MATHEMATICS C1-C4 and FP1-FP3

Transcription

1 MS WELSH JOINT EDUCATION COMMITTEE. CYD-BWYLLGOR ADDYSG CYMRU General Certificate of Education Advanced Subsidiary/Advanced Tystysgrif Addysg Gyffredinol Uwch Gyfrannol/Uwch MARKING SCHEMES SUMMER 6 MATHEMATICS C-C and FP-FP

2 INTRODUCTION The marking schemes which follow were those used by the WJEC for the 6 examination in GCE MATHEMATICS. They were finalised after detailed discussion at examiners' conferences by all the examiners involved in the assessment. The conferences were held shortly after the papers were taken so that reference could be made to the full range of candidates' responses, with photocopied scripts forming the basis of discussion. The aim of the conferences was to ensure that the marking schemes were interpreted and applied in the same way by all examiners. It is hoped that this information will be of assistance to centres but it is recognised at the same time that, without the benefit of participation in the examiners' conferences, teachers may have different views on certain matters of detail or interpretation. The WJEC regrets that it cannot enter into any discussion or correspondence about these marking schemes.

3 MATHEMATICS C. (a) increase in y Gradient of AC(BD) increase in x Gradient AC Gradient BD ½ Gradient AC Gradient BD AC and BD are perpendicular (b) A correct method for finding the equation of AC(BD) Equation of AC : y (x ) (or equivalent) (f.t. candidate s gradient of AC) Equation of AC : x y (convincing) Equation of BD : y ½(x + ) (or equivalent) (f.t. candidate s gradient of BD) (c) An attempt to solve equations of AC and BD simultaneously x, y (c.a.o.) (d) A correct method for finding the length of AE AE 5. (a) 5 (5 )( ) + ( + )( ) Numerator: Denominator: 5 + Special case If not gained, allow for correctly simplified numerator or denominator following multiplication of top and bottom by a + b (b) Removing brackets 6 ( + )( ) (c.a.o.)

4 dy. (a) An attempt to find dx dy x dx dy Value of at A dx dy (f.t. candidate s ) dx Equation of tangent at A: y (x ) (or equivalent) (f.t. one error) (b) Gradient of normal Gradient of tangent Equation of normal at A: y ½(x ) (or equivalent) dy (f.t. candidate s numerical value for ) dx. (a) An expression for b ac, with b ±, and at least one of a or c correct b ac k(k ) b ac (k )(k + ) Putting b ac m k, (f.t. one slip) (b) a b Least value (f.t. candidate s b) 5. (a) Use of f () 8p + q 6 Use of f () 7p 9 + q 6 Solving simultaneous equations for p and q p, q (c.a.o.) Special case assuming p Use of one of the above equations to find q q Use of other equation to verify q (b) Dividing f (x) by (x ) and getting coefficient of x to be Remaining factor x + ax + b with one of a, b correct f (x) (x )(x + )(x + ) (c.a.o.)

5 6. (a) (a + b) a + a b + 6a b + ab + b An attempt to substitute x for a and ± for b in r. h. s. of above expansion Required expression 8x 6x x 8x ( 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 5 7. (a) y + δy (x + δx) (x + δx) + Subtracting y from above to find δy δy xδx + (δx) δx Dividing by δx, letting δx and referring to limiting δy value of δx dy x dx (b) Required derivative 7 x x +, 8. (a) An attempt to collect like terms across the inequality 7 x > 6 (b) An attempt to remove brackets x + 6x + 8 < Graph crosses x-axis at x, x Either: < x < Or: < x and x < Or: (, )

6 9. (a) Translation along y-axis so that stationary point is (, a), a, 8 Correct translation and stationary point at (, ) (b) Translation of units to left along x-axis Stationary point is (, ) Points of intersection with x-axis are (, ) and (, ) Special case Translation of units to right along x-axis with correct labelling. dy x 6x 9 dx dy Putting derived dx x, (both correct) (f.t. candidate s dy) Stationary points are (, 7) and (, 5) (both correct) (c.a.o) A correct method for finding nature of stationary points (, 7) is a maximum point (f.t. candidate s derived values) (, 5) is a minimum point (f.t. candidate s derived values)

7 MATHEMATICS C. h (correct formula h ) Integral [ ( ( values) + 8)] ( values) (F.T. one slip) S. Case h 8 8 Integral [ ( 5 + (correct formula h 8) )] (all values) (F.T. one slip). (a) x 58 º, 8 º, (b) x 6º, º, º, 66º (any value) x º, º, º,, (c) ( sin ) + sin (correct use of cos sin ) sin sin (attempt to solve quad in sin correct formula or ( sin + ) (sin ) (a cos + b) (c sin + d) with ac coefft. of sin bd constant term) sin, º, º (º) (º). (a) Area x 8 sin 5º x 8 sin 5º (correct equation) x 5 sin5 (C.A.O.) 5

