Contents. Page. Mark Schemes C1 3 C2 9 C3 15 C4 21 FP1 27 FP2 31 FP3 35 M1 39 M2 45 M3 51 S1 57 S2 61 S3 65

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1 GCE AS/A MATHEMATICS Specimen Assessment Materials Contents Page Mark Schemes C C 9 C 5 C FP 7 FP FP 5 9 M 5 M 5 S 57 S 6 S 65

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3 GCE AS/A MATHEMATICS Specimen Assessment Materials WELSH JOINT EDUCATION COMMITTEE General Certificate of Education Advanced Level/Advanced Subsidiary CYD-BWYLLGOR ADDYSG CYMRU Tystysgrif Addysg Gyffredinol Safon Uwch/Uwch Gyfrannol MATHEMATICS C Pure Mathematics Specimen Mark Scheme 005/006

4 GCE AS/ MATHEMATICS Specimen Assessment Materials. (a) Gradients: AB 0, BC 6 0, 0 CA 6. 0 (b) gradient AB gradient AC = (o.e.) (mm = ) (or Pythagoras Theorem) AB r AC giving BÂC = 90 (c) Equation of AB is y ( ) (y y = m ( )) so that y 8 = or y = 0 () (convincing) (d) Coordinates of D: (,8)., AD = ( ) ( 8) (correct formula) 5. ( ( 5 5 )( )( 5 5 ) ) = = (rationalise) (numerator unsimplified) (denominator unsimplified) = 0

5 GCE AS/ MATHEMATICS Specimen Assessment Materials 5. Equate ys 7 giving (standard form of quad ( ) 0 and attempt to solve) so that, y The line y 7 is a tangent to the curve y at the point (, ) B (Allow for (, ) is point of intersection of line y 7 and the curve y ). Condition for distinct real roots is (Use of b ac ) 6 k ( k ) 0 k k > 0 k k < 0 (convincing) giving ( k )( k ) < 0 (fied points) then k or (,). (any method) 5. (a) ( 6) 5 = [( ) 9] 5 = ( ) 7 (a), (b), (c) (a, b, c need not be displayed). (b) Least value of 5 7 Corresponding value of (c) (shape) (location of minimum)

6 GCE AS/ MATHEMATICS Specimen Assessment Materials 6 6. (a) 8k 6 0 (use of factor theorem) (correct) so that k (convincing) (b) 8 = ( )( ) = ( )( )( ). Roots are,,. (c) Remainder = () 8() 9 (use of remainder theorem) = (a) ( ) () () () 6( ) () (5 terms, correct) ( )() A (unsimplified, for each error) (simplified) (b) n C () so that n( n ). = 5 = 5 n n = 0 (standard form of quadratic and attempt to solve) n

7 GCE AS/ MATHEMATICS Specimen Assessment Materials 7 8. (a) y y ( ) ( ). (o.e.) y ( ) ( ) (o.e.) ( ). (o.e.) dy lim y lim ( ) d 0 0 (o.e.). (b) 5 ( ) (attempted differentiation of =. 5 (attempted differentiation 9. dy d = 5 at (,). of ) Equation is y 5( ) (for using gradient)

8 GCE AS/ MATHEMATICS Specimen Assessment Materials 8 0. (a) d y 6 9. d d y 0 ( ) 0. d giving =, When =, y =, =, y = 8, d d y 6 6. (any method) =, d y d, d y d Ma. point at (, 8) Min. point at (, ). (b) (general shape) (Position of Ma) (Position of Min)

9 GCE AS/A MATHEMATICS Specimen Assessment Materials 9 WELSH JOINT EDUCATION COMMITTEE General Certificate of Education Advanced Level/Advanced Subsidiary CYD-BWYLLGOR ADDYSG CYMRU Tystysgrif Addysg Gyffredinol Safon Uwch/Uwch Gyfrannol MATHEMATICS C Pure Mathematics Specimen Mark Scheme 005/006

