4755 Mark Scheme June 2013

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Aldous Douglas
6 years ago
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1 4755 Mark Scheme June 0 Evidence of comparing coefficients, or multiplying out x x 5 x A x B x C D M the RHS, or substituting. May be implied by A = or D = - 4 Comparing coefficients of x, A B Comparing coefficients of x, B A 0 B 4 B Comparing coefficients of x, C B 0 C B Comparing constants, D C 0 D 4 B Unidentified, max 4 marks. [5] z is a root z is a factor. M Use of factor theorem, accept z, z z z b z c z 9 z z 0 M Attempt to factorise cubic to linear x quadratic Other roots when z 6 z 0 0 M A Compare coefficients to find quadratic (or other valid complete method leading to a quadratic) Correct quadratic z M Use of quadratic formula (or other valid method) in their quadratic j o r j oe for both complex roots FT their -term quadratic A provided roots are complex. OR 9, 5, o r M Two root relations (may use ) 6, 0 M leading to sum and product of unknown roots z 6 z 0 0, M and quadratic equation A which is correct z M Use of quadratic formula (or other valid method) in their quadratic j o r j oe For both complex roots FT their -term quadratic A provided roots are complex. or roots must be complex, so a b j, a 6, 9 b 0 z j, z j M A [6] SCM0B if conjugates not justified

2 4755 Mark Scheme June 0 (i) 4 p 0 M Any valid row x column leading to p (ii) 4 (i) p B [] x 9 y N 5 M z z 5 c o s j s in 4 4 M A A M Attempt to use N Attempt to multiply matrices (implied by x result) One element correct All correct. FT their p May be implied Correct solution by means of simultaneous equations can earn full marks. M elimination of one unknown, M solution for one unknown A one correct, A all correct 5 5 j A oe (exact numerical form) []

3 4755 Mark Scheme June 0 4 (ii) 5 5 z z j j Attempt to add and subtract z and their z - may be M 5 5 implied by Argand diagram z z j j z z z B For points cao, - each error dotted lines not needed. z 5 n z z n 4 4 M For splitting summation into two. Allow missing /4 r r r 4 r 4 r 4 r n 4 n M Write out terms (at least first and last terms in full) A Allow missing /4 4 4 n M A Cancelling inner terms; SC insufficient working shown above,mm0ma (allow missing /4 ) Inclusion of /4 justified 4 n n 4 4 n 4 n A [6] Honestly obtained (AG)

4 4755 Mark Scheme June 0 6 x w w x M x 5 x x 6 0 w w w M 7 w w w 4 5 w w 9 w 5 0 A Substituting Correct 7 w 6 w 8 0 w A FT x w, w 9 w 4 w 6 0 w 9 0 A cao, - each error OR In original equation 5,, 6 New roots,, MA all correct for A, 9 M At least two relations attempted 7 9 A Correct - each error FT their 5,,6 Fully correct equation A Cao, accept rational coefficients here [7]

5 4755 Mark Scheme June 0 7 (i) Vertical asymptotes at x and x occur when M Some evidence of valid reasoning may be implied b x x a 0 7 (ii) a an d b A A Horizontal asymptote at x x cx c Valid reasoning seen y so when x gets very large, A M Some evidence of method needed e.g. substitute in large values with result Large positive x, y fro m b e lo w Large negative x, y fro m a b o v e A Both approaches correct (correct b,c) x x y B B LH branch correct RH branch correct Each one carefully drawn.

6 4755 Mark Scheme June 0 7 (iii) x x x x M x x Or other valid method, to values of x (allow valid solution of inequality) 0 x x x o r x A Explicit values of x (, ) (, ) y x From the graph x x for x o r x B B FT their x=, provided >/.

7 4755 Mark Scheme June 0 8 (i) n n n r r r r n M Split into separate sums r r r n n n n n n M Use of at least one standard result (ignore rd term) 6 A Correct n[ n n n 6 ] M Attempt to factorise. If more than two errors, M0 6 n[ n 8 ] 6 n [ n 4 ] 8 (ii) When n =, A Correct with factor n oe n n n Answer given n r r r 0 a n d n n n So true for n = B Assume true for n = k E Or if true for n=k, then k r r k k k r k r r k k k k k M* Add (k + )th term to both sides r k k 4 k k k k k [5]

8 4755 Mark Scheme June 0 k k k Mdep * k k k A Attempt to factorise a cubic with 4 terms Or = k k k n ( n )( n ) where n k ; or target seen But this is the given result with n = k + replacing n = k. Therefore if the result is true for n = k, it is also true for n = E Depends on A and first E k+. Since it is true for n =, it is true for all positive integers, n. E Depends on B and second E [7] 9 (i) Q represents a rotation B 90 degrees clockwise about the origin B Angle, direction and centre 9 (ii) 0 x 0 M P, A Allow both marks for P(-, ) www 9 (iii) 0 x y 0 y y [] [] M Or use of a minimum of two points l is th e lin e y x A Allow both marks for y x www [] 9 (iv) 0 x y 6 Use of a general point or two different points leading to M 6 0 y y 6 6 n is th e lin e y 6 B y=6; if seen alone MB []

9 4755 Mark Scheme June 0 9 (v) d et M 0 M is singular (or no inverse ). B www The transformation is many to one. E Accept area collapses to 0, or other equivalent statements [] 9 (vi) R Q M M Attempt to multiply in correct order 0 x y Or argue by rotation of the line y x 0 y y q is th e lin e y x A y x SC B following M0 []

Separate sum (may be implied) ( 1)(2 1) ( 1) 6 n n n n n A1,A1 1 mark for each part oe

4755 Mark Scheme June 04 n n n (i) (ii) 0 0 (iii) r( r ) r r Separate sum (may be implied) ( )( ) ( ) 6 n n n n n A,A mark for each part oe ( )[( ) 6] 6 n n n nn ( )(linear factor) ( )( 5) 6 n n n A Or