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MATHEMATICS – SEC LEVEL MS06 MARKING SCHEME – PAPER IIA MAY 2006 SESSION Question Number 1(a) (b) Answer Mark 4.5% --- Lm382.5 100% --- 382.5 ÷ 4.5 × 100 = Lm8500 M1 3km/h --- 30min track is 1.5km long 5km --- 60min 1.5km --- 60 ÷ 5 × 1.5 = 18min Notes A1 Dividing by 4.5 and multiplying by 100 [or use formula and substitute values] Accept 8500 [or 850 000c, unit given] M1 1.5 (or equivalent) M1 A1 Dividing by 5 and multiplying by 1.5 Total: 5 marks 2 AD̂E = AB̂E (angles on same arc) ED̂F = CB̂E (Exterior angle = interior opposite angle of a cyclic quadrilateral) AB̂E = CB̂E (BE bisector) AD̂E = ED̂F ; hence DE bisects AD̂F M1 M1 Reason must be given Reason must be given A1 M1 No need for reason Up to AD̂E = ED̂F enough Total: 4 marks 3(i) f(−6) = 2 − −36 = 2 + 2 = 4 (ii) y = 2 − 3x x 3 M1 = 2− y Making x 3 subject [not − 3x ] M1 A1 Multiplying by 3 (or LCM method) Or equivalent (using correct function notation) M1 Substitution of both functions [×3] 9(2 − x) + 6 − x = 4 18 − 9x + 6 − x = 4 18 + 6 − 4 = 9x + x 20 20 = 10x; x = 10 M1 Multiplying by common denominator [or LCM method] x=2 A1 x = 3(2 − y) f−1(x) = 3(2 − x) (iii) B1 f−1(x) + f(x) = 4 3 3(2 − x) + 2 − x 3 = 4 3 M1 M1 Collecting like terms Making x subject No ft marks awarded here Total: 9 marks Page 1 of 5 MATHEMATICS – SEC LEVEL 4(i) (ii) AĈB common A1 CÊD = CB̂A (or CD̂E = CÂB) (corresponding angles) A1 Reason must be given M1 Establish relationship between corresponding sides x+1=6 M1 Simplification [removing denominators and multiplication] x=5 A1 AB = 6(5) = 30cm BC = x + 1 = 6cm DE = 2x = 2(5) = 10cm A1ft A1ft A1ft ft for incorrect value of x in (ii) ft for incorrect value of x in (ii) ft for incorrect value of x in (ii) M1 M1 Substitution Making CD subject from previous statement DE CE 2x 2 (iii) (iv) MS06 = = AB CB 6x x +1 CD CE = AC CB CD 2 = 27 6 CD = 2 × 27 ÷ 3 CD = 9cm A1 Total: 11 marks 5(a) x 3 − 7 x − 24 = 0 [x = 3.5] 3.53 − 7(3.5) − 24 = −5.625 [x = 3.6] 3.63 − 7(3.6) − 24 = −2.544 [x = 3.7] 3.73 − 7(3.7) − 24 = 0.753 [x = 3.66] 3.663 − 7(3.66) − 24 = −0.59 Hence root lies between 3.66 and 3.7 x = 3.7 (1 d.p.) (b)(i) Line A: y = 3 Line B: gradient =1; intercept = 0 y=x Line C: gradient = 52 ; intercept = 3 y= (ii) 2 5 x+3 y≥3 y≥x y ≤ 52 x + 3 [ft from equation] M1 M1 M1 Substituting any value 3 < x < 4 Substituting to obtain a better approximation Substituting to obtain a better approximation A1 Conclusion B1 M1 A1 [Or equivalent] [Or equivalent] M1 A1 [Or equivalent] B1 B1 A1ft [Or equivalent] [Or equivalent] [ft for incorrect equation in (i)] Total: 12 marks Page 2 of 5 MATHEMATICS – SEC LEVEL 6(i) (ii) MS06 A = −5 C = 10 E = 23 or 0.66 or 0.67 B = −10 D=5 F = 12 [Refer to graph] B1 B1 B1 For any two values For any other two values For remaining two values A1 M4 For scale on both axes 1 mark for each 3 points plotted correctly according to table of values of student Overall shape obtained [Ends must not be joined] Accept 0 A1 (iii) (iv) y will approach 0 B1 when x = 0: y = 3 when