10 GCE AS/ MATHEMATICS Specimen Assessment Materials 0. Integral 05 [05 + ( (correct formula) ) ( values) + 0] (other values) 0575 (F.T. one slip). (a) 6 ( cos ) + cos 5 = 0 (correct elimination of sin ) 6 cos cos = 0 (standard form of quadratic and an attempt to solve) giving cos =,. = 60, 00, 095, 505 (60, 095), (b) = 5, 5, 95 = 5, 05, 65.,,. (a) By the Sine Rule, 0 sin 5 = sin C (correct Sine Rule) sin C = sin 5 0 C = 58, (correct to the nearest degree),

11 GCE AS/ MATHEMATICS Specimen Assessment Materials (b) When C = 58, B = 80 (5 + 58) = 77 Sine Rule, 0 sin 5 = AC sin 77 (any correct method) (correct B) so that AC = 0sin 77 sin 5 8cm When C =, B = 80 (5 + ) = Sine Rule, 0 sin 5 = AC sin so that AC = 0sin sin 5 cm. (a) Sn = a + a r + + a r n + a r n rsn = a r + a r + + a r n + a r n + a r n Sn rsn = a a r n Sn = a( r r n ) (convincing) Sum to infinity = a r (b) a r a = a ( k a, k = or ) r (correct) r = (eliminate a and attempt to solve) giving r = 5. (a) a + d = a + 5d = 7 d =, a = (attempt to solve) (b) S8 = 8 [ + 7 ] = 00

12 GCE AS/ MATHEMATICS Specimen Assessment Materials C (attempt to integrate ) = + + C (attempt to integrate ) 7. (a) Coordinates of A = (equate ys) + 6 = 0 = A (, 8) Coordinates of B are (6,0) (standard form of quadratic and attempt to solve) (b) Integral = 0 d 6 ( )d (sum up difference or definite integrals) (all correct) = 0 6,, 6 = ( ) (use of limits) 6 = (C.A.O.) 8. AÔB =, = so that = shaded area = r ( sin ) (sector area) (triangle area) (difference of areas) 958cm

13 GCE AS/ MATHEMATICS Specimen Assessment Materials 9. (a) Let = a y, y = loga so that n = (a y ) n = a ny and loga n = ny = n loga (b) (y + ) log = log y = log log log (attempt to isolate y) (a) (5, 7); 5, (b) Gradient of radius = 7 = 5 (correct method of finding gradient of radius) Gradient of tangent = Equation of tangent is y = ( ) (c) (i) CQ = ( 5) + ( 7) = = 00 (attempt to find CQ) (ii) CQ radius and Q lies outside circle For circles touching eternally, sum of radii = CQ (convincing) Radius of required circle = Equation of circle is ( ) + (y ) = 5 (o.e.)

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15 GCE AS/A MATHEMATICS Specimen Assessment Materials 5 WELSH JOINT EDUCATION COMMITTEE General Certificate of Education Advanced Level/Advanced Subsidiary CYD-BWYLLGOR ADDYSG CYMRU Tystysgrif Addysg Gyffredinol Safon Uwch/Uwch Gyfrannol MATHEMATICS C Pure Mathematics Specimen Mark Scheme 005/006

16 GCE AS/ MATHEMATICS Specimen Assessment Materials (attempt to find values) change of sign of + 0 indicates that there is a root between 0 and. (correct values and conclusion) o = 0, = 097, = () = 09896, = () Change of sign indicates presence of root which is 0989, correct to five decimal places. Integral 0 5 [ (65) (formula with + ( )] h = 05) ( values) ( other values) 56 (C.A.O.). 9 > 5 > 9., > > giving 7 > > or (,7) (7 > ) ( > )

17 GCE AS/ MATHEMATICS Specimen Assessment Materials 7. (a) d d d d y y y y y y d dy d d y y y y y y y d d (all correct) (b) t t t y d d t t y y d d / d d d d t y t y d d d d d d d d (correct formula) 9 t t t, (convincing) 5. cosec = 7 cosec (correct elimination of cot ) cosec + cosec 8 = 0 (standard form of quadratic and attempt to solve) giving cosec =,, so that sin = ½, ¼ = 0, 50, 95º, 55º (0, 50) (95) (55)