8 (b) BC cos º (o.e.) BC (a) S n a + a + d a + (n ) d + a (n )d (at least terms one at each end) S n a + (n )d + a + (n )d a + d + a S n a + (n )d + a + (n )d a + (n )d + a + (n )d n[a + (n )d] S n n [a + (n )d] (convincing) (b) (i) [a + 9d] 5 [a + 9d] 6 a + 9d 5 () a + 9d 8 () Solve (), (), d (reasonable attempt to solve equations) a (both) C.A.O. (ii) 5 th term + (n ) (n 5) (correct ) 5 5 (F.T. derived values) 9 5. (a) ar 9 ar (ar kar, k 9, 9 ) (correct) 9r (F.T. value of k) r ± (F.T. value of k, r ± ) 6

9 a (b) (use of correct formula) a 8 (F.T. derived r) Third term (F.T. r) Al 7 6. x + x + 5x( + C),, 7. (a) 7 + x x x + (equating ys) x x 6 (correct attempt to solve quad) (x )(x + ) x, B (, ) (b) Area ( 7 + x x ) dx (use of integration to find areas) ( x + ) dx m (6 + x x ) dx (simplified) 6 x x + x B ( correct integrations) ( + ) (use of limits) 7 (C.A.O.) 7

10 8. (a) Let log a x p (props of logs) x a p x n (a p ) n a pn (laws of indices) log a x n pn n log a x (convincing) (b) ln 5 x + ln6 (taking logs) (x + ) ln 5 ln6 (correct) x ln 5 ln 6 ln 5 ln6 ln5 x (o.e.) m (reasonable attempt to isolate x) ln5 78 (C.A.O.) 7 9. (a) Centre (, ) Radius + 8 (use of formula or std form) (answer) (b) DP 9 (o.e.) PT DP (radius) 9 9 PT (F.T. coords of centre) (use of Pythagoras) (convincing) (c) Equation of circle is (use of x + y + gx + fy + c (x ) + (y 6) or (x ) + (y 6) any +ve no) or x + y 8x y + (either) 8. (a) x A (correct equation) 8, (convincing) (b) Area sin (sector) 85 (Δ) (sector Δ) (C.A.O.) 8 8

11 MATHEMATICS C. h 5 (h 5 correct formula) Integral.5 [ ( ) ( values) + ( 6676)] ( values) 58 (F.T. one slip). (a) a b 5º, for example (choice of values) cos (a + b) cos a + cos b + (for correct demonstration) ( cos (a + b) cos a + cos b) [other cases, cos º + cos º cos º 999 cosº + cos º cos 5º 996] (b) 7 ( + tan ) tan + tan (substitution of sec + tan ) tan + tan 6 (attempt to solve quad (a tan + b)(c tan + d) ( tan )(tan + ) with ac coefficient of tan bd constant term, tan, or formula) 56 º, 6 º, 6 6º, 96 6º (56 º, 6 º) (6 6º, 96 6º) Full F.T. for tan t, marks for tan, mark for tan +, + 8 9

12 . (a) dy dx cos t (attempt to use sin t ( sin t) dy dx y& ), x (kcos t, k,,, ) cos t, C.A.O. sin t (b) dy x + x dx dy + xy + y dx (x + xy) dy (x + xy) dx x + y dy (y ) dx (x, ) (C.A.O.) 8. (a) (i) e x x a a e a (ke x, k,) e x x (F.T. one slip (in k)) e a (ii) a (9 a) e a a 9 a e a a (convincing) (b) a f(a) 6 change of sign (attempt to find values or signs) 6 indicates presence of root (between and ) (correct values or signs and conclusion) a, a 79569, a 898, a 878 (a ) a 8 ( 885) (a to 5 places, C.A.O.) Try 8, 85 a f(a) 85 8 (attempt to find values or signs) 85 (correct values or signs) Change of sign indicates root is 8 (correct to 5 decimal places)

13 k 5. (a) (i) + (x) + 6x, + (x) k (Allow for ) (k ) + x x f ( x) (ii) x, f ( x) + x + x + x (f(x) x) (iii) x e x + x e x (x f(x) + e x g(x)) (f(x) ke x, g(x) x) (all correct) (b) (sin x)( sin x) (cos x)(cos x) sin xf ( x) cos xg( x) sin x sin x (f(x) sin x, g(x) cos x) sin x cos x sin x (sin x + cos x) sin x cosec x sin x (convincing) 6. (a) 5 x x ± 5 (both) (F.T. a x b) (b) 7x 5 x 7 8 7x 5 (7x 5 ) x 7 7. (a) (i) 7 5(5x + ) (+C) (o.e.) k ( 5x + ) 7 k 5 5 (ii) ln 8x + 7 (+C) (o.e.) (k ln 8x + 7 ) (k (o.e.))

