4753 Mark Scheme June 2017 Question Answer Marks Guidance. x x x A1 correct expression, allow. isw

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1 475 Mark Scheme Jne 7 Qestion Answer Marks Gidance (5 ) (5 ) d Chain rle on (5 ) (5 ) ( 6 )( )(5 ) crect epression, allow ( 6 )( ) o.e. (5 ) isw cao isw (5 ) inverted v shape throgh (, ), (, ) and (, ) crect domain ( ) qotient ( prodct) e.g. (5 )..(5 )( 6 ) 4 (5 ) [mst have crect denom f ] v v in QR is M (i) = ln( ) = ln( ) e = = e [so f () = e ] domain < ln range < < (ii) f( ) = ln ( + ) fg() = ln( ) ln( ) + ln ( + ) = ln ( )( + ) = ln( ) e = = e f () = e allow <.69 better, < < ln < f () <. Mst se f domain, f () f range. soi e.g. from ln( ) + ln( + ) = can interchange and at an stage ln is B, ln < < ln is B allow (, ) bt not [, ]. If not labelled, take ineqalit with as domain and with f () as range mst inclde brackets mst inclde brackets 7

2 475 Mark Scheme Jne 7 4 (i) d d d o.e. (ii) when =, = 8, d 4 cao [] 5 (i) Initial temperatre is.5 [ C] boiling point is 8[ C] [] 5 (ii) = ( e k ), e k = 9.5/69.5 k = ln(.794 ) k = ln(.794 ) =.9 5 (iii) 79 = ( e kt ) e kt = = d d ( ) seen crect eqation mst simplif / (/) = sbstitting both = and = 8 into their d/d NB check power of is crect in part (i) re-arranging and taking lns (crectl) art. ln(9/) o.e. sbstitting = their (8 ) into the eqn and rearranging f e kt taking lns crectl t = ln(.488 )/.9 [=.879 ] = mins cao 6 Sppose the polgon has n sides. Then 8(n ) = 55n sm of et angles = 6 so 5n = 6 5n = 6 [ n = 4.4] 7/5 which is impossible as n is an integer So no reglar polgon has interi angle 55 When n = 4, int angle = 8/4 = 54.9 When n = 5, int angle = 8/5 = 56 So no n which gives an interi angle 55. cao clear statement of conclsion accept 54 mark final answer Trial and err: e.g. t =, = t =, = 79.4, so mins SCB 8

3 475 Mark Scheme Jne 7 7 (i) sin = sin cao cos d cao [] 7 (ii) d dt d dt When =, arcsin 6 so cos d 6 sin = ½ cos sin = ( ¼ ) = d d. d 4 when =, d d., d d t dt [5] o.e. sbstitting = into = arcsin ½ /6. d ( ¼ ) soi, e.g. d dt condone soi e.g. b d dt mst be eact, bt isw if approimated 9

4 475 Mark Scheme Jne 7 8 (i) 8 (ii) 8 (iii) (A) (iii) (B) (, ), (, ) [] d ( sin )( sin ) cos ( cos ) ( sin ) when d ( sin )( sin ) cos ( cos )= sin =, o.e. 6 [7] cos [] = ln( sin ) ft sin let = sin, d/d = cos = d = = /; = = ½ (isw) crect qotient prodct rle crect epression (isw) setting (onl) their nmerat to zero se of sin + cos = mst be eact, isw crect integral and limits c ln ( sin ) c = [d ] (igne limits) [d ] Igne increct labelling denom mst be crect at some stage missing brackets ma be inferred from sbseqent wk not denominat withhold if denom is set to zero ft their /, not 9, limits ma be implied from sbseqent wk = sin, d/d = cos [d ] = ln [ ln ] (igne limits) ln [ ln( )] = ln ln (½) is A, isw after ln not ln ln k cos [] ln sin ln ln( sin k) = ½ ln ln ( sin k) = ½ ln = ln sin k = sin k = k = arcsin( )* arcsin( ) cos [] ln( sin ) sin = ln ln( + )= ln ln = ln ½ ln = ½ ln * arcsin( ) cao [5] eqating integral from to k from k to / to ½ their area o.e. e.g. ln( sin k) = ½ ln eliminating logarithms crectl o.e. e.g ( sin k) = NB AG SC: verifing (ma marks ot of 5): attempt to find integral from to arcsin( ) crect epression N.B AG eqating integral from to k to integral from k to / ln ln( sin k) = ln( sin k) dep first from arcsin( ) to /

5 475 Mark Scheme Jne 7 9 (i) 9 (ii) 9 (iii) ( ) f( ) ( ) e e f( ) Rotational smmetr of der two abot the igin. f ( ) e ( ) e 4 f() = when e e = 4 =,.5,.5 =,.4..4 So (, ), (.,.4), (.,.4) 9 (iv) (A) let t =, dt/d = [ d = ½ dt] o.e. t e [] e [] te [d t] 9 (iv) (B) d 4 t e k te dt let = t, v=e t, =, v = e t (.,.4) 4 4 t t [ k] t( e ) ( e )dt * dep Adep [7] dep [] [] sbstitting f in f() ( ) mst have f( ) ( ) e f point half-trn (8) smmetr abot O prodct rle crect epression their deriv = taking ot dividing b e dep * =.5 o.e. dep * crect shape f with TPs, throgh O, reasonable half trn smmetr cods of TPs and stationar infleion at igin shown dep (,) given as a stationar point in part (ii) k = ½ t t crect parts on t e [ dt] kt e [ dt] igne limits, ft their k t t 4 limits mst be crect here, ft their k [ k] e te 5 e 4 4 e e 4 (.,.4) cao oe bt mst evalate e = and combine e 4 terms at least once allow description of smmetr, e.g. fits its otline if rotated etc consistent with their derivatives - condone deriv of e is e f mst be terms mst be terms Allow SC if both -cods crect one point crect (dep *) need not show stationar infleion at O. igne shape otside condone plotting beond [, ] provided shape is crect ft their k, condone v = e t

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PhysicsAndMathsTutor.com Qestion Answer Marks (i) a = ½ B allow = ½ y y d y ( ). d ( ) 6 ( ) () dy * d y ( ) dy/d = 0 when = 0 ( ) = 0, = 0 or ¾ y = (¾) /½ = 7/, y = 0.95 (sf) [] B [9] y dy/d Gi Qotient (or prodct) rle consistent