18 GCE AS/ MATHEMATICS Specimen Assessment Materials 8 6. (a) e cos + e sin (e f() + g() sin ) (f() = cos ), (g() = ke ) (k = ) ( )() ( )() 0 (b) ( ) ( ) ( ) f ( ) ( ( ) ) g( ) (f() =, g() = ) (simplification) (c) 8 sec ( + ) (f() sec ( + ) (f() = 8) 7. (a) (i) ¼ e + (ke + ) (k = ¼) (ii) ln 6( 7) (k ln + ) (k = ½) A k ( 7) k 6 (b) cos 0 cos ( for ½) (k cos, k = ± ½, ) (k = ½)

19 GCE AS/ MATHEMATICS Specimen Assessment Materials 9 8. (a) y = tan so that tan y = then sec y = d dy dy d sec y tan y (b) f ( ) f ( ) (c) d d d (attempt to rewrite) (correct) 9. tan ln( ) C, (Asymptote y = ) (shape) (0, 5) (shape) (0,)

20 GCE AS/ MATHEMATICS Specimen Assessment Materials 0 0. (a) (i) Let y = ln ( ) + (attempt to isolate ) y = ln ( ) giving e y = y e so that f e ( ) (ii) domain [,), range [,), (b) gh() = ( + ) +, hg() = ( + ) +., ( + ) + = [( + ) + ] = and = (unsimplified) (attempt to solve)

21 GCE AS/A MATHEMATICS Specimen Assessment Materials WELSH JOINT EDUCATION COMMITTEE General Certificate of Education Advanced Level/Advanced Subsidiary CYD-BWYLLGOR ADDYSG CYMRU Tystysgrif Addysg Gyffredinol Safon Uwch/Uwch Gyfrannol MATHEMATICS C Pure Mathematics Specimen Mark Scheme 005/006

22 GCE AS/ MATHEMATICS Specimen Assessment Materials = ( ) 5 ( ) 5 = = 5 Valid for. (a) Substitution of appropriate value of Convincing conclusion 9 (b) ( sin ) = sin (correct elimination of cos ) 6 sin sin = 0 (Standard form of quadratic and attempt to solve) ( sin ) ( sin + ) = 0 so that sin =, = 8º, 8º, 0º, 0º (8º, 8º) (0º) (0º)

23 GCE AS/ MATHEMATICS Specimen Assessment Materials. R cos = 5, R sin = (both) R, = 866º (any method of funding R or ) (R) (8.7º) sin ( + 87º) = + 866º = 506º, 879º = º, 9º (any value) (both). (a) Let A B C (correct form) + + A ( ) + B ( ) + C so that A =, B =, C = 6 m (clearing fraction and attempt to find constants) A ( constants) ( constants) (b) d ln 6ln, ( may be omitted) 5. dy a dy dy d / d at t d dt dt For normal at P, gradient = p y ap = p ( ap ) giving p + y ap ap = 0 (convincing) Q: y = 0, = a + ap R: y = 0, = ap QR = a

24 GCE AS/ MATHEMATICS Specimen Assessment Materials 6. (a) d k dt d Now at t = 0, =, 0 06 dt (use of data) 0.06 Then k 0 0 d dt 0 0 (b) d 0 0dt (separation of variables and integration of r.h.s.) ln = 00t + A t = 0, = A = ln ln = -00t + ln 0.0t = ln ln = ln t ln ln (c) 5, t ln 66 years

25 GCE AS/ MATHEMATICS Specimen Assessment Materials 5 7. Volume e d n e e d ln (parts, correct choice of u, v) e e d ln m (division) e e 6 ln e 6 8. (a) d cos d cos 0 0 (a + b cos ) (a = ½, b = ½) 0 sin (convincing) (b) = tan, d = sec d (substitution for d) = 0, = 0, =, = (limits) 0 0 d 8sec sec 7 d 9 9 tan sec 7 (unsimplified) ( + tan = sec ) d cos (simplified) 8

26 GCE AS/ MATHEMATICS Specimen Assessment Materials 6 9. (a) Where the lines intersect, i + j + (i + j + k) = i + j + tk + μ (i + j + k) (attempt to equate i, j, k terms) + = + μ, + = + μ, (correct) = t + μ. Then = μ = (attempt to solve) t = (, μ) (convincing) Position vector of point of intersection is i k (b) i j k. i j k cos (identification of i j k i j k appropriate vectors) (scalar product) Now (i + j + k). (i + j + k) (correct method) = + + = 5 i + j + k = 6 (for one) i + j + k = 6 5 cos 6 º.