14 (b) 6 sin x (k sin x, k,,) (k ) sin sin (use of limits, F.T. allowable k) 6 (C.A.O.) 8 8. (a) f (x) + x f (x) > since ( > and) Least value when x and is x > (b) Range of f is [o, ) (c) (x x ) + x (x + x ) x + 8 x x ± (accept ) x since domain of f is x + 8 (correct composition) (writing equations and correct expansion of binomial) (C.A.O.) (F.T. removal of ve root) 8 9. y f (x) (,) (correct behaviour for large +ve, ve x) y f (x) (, ) (behaviour for ve x, must approach ve value of y) (greater slope even only if in st quadrant) 5

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

16 MATHEMATICS C. (a) Let x + A + ( x ) ( x + ) x + B C + x ( x ) (Correct form) x + A(x ) + B(x ) + C(x + ) (correct clearing of fractions and attempt to substitute) x C(6), C x 6 A(6), A (two constants, C.A.O.) Coefft of x A + B, B (third constant, F.T. one slip) (b) f ( x) (first two terms) ( x + ) ( x ) ( x ) (third term) f ( ) + (o.e.) (C.A.O.) x + 6y dy + xy dy y dx dx (y dy ) dx (6y dy + xy ) dx dy (C.A.O.) dx Gradient of normal dy (F.T. candidate's ) dx Equation is y (x ) (F.T. candidate's gradient of normal) 5

17 . + ( cos ) cos (correct substitution for cos ) 6 cos cos (correct method of solution, ` (a cos + b)(cos + d) with ac coefft of cos, bd constant term, or correct formula) ( cos + ) ( cos ) cos, 9 5º, 5 5º, 6º, º (9 5º) (5 5º) (6º, º) 6 Full F.T. for cos +, marks for cos, mark for cos +, + Subtract mark for each additional value in range, for each branch.. (a) R cos α, R sin α R 5, α tan 6 87º (reasonable approach) (or 6 9º, 7 º) (α) (R 5) (b) Write as 5sin( x o ) + 7 o (attempt to use sin (x + α) ± ) Greatest value (F.T. one slip) 5 5. Volume sin xdx cos x ( ) dx (a + b cos x) (a, b ) x sin x x sin x ( ) (F.T. a, b)) ( 67, accept 7, sig. figures) (C.A.O.) 5 5

18 6. (a) dy dx t t t y& x& (simplified) Equation of tangent is y p p x p y p p x + p p x + y p (y y m(x x )) o.e. (convincing) (b) y, x p (o.e.) x, y p ( p ) p PA + p p p + (o.e.) (correct use of distance PB PA, PB PA formula in context) (one correct simplified distance C.A.O.) (convincing) 9 x x 7. (a) x ln xdx ln x. dx ( f ( x) ln x g( x). dx) x x x x ln x dx x x ln x ( + C) (C.A.O.) (b) u sinx +, dx cosxdx (limits are, ) x (f(x)g(x)) (f(x) g(x) ) du u (a ) a du with a ±,, u a (, allowable a) u u (o.e.) (F.T. allowable a or one slip) 9 6

19 dx 8. (a) k x dt (b) (c) dx kdt x (attempt to separate variables, allow similar work) x kt + C (unsimplified version, allow absence of C) t, x 9, 9 C (correct attempt to find C) C 6 kt 6 x (convincing) t, x gives k (o.e.) (attempt to find k) Tank is empty when 6 t, t 6 (mins.) 8 9. (a) OP OA + λ AB (correct formula for r.h.s. and method for finding AB) i + j + k + λ (i + 5j k) (AB) r i + j + k + λ (i + 5j k) (must contain r, F.T. AB) (b) (Point of intersection is on both lines) Equate coefficients of i and j + λ + μ (attempt to write equations using candidate's equations, one correct) + 5λ + μ (two correct, using candidate's equations) λ, μ (attempt to solve equations) (C.A.O.) (Consider coefficients in k) p μ λ (use of equation in k to find p) p μ + λ (F.T. candidate's λ, μ) (c) b. c (i + 8j k).(i j k) (correct method) (correct) b and c are r vectors (C.A.O.) 7

