MATHEMATICS C1-C4 and FP1-FP3

Transcription

1 MS WELSH JOINT EDUCATION COMMITTEE. CYD-BWYLLGOR ADDYSG CYMRU General Certificate of Eucation Avance Subsiiary/Avance Tystysgrif Aysg Gyffreinol Uwch Gyfrannol/Uwch MARKING SCHEMES SUMMER 6 MATHEMATICS C-C an FP-FP

2 INTRODUCTION The marking schemes which follow were those use by the WJEC for the 6 eamination in GCE MATHEMATICS. They were finalise after etaile iscussion at eaminers' conferences by all the eaminers involve in the assessment. The conferences were hel shortly after the papers were taken so that reference coul be mae to the full range of caniates' responses, with photocopie scripts forming the basis of iscussion. The aim of the conferences was to ensure that the marking schemes were interprete an applie in the same way by all eaminers. It is hope that this information will be of assistance to centres but it is recognise at the same time that, without the benefit of participation in the eaminers' conferences, teachers may have ifferent views on certain matters of etail or interpretation. The WJEC regrets that it cannot enter into any iscussion or corresponence about these marking schemes.

3 MATHEMATICS C. (a) increase in y Graient of AC(BD) increase in Graient AC Graient BD ½ Graient AC Graient BD AC an BD are perpenicular M M (b) A correct metho for fining the equation of AC(BD) M Equation of AC : y ( ) (or equivalent) (f.t. caniate s graient of AC) Equation of AC : y (convincing) Equation of BD : y ½( ) (or equivalent) (f.t. caniate s graient of BD) (c) An attempt to solve equations of AC an BD simultaneously M, y (c.a.o.) () A correct metho for fining the length of AE M AE. (a) ( )( ) ( )( ) M Numerator: Denominator: Special case If M not gaine, allow B for correctly simplifie numerator or enominator following multiplication of top an bottom by a b (b) Removing brackets M B 6 B ( )( ) (c.a.o.)

4 y. (a) An attempt to fin y M y Value of at A y (f.t. caniate s ) Equation of tangent at A: y ( ) (or equivalent) (f.t. one error) (b) Graient of normal Graient of tangent M Equation of normal at A: y ½( ) (or equivalent) y (f.t. caniate s numerical value for ). (a) An epression for b ac, with b ±, an at least one of a or c correct M b ac k(k ) b ac (k )(k ) Putting b ac m k, (f.t. one slip) (b) a B b B Least value (f.t. caniate s b) B. (a) Use of f () M 8p q 6 Use of f () M 7p 9 q 6 Solving simultaneous equations for p an q M p, q (c.a.o.) Special case assuming p Use of one of the above equations to fin q q Use of other equation to verify q M (b) Diviing f () by ( ) an getting coefficient of to be M Remaining factor a b with one of a, b correct f () ( )( )( ) (c.a.o.)

5 6. (a) (a b) a a b 6a b ab b B An attempt to substitute for a an ± for b in r. h. s. of above epansion M Require epression ( terms correct) (all terms correct) A (f.t. one slip in coefficients of (a b) ) (b) Either: n( n ) k ( k,) Or: n C n M 7. (a) y δy ( δ) ( δ) B Subtracting y from above to fin δy M δy δ (δ) δ Diviing by δ, letting δ an referring to limiting δy value of δ M y (b) Require erivative 7 B, B 8. (a) An attempt to collect like terms across the inequality M 7 > 6 (b) An attempt to remove brackets M 6 8 < Graph crosses -ais at, B Either: < < Or: < an < Or: (, ) B

6 9. (a) Translation along y-ais so that stationary point is (, a), a, 8 M Correct translation an stationary point at (, ) (b) Translation of units to left along -ais M Stationary point is (, ) Points of intersection with -ais are (, ) an (, ) Special case Translation of units to right along -ais with correct labelling B. y 6 9 B y Putting erive M, (both correct) (f.t. caniate s y) Stationary points are (, 7) an (, ) (both correct) (c.a.o) A correct metho for fining nature of stationary points M (, 7) is a maimum point (f.t. caniate s erive values) (, ) is a minimum point (f.t. caniate s erive values)