y = 0: 4x + 3 = 0 4x = −3; x = − 34 [Refer to plotting of straight line] (v) From graph: x = −1; x = 1 4 M1 For attempting to obtain the coordinates of at least 2 points on line [e.g. table of values] A1 A1 A1 For coordinates of x intercept For coordinates of y intercept For correct plotting of line on same axes [Correct values only] M1 Use graph to determine x coordinate of points of intersection [or algebraic method – obtaining quadratic equation] A1 Accept either exact values only for algebraic solution or −1.1 ≤ x ≤ −0.9 and 0.2 ≤ x ≤ 0.4 for graphical solution [Accept in coordinate form – ignore y coordinate] Total: 16 marks 7(i) Area ∆PBS = 1 2 × 6× 6 2 = 18cm (ii) Volume = 1 × 18 × 6 3 M1 A1 Using 6cm for base [or height] M1 A1 For formula and substitution using ans. (i) [accept verification method] B1 M1 For 8 × volume of 1 pyramid = 36cm3 (iii) Volume of cube = 123 = 1728cm3 Volume of 8 pyramids × 36 = 8 × 36 = 288cm3 Hence 1728 − 288 = 1440cm3 M1 Subtracting volume of 8 pyramids from volume of cube A1 (iv) 6 + 8 = 14 faces B1 B1 8 14 Total: 10 marks Page 3 of 5 MATHEMATICS – SEC LEVEL 8(i) a=9 c=2 e=2 (ii) P(B, B, B) = = (iii) (iv) b=7 d=8 f=8 B3 1 mark for each 2 correct values 7 10 × 96 × 85 M1 A1 For multiplying respective probabilities Or equivalent [accept 3 d.p or more] 7 10 × 96 × 83 M1 A1 Multiplying the 3respective probabilities Or equivalent M1 Using P(A) = 1 − P(A*) M1 A1 For subtracting P(B, B, B) from 1 7 24 P(B, B, Y) = = MS06 7 40 P(at least 1 Y) = 1 − P(No Yellow) = 1 − P(B, B, B) 7 = 1 − 24 = 17 24 Total: 10 marks 9(i) In ∆ABC: AC2 = 232 + 262 AC = (ii) 1205 = 34.71 = 34.7 cm 23 − 18 = 5cm AD2 = 52 + 262 AD = 701 = 26.47 = 26.5 cm tan 35 = DY AD DY = 701 × tan 35 (iii) (iv) M1 A1 For Pythagoras and substitution B1 M1 A1 For (23 − 18) or 5 For Pythagoras and substitution M1 For using tangent ratio and substitution, including AD Accept also 18.6cm [ft for incorrect AD in (ii)] DY = 18.54 = 18.5cm A1ft In ∆ACX: AĈX = 90 AX2 = AC2 + CX2 = 34.712 + 18.542 AX = 39.35 = 39.4cm M1 M1 A1ft For using ∆ACX as a right angled ∆ at C For Pythagoras and substitution [ft for incorrect AC and/or CX in (i)/(ii)] M1 For any trigonometric ratio and respective substitution [ft for incorrect CX and/or AC in (i)/(ii)] .54 tan XÂC = CX = 18 AC 34.71 XÂC = 28.11 = 28.1 A1ft Total: 12 marks Page 4 of 5 MATHEMATICS – SEC LEVEL MS06 10(a) x x +1 − x ( x2+1) + 2 x = x ( x )− 2 + 2( x +1) x ( x +1) = x 2 − 2+ 2 x + 2 x ( x +1) ( ) = xx (xx++12) = M1 M1 x+2 x +1 M1 M1 LCM and all numerators correct [If denominator is crossed out award first M1 only and no other marks] Expand all numerators [do not award for trivial cases due to incorrect previous step] Factorise numerator [do not award if there is no common factor due to previous errors] Dividing numerator and denominator by common factor A1 5th term = 5(5) + 3 = 28 B1 (ii) 2n + 4 = 3(28) 2n = 80 n = 40 M1 M1 A1 For establishing relationship Making 2n subject (iii) 216 = 2n + 4 2n = 212 n = 106 M1 For establishing relationship (b)(i) A1 Total: 11 marks Page 5 of 5