20 . x + 8 x x x x x + + K x 5 Expansion is valid for x < 8 (only for conditions on x ) (o.e.) (expression must involve ) (convincing) k b b k one or other of b has a factor these statements b has a factor a and b have a common factor Contradiction ( is irrational) (if and only if previous B gained) (depends upon previous B being gained) 8

21 MATHEMATICS FPl. Sum fr + )(r +s)i r' + afr' * si, rl r7 rl n'(n +)z, 6n(n +l)(n +), 5n(n +) 6 n(n!) gz t- n -r gn+ + o) n(n +),rz +9n +) n(n +)(n + )(n + 7) A + [Award 6 for a correct unfactorised answer] MlA]. J@ + h) - tu) : -r- -,-- (x + h) - x - _ x--(x+h)+ l(x + h) -(x') _ -h [(x + h) -](x -) im -h f(x) h --> hlz(x + lt) -Qx - ) _L " (x *). Cross multiplying,,:(z+x+d : * iz +Z + i z(t + ): -( + i) +i z -- + i - ( + ( - i) (+ ixl - i) i Mi Altemative solutions: Candidates who repiace zby x + iy and multipiy the LHS by x + - iy to obtain x(x+)+ yz (x +l) + yz _. (x +) + y - Award t9

22 No further progress is possible at this level. Candidates who cross multiply and then replace z by x + iy to obtain 5. x * iy x + -y + i(zy + x +) x-y:-,x*y:- Solving these equahons (r) (b) Let the roots be a,a anda. Then 7a -:_5.6 u-l The roots are, and 8. 7a:-p givesp:* 8at -qgivesq-6 l-q+ )" - f tl L )") (a) A+tuI:l-l tt I I I (b) Det: (I- )(x(7 + 7") + 6)+ (-7 -)"+7)+(6+7).") :f+*+?," The matrix is singuiar when this equals zero, ie, when ),: or * +}"+: giving )"-,-. Award Award Award m AI m 6. (a),,[i? [oo [ol T,: l '[oo lo r lr o [oo l-o, o I -il,l r.l l 'l. ollr o - rl oll o l,]lo o.l l -rl,] (b) Fixed points satisfy [o illxl [rl l, o -rllrllrl [o o,][i] Lil giving y*l:xandx-l:y The required equation is y : * - r.

23 The statement is true for n: since 9-5 : is divisibleby Let T, 9' - 5' and assume that Tn is divisible by. Consider It follows that To*, is divisible by, Tt*, 9t+l -5/'+ : 9k.g - 5r.5 :9t.8-5k.+9r -5r) m It follows by induction thal 9' - 5'is divisible by for all positive integers n. 8. Using reduction to echelon form, [t zll il xl [t:l rl,rl l s :ll"r,l [o,,)",),,] [t )l x [tgl lo ',ll,ll,el rt forows rnr,,!or, i*l \.') (a) (b) lny - xln(x) I dy -h(x)- ydx At the stationary point, ln(x) - so r e-'.:os andy: (e-;-''' (.) Differentiating again, ml d'y t(ay\' ydx' y'(dr) -- x whence the printed result. AG At the stationary poirrt, E so clx dv v _!_<l dxz x It is therefore a maximum. AI. u t iv : (x + iy) x' - y' + Zxyi Lrx'-y';vzxy Substituting!: x - \, Ml ux'-(x-)'x-l v Zx(x -l) Substituting from first into second, (r' -t) Mi t

24 MATHEMATICS FP. (i) f is continuous because f(x) passes through zero from both sides around x. (ii) For x, f ( x) cos x when x and for x <, f ( x). So f is continuous.. Converting to trigonometric form, i cos(/) + isin(/) B Cube roots cos( / 6 + n / ) + isin( /6 + n/) (si) n gives cos(/6) + isin(/6) + i n gives cos( 5 / 6) + isin(5/6) + i n gives cos( / ) + isin(/) i [FT on candidate s first line but award a max of marks for the cube roots of ]. (a) f ( x) (x + ) x ( + x ) < for all x > (cso) (b) f is odd because f(-x) f(x) B (c) The asymptotes are x and y. (d) Graph G

