June Core Mathematics C1 Mark Scheme

Transcription

1 June 005 Mark. Penalise () 8 64 or ( a) or 8 or Allow 8 M = 4 or 0.5 A () () M for understanding that - ower means recirocal 8 4 is M0A0 and - is MA0 4. dy 6 8x x n x n ( 6x 4x 6x ) 4x 0 both ( 6x is OK) c M A () M A A () (5) In and M is for a correct ower of x in at least one term. This could be 6 in or c in st 6x A for one correct term in x : or 4x (or better simlified versions) nd A for all terms as rinted or better in one line. N.B. MA0A is not ossible. SC. For integrating their answer to art just allow the M if +c is resent June 005 Advanced Subsidiary/Advanced Level in GCE Mathematics

2 . x 8x 9 ( x 4) 45 ( x 4) ( x 4) 6 ( 9) ( x 4) 45 M A A () ALT Comare coefficients 8 a equation for a a 4 AND a b 9 b 45 M A A () ( x 4) 45 (follow through their a and b from ) x 4 45 c 4 x 4 5 d ( OK) M A A () (6) M for ( x 4) or an equation for a (allow sign error 4 or 8 on ALT) sta for ( x 4) 6( 9) can ignore -9 or for stating a 4 and an equation for b nd A for b 45 Note MA0 A is ossible for ( x 4) 45 N.B. On EPEN these marks are called MA but aly them as MAA M for a full method leading to x 4... or x... (condone x 4 = n ) N.B. ( x 4) 45 0 leading to ( x 4) 45 0 is M0A0A0 A for c and A for d N.B. M and A for c do not need (so this is a secial case for the formula method) but must be resent for the d mark) Note Use of formula that ends with scores M A A0 (but must be 5) i.e. only enalise non-integers by one mark. June 005 Advanced Subsidiary/Advanced Level in GCE Mathematics

3 4. Shae Points () M - and 4 max for shae: grahs must have curved sides and round to. Don t enalise twice. (If both grahs are really straight lines then enalise B0 in art only) st for shae through (0, 0) and ( (k,0) where k >0) nd for max at (, 5) and 6 labelled or (6, 0) seen Condone (5,) if and 5 are correct on axes. Similarly (5,) in A A () (5) 5. M for shae NOT through (0, 0) but must cut x-axis twice. st A for - and 4 labelled or (-, 0) and (4, 0) seen nd A for max at (, 5). Must be clearly in st quadrant x y and sub ( y) y 9 5y 4y 8( 0) i.e. ( 5y 4)( y ) 0 ( y ) or 4 (o.e.) (both) 5 M A M A y x 4 5 ; 4 y 5 x (o.e) 5 MA f.t. (6) st M Attemt to sub leading to equation in variable Condone sign error such as y, x = -(+y) enalise st A only st A Correct TQ (condone = 0 missing) nd M Attemt to solve TQ leading to values for y. nd A Condone mislabelling x = for y = but then M0A0 in art (c). rd M Attemt to find at least one x value (must use a correct equation) rd A f.t. f.t. only in x y (sf if not exact) Both values. N.B False squaring. (e.g. x 4y ) can only score the last marks. June 005 Advanced Subsidiary/Advanced Level in GCE Mathematics

4 6. 6x > 5 x 8x > x > or 0.5 or 4 8 M A () ( x )( x ) (> 0) Critical values x, (both) M A Choosing outside region x > or x < M A f.t. (4) ( c ) x > or < x < 4 [ 4 (, ) or (, ) is OK] f.t. f.t. () (8) M Multily out and collect terms (allow one sli and allow use of = here) st M Attemting to factorise TQ x... nd M Choosing the outside region nd A f.t. f.t. their critical values N.B.(x>, x > is M0A0) (c ) f.t. their answers to and st a correct f.t. leading to an infinite region nd a correct f.t. leading to a finite region Penalise or once only at first offence. For < x < q where > q enalise the final A in. e.g. (c) Mark x > 4 x > 4 <x < <x < B0 x >, x > x > B0 June 005 Advanced Subsidiary/Advanced Level in GCE Mathematics

