C3 June 2005 Mark Scheme

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maths c3 edexcel june 05
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  6665 Core C3 Mark Scheme (Post standardisation) Question Number SchemeMarks1 (a)Dividing by cos  θ  2 : θ θ θ θ θ  22222 cos1coscoscossin ≡+ M1Competion  : 1 ! tan  θ θ   22 sec ≡  (no errors seen) 1 (2)(b)#se o$ θ θ   22 sectan1  =+ : 1sec)1(sec2  2 =+−  θ θ  M1%  &'secsec2  2 =−+  θ θ  actorising or soving: &)1)(sec'sec2(  =−+  θ θ  M1% sec  = θ    * 2'  or sec  (1 = θ     & = θ   +1 = θ   cos  * '2  , -.1'1 1  = θ  /M1 1 2 22-.2 θ   = / 10 ()% 1$t $or = 2 θ  '&/ * 1 θ    [8] 1  6665 Core C3 Mark Scheme (Post standardisation) Question Number SchemeMarks2 (a)(i)  x x x x  2tan2sec2cossin  + M1 1 1 (') or 'sin22sec2tan2  xxx +   %M1 $or  sin  x]  (ii)'(  )1()2n  1 2  x  x x  ++ +1M1 1 (')% +1 $or '(  2 n2)  xx + (b)Di$$erentiating numerator to obtain 1&  x – 10  Di$$erentiating denominator to obtain 2(  x 1) #sing 3uotient rue $ormua correcty:4o obtain   522 )1()1(2)61&7()1&1&()1( dd −−+−−−− =  x x x x x x  x y Simpi$ying to $orm 522 )1()61&7()1(7)%1(2 −+−−−−  x x x x x 8 ' )1(- −−  x 9(c.s.o.)' (a) 2')1)(2( 17 +−−++  x x x x +1 8 )1)(2( )1('17 −+−−+  x x x x M1M1 $or combining $ractions even i$ the denominator is not oest common8 )1)(2( 52 −++  x x x  8 )1)(2( )2(2 −++  x x x  8 12 −  x   9 M1 1 cso (5)M1 must have inear numerator (b)  y  8 12 −  x   ⇒   2 =−  y xy   ⇒    xy  8 2 !  y M1 1$  *1 (  x ) 8  x x + 2  o.e. 1 (')$g(  x ) 8 52 2 +  x  (attempt)% 1;;2 −  g  M1Setting 52 2 +  x  8 51  and $inding  x 2  8 <,  x  8 ±  2 M1, 1 (') [10] 2  6665 Core C3 Mark Scheme (Post standardisation) Question Number SchemeMarks 4 (a) $  ′  (  x ) 8 ' e  x  *  x 21 M1 1 1(')1'e2  x  x − 8 & M1 ⇒   1e  = α  α   ⇒  α  α   − =  e 1  (9) 1 cso (2)(c)  ....1557.&...=1527.&..=17-.&...=&1'.& 5'21  ====  x x x x M1 1 (2)%M1 correcteastat 1  x = 1 a correct to 5 d.p.(d)#sing $  ′ (  x ) 8 ' e  x  *  x 21 ith suitabe intervae.g. $  ′  (&.15527) 8 * &.&&&>M1 $  ′  (&.155'7) 8 ! &.&&2(1) ccuracy (change o$ sign and correct vaues) 1   (2) [9] 3  6665 Core C3 Mark Scheme (Post standardisation) Question Number SchemeMarks7 (a)cos 2  A  8 cos 2    A  * sin 2    A   (   ! use o$ cos 2    A ! sin 2    A  )1 ≡   M1 = (1 * sin 2    A ), * sin 2    A   = 1  – 2 sin 2    A  (9) 1 (2)(b)  2 2sin2'cos2'sin'5sincos,'(12sin)'sin' θ θ θ θ θ θ θ  − − + ≡ − − − +  +1, M1 θ θ θ θ   sin'sin1cossin5  2 −+≡ M1 )'sin1cos5(sin  −+≡  θ θ θ  (9) 1 (5)(c)  α θ α θ θ θ   sincoscossinsincos5  R R  +≡+ Compete method $or  R  (may be impied by correct anser)% =5  222 +=  R   =5sin  = α   R   cos  = α   R M1  R 8 72  or >.21 1Compete method $or α  , 7--.& = α   (ao ''.>?)M1 1 (5)(d)sin θ    (  θ  cos5 ! 'sin  − θ  ) 8 &M1 & = θ   +1sin(  )7--.& + θ   8 ..51&.& 72' = (25.?)M1 7--.& + θ   8 (&.5261)= 2.>127 %or >.'' + θ  / 8 (25.?)= 177.5/dM1 12.2 = θ    cao 1 (7) [15] 4
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