25 . (a) Completing the square, {( x ) } ( y + ) + ( x ) ( y + ) The centre is (, ) (b) In the usual notation, a, b. ( e ) e Foci are ( ±, ), Directrices are x ± (FT) 5. Putting t tan(/) and substituting, ( t ).t + + t + t t t + 8t + t t t t 6t + 9t Either t giving / n or n Or t giving /.5 + n.5 + n (Accept degrees) 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 Consider k + (cos + isin) (cos k + isin k)(cos + isin) cos k cos sin ksin + i(sin kcos + sin cos k) cos( k + ) + isin( k + ) Thus, true for n k true for n k + and since true for n, proof by induction follows. (b) cos5 5 + isin5 (cos + isin) 5 cos + 5cos (isin) + cos (isin) 5 + cos (isin) + 5cos(isin) + (isin) 5 cos + 5icos sin cos sin 5 icos sin + 5cossin + isin

26 Equating imaginary parts, 5 sin5 5 cos sin cos sin + sin 5 5 sin ( sin ) sin ( sin ) + sin sin sin + 5sin sin + sin + sin 5 6sin sin + 5sin 7. (a) x A Bx + C Let + ( x + )( x + ) x + ( x + ) x A ( x + ) + ( x + )( Bx + C) x gives A / Coeff of x gives B / Constant term gives C / x (b) Integral dx ( x ) ( x ) ( x ) x ln( x + ) + ln( x + ) + arctan 8 ln 5 + ln + arctan + ln ln8 arctan cao 8. (a) The line meets the circle where x + m ( x ) ( + m ) x m x + m Sum of roots x coordinate of M m + m AG Substitute in the equation of the line. m y m + m m + m AG (b) Dividing, x x m or m y y Substituting, x / y y + x / y xy x + y whence x + y x [Accept alternative forms, e.g. y x( x) ] 5

27 MATHEMATICS FP x x e e. (a) sinh x + + (e (e x x + e + ) ) + e coshx x x (b) Substitute to obtain the quadratic equation sinh x sinh x + sinhx,/ x 88, 8 cao. x dt t tan dx + t (,/) (,) dt /( + t ) I + ( t ) /( + t ) dt t ln ln + t t + or or tanh tanh t (.6) cao. (a) f() si f ( x) sec x tan x tan x, f ( ) sec x si f ( x) sec x, f ( ) si f ( x) sec x tan x, f ( ) si f ( x) sec x tan x + sec x, f ( ) si [This expression has several similar looking forms, eg 6 sec x tan x + sec x ] The series is x x f ( x) (b) Substituting the series gives x + 6x Solving, x.95, x.85 6

28 7. (a) sin d d, cos d d + y x sin cos cos d d d d y x ( + cos cos AG (b) Arc length d cos sin (c) CSA d ).cos cos ( + d.cos cos or d cos cos d cos + d cos 8 or )d cos (cos cos + + dsin sin 6 or d cos d cos 6 + sin sin 6 or sin d sin + ( 5) 5. (a) os d c I n n [ ] d. cos cos + n n n n d os c n n + dsin + n n n [ ] d sin ) ( sin + n n n n n n ) ( n n I n n

29 (b) I I ( I ) + sind + [ cos ] (a) Area sinh d (cosh )d sinh sinh ( 5) (b) (i) Consider x rcos sinhcos dx cosh cos sinh sin d At P, coshcos sinhsin so tanh cot (ii) The Newton-Raphson iteration is (tanh cot ) x x (sec h + cos ec ) Starting with x, we obtain x 98 GCE M/S Mathematics C-C & FP-FP (June 6)/JD 8

30 MS WELSH JOINT EDUCATION COMMITTEE. CYD-BWYLLGOR ADDYSG CYMRU General Certificate of Education Advanced Subsidiary/Advanced Tystysgrif Addysg Gyffredinol Uwch Gyfrannol/Uwch MARKING SCHEMES SUMMER 6 MATHEMATICS -M and S-S

31 INTRODUCTION The marking schemes which follow were those used by the WJEC for the 6 examination in GCE MATHEMATICS. They were finalised after detailed discussion at examiners' conferences by all the examiners involved in the assessment. The conferences were held shortly after the papers were taken so that reference could be made to the full range of candidates' responses, with photocopied scripts forming the basis of discussion. The aim of the conferences was to ensure that the marking schemes were interpreted and applied in the same way by all examiners. It is hoped that this information will be of assistance to centres but it is recognised at the same time that, without the benefit of participation in the examiners' conferences, teachers may have different views on certain matters of detail or interpretation. The WJEC regrets that it cannot enter into any discussion or correspondence about these marking schemes.

