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

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1 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 ( )( n 5) / 6 only, ie /6 must be a factor [5], [] [], [] mark for each column. Must be a matrix Condone lack of brackets throughout mark for each column (no ft). Must be a matrix 5

2 4755 Mark Scheme June 04 j z is also a root Either z ( j) z ( j) ( z ) j) ( z ) j Condone (z+ + j)(z+ j) z 4z A Correct quadratic 4 z z z z z z z z A z z z z So the other roots are and A,A mark for each root, cao Or Valid method to find the other quadratic factor. Correct quadratic j j 5 oe Sum of roots with substitution of roots j for α and β ( j)( j) and A Attempt to obtain equation in using a root relation and j Eliminating or leading to a quadratic equation Correct equation obtained So the other roots are and. A,A mark for each, cao If, guessed from and give A A for these equations and AA for the roots. SC factor theorem used. for substitution of z (or ) or division by (z + ) (or by z ), A if zero obtained, for the root stated to be (or ). For the other root, similarly but AA Max [7/7] Answers only get M0M0, max [/7] 6

3 4755 Mark Scheme June 04 4 n n Split to partial fractions. Allow missing rr r 5 rr r 5 A n 5 n 5 n5 5 n5 A [5] Expand to show pattern of cancelling, at least 4 fractions All correct, allow missing, condone r Cancel to first minus last term must be in terms of n. oe single fraction 7

4 4755 Mark Scheme June 04 5 Either y y y y x x * Change of variable, condone,. y y y 9 0 dep* A Substitute into cubic expression Correct Correct coefficients in cubic expression (may be fractions) Aft ft their substitution (- each error) y 6y y 4 0 A cao. Must be an equation with integer coefficients Or 9 Let new roots be k, l, m then k l m 6 kl km lm 9 6 klm y y y A Aft All three root relations, condone incorrect signs All correct Using (α-) etc in k, kl, klm using,, One each for 6,, 4, ft their, at least two attempted, and,, A cao. Must be an equation with integer coefficients 8

5 4755 Mark Scheme June 04 6 When n =, n Condone eg and, so true for n = n Assume true for n = k Sum of k + terms E Assuming true for k, (some work to follow) If in doubt look for unambiguous if then at next E Statement of assumed result not essential but further work should be seen NB last term = sum of terms seen anywhere earns final E0 k Adding correct (k + )th term to sum for k terms k k k kk k k k k k k k k k k k k A Combining their fractions Complete accurate work n which is n with nk May be shown earlier Therefore if true for n = k it is also true for n = k. E Dependent on A and previous E. Since it is true for n =, it is true for all positive integers, n. E Dependent on and previous E E0 if last term = sum of terms seen above 9

6 4755 Mark Scheme June 04 7 (i) 5 5 0, Allow for both x 0 and y seen 6 6 5, 0, 5, 0 (both) Allow ( 5,0) or for both y 0 and x 5 seen 7 (ii) a = 7 (iii) y = 0 x = -, x = Must be two equations y [] [] Two outer branches correctly placed Inner branches correctly placed Correct asymptotes and intercepts labelled 5 5/6 5 x For good drawing. Dep all marks above Look for a clear maximum point on the right-hand branch, ( not really shown here). Condone turning points in 5 x, y 0 x = x = ½ x = [4] (iv) x 5, x, x 5 B [] One mark for each. Strict inequalities. Allow.4 for 5 (if B then if more than inequalities) 0

7 4755 Mark Scheme June 04 8 (i) 4 w π arg w arctan π π w 4cos jsin A Accept [] π 4,,.05 rad, in place of π, or π 4e j

8 4755 Mark Scheme June 04 8 (ii) Im w Circle, or arc of circle, centre the origin Radius 4 Re π π Half line from origin angle with positive real axis 4 or acute angle labelled as π/ Use of negative Im axis clearly indicated Correct region indicated. Dependent on first 4 B marks Ignore placing of w. Maximum z w = 4 = 7.7 ( s.f.) = 6 w at intersection of π line and circle (dep st B marks) Maximum z w indicated by chord on diagram oe or sight of 4 j ( j) oe Valid attempt to calculate maximum z w A allow 6 oe (accept s.f. or better) [9]

9 4755 Mark Scheme June 04 9 (i) 5 multiply second row of A with first column of B 5 0 A Correct 9 (ii) 5 9 (iii) A Correct When, 5 A [] [] Attempt to multiply relevant row of A with relevant column of B. Condone use of BA instead Multiplication of B by A Correct elements in matrix and correct γ. A does not exist when ft ft their 0. Condone 9 (iv) x 5 y z x 4, y 6, z [],(γ ) using in both their Set-up of pre-multiplication by their x A ( using ) soi need not be fully evaluated A [5], or by B cao A for each explicit identification of x, y, z in a vector or a list. (- unidentified) Answers only or solution by other method, M0A0