27 GCE AS/A MATHEMATICS Specimen Assessment Materials 7 WELSH JOINT EDUCATION COMMITTEE General Certificate of Education Advanced Level/Advanced Subsidiary CYD-BWYLLGOR ADDYSG CYMRU Tystysgrif Addysg Gyffredinol Safon Uwch/Uwch Gyfrannol MATHEMATICS FP Further Pure Mathematics Specimen Mark Scheme 005/006

28 GCE AS/ MATHEMATICS Specimen Assessment Materials 8. If - i is a root, so is + i. Also, 5 s a factor of the quartic. Using long division, ) ( The other roots are 0, ie i.. f( + h) - f() = ( h) = h ( ) ( h) = h( h h ) ( h) lim ( h h ) f () h 0 ( h) =. 8 6i (8 6i)( + i) i ( i)( + i) 0 0i i 5 (a + ib) (a ib) = + i a = and b = a = and b = /. (a) T n n n n ( ) ( n ) = n n n n n = n + (AG) n n n (b) Sn r r r r r n( n )( n ) n( n ) n 6 n ( n n n ) n ( n n )

29 GCE AS/ MATHEMATICS Specimen Assessment Materials 9 5. Assume the proposition is true for n = k. k Letting T k 7, consider k k T T 7 ( 7) m k k k ( 9 ). 8 So, if the proposition is true for n = k, it is also true for n = k +. It is true for n = since T 6 which is divisible by 8. The result is therefore proved by mathematical induction [The final is for a concluding statement plus satisfactory lay-out] 6. lnf() = ln( ) ln( ) f ( ).. f( ) ( ) ( ) Putting =, f () / f () = 8 (0.) and ( ) = ( ) = Required equation is + = 0

30 GCE AS/ MATHEMATICS Specimen Assessment Materials 0 8. (a) Det = (0 ) (5 ) +( ) = 6 Matri is singular when =. (b) (i) Using reduction to echelon form, 0 y 0 0 z Comparing the second and third rows, = for consistency. (ii) Put z = a so that y = a and = a. 9. (a) 0 Matri of translation = Matri of rotation = T - Matri = = (b) Consider the image of the point ( a, a ). This is given by 0 a a 0 a a 0 0 So, a, y a. Eliminating a, the required equation is ( y ).

31 GCE AS/A MATHEMATICS Specimen Assessment Materials WELSH JOINT EDUCATION COMMITTEE General Certificate of Education Advanced Level/Advanced Subsidiary CYD-BWYLLGOR ADDYSG CYMRU Tystysgrif Addysg Gyffredinol Safon Uwch/Uwch Gyfrannol MATHEMATICS FP Further Pure Mathematics Specimen Mark Scheme 005/006

32 GCE AS/ MATHEMATICS Specimen Assessment Materials. For the continuity of g at =, + b = + a Since g( ) b for < and a for >, it follows that + b = a The solution is a =, b =.. The equation can be rewritten sin + sincos = 0 sin(cos -) = 0 Either sin = 0 from which = 0, 60, 0, 80. Or cos = / from which = 0, 50. A ( each error or etra root in the range). Converting to trigonometric form i = r(cos + i sin ) where r and = tan, / (cos sin )) / r i r (cos isin ) ( So ( i) / / = ) (cos i sin ), ( = 50 00i / or = ) (cos i sin ) ( = i / or = ) (cos i sin ) ( = 0 5i. (a) ( ) ( ) f ( ) ( ) ( ) This is < 0 for all in the domain so f is strictly decreasing. (b) The range is (,). (c) (i) f() = 7/ and f() = so f(s) = [,7/]. (ii) f() = = (from above) and f() = = 5/. So f - (S) = [5/,].