32 MATHEMATICS. NL applied to object 6g R 6a o.e. Accelerating a R 6g 6 8N c.a.o. Constant speed a R 6g 58 8N Decelerating a R N c.a.o.. (a) Use of v u + at with u, a, v 5 6 t s (b) (c) 6 + 6T + 6, 6T + 8 f.t.(a) T s c.ao.. (a) R g F limiting friction μ R used g 58 8N NL dim. correct T F a a ms - c.a.o. (b) T < limiting friction F T 5N

33 . NL applied to A and B.g T a NL applied to B T.8g 8a Adding.g.8g.9a m a 6ms - c.a.o. T 8 ( ) N c.a.o. 5. (a) Using s ut + at with s, u 5, a (±) 9 8 o.e. 5t 9 8 t t.5s v 5ms (b) Using v u + as with v, u 5, a (±) 9 8 o.e s s 8 (65)m c.a.o. (c) Using v u + at with u 5, a (±) 9 8, t o.e. v f.t. (a) if used v 7 5 Speed 7 5ms - Direction is downwards 6. Moments about C all forces dim. correct g 5 + g R D A R D 6 7N (6.g) c.a.o. Resolve R C + R D g g R C g 96N f.t. R D

34 7. (a) Conservation of momentum used + 6 V A + 6 V B V A + 6V B Restitution used V B V A ( ) V A + V B 6 Adding 7V B 8 dep. on both Ms V B ms - c.a.o. V A ms - c.a.o. (b) After B collide with wall, it is moving with speed V B ms - Since V B and VB towards A V A, B will not catch up with A f.t. (c) I 6 ( + ) NS f.t. V B, V B 8. (a) Area dist. from AB dist. from AE ABDE 6 5 BCD 8 Lamina 78 x y (areas) Moments about AB x f.t. cand's values x c.a.o. Moments about AE xy f.t. cand's values 86 y c.a.o. (b) AX cm f.t. x

35 9. μ to plane R 5g cos 5º Limiting friction F μr used x 5g cos 5 º si 5 99 N Max T when body on point of moving up plane T 5g sin α + F f.t. F T 5 N c.a.o. Min T when body on point of moving down plane T 5g sin α F f.t. F T 8 8 N c.a.o.

36 MATHEMATICS M. sin α 8 (a) P 5 v NL to whole system 5 g sin α 5 a A a a 5 ms - (b) NL applied to trailer T A T 7N. (a) r A (i k) + t ( i j 5k) ( t)i + ( t)j + ( 5t)k r B (7i + 9j 6k) + t(i 8j 5k) (7 + t)i + (9 8t)j + ( 6 5t)k (b) When t r A r B ( 9)i + ( + 7)j + ( + 6)k i + j k Distance between A and B + + m 5

37 . (a) s ( t t )dt 6t t (+C) m When t, s 6 + C o.e. C 5 s 6t t 5 m (b) a dv dt attempted a 6t (c) Power F.v. used F ma ( 9) 9 Power 9 ( 5 5 ) 9 5 5W f.t. F. Initial energy PE mgh used J si Final EE ( 8) λ J Conservation of energy 5v v 6 ms - 6

38 5. (a) Let u be initial horizontal and vertical velocities (v sin/cos 5) o.e. t be time to hit target. Horizontal motion ut Vertical motion Using s ut + at with s 6, u u (c), a (±) ( 9 8) u u ms m (subst.) Speed of projection u 6 ms - t u s convincing (b) Using v u + at with u (c), a (±) 9 8, t v 9 8 f.t. u 9 f.t. u Resultant speed ( 9 ) + ( c) 9 Direction of motion tan ( c ) 8 ms f.t. u 7 5º [No if +9. used] i.e. angle above horizontal 7

39 6. (a) r cos t i + sin t j v d (r) dt used sin t i + cos t j A (b) Consider v.r ( sin t i + cos t j).(cos t i + sin t j) sin t cos t + sin t cos t dot product v is perpendicular to r for all values of t (c) Speed of P v si ( sin t ) + (cos t) 9(sin t + cos t) c.a.o. 7. (a) Resolve T cos mg T N c.a.o. (b) NL towards C T sin mrω (a rω ) ω ω 86 rads - f.t. T 8

40 8. (a) At lowest point T mg mv r (ma) T mg, v u, r l mg mu l u gl m u gl convincing (b) Conservation of energy mu mv mgl ( cos ) (KE) (all correct) v gl gl ( cos ) gl cos gl (c) At max, v used cos c o 6 c.a.o. (d) At general mv T mg cos l (ma) mv r Subst. for v and r l T l m gl cos + mg cos l m gl m mg cos mg T mg cos 8 º, 8º 9