7 MATHEMATICS C. h M (correct formula h ) Integral [ 79 ( 7997 B ( values) 8)] B ( values) (F.T. one slip) S. Case h 8 8 Integral [ 79 ( M (correct formula h 8) )] B (all values) (F.T. one slip). (a) 8 º, 8 º B, B (b) 6º, º, º, 66º B (any value) º, º, º B, B, B (c) ( sin ) sin M (correct use of cos sin ) sin sin M (attempt to solve qua in sin correct formula or ( sin ) (sin ) (a cos b) (c sin ) with ac coefft. of sin b constant term) sin, º, º B (º) B (º). (a) Area 8 sin º B 8 sin º B (correct equation) sin B (C.A.O.)

8 (b) BC 8..8 cos º (o.e.) B B BC 8 B 6. (a) S n a a... a (n ) a (n ) B (at least terms one at each en) S n a (n ) a (n )... a a S n a (n ) a (n )... M a (n ) a (n ) n[a (n )] S n n [a (n )] (convincing) (b) (i) [a 9] B [a 9] 6 B a 9 () a 9 8 () Solve (), (), M (reasonable attempt to solve equations) a (both) C.A.O. (ii) th term (n ) (n ) M (correct ) (F.T. erive values) 9. (a) ar 9 ar M (ar kar, k 9, 9 ) (correct) 9r (F.T. value of k) r ± (F.T. value of k, r ± ) 6

9 a (b) M (use of correct formula) a 8 (F.T. erive r) Thir term (F.T. r) Al 7 6. ( C) B, B, B 7. (a) 7 M (equating ys) 6 M (correct attempt to solve qua) ( )( ), B (, ) (b) Area ( 7 ) M (use of integration to fin areas) ( ) m (6 ) B (simplifie) 6 B ( correct integrations) ( ) M (use of limits) 7 (C.A.O.) 7

10 8. (a) Let log a p B (props of logs) a p n (a p ) n a pn B (laws of inices) log a n pn n log a B (convincing) (b) ln ln6 M (taking logs) ( ) ln ln6 (correct) ln ln 6 ln ln6 ln (o.e.) m (reasonable attempt to isolate ) ln 78 (C.A.O.) 7 9. (a) Centre (, ) B Raius 8 B (use of formula or st form) B (answer) (b) DP 9 (o.e.) PT DP (raius) 9 9 PT B (F.T. coors of centre) M (use of Pythagoras) (convincing) (c) Equation of circle is M (use of y g fy c ( ) (y 6) or ( ) (y 6) any ve no) or y 8 y (either) 8. (a) 8 B A 8 B 8 8 B (correct equation) 8, B (convincing) (b) Area sin B (sector) 8 B (Δ) M (sector Δ) (C.A.O.) 8 8

11 MATHEMATICS C. h M (h correct formula) Integral. [ 86 ( ) B ( values) ( 6676)] B ( values) 8 (F.T. one slip). (a) a b º, for eample B (choice of values) cos (a b) cos a cos b B (for correct emonstration) ( cos (a b) cos a cos b) [other cases, cos º cos º cos º 999 cosº cos º cos º 996] (b) 7 ( tan ) tan tan M (substitution of sec tan ) tan tan 6 M (attempt to solve qua (a tan b)(c tan ) ( tan )(tan ) with ac coefficient of tan b constant term, tan, or formula) 6 º, 6 º, 6 6º, 96 6º B (6 º, 6 º) B (6 6º, 96 6º) Full F.T. for tan t, marks for tan, mark for tan, 8 9

