GRADE 12 LEARNER SUPPORT PROGRAMME

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1 Province of the EASTERN CAPE EDUCATION Steve Vukile Tshwete Education Complex Zone 6 Zwelitsha 5608 Private Bag X00 Bhisho 5605 REPUBLIC OF SOUTH AFRICA CHIEF DIRECTORATE CURRICULUM MANAGEMENT GRADE LEARNER SUPPORT PROGRAMME REVISION AND REMEDIAL TEACHING INSTRUMENT: ANSWERS SUBJECT: MATHEMATICS SECOND PAPER June 009 This document consists of pages. Strictly not for test/examination purposes

2 MATHEMATICS SECOND PAPER (MATH) (MEMO 06/09) QUESTION P ; = P(0; ) P lies on the y-axis because x = 0 corr. subst. midpt. formula coordinates of P conclusion. ( 6) M LM = = = = 5 ( ) corr. subst. midpt. formula (). y = mx + c form. straight y = x + line equal gradients y-intercept.4 Substitute N(; -) into y = -x + LHS = - RHS = -() + = - LHS = RHS N(; -) lies on y = -x + substitution in equation result conclusion.5 LM = = ( 5 + ) + ( + ( 4) + (4) 6) dist. formula substitution = = = 4.6 PN = LM ( 0) + ( + = = 8 =. = 4 = PN ) substitution simplification simplest surd form (4) [8]

3 (MEMO 06/09) MATHEMATICS SECOND PAPER (MATH) QUESTION. y-intercept. x + y - = 0 y = x + 6 y = 6 K(0; 6) RK = KO + OR (pyth)\ = RK = 5 RK = 5 = y = - x + 6 stand. form of equation value of y Pythagoras (4) M KR = - M KR.M NR = - M NR = y = x + c (radius tangent) (4; 0) : 0 = (4) + c or y 0 = (x 4) c = 8 - = y 8 = x - y = x - y = x - 8. NR = NP (radii) NR = NP (x - 4) + (y - 0) = (x - ) + (y + 5) x - 8x y = x - 6x y + 0y + 5-0y = x + 8 () Eq NR : y = x - 8 () () - () -y = 6 [ (-)] y = - in () -0(-) = x + 8 x = 0-8 x = x = N(; -).4 (x - x N ) + (y - y N ) = (x N - x P ) + (y N - y P ) (x - ) + (y + ) = ( - 4) + (- - 0) x - x + + y + 4y + 4 = x + y - x + 4y - 8 = 0 prod. gradients - gradient NR equation with perp. Gradient subst. of point R equation in any form equal radii substitution both sides simplification both sides equation () value of y (5) value of x (8) equate distances substitution simplification (5) []

4 4 MATHEMATICS SECOND PAPER (MATH) (MEMO 06/09) QUESTION.. Scale factor of enlargement is ().. Area of ABCD = 6 units Area of A'B'C'D' = 4 units (4).. Four times ()..4 The shapes are similar as the shape is preserved. The shapes are not congruent, as the size is not preserved in an enlargement. statement & reason statement & reason ().. (See diagram) plotting of points.. (See diagram) P (- ; 5), Q (- ; ), R (-5 ; ) correct coordinates for Δ P Q R.. (See diagram) K(5 ; -), L( ; -), M(; -5) correct coordinates of Δ KLM P y R Q Q R P - L - K M x..4 P (- ; ), Q (- ; -), R (- 4 ; -) coordinates []

5 (MEMO 06/09) MATHEMATICS SECOND PAPER (MATH) 5 QUESTION 4 4. A(-; - θ A (-; - AO = (-) + ) = + AO = 4 AO = cos θ = - 4. A ( ; ) AỐX = = 40 use of Pythagoras or distance formula value of AO value of cos θ both coordinates [6]

6 6 MATHEMATICS SECOND PAPER (MATH) (MEMO 06/09) QUESTION tan( 5 ).cos 585 = tan 45.cos 45 =.. =. = 4 sinθ cos θ cos θ tan θ = cos θ sin θ cos θ = cos θ sin θ cos θ cos θ cos θ = cos θ cos θ =.cos θ cos θ = cos θ - cos θ = 0 special angles ratios (5) identity simplification of fraction simplification of fraction answer (4) [9]

7 (MEMO 06/09) MATHEMATICS SECOND PAPER (MATH) 7 QUESTION 6 6. cos(x +,4 ) = -0,4 x +,4 = ,49 x = 8,09 reference angle quadrant 6... sin x. tan x. tan x sin x + = 0 tan x( sin x ) ( sin x ) = 0 tan = or sin x = x = k or x = , k Ζ x = k, k Ζ common factor common factor values of tan x and sin x values for x, k specified (8) [] QUESTION 7 7. y f g x f g shape y-intercept x-intercept turning points (8)

8 8 MATHEMATICS SECOND PAPER (MATH) (MEMO 06/09) () () 7. sin x + cos( x + 0 ) = sin x + cos( x + 0 ) = x = 50 () 7.4 g:x cos x + cos x + () [4] QUESTION In ACB: AĈB = 80 - (x + y) angles of triangle AB BC = sin C sin x BC. sin[80 (x + y)] AB = sin x BC.sin(x + y) = sin x AĈB sine rule substitution 8.. MN = AB, AMNB is a rectangle h = tan z BC h BC = tan z properties of rect. tan ratio value of BC BC. sin(x + y) MN = MN = AB sin x = h sin(x y) x + tan z sin x subst of BC h.sin(x + y) = tan z.sin x (4)

9 (MEMO 06/09) MATHEMATICS SECOND PAPER (MATH) 9 8. h sin(x + y) MN = tan z.sin x h sin(5, + 7, ) 70 = tan 4. sin 5, 70. tan 4. sin 5, h = sin 79,6 h = 50,7 m substitution subject of formula 8. In ABC: AC = AB + BC - AB.BC cos ABˆ C = (70) + (48,) -(70)(48,)cos7, = 4,06 m AC = 4,99 m Perimeter of ABC = 70 m + 48, m + 4,99 m = 5,9 m cos formula substitution AC (5) 8.4 Area : Δ ABC = AB.BC sin ABˆ C =.70.48,. sin 7, = 775, m number of mice = 775, = 77 mice 0 (accept 78) area rule substitution area (5) [0]

10 0 MATHEMATICS SECOND PAPER (MATH) (MEMO 06/09) QUESTION Lower quartile Q = Upper quartile Q = 6 Inter-quartile range = Q Q = 6 = box & whiskers and 9 and 6 (4) 9.. The dispersion of the data is skewed to the right () 9.. Height in cm Frequency Cumulative Frequency 0 h< h< h< h< h< h< cum. frequency () 9.. Ogive representing heights Cumulative frequency Heights in cm 8 y-co-ords of points x-co-ords of points curve (upper limits of intervals) girls are shorter than 8 cm (accept 55-65) answer ()

11 (MEMO 06/09) MATHEMATICS SECOND PAPER (MATH) Mean mass: 04 x = = 8 8 answer only via calculator: full marks Mass in x x ( x x ) kg = 04 = x = = 8 8 formula for mean x x ( x x ) 684 () (5) 9.. n ( ) x x 684 i = 85,5 i= Variance = = 8 n SD.. = 85,5 = 9.5 only via calculator: full marks formula of variance variance form. st. dev. st. deviation (4) 9..4 interval (8,75;47,5) () 5 people [9] TOTAL: 50