48 45 0 The quadratic row /kwadratiese ry value of a value of b value of c inequality /ongelykheid factorization faktorisering (4) n 48 or n 45 T 46 ALTERNATIVE: n n n 57 n 60 0 select correct value of n /selekteer korrekte waarde () inequality /ongelykheid n a b b 4ac 8649 n n 48 or/of n 45 T 46 formula /formule select correct value of n /selekteer regte waarde () [] QUESTION /VRAAG.. The first term a p and the constant difference d q T p 9( q p) 9q 8 p 0 p p and ( q p) 9q 8 p ()
Mathematics P 7 LimpopoDoE/September 08.. n S n a ( n ) d S0 0(a 9 d ) S0 0 p 90 ( q p) =90 q 70 p correct substitution into korrekte waarde answer /antwoord ().. T T4 67 a d a d 67 a5d 67...() T 4 a 0d 4 a 0 d 4...() Substitute () into () : ( 0 d 4) 5d 67 40d 8 5d 67 5d 75 d 5.. a 0d 4 a 0 ( 5) 4 96 S a 0 d (96 ) 0 ( 5) 966 a 0d 4 a 0( 5) 4 96 S a l 96 4 966 equation/vergelyking equation/vergelyking equating : Equation = equation value of d (4) value of a correct substitution in korrekte waarde answer/antwoord value of a correct substitution in korrekte waarde answer/antwoord ()
Mathematics P 8 LimpopoDoE/September 08. a ; r Sequence is: ; ; ; ; 9 7 Sequence of negative numbers: geometric sequence /meetkundige ry ; ; a ; r 7 9 S ; a r 9 9 8 7 8 8 sequence of negative numbers /ry met negatiewe getalle value of r and a correct substitution in korrekte formule correct answer /korrekte antwoord (5) [6] QUESTION 4/VRAAG4 4. x y y - intercept: x 0 f 0 (0) C(0;) g( x) x q C(0;) (0 ) q q q substitute C(0;) /vervang q () 4. g( x) ( x ) answer /antwoord () D(; ) 4. t answer/antwoord ()
Mathematics P 9 LimpopoDoE/September 08 4.4 y f : x y log x answer/antwoord () 4.5 form of the graph /vorm van die grafiek x intercept/afsnit any other coordinate on the graph /enige ander koordinaat op die grafiek () 4.6 g( x) x g( x ) ( x ) h( x) x + and answer/antwoord () 4.7 Domain: x 0 or x 0 answer /antwoord () [] QUESTION 5/VRAAG 5 5. x y answer/antwoord answer/antwoord () 5.. A : x 0 4 y 0 y y A(0 ; ) answer /antwoord ()
Mathematics P 0 LimpopoDoE/September 08 5.. y 0 4 0 x ( x ) 4 x 4 x 6 D( 6 ; 0) equating to 0 /gelykstel aan simplification /vereenvoudiging answer/antwoord () 5.. 4 x x 4 ( x ) (x )( x ) equating /gelykstelling B : 4 x 6x 4 0 x 6x standard form /std vorm 0 x( x ) x 0 or x g( ) ( ) B( ; ) value of x value of y (4) 5. Average gradient: y y m x x 0 6 0 applying /toepassing y y x x answer /antwoord () 5.4 g( x) x / g ( x) f ( x). g( x) 0 / g ( x) 0, x R f ( x) 0 for g / ( x ) explaining /verduidelik / g x answer of f x 0 /antwoord () [5] 0
Mathematics P LimpopoDoE/September 08 QUESTION 6/VRAAG 6 6. 6.. 6.. 500 i 5 =5 546, i 5 0, 447056 ( i) 0, 447056 5 5 5 i 0,8500000 i 0,499998898 i 5% im m i eff ( ) m 0,08 i ( ) eff ieff 0,08999 00 ieff 8,99 8, 60 Deposit paid = 800 000 R 080 000 00 Deposito betaal 40 Balance owed = 80 000 R70 000 00 Therefore/dus : R70 000 was financed by the bank was paid by the bank. Was gefinansier deur die bank /betaal deur die bank. correct substitution in korrekte formule applying 5 /toepassing value of i () m correct substitution in korrekte formule simplification/vereenvoudig answer/antwoord (4) deposit paid/deposito betaal balance owed /balans verskuldig 8 x 0,08 00 70 000 8 00 48 (0,546470496 ) x 884,6 x 9 correct substitution in korrekte formule n 9 answer of R8 84,6 (5) /antwoord