12 . (a) y cos t M (attempt to use sin t B ( sin t) y y& ), B (kcos t, k,,, ) cos t, C.A.O. sin t (b) y y y y B ( y) y ( y) y y B (y ) B (, ) B (C.A.O.) 8. (a) (i) e a a e a M (ke, k,) e (F.T. one slip (in k)) e a (ii) a (9 a) e a a 9 a e a a B (convincing) (b) a f(a) 6 change of sign M (attempt to fin values or signs) 6 inicates presence of root (between an ) (correct values or signs an conclusion) a, a 7969, a 898, a 878 B (a ) a 8 ( 88) B (a to places, C.A.O.) Try 8, 8 a f(a) 8 8 M (attempt to fin values or signs) 8 (correct values or signs) Change of sign inicates root is 8 (correct to ecimal places)

13 k. (a) (i) M () 6, () k (Allow M for ) (k ) f ( ) (ii) M, f ( ) (f() ) (iii) e e M ( f() e g()) (f() ke, g() ) (all correct) (b) (sin )( sin ) (cos )(cos ) sin f ( ) cos g( ) M sin sin (f() sin, g() cos ) sin cos sin (sin cos ) sin cosec sin (convincing) 6. (a) B ± B (both) (F.T. a b) (b) B 7 M (7 ) 7 7. (a) (i) 7 ( ) (C) (o.e.) M k ( ) 7 k (ii) ln 8 7 (C) (o.e.) M (k ln 8 7 ) (k (o.e.))

14 (b) 6 sin M (k sin, k,,) (k ) sin sin M (use of limits, F.T. allowable k) 6 (C.A.O.) 8 8. (a) f () f () > since ( > an) Least value when an is B > B (b) Range of f is [o, ) B B (c) ( ) ( ) 8 ± (accept ) since omain of f is 8 B (correct composition) M (writing equations an correct epansion of binomial) (C.A.O.) B (F.T. removal of ve root) 8 9. y f () B (,) B (correct behaviour for large ve, ve ) y f () B (, ) B (behaviour for ve, must approach ve value of y) B (greater slope even only if in st quarant)

15 . (a) Let y M (attempt to isolate, y ) y y f - () (F.T. one slip) (b) omain [, ), range [, ) B, B (c) B (parabola y ) B (relevant part of parabola) B (y f(), F.T. graph of y f ()) 8

16 MATHEMATICS C. (a) Let A ( ) ( ) B C ( ) M (Correct form) A( ) B( ) C( ) M (correct clearing of fractions an attempt to substitute) C(6), C 6 A(6), A (two constants, C.A.O.) Coefft of A B, B (thir constant, F.T. one slip) (b) f ( ) B (first two terms) ( ) ( ) ( ) B (thir term) f ( ) (o.e.) B (C.A.O.) y y y y y B (y y ) B (6y y y ) y B (C.A.O.) Graient of normal y M (F.T. caniate's ) Equation is y ( ) (F.T. caniate's graient of normal)

17 . ( cos ) cos M (correct substitution for cos ) 6 cos cos M (correct metho of solution, ` (a cos b)(cos ) with ac coefft of cos, b constant term, or correct formula) ( cos ) ( cos ) cos, 9 º, º, 6º, º B (9 º) B ( º) B (6º, º) 6 Full F.T. for cos, marks for cos, mark for cos, Subtract mark for each aitional value in range, for each branch.. (a) R cos α, R sin α R, α tan 6 87º M (reasonable approach) (or 6 9º, 7 º) (α) B (R ) (b) Write as sin( o 6 87 ) 7 o M (attempt to use sin ( α) ± ) Greatest value 7 (F.T. one slip). Volume sin B cos ( ) M (a b cos ) (a, b ) sin sin ( ) (F.T. a, b)) ( 67, accept 7, sig. figures) (C.A.O.)