33 GCE AS/ MATHEMATICS Specimen Assessment Materials 5. Using de Moivre s Theorem, z cos n i sin n n z cosn i sin n So, n z n z isin n 5 z z 5z 0z z 5 z z z 5 z 5 z 0 z 5 z z z 5 ( isin ) isin5-0isin + 0isin whence sin sin sin sin (a) Completing the square, ( y ) 8( ) (i) Verte is (, ). (ii) Focus is (, ) (b) (i) For the point P, ( y ) 6p and 8( - ) = 6 p. confirming that P lies on the parabola. (ii) (iii) dy d p p Equation of tangent is y p p p ( ) This passes through the origin if p p p p p 0 p 8 In view of line in (ii), Gradients of tangents =

34 GCE AS/ MATHEMATICS Specimen Assessment Materials 7. (a) (i) y G (shape) G (asymptotes) (ii) The asymptotes are = - and y = 0. (b) A B C Let ( )( ) ( ) ( ) = - gives A 5. A( ) ( )( B C) ( )( ) Coeff of gives B. 5 Const term gives C 5. (c) Int = 5 0 d d d 5 0 ( ) ( ) 5 0 ( ) = ln( ) tan ln( = ln tan (ln 5 ln ) = 0.6

35 GCE AS/A MATHEMATICS Specimen Assessment Materials 5 WELSH JOINT EDUCATION COMMITTEE General Certificate of Education Advanced Level/Advanced Subsidiary CYD-BWYLLGOR ADDYSG CYMRU Tystysgrif Addysg Gyffredinol Safon Uwch/Uwch Gyfrannol MATHEMATICS FP Further Pure Mathematics Specimen Mark Scheme 005/006

36 GCE AS/ MATHEMATICS Specimen Assessment Materials 6. sinh sinh leading to sinh sinh 0 (sinh )(sinh ) 0 sinh ( ) ln( 5 ) or sinh ( ) ln( ). (a) y cot y GG / The one point of intersection confirms that there is only one root in (0,/). (b) The Newton-Raphson iteration is cot cosec The iterates are correct to 6 decimal places. Let f() = cot f( ) = negative f(0.9585) = positive So the root is correct to 6 decimal places.

37 GCE AS/ MATHEMATICS Specimen Assessment Materials 7 y. (a) Area = y 0 d d d = 9 d 0 (b) u 9,du 6 d [0,] [,0] 0 Area =. 6 ud u = 8 u / 0 = ( ) /. I e d sin 0 = / / e sin sin. e d 0 0 = e J / J e d( cos ) 0 = / / e e cos cos. d 0 0 = - I Substituting, I e ( I) so I 5 5 e It follows that J 5 5 e 5. (a) f() = ln( + sin) : f(0) = 0 cos f ( ) sin : f (0) = f ( ) : f (0) = - sin cos f ( ) ( sin ) The series is 6 / (b) Int 6 0 = = 0.05

38 GCE AS/ MATHEMATICS Specimen Assessment Materials 8 6. (a) C (b) Area = ( cos ) d 0 = ( cos cos ) d 0 = cos cos d 0 = sin sin 0 = GG (c) The curves meet where cos = - cos cos cos 0 cos = giving (0.9, 0.675) and (0.9, ) The curves also intersect at the pole (origin). 7. (a) sin n sin( n ) cos( n ) sin The given result follows by division by sin. (b) Using this result, sin n sin( n ) d cos(n ) d 0 sin 0 In In sin( n ) n 0 = 0 (c) (i) If n is even, (ii) I n If n is odd, I n I 0 0 = 0 I 0 = 0 =

39 GCE AS/A MATHEMATICS Specimen Assessment Materials 9 WELSH JOINT EDUCATION COMMITTEE General Certificate of Education Advanced Level/Advanced Subsidiary CYD-BWYLLGOR ADDYSG CYMRU Tystysgrif Addysg Gyffredinol Safon Uwch/Uwch Gyfrannol MATHEMATICS Mechanics Specimen Mark Scheme 005/006

