GCE Mathematics (MEI) Advanced Subsidiary GCE Unit 475: Concepts for Advanced Mathematics Mark Scheme for June 011 Oxford Cambridge and RSA Examinations
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475 Mark Scheme June 011 SECTION A 1 ½x 4 + x F[5] F[] [=7.5 14] =1.5 o.e. accept unsimplified at least one term correctly integrated, may be implied by ignore + c condone omission of brackets 1.5 unsupported scores 0 0.05, 000, 1.5 10-6 or 1 1, 000, 0 800000 o.e. B for two correct divergent allow alternate terms tend to zero and to infinity o.e. do not allow oscillating, getting bigger and smaller, getting further apart (i) m = 1+ 4.1 1+ 4 s.o.i 4.1 4 9. 9 grad = s.o.i 4.1 4 0.15 cao (ii) selection of value in (4, 4.1) and 4 or of two values in [.9, 4.1] centred on 4 no marks for use of Chain Rule or any other attempt to differentiate SC for 0.. appearing only embedded in equation of chord allow selection of 4 and value in (.9, 4) answer closer to 1/ than 0.15(...) 4 6 = ab and.6 = ab log6 = loga + logb and log.6 = loga + logb a = 10, b = 0.6 c.a.o. A each; if M0 then B for both, for one 1
475 Mark Scheme June 011 5 [ dy dx = ] x c.a.o. substitution of x = ½ in their dy dx 1 grad normal = their4 [= 4] must see kx their 4 must be obtained by calculus when x = ½, y = 4 ½ o.e. 1 1 1 y 4 )i.s.w = 4( x y = + 1 x 4 5 o.e. 4 8 6 1 dy 6x dx = y = kx x + c o.e. y = 4x x+ c o.e. M for k x and for -x + c 1 6 x is a mistake, not a misread y = need not be stated at this point, but must be seen at some point for full marks correct substitution of x = 9 and y = 4 in their equation of curve y = 4x x 8 6 dep dependent on at least already awarded allow for c = 86 i.s.w. if simplified equation for y seen earlier must see + c
475 Mark Scheme June 011 7 sinθ sinθ cosθ = cos θ 1 = 0 and sin θ = 0 [θ = ] 0, 180, 60, [θ = ] 60, 00 may be implied by cos θ - 1 = 0 or better or, if to advantage of candidate B4 for all 5 correct B for 4 correct B for correct for correct if 4 marks awarded, lose 1 mark for extra values in the range, ignore extra values outside the range if extra value(s) in range, deduct one mark from total do not award if values embedded in trial and improvement approach 8 log p = log s+ log t n log p = log s+ nlog t p log log p log s s [ n = ] or log t log t [base not required] p n or t s = nlog t log p = s as final answer (i.e. penalise further incorrect simplification) or A for p [ n = ]log t [base t needed ] following first s 9 log16 ½ or [ ] log5 s.o.i. if a = 10 assumed, x = 1 c.a.o. scores B www log(4 75) or log 75 5 s.o.i. x = 1 www 10 t 1 = sin θ t = sin θ x = 4 75 5 www www implies no follow through e.g. sin(θ + 60) = sin θ + sin60 = sinθ B0 Section A Total: 6
475 Mark Scheme June 011 SECTION B 11 (i) 00 - πr = πrh 00 π r h = o.e. π r substitution of correct h into V = πr h V = 100r πr convincingly obtained 100 = πr + πrh 100r = πr + πr h 100r = πr + V V = 100r πr or V for h = π r for 00 = πr V + πr π r sc for complete argument working backwards: V = 100r πr πr h = 100r πr πrh = 100 πr 100 = πrh + πr 00 = A = πrh + πr sc0 if argument is incomplete for 00 = πr + V r for V = 100r πr convincingly obtained 11 (ii) dv = 100 πr dr dv = 6πr dr B for each term allow 9.4(.) r or better if decimalised -18.8( ) r or better if decimalised 4
475 Mark Scheme June 011 11 (iii) their dv dr = 0 s.o.i. r =.6 c.a.o. A must contain r as the only variable 100 for r = ( ± ) π ; may be implied by.5 V = 17 c.a.o. deduct 1 mark only in this part if answers not given to sf, there must be evidence of use of calculus 5
475 Mark Scheme June 011 1 (i)(a) 90 B for 500 11 10 1 (i)(b) 4 S 4 = ( 500 + (4 1) 10) o.e. i.s.w. B nothing simpler than 1(1000 + -10) or 4 (1000 0) or 1( 500 0) if B not awarded, then for use of a.p. formula for S 4 with n = 4, a = 500 and d = -10 condone omission of final bracket or ()-10 if recovered in later work if they write the sum out, all the terms must be listed for marks 4 or ( ) S 4 = 500 + 70 o.e. i.s.w. [=940] (answer given) or for l = 70 s.o.i. 1 (1000 0) or 1 770 on its own do not score 1 (ii)(a) 68.(...) or 68.4 B for 460 0.98 11 1 (ii)(b) J 0 = 10 M 0 = 1.6(...), 1.4, 1., 1.7 or 1 B B for all 4 values correct or B for values correct or for values correct values which are clearly wrongly attributed do not score J 19 = 0 M 19 = 19.76(...), 19.8 or 19.7 1 (ii)(c) 887 to 887.06 B for S 4 4 ( ) 460 1 0.98 = 1 0.98 o.e. 1 a (ii)(d) (1 0.98) 480.97 to 480.98 4 (1 0.98 ) = 940 o.e. f.t. their power of 4 from (ii)c 6
475 Mark Scheme June 011 1 (i) arc AC =.1 1.8 =.78 c.a.o. area = their.78 5.5 = 0.79 or 0.8 i.s.w. dep* 10 π.1 60 dependent on first 10 or better.78 must be seen but may be embedded in area formula 1 (ii) BD =.1 cos (π 1.8) or.1cos1.(4159..) or.1sin0.(9 ) r.o.t to 1 d.p. or more = 0.48 M for BD cos( π 1.8) = o.e..1 allow any answer which rounds to 0.48 M for BD =.1 cos 76.8675 or.1sin1.14 rounded to or more sf or M for CD =.045... r.o.t. to s.f. or better and BD = (.1.045 ) 1 (iii) sector area =.969 M for ½.1 1.8 or equivalent with degrees for first two Ms N.B. 5.5.969 = 1.895 so allow M for 1.895 triangle area = 0.487 to 0.491 M for ½.1 their 0.48 sin (π 1.8) or ½ their 0.48.045.. r.o.t. to s.f. or better may be sin 1.8 instead of sin (π 1.8) N.B. 5.5 area =.6785 to.7005 so allow M for a value in this range 4.5 allow any answer which rounds to 4.5 Section B Total: 6 7