18 6. (a) y t t M t y& & (simplifie) Equation of tangent is y p p p y p p p p y p M (y y m( )) o.e. (convincing) (b) y, p (o.e.) B, y p B ( p ) p PA p p p (o.e.) M (correct use of istance PB PA, PB PA formula in contet) (one correct simplifie istance C.A.O.) (convincing) 9 7. (a) ln ln. M ( f ( ) ln g( ). ) ln ln ( C) (C.A.O.) (b) u sin, cos (limits are, ) (f()g()) (f() g() ) u u M (a ) a u with a ±,, u a (, allowable a) u u (o.e.) (F.T. allowable a or one slip) 9 6

19 8. (a) k B t (b) (c) kt M (attempt to separate variables, allow similar work) kt C (unsimplifie version, allow absence of C) t, 9, 9 C M (correct attempt to fin C) C 6 kt 6 (convincing) t, gives k (o.e.) M (attempt to fin k) Tank is empty when 6 t, t 6 (mins.) 8 9. (a) OP OA λ AB M (correct formula for r.h.s. an metho for fining AB) i j k λ (i j k) B (AB) r i j k λ (i j k) (must contain r, F.T. AB) (b) (Point of intersection is on both lines) Equate coefficients of i an j λ μ M (attempt to write equations using caniate's equations, one correct) λ μ (two correct, using caniate's equations) λ, μ M (attempt to solve equations) (C.A.O.) (Consier coefficients in k) p μ λ M (use of equation in k to fin p) p μ λ (F.T. caniate's λ, μ) (c) b. c (i 8j k).(i j k) M (correct metho) 6 8 (correct) b an c are r vectors (C.A.O.) 7

20 K B 6 6 B Epansion is vali for < 8 B (only for conitions on ) 8 6 (o.e.) B (epression must involve ) 6 B (convincing) 8. k b B b k one or other of b has a factor these statements B b has a factor a an b have a common factor Contraiction ( is irrational) B (if an only if previous B gaine) B (epens upon previous B being gaine) 8

21 MATHEMATICS FP. Mo B Arg tan 6 ( ) B i (cos isin ) 6 6 B Arg of ( n i) 6 M Power is real when n is an integer multiple of 6 Smallest n 6.. Det. λ λ(λ - ) Mm λ λ EITHER ( / ) / 9) λ M > for all λ OR ' b ac' M So eterminant not equal to zero for any real λ.. f( h) - f() M ( h) [ ( h) ] [ ( h) ]( ) m h( h) [ ( h) ]( ) lim ( h) ( ) h [ ( h) ]( ) M ( ) f. 7i ( 7i)( i) i ( i)( i) M 8 i 9 i ( iy) ( iy) 9 i M Equating real an imaginary parts m an y 9

22 . (a) T B T B T M A Note: Multiplying the wrong way gives Awar M (b) Fie points satisfy y y M giving y an y Fie point is (/, /). M FT from wrong answer to (a) the incorrect matri above leas to (/,/). 6. (a) Assume the proposition is true for n k, that is ) ( ) ( k r k r B Consier ) ( ) ( k k r k r M ( ) k So, if the proposition is true for n k, it is also true for n k. (b) Since, P is false for n is therefore false. B

23 7. (a) Using reuction to echelon form, y 6 λ z μ M y 6 λ z μ The solution will not be unique if λ. (b) The equations are consistent if μ. B Put z α y α 6 6 α M M 8. (a) Let the roots be α, α, α. M Then α(α ) α(α ) (α )(α ) 7 m α 6 α The roots are, an. (b) p, q 6 BB 9. (a) lnf() ln( ) M f ( ) ln( ) f ( ) m ( ln( ) ) At the stationary point, ln() M e so e (.7) an y e ( ) (b) We now nee to etermine its nature. We see from above that For < e, f ( ) > an for > e, f ( ) < M Showing it to be a maimum.