5 7. ( x) 9 6 x x 9x 6 x by x M A c.s.o. () (9x 6 x ) 9x 6x x ( c) M A//0 use y and x : 8 6 c So y 8x 6x x - c = - M A c.s..o. Af.t. (6) (8) M Attemt to multily out ( x). Must have or 4 terms, allow one sign error A cso Fully correct solution to rinted answer. Penalise invisible brackets or wrong working st n M Some correct integration: x x A At least correct unsimlified terms A All terms correct (unsimlified) n Ignore + c nd M Use of y and x to find c. No + c is M0. Ac.s.o. for -. (o.e.) Award this mark if c " stated i.e. not as art of an exression for y Af.t. for simlified x terms with y and a numerical value for c. Follow through their value of c but it must be a number. June 005 Advanced Subsidiary/Advanced Level in GCE Mathematics

6 8. y ( 4) y ( 4) ( x 9) or x 9 y x 0 (o.e.) (condone terms with integer coefficients e.g. y+=x) M A A Equation of l is: y x (o.e.) Solving l and l : 6x x 0 is oint where x, y 6 x or or y y x M () A Af.t. ( x) (4) (c ) ( l is y x 7 ) C is (0, -7) or OC = 7 Area of OCP OC x, or f.t. M Ac.a.o. ALT By Integration: M for l l, x P 0 ft for correct integration (follow through their ), then Acao. l () (0) (c ) MR M for full method to find equation of l sta any unsimlified form M Attemt to solve two linear equations leading to linear equation in one variable nd A f.t. only f.t. their x or y in y x N.B. A fully correct solution by drawing, or correct answer with no working can score all the marks in art, but a artially correct solution by drawing only scores the first. f.t. Either a correct OC or f.t. from their l M for correct attemt in letters or symbols for OCP A c.a.o. 7 scores M A0 (x-axis for y-axis) Get C = (, 0) Area of OCP OC y 6 = 6 (B0MA0) June 005 Advanced Subsidiary/Advanced Level in GCE Mathematics

7 9 ( ) a ( a d) a ( n ) d ( S ) a ( n ) d a S a ( n ) d a ( n ) d either S na ( n ) d n S a ( n ) d S M dm A c.s.o (4) ( a 49, d ) u 49 0( ) = 09 M A () (c ) n S n 5000 n 50n (*) S n 49 ( n )( ) n( 50 n) M A A c.s.o () (d) ( n 00)( n 50) 0 n 50 or 00 M A//0 () (e) u < 0 n 00 not sensible 00 f.t. () () requires at least terms, must include first and last terms, an adjacent term and dots! There must be + signs for the (or at least imlied see sniet 9D) st M for reversing series. Must be arithmetic with a, n and d or l. (+ signs not essential here) nd dm for adding, must have S and be a genuine attemt. Either line is sufficient. Deendent on st M (NB Allow first marks for use of l for last term but as given for final mark ) M for using a = 49 and d = in a ( n ) d formula. (c) M for using their a, d in Sn A any correct exression Acso for utting S n =5000 and simlifying to given exression. No wrong work NB EPEN has MA here but aly marks as MAA as in scheme (d) M Attemt to solve leading to n... A//0 Give AA0 for correct value and AA for both correct (e) f.t. Must mention 00 and state u 00 < 0 (or loan aid or equivalent) If giving f.t. then must have n 76. June 005 Advanced Subsidiary/Advanced Level in GCE Mathematics

8 0 x, y ( =0 is OK) () (c) dy x 4 x 8 ( x 8x 8) dy When x =, m 7 Equation of tangent: y 0 7( x ) y 7x dy m gives x 8 x 8 7 ( x 8x 5 0) ( x 5)( x ) 0 x () or 5 5 M A M M A c.a.o M M A (5) y y 5 or M A (5) () st M some correct differentiation ( x n x n for one term) st A correct unsimlified (all terms) nd M substituting x P ( ) in their dy clear evidence rd M using their m to find tangent at dy.the m must be from their at x P( ) Use of 7 here scores M0A0 but Could get all Ms in Part (c). (c) st dy M forming a correct equation their = gradient of their tangent nd dy M for solving a quadratic based on their leading to x =... The quadratic dy could be simly =0. rd M for using their x value (obtained from their quadratic) in y to obtain y coordinate. Must have one of the other two M marks to score this. MR For misreading (0, ) for (, 0) award B0 and then MA as in scheme. Then allow all M marks but no A ft. (Max 7) June 005 Advanced Subsidiary/Advanced Level in GCE Mathematics