PREPARATORY EXAMINATION

Transcription

1 NATIONAL SENIOR CERTIFICATE GRADE MATHEMATICS P PREPARATORY EXAMINATION 008 MEMORANDUM MARKS: 50 TIME: ours Tis memorandum consists of pages.

2 Matematics/P DoE/Preparatory Eamination 008 QUESTION ( 5)( + ) 0 5 or ( + ) + 8 8( + ) Standard form Factors Bot solutions LCD: (+) () Standard form 0 ( ) 0 0 or Factors 0 and Bot values (). y.() y + y + 7.() y From (): y Substitute () into () y + y(y ) + (y ) y 7 y + y ( 5) ± y y 0,6 (,5),7 5y 6 0,8 y + y 5 ± 9 y y,5 or or ( 5) (7) ( 0,6) (,,7;,5) (,8; 0,6) () 7 y (7)( 6) substitution standard form simplification bot values bot y values (7)

3 Matematics/P DoE/Preparatory Eamination 008. ( )( + ) ( + 5)( ) 0 5 or standard form factors Solutions of ( mark eac) correct inequality signs OR must be stated 5 [9] QUESTION A P( + i). ( ) n n (.06) n log(.06) n log(.06) n formula substitution log log log log n log(.06) answer. n 5.66 n.. R Answer.. n P [ ( + i) ] i R067,7 80 [ ( + 0,05) ] 0.05 formula i substitution answer ()

4 Matematics/P DoE/Preparatory Eamination A ( + i) ( + 0,05) ,57 F i n 067,7 0,05 50 n [( + i) ] 50 [( + 0,05) ] 779,69 Balance A F 607,57 779,69 R597,88 formula substitution k 50 answer () QUESTION. answer. Te sum of 's is given by sum of s S 5 ( 5) 5 Te sum of te oter numbers in te sequence is given by n S 5 [a + ( n ) d] 5 [ + (5 )] 950 formula substitution Te required sum is answer ()

5 Matematics/P 5 DoE/Preparatory Eamination 008 QUESTION. T6 8 T 5 7. Solution st differences: 5, 7, 9;, nd differences:,,, ince te nd differences are constant T p ap + bp + a a c mark for eac term () quadratic equation value of a T p p + b + c b + c 6 + b + c b + c b 0 c T p + p + bp + c. p + 67 p 65 p ±5 p 5 value of b value of c equation making n te subject two roots coosing te rigt answer [0] QUESTION r 50 a R0, n a( r ) S n r 0 0,5( ) S0 S 5,5 0 Total amount R5, Yes; R5,50+R90R80,50 e will ave enoug money to buy te boots. value of r value of a formula substitute answer answer reason CA (5) ()

6 Matematics/P 6 DoE/Preparatory Eamination T ( ) T 8( ) ( ) r value of r 5.. To converge: < < < < 0 < < S a r statement bot values and inequality signs correct formula () 8( ).5 ( ) S S 8 8 substitution answer () []

7 Matematics/P 7 DoE/Preparatory Eamination 008 QUESTION 6 6. a() a 6. y 6. one value of a y y asymptote asymptote asymptotes sape coordinate () [6] QUESTION y y y log and y Correct base 7.. Yes, only one corresponding y value for every value yes reason 7.. y y y ( ) () () ()

8 Matematics/P 8 DoE/Preparatory Eamination y k k g( ) a m ( ) 0 ( ) 6 (6;0) -intercept m(0) 0 m(0) (0; ) Is te y-intercept substitution of ( ; ) value for a - Substitution -intercept y-intercept () () () []] QUESTION 8 8. answer < 80 or ( 80 ;80 ] inequality 8. g ( ) sin( 60 ) answer 8. 0 ;90 ; 80 Any two () () () () [7]

9 Matematics/P 9 DoE/Preparatory Eamination 008 QUESTION f ( ) f ( + ) f ( ) f '( ) lim 0 ( + ) ( ) f '( ) lim 0 + f '( ) lim 0 f '( ) lim 0 ( + ) f '( ) lim 0 f '( ) f ( + ) f ( ) Average gradient ( + ) ( ) ( + + ) + + () Formula Substitution Simplification Common factor 0 formula value of answer (5) () y + dy d 9.. y y dy d epanding one per term te two fractions per term answer () () [6]

10 Matematics/P 0 DoE/Preparatory Eamination 008 QUESTION 0 0. y intercept is (0;) ( )( + ) 0 ( )( + )( ) So intercepts are (;0) and (-;0) y intercept equation eac intercept () 0. f / ( ) ( + 5)( ) 0 5 ; 5 56 f ( ) ; f () Turning points: (;0) and ( ; ) 7 derivative zero factors bot values eac y value (5) 0. 5 ;9, 8 9 y sape intercepts y intercept Turning point ()

11 Matematics/P DoE/Preparatory Eamination / f ( ) + 5 / f () f () y + c 5. + c; c 7 y 7 Turning points: (;0) and ( ;9,8) 0.6 f ''( ) Te concavity canges as increases troug an inflection is at value of gradient y value equation () One mark per coordinate () 0 value statement conclusion () [] QUESTION. Incorrect formula was given in Q paper ( + )( ) A( ) A( ) A( ) + +. da + d 0 + Please consider te working wit te incorrect formula. A A,5 units 8 Incorrect formula was given in Q paper RHS simplification derivative 0 answer Substitution answer () () ()

12 Matematics/P DoE/Preparatory Eamination 008 QUESTION. <50 y <0 >0 y >0 + y 00, y N 0 constraint constraint constraint implicit constraint (6). Please consider alternatives for all questions. y Classic Sandals feasible region Elegance Sandals one mark per constraint CA 6. (80 ;0) 80 Elegance and 0 Classic searc line value y value. P y equation and y.5 P 60(80)+00(0) P R6 800 substitute values from. R6 800 () () () [8]