NATIONAL SENIOR CERTIFICATE GRADE 11

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1 NATIONAL SENI CERTIFICATE GRADE MATHEMATICS P NOVEMBER 007 This memorandum consists of 0 pages.

2 Mathematics/P DoE/November 007 QUESTION..(a) ( + )( ) = + + = + multiplication / ( + )( ) + ( ) = 0 transposing + = 0 ( )( + ) = 0 equating to zero ( + )( ) = 0 = or = factors = or = -values ()..(b) + < 0 factors ( + )( ) < 0 s < < ().. + =. + = + = 0 ± = ± = () () ()( ) = 0, or =, = y + y = 5y ( ) + = ( ) y y 5 y y ( ) 9 6y + y + y = 5y 5y 8 y + y + y = 5y 5y 9y 7y+ 8= 0 y y+ = 0 = 0 ( y )( y ) y = or y = = or = = + = ± = + = ± = 0, or =, standard form formula s making the subject of the formula substitution multiplication simplification standard form y values values making y the subject of the formula substitution multiplication simplification (5) standard form (9)

3 Mathematics/P DoE/November 007 y = y values + y = 5y + = 5 values ( ) ( ) ( ) = = = 0 + = 0 = 0 ( )( ) = or = y = or y =

4 Mathematics/P DoE/November 007. f( ) = ( ) ( ) = + + = + ( ) ( ) = + = f ( ) = = + = ( ) f ( ) = ( ) = ( ) substitution multiplication completing the square ( + ; -) () [5] QUESTION = = 8 M = (,5) M = + 5 M = is rational 6 (,5) 5 8 simplification M = 6 5 () > 0 5 > 0. Yes = = 8.(.5) = 8.(0) the sum of the digits is = + 5 > 0 () 5 > 0 () yes (eponential law) 8 grouping bases with same eponents sum of digits (5) [5]

5 Mathematics/P 5 DoE/November 007 QUESTION. 9. T n = an + bn + c From ( ; ) = a + b + c c = a b c = () From ( ; 5) 5 = a + b + c 5 = a + b + a b = a + b (i) (i) From (; ) = 9a + b + c = 9a + b + a b 0 = 8a + b (ii) Solving (i) and (ii) simultaneously: 8 = 6a + b 0 = 8a + b = a a = b = c = (ii) a b c T n = n + n T n = an + bn + c From ( ; ) = a + b + c (i) From ( ; 5) 5 = a + b + c (ii) From (; ) = 9a + b + c (iii) (ii) (i) a + b = (iv) (iii) (ii) 5a + b = 6 (v) (v) (iv) a = a = b = c = T n = n + n. P n = n + n P 00 = = P n = 00(0) P n = eqn s (i),(ii),(iii) eqn (iv) eqn (v) a = b = c = (7) () [0]

6 Mathematics/P 6 DoE/November 007 QUESTION. ; ; ; ; 8 ; 6 Answer Height reached during the 6 th bounce = 6. st bounce : =. nd bounce : =. rd bounce: =. n th bounce : n = ( n)... = = = = = 0 = (for st bounce nd bounce and rd bounce) nth bounce () ste bounce: = de bounce: = 0 de bounce: =. de bounce:.. n de bounce : n = n = n = n = = n 9 equating eponents () during the th bounce () [0]

7 Mathematics/P 7 DoE/November 007 QUESTION 5 5. A = P( i) 500 = 86000( i) n 0.5 = ( i) formula 500 substitution 0.5 = i i = i = percentage rate = 9,9% answer answer as percentage (5) 5.. ( m) i + i = + m 0, + i = + + i =,685 i =,68% m Formula Substitution Simplification,68% () 5.. T 0 T T T T T 5 () 0. i = 0 i =, = 0.0 () 0. i = 0 i =, = Original investment 0, = P =, P P = R 9 59, 0, + i net i. formula substitution answer (6)

8 Mathematics/P 8 DoE/November y plotting of points E (in billions of rands) (one mark for plotting one or two correct points; two marks for plotting three or four correct points) () 0.5 O t (in years) 5.. The ependiture is increasing by 0,5 billion rand each year. The graph is a straight line Linear appreciation 0,5 per year straight line () 5.. E = 0,5t + 0,5t () 5.. E = 0,5(8) + E = R 6 billion () []

9 Mathematics/P 9 DoE/November 007 QUESTION 6 6. P( ; 8) is the turning point f ( ) = a( ) + 8 f ( 0) = 6 a + 8 = 6 a = f ( ) = ( ) + 8 f ( ) = ( f ( ) = + ) substitute Turning point substitute (0; 6) a + 8 = 6 a = multiplication Shift the parabola one unit to the left Then f ( + ) = a + 8 f ( 0) = 6 for = 6 = a ( ) + 8 a = f ( + ) = + 8 f ( ) = f ( ) = ( ) f ( ) = f (( ) + ) f () f () f () f () f () f () Ave gradient = or = = = = y = m 0 = m m = m = y = = = = 0 (6 + 7)( ) = 0 7 = or = 6 (; 0) m equating standard form factors (6) () () (6)

10 Mathematics/P 0 DoE/November 007 C 5 ; 6 8 coordinate 6.5 The parabola is reflected about the y-ais reflected y-ais () 6.6 h ( ) = + 6 () [] QUESTION 7 7. R ; 8 answer f ( ) = = = 8 + ( 8) = = 6 8 f ( 0) = p = + 8 p = + y = 0 answer p = answer () 7. = 8 and y = answers () 7.5 y substitute = 0 () () shape y-intercept -intercept asymptotes () []

11 Mathematics/P DoE/November 007 QUESTION 8 8. f ( ) = + a. f ( 0) = 0 + a() 0 = 0 a = 8. f( 5) = + ( )() 5 = () = 0, = + = = 5 f ( 0) = 0 () () = = = 8. h ( ) = f ( ) = () () [9] QUESTION 9 9. Period = 60 = 80 formula () 9. y [-; 0] () 9. Amplitude = () [6]

12 Mathematics/P DoE/November 007 QUESTION y 7 + 6y 0 y 6 constraint constraint constraint (5) 5 Sketching + 9y 7 + 6y 0 feasible region Feasible Region P = y coefficient of coefficient of y 0. At point A (; ): P = 0() + 50() = 90 At point B (; 0): P = 0() + 50(0) = 90 At point C (7,5; 0): P = 0(7,5) + 50(0) = 5 At point D (6; ) : P = 0(6) + 50() = 0 of points A, B, C and D (8) () Maimum value of P is at D (6;) = 6 i.e. 6 liters of Laughter y = i.e. liter of Joy 0.5 Ma profit = 0(6) + (50) = R 0 - answer y-answer () () []

KwaZulu-Natal Department of Education REPUBLIC OF SOUTH AFRICA MATHEMATICS SEPTEMBER PREPARATORY 2018 MEMORANDUM NATIONAL SENIOR CERTIFICATE GRADE 12

Education KwaZulu-Natal Department of Education REPUBLIC OF SOUTH AFRICA MATHEMATICS SEPTEMBER PREPARATY 08 MEMANDUM NATIONAL SENI CERTIFICATE GRADE MARKS: 50 TIME: hours This memorandum consists of 4