40 GCE AS/ MATHEMATICS Specimen Assessment Materials 0. (a) Attempt to apply NL to lift, dimensionally correct equation 600g T = 600a a = 0., T = 600(9.8 0.) = 560 N (b) a = 0, T = = 5880 N. (a) Attempted use of s ( u v) t with u = 0, v = 5, s = = 0.5(0 + 5)t time taken during acceleration = t = s (b) (vt graph) G (st points correct) G (last part correct) (c) Use of distance = area under graph Distance = = 095 m. (a) Use of v u as with u = 0, v =., a = 9.8, s = h = h h = 0 m (b) Use of v u at with u = 0, v =., a = 9.8 = 9.8t 0 t = s 7 (c) Required speed = 0.6 = 8. ms -

41 GCE AS/ MATHEMATICS Specimen Assessment Materials. (a) Attempt at restitution equation 8 7 = - e(6 0) e = 0.5 (b) Attempt at use of conservation of momentum, dimensionally correct equation 9 8 m = (9 + m) 5 m = kg 5.

42 GCE AS/ MATHEMATICS Specimen Assessment Materials (a) NL applied to each particle, dimensionally correct equation For A, T = a For B, g T = a Substitute for T, g a = a m 9 8 a = 7 =. ms - T =. = 6.8 N (b) Vertically for B, T = g Vertically for A, R = g Horizontally for a, F = T = g Use of F R F g = 0.75 R g 6. (a) (b) Attempt to take moments about C/B to obtain dimensionally correct equation P + 5g 0.6 = 0 P = g = 9. N Attempt at second equation with P and Q P + Q = 5g (o.e.) Q = 5g g = g = 9.6 N Attempt to take moments about C, dimensionally correct equation -P + 5g 0.6 = W 0. Plank does not tilt iff P = g 0.W 0 m g Greatest W = 0. = 5g = 7 N 7. (a) Resolve perpendicular to plane R = g cos0 F = 0.6 R F = cos0 Attempt to apply NL // to plane, dimensionally correct 50 F g sin0 = a A ( each error) a =.6(8) ms -

43 GCE AS/ MATHEMATICS Specimen Assessment Materials (b) e.g. Body modelled as particle No air resistance 8. (a) attempt at use of Impulse = change of momentum - = 0.5(v 5) v = - (b) Attempt at use of Impulse = force time - = F 0. Magnitude of force = 0 N 9. (a) Area Dist of c of m Dist of c of m from AE from AB DEF 6 0 Areas ABCF Distance from AE Lamina 8 y Distances from AB Take moments 8 = 6 + = 75 8y = y = 85 (b) Required angle identified 75 tan = = 08

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45 GCE AS/A MATHEMATICS Specimen Assessment Materials 5 WELSH JOINT EDUCATION COMMITTEE General Certificate of Education Advanced Level/Advanced Subsidiary CYD-BWYLLGOR ADDYSG CYMRU Tystysgrif Addysg Gyffredinol Safon Uwch/Uwch Gyfrannol MATHEMATICS M Mechanics Specimen Mark Scheme 005/006

46 GCE AS/ MATHEMATICS Specimen Assessment Materials 6. (a) dv Use of a dt a = 6t + 0t (b) Use of v d t = t + t 5 + c Use of t = 0, = - m = t + t 5 when t =, = = 69 m. (a) P Use of F 5 NL F 600 = 0 F P = = 5000 W (b) 6000 F 000 F g sin = 8000a a a = 0.75 ms -. (a) Tension in string, T = T = N (b) Energy stored in string = 0 7 =.5 J