41 MATHEMATICS M. (a) dv a v dx used B [ B( x + A) ] x + A B ( x + A) (b) t, v, x B A t, a 6, x B 6 A A 6 A A 9 B 8 as required convincing (c) v dx 8 dt x + 9 x ( 9 ) x + dx 8 dt + 9x 8t + C t, x C f.t. minor error 6t x + 8x t x ( x + 8) 6. Auxiliary equation m + m + ± m ± i C.F. - is x e -t (A sin t + B cos t) For P.I. try x at + b dx a dt a + (at + b) 5t a 5 comp. coeff. a

42 a + b b both c.a.o. General solution is x e -t (A sin t + B cos t) + t When t, x, B dx dt used B 6 f.t. a.b. dx e -t (A sin t + B cos t) + e t ( A cos t B sin t) + dt B + A + A c.a.o. x e t (sin t + cos t) + t. (a) NL v g a dv dx dv a v dx dv v (9 + v ) dx convincing (b) vdv + dx (9 v ) sep. var. ln (9 + v ) x + C t, v 7, x m ln (9 + 7 ) C C ln (68) f.t. minor error x ln (68) ln (9 + v ) 68 ln 9 + v At greatest height, v m 68 x ln 9.5 m c.a.o. (c) Speed of ball when it returns to O is less than 7 ms because energy used (lost) in overcoming air resistance.

43 . (a) Period ω ω Using v MAX aω with v MAX, ω a a 6 m c.a.o. (b) Using v ω (a x ) with ω (c), a 6 (c), x 8 v (6 8 ) f.t. a, ω v ms f.t. a, ω (c) Let x distance from O, y, p is at O x 6 sin t 8 6 sin t 8 sin t 8 6 t sin ( 8) f.t. a, ω 59s f.t. a, ω (d) Max acceleration when x a Max acceleration ω a 6 8 ms f.t. a, ω (e) Distance travelled oscillations (a) 7m f.t. a

44 5. (a) Impulse change in momentum Apply to A I v Apply to B I cos α + v cos α + 8 cos α + cos α cos α α 6º (b) sin α u u Speed of b + ms tan 7 89º

45 6. (a) Moments about B dim correct, attempted equation 7 5g cos 6º + 75g x cos 6º S 8 sin 6º A Resolve R 7 5 g + 75 g 5 g Resolved S F F μr S μ. 5g Substitute S into moment equation and μ 5 m x (75 g cos 6º) 5 5 g 8 sin 6º 7 5 g cos 6º x c.a.o. (b) Substitute x 8 and s μ. 5g (c) into moment equation M μ. 5g 8 sin 6º 7 5g cos 6º + 75g 8 cos 6º + 75 μ. 5 8 c.a.o. (c) Person modelled as particle/ Ladder modelled as rod

46 MATHEMATICS S. (a) P(A eats red sweet) (b) P(B eats red sweet) [Method must be shown, award for either or ] (c) P(C eats red sweet) or (no working required) [FT on answers to (a) and (b) - Special case award /5 for writing down the correct answers with no working]. (a) P(A B) P(A) + P(B) P(A B).5 P(A).P(B) or P(A B) / or P(B A) / A and B are not independent (b) P(exactly one of A,B) P( A B ) + P( A B) [Award M if independence assumed having stated not independent in (a). If independence stated in (a), FT partially bearing in mind that the problem is now easier] OR P(exactly one of A,B) P(A B) P(A B) (a) Var(X) E(Y) Var(Y) 6 (b) Because Y can only take the values 8,, etc 5

47 . (a) (i) P(X > ) 658 (or 7) (ii) P(X 5) 8 77 or (cao) (b) (i) P(Y 5) e. 5 5! (ii) P(Y < ) e (cao) 5. (a) P(DY) (b) (i) P(A DY) 85 (ii) 5 5 P(B DY) 85 6 P(C DY) 69 9 (si) So most likely to have come from Farm B. 6. (a) (i) B(5,.) (ii) Mean 5 Var 5 8 SD 8 ( ) (iii) P(8 X ) 89 9 or (cao) [For candidates summing probs award for an expression involving binomial probs and for the answer] (b) X is approx Po(). P(X < ) 579 (or 5) 7. (a) Sum of probabilities 5k 5k so k 5 (b) 5 E(X) E(X 5 ) ( 5) Var(X)

48 (c) The possibilities are (,5), (,),(,). 5 P(Y 6) (a) (i) P( 5 X 5) F( 5) F( 5) ( 5 + 5) ( 5 + 5) 7 ( 9) (ii) The median satisfies ( m + m) m + m + 5 m 68 [Special case for candidates integrating F(x):- For obtaining m + m 6 ] (b) (i) f ( x) F ( x) ( x + ) (ii) E(X) x(x ) dx + [FT on candidate s f(x) from (i). If candidates use F(x) instead of f(x), not having obtained an answer in (i), award ] x 7 + x ( 58) 7