Mathematics P LimpopoDoE/September 08 6.. 8 A 74800 00 A R90 667, 85 6 8 00 F 8 84,6 8 00 F= R58 4,887 Balance of the loan = R 90 667,85 - R5 84,8874 9 =R 56 46 ALTERNATIVE/ALTERNATIEF correct substitution in korrekte formule A = R90 667,85 correct substitution in korrekte formule difference/verskil balance/balans (5) P n x[ ( i) ] i 8 8 00 P = 8 84,6 8 00 Balance of the loan: P = R56 46 /korrekte formule n = 8 8 84,6 correct substitution balance/balans (5) [7]
Mathematics P LimpopoDoE/September 08 QUESTION 7/VRAAG7 Penalty for notation in 7. or 7.- only once. 7. f ( x h) f ( x) lim h0 h lim h0 lim h0 lim h0 h0 ( x h) x ( x hx h ) x x 6hx h x 6hx h lim h0 h h(6x h) lim h0 h lim(6x h) 6x h h h subst. in formula simplifying factorising 6 x (5) 7. y (4x x 9) y 4x x 9 dy 8x dx simplifying 8x () 7. f ( x) x x f ( x) x x x x x / f ( x) x x / f 4 4 4 simplifying determining derivative of each fraction /bepaal eerste afgeleide van elke term substitution of 4 answer (4) []
Mathematics P 4 LimpopoDoE/September 08 QUESTION 8/VRAAG 8 8. g ( x) ( x )( x )( x 5) 0 x or x 5 A ( ; 0) and D(5 ; 0) 8. / g ( x) x 6x 9 0 : x x 0 ( x )( x) 0 x or x g() C(; ) and A( ;0) 8. EC: y x x 9x 5 0 x x 9x 5 0 x x 9x 7 0 ( x )( x )( x ) x or x x coordinate 8.4. / g ( x) x 6x 9 8.4. // g x x g ( ) 6 6 0 x at E : x () () () 9() 5 6 point of inflection: (; 6) xa + xc y x A y y C and 0 x y 6 point of inflection: ( ; 6) Gradient of tangent: g / () () 6() 9 Point of inflection (; 6 ) y x c Equation of tangent: 4 c 6 () c y x 4 A(-; 0) D(5 ; 0) first derivative = 0 /eerste afgeleide C(;-) () A(-;0) () equation of tangent /vergelyking van raaklyn equating /gelykstelling value of x second derivative=0 /tweede afgeleide value of x value of y average value of x value of y () () first derivative /eerste afgeleide value of gradient gradient substitution of coordinates of the point of inflection equation of tangent (4)
Mathematics P 5 LimpopoDoE/September 08 QUESTION 9 /VRAAG9 9. Area shaded= 8x x 6 = 8 x 4 x 9 = 8 5 x x 9 x formula /formule factorization (4) /faktorisering 9. 5 A = 8 x x 9 Maximum area: 04 Dx ( A) 8 x 0 9 04 x 7 7 x 0, m 04 first derivative = 0 /eerste afgeleide simplifying /vereenvoudiging x value /waarde (4) 9. 5 maximum area = 8(0,) (0,) = 0,88 9 answer /antwoord () m [9] QUESTION 0/VRAAG 0 0. For an odd product,the two numbers must be odd. Vir n onewe produk moet albei getalle onewe wees First draw: 4 odd out of 7 cards Eerste trekking: 4 onewe van 7 kaarte Second draw: odd out of 6 Tweede trekking: onewe van 6 4 P(odd product) = 0, 7 6 7 4 P(onewe produk)= 0, 7 6 7 4 7 6 7 0r 0, ()
Mathematics P 6 LimpopoDoE/September 08 0. S P(A) P(B) 0, and 0,5 0, 0,5 0,5 0,5 0.4. 0.4.. 0, P(A and B / ) = 0,55 0,5 = 0, 0, () 0,55-0,5 0, () P(A or B / ) = 0,5 = 0,65 answer /antwoord () [9] QUESTION /VRAAG. 6! = 6 5 4 70 answer /antwoord ().! 5! 0 40! 5! 40 (). Left/links : 5! Right /regs: 5! 5! 5! 6! 0 0 70 40 70 Left/links : 6 5! 5! 6! answer /answer (4) Right /regs: 6 6 + 6 6 [8] TOTAL 50