47 GCE AS/ MATHEMATICS Specimen Assessment Materials 7 (c) Attempt at conservation of energy (at least terms) Gain in potential energy = sin = =.76 J Gain in kinetic energy = v v =.5 v =.(088) ms -. (a) y = ut gt 0 = 5 sin 0t t = 5 seconds 9 8t, (b) AB = 5 cos 0 5 FT student's time = 50 m (c) Greatest Height = 5 sin (5) FT student's time = 7656 m (d) y = 5 sin 0 98 = 75 cao = 5 cos 0 = 8 cao Speed = ( 7 5) ( 8) 5 m/s, Angle with horizontal = tan 7 5 ( 9, 8 5. (a) a = ((i j) (i + j)) a = i j (b) Use of v = a dt v = ti tj + c When t = 0, v = i + j m c = i + j So v = (t + )i (t )j (o.e.) (c) Use of a.v = 0 (t + ) ( t) = 0 t = 0.6

48 GCE AS/ MATHEMATICS Specimen Assessment Materials 8 (d) Use of r = v dt r = t i - t j + ti + tj + c when t = 0, r = i j, c = i j r = ( t + t + )i + (-t + t )j m (e) when t =, r = ( + + )i + (- + )j r = 6i j r = 6 r = 7 6. (a) Attempt at conservation of energy mu = mv + mg(a - acos) A (- each error) v = u + ag( - cos) (b) Attempt at NL towards centre R mg cos = m acceleration mv R mg cos = a m R = mg cos - (u + ag agcos) a m R = mg cos - mg + mgcos - mu R = mg(cos - ) - a mu a (c) R = 0 when ball bearing leaves bowl (cos - ) = 0 cos = 0.98(7755) = 0.5()

49 GCE AS/ MATHEMATICS Specimen Assessment Materials 9 7. (a) Resolve vertically. T cos0 = mg T = =.6 cos0 (b) NL towards centre mu T sin0 = 0 8sin 0 L.H.S. R.H.S. Substitute for T u =.50(979) ms - m (c) Tension is constant throughout the rope.

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51 GCE AS/A MATHEMATICS Specimen Assessment Materials 5 WELSH JOINT EDUCATION COMMITTEE General Certificate of Education Advanced Level/Advanced Subsidiary CYD-BWYLLGOR ADDYSG CYMRU Tystysgrif Addysg Gyffredinol Safon Uwch/Uwch Gyfrannol MATHEMATICS M Mechanics Specimen Mark Scheme 005/006

52 GCE AS/ MATHEMATICS Specimen Assessment Materials 5. (a) NL applied to particle gv a g 9 g Magnitude of retardation = (6 v ) 6 (convincing) dv g (b) 6v dt 6 dv 6 g dt 6 v v 6 tan gt c 6 6 When t = 0, v = 6 m C = 6 At maimum height, v = 0 t 0 8s g dv g (c) v 6 v d 6 vdv 6 g d 6 v 0 ln(6 + v ) = -g + c When t = 0, v = 6, = 0 m c = 8 ln(7) At maimum height. v = 0 8 = ln() =.7 m 9 8. (a) Ma. speed = a a = 8 = 8 5 Period = 5 (F.T.) Time for 9 oscillations = 5 (or 5.s)

53 GCE AS/ MATHEMATICS Specimen Assessment Materials 5 (b) Using v ( a ) with 8 a 5,, o.e. 5 v 6 (5 ) v = 8ms (c) a = = 6 = 768ms (d) = 5sin (6t) When =, ta = 09 When = 6, tb = 050 Required time = tb ta (overall strategy) = 085 (C.A.O.). Auiliary equation m + 8m + = 0 (m + )(m + 6) = 0 m = -, -6 Complementary function = Ae -t + Be -6t For particular integral, try = at + b d a dt Substitute into differential equation 8a + (at + b) = t + 0 Compare coefficients m a = a = 8a + b = 0 b = both values General solution = Ae -t + Be -6t + t + When t = 0, = 0 0 = A + B + d When t = 0, dt t e 6 e t A B dt -A 6B + = Solving A =, B = 0 General solution is = e t +t+ When t =, = 98 m