24 . z w z wz w z z(w ) w w z w Since z, it follows that M - w w - M ( u ) v ( u ) v 9 6u u v u u v leaing to u. Straight line parallel to the v-ais (passing through (,)). B ALTERNATIVE METHOD iy iy u iv. M iy iy ( )( ) y ( ) Equating real an imaginary parts, y u y y v y iy( ) y Putting y, M u M which is a straight line parallel to the v-ais (passing through (,)). B

25 MATHEMATICS FP. (i) f is continuous because f() passes through zero from both sies aroun. BB (ii) For, f ( ) cos when an for <, f ( ). M So f is continuous.. Converting to trigonometric form, i cos(/) isin(/) B Cube roots cos( / 6 n / ) isin( /6 n/) (si) M n gives cos(/6) isin(/6) i M n gives cos( / 6) isin(/6) M i n gives cos( / ) isin(/) i [FT on caniate s first line but awar a ma of marks for the cube roots of ]. (a) f ( ) ( ) M ( ) < for all > (cso) (b) f is o because f(-) f() B (c) The asymptotes are an y. BB () Graph G

26 . (a) Completing the square, {( ) } ( y ) M ( ) ( y ) The centre is (,) (b) In the usual notation, a, b. M ( e ) e Foci are ( ±,), Directrices are ± (FT) BB. Putting t tan(/) an substituting, M ( t ).t t t t t 8t t t t t 6t 9t Either t B giving / n or n B Or t B giving /. n B. n (Accept egrees) B 6. (a) Putting n, LHS cos isin RHS so true for n. Let the result be true for n k, ie k (cos isin) cos k isin k B M Consier k (cos isin) (cos k isin k)(cos isin) M cos k cos sin ksin i(sin kcos sin cos k) cos( k ) isin( k ) Thus, true for n k true for n k an since true for n, proof by inuction follows. (b) cos isin (cos isin) M cos cos (isin) cos (isin) cos (isin) cos(isin) (isin) cos icos sin cos sin icos sin cossin isin

27 Equating imaginary parts, sin cos sin cos sin sin M sin ( sin ) sin ( sin ) sin sin sin sin sin sin sin 6sin sin sin 7. (a) A B C Let ( )( ) ( ) A ( ) ( )( B C) gives A / Coeff of gives B / Constant term gives C / M (b) Integral ( ) ( ) ( ) M ln( ) ln( ) arctan 8 ln ln arctan ln ln8 arctan 8 8. cao 8. (a) The line meets the circle where m ( ) M ( m ) m m Sum of roots coorinate of M M m m AG Substitute in the equation of the line. m y m m M m m AG (b) Diviing, m or m y y M Substituting, / y y / y M y y whence y [Accept alternative forms, e.g. y ( ) ]

28 MATHEMATICS FP e e. (a) sinh M (e (e e ) ) e cosh (b) Substitute to obtain the quaratic equation M sinh sinh sinh,/ M 88, 8 cao. t t tan t B (,/) (,) B t /( t ) I ( t ) /( t ) M t t ln ln t t or or tanh tanh t M (.6) cao. (a) f() si B f ( ) sec tan tan, f ( ) sec si BB f ( ) sec, f ( ) si B f ( ) sec tan, f ( ) si BB f ( ) sec tan sec, f ( ) si BB [This epression has several similar looking forms, eg 6 sec tan sec ] The series is f ( )... B (b) Substituting the series gives 6 M Solving,.9, M.8 6

29 7. (a) sin, cos y BB sin cos cos y M ( cos cos AG (b) Arc length cos M sin (c) CSA ).cos cos ( M.cos cos or cos cos cos cos 8 or ) cos (cos cos sin sin 6 or cos cos 6 M sin sin 6 or sin sin ( ). (a) os c I n n M [ ]. cos cos n n n n os c n n sin n n n M [ ] sin ) ( sin n n n n n n ) ( n n I n n

30 (b) I I B ( I ) B sin M [ cos ] 8 6. (a) Area sinh M (cosh ) sinh sinh ( ) (b) (i) Consier rcos sinhcos cosh cos sinh sin M M At P, coshcos sinhsin so tanh cot M (ii) The Newton-Raphson iteration is (tanh cot ) M (sec h cos ec ) Starting with, we obtain M 98 GCE M/S Mathematics C-C & FP-FP (June 6)/JD 8

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