49 MATHEMATICS S. 6 6 x ( 6 6) SE of x ( 6) 95% conf limits are 6 6 ±.96 6 [ correct form, 96] giving [6,6 ] Yes because 6 is within the interval.. ( b a) Variance ( b a) 6 b a 6 AG Mean a + b b + a Solving, a 7, b.. (a) (i) 8 z ; z Prob or (cao) (ii) Reqd weight [ for 5 ± zσ] (b) X - Y is N(5,7 ) We require P(X - Y > ) z 5 ( ± ) Prob.969 (cao). (a) Prob of crash on a computer 8 e 595 Prob of crash on each of 5 computers (b) Distribution of Total is Po(). 5 P(Total 5) e. 5! 56 (cao) 8

50 5. (a) : μ versus H : μ > (Accept μ ] H In 5 days, number of passengers Y is Poi() under H. si [A for normal approx] p-value P( Y 8 ) 6 We cannot conclude that the mean has increased. (b) Under H the number of passengers in days is Po() N(,) 79 5 z 55 Either p-value 59 or CV 6 [No cc gives z 58, p 9, wrong cc gives z 6, p 5] We conclude at the % level that the mean has increased. 6. (a) (i) X is B(5,p) (si) Sig level P(X p ) 5 (cao) (ii) We require P(X 5 p ) 55 (b) Under H, X is now B(5, ) N(,) 85 5 z p-value 9 [No cc gives z 7, p 85, wrong cc gives z, p 79] Insufficient evidence to support the agent s belief. (oe). 7. (a) H : μ A μ B versus H : μ A μ B (b) 5 x A x B The appropriate test statistic is x y TS σ + m n (cao) Prob from tables p-value (i) Accept H (or the fuel consumptions are the same) at % SL (ii) Accept H (or the fuel consumptions are not the same) at 5% SL 9

51 8. (a) The mean of a large (random) sample from any distribution is (approximately) normally distributed. (b) 5 E ( X ).5, Var( X ) (si) 6 5 z 7 5 / 6 Prob 98

52 MATHEMATICS S. (a) The possible combinations are given in the following table. Combination [Accept a table with, and repeated] The sampling distribution of the mean is Mean / 5/ 7/ 8/ Prob (correct means) (correct probs) [Award for correct probs] B The sampling distribution of the median is Median Prob... (correct medians) (correct probs) B [For each table award for correct probs with replacement] 98. (a) p ˆ 5 (b) (c) SE... Approx 9% confidence limits are 5 ± 65 giving [ 9, 8]. (d) We are using a normal approximation to a binomial situation. The standard error and/or p are estimated and are not exact.

53 . x ; x 6 A B A B s s [Accept division by n giving.. and..] SE + ( 7) 5 95% conf lims are 6 ± 96 7 giving [, ] (cao). (a) UE of μ 8 5 x UE of σ 9 9 5(666..) (b) : μ 9 versus H : μ 9 H Test statistic / 97 DF 9 Crit value 6 Insufficient evidence to reject the farmer s claim at the 5% level. [Award the final only if t used] 5. (a) Σx 75, Σy 89, Σxy 7 5, Σx 75 B [ error] (b) b a 6 [Note S 7 5, S 55 5] xx xy (c) (i) Est resist ( 5) SE / 6 7( ) (ii) 95% confidence limits are 7 5 ± 96 7 giving [7,8 ] [Accept 7 9 as the upper limit]

54 (d) Test stat σ / ( Σx b β ( Σx) / n) / (75 75 / 6 (Accept, 5) EITHER p-value OR Crit value 576 The value is consistent with his prediction at the % level. 6. (a) E(U) a(μ + μ) μ if a 5 E(V) b( μ + μ) μ if b Var(V) ( σ + σ ) σ 6 6 V is the better estimator (since it has the smaller variance). (b) Var(U) ( σ σ ) σ (c) (i) μ + k μ E(W) μ + k [AG so must be convincing] (ii) σ + k σ ( + k ) Var(W) σ ( + k) ( + k) (iii) ( W ) d 6k( + k) ( + k)( + k ) Var( ) dk ( + k) For minimum variance, 6k ( + k) ( + k)( + k ) k ( + k) ( + k ) k [Award full marks if correct answer given with no/partial working] GCE M/S Mathematics -M & S-S (June 6)/JD

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