# C3 June 2005 Mark Scheme

View again

All materials on our website are shared by users. If you have any questions about copyright issues, please report us to resolve them. We are always happy to assist you.
Information Report
Category:

## Documents

Published:

Views: 5 | Pages: 6

Share
Related documents
Description
maths c3 edexcel june 05
Tags

## MathematicaL Analysis

Transcript
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
Recommended

5 pages

5 pages

15 pages

10 pages

17 pages

19 pages

9 pages

7 pages

4 pages

7 pages

4 pages

7 pages

9 pages

9 pages

### C3 January 2006 Mark Scheme | Trigonometric Functions

6 pages

View more...

#### YES on 38 Political issue inq form (13451507812376)_.pdf

We Need Your Support
Thank you for visiting our website and your interest in our free products and services. We are nonprofit website to share and download documents. To the running of this website, we need your help to support us.

Thanks to everyone for your continued support.

No, Thanks