54 GCE AS/ MATHEMATICS Specimen Assessment Materials 5. (a) Resolve horizontally S = Tsin Moments about A, dimensionally correct 5g + T = S 8 Substitute for S = 08T 5g + T = 6T T = 875 N S = 7 N Resolve vertically R = 5g + T cos R = = 575 N m (b) Using T = 000 N and adding etra term in moment equation + T + 80g cos = 6 T 70 = (6 ) 000 = 97 m Man will not reach top of ladder (c) Any reasonable assumption (e.g. man is a particle, ladder is a rigid rod)

55 GCE AS/ MATHEMATICS Specimen Assessment Materials (a) v = 7 cos =. ms - For B, -J = 5 7 sin - 5u For A, J = u -J = 8 5u = -u u = ms - Speed of A = ms - Speed of B = Speed of B =.6 58 ms - J = u = 8 Ns (b) Loss in K.E. = Initial K.E. - Final K.E. Loss in K.E. = m Loss in K.E. =. J

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57 GCE AS/A MATHEMATICS Specimen Assessment Materials 57 WELSH JOINT EDUCATION COMMITTEE General Certificate of Education Advanced Level/Advanced Subsidiary CYD-BWYLLGOR ADDYSG CYMRU Tystysgrif Addysg Gyffredinol Safon Uwch/Uwch Gyfrannol MATHEMATICS S Statistics Specimen Mark Scheme 005/006

58 GCE AS/ MATHEMATICS Specimen Assessment Materials 58. (a) Prob = 6 (or ) (b) P( reds) = = P( blues) = = P( yellow) = 0 P( same colour) = Sum = 5 8 ( or 9 ). E(Y) = + = 7 Var (X) = Var (Y) = 6 = 6 SD(Y) = 8. (a) P( A B) P( A) P( B) P( A B) = = 0.8 (or = P( A B) 08. ) (b) (i) Prob = P( A) P( B) P( A ) P( B) = = 0.5 (ii) P( A and B) = = 0.8. (a) (i) P(X ) = (or - 0.5) (ii) P(X = 6) = (or ) = (b) (i) P(Y = ) = e (ii) P(Y ) = - P(Y ) = - e. (. ) = (a) P(White) = = 0.85 (b) P(AWhite) =

59 GCE AS/ MATHEMATICS Specimen Assessment Materials (a) Mean = = 8 ; SD = (b) (i) 0 8 P ( X 8) (or using tables or ) (ii) Y is B(0,0.) P( X 7) P( Y 6) = = 0. m (or ) 7. (a) Mean = =.6 E( X ) = = 8.0 Var = =. (b) E X (c) Possibilities are (,), (,), (,). (si) Prob = = (a) E(X) = (b) (i) F() = ( 8 ) d 0 = =. = y y = ( 8 y y ) d y 0 0 (ii) P(X ) = - F() = (iii) The median satisfies F(m) = 0.5, ie m m 6 leading to the given equation. Solving, m giving m =.9

60

61 GCE AS/A MATHEMATICS Specimen Assessment Materials 6 WELSH JOINT EDUCATION COMMITTEE General Certificate of Education Advanced Level/Advanced Subsidiary CYD-BWYLLGOR ADDYSG CYMRU Tystysgrif Addysg Gyffredinol Safon Uwch/Uwch Gyfrannol MATHEMATICS S Statistics Specimen Mark Scheme 005/006

62 GCE AS/ MATHEMATICS Specimen Assessment Materials (a) (i) z = 5 P(overflow) = (ii) = = 88. (b) T is N(980,5) z = m Prob = (a) Distribution of T is Poi(.5). Prob = e (b) T is now Poi(5) N(5,5) z =. 8 5 Prob = 0.0. (a) % confidence limits are giving [7.0, 76.0]. (b) A 95% confidence interval is an interval determined by a method which would ensure that the parameter lies within the interval 95% of the time. B (Allow if not completely convinced). (a) P(R 6 ) P(R > 6) = ( 0 6 ) ( 0 ) = (b) (i) The density of R is f(r) = /6 (si) 0 E(A) = r. d r 6 0 = r 8 = 5

63 GCE AS/ MATHEMATICS Specimen Assessment Materials 6 (ii) E 0 A r. dr r = = 99. Var(A) = = The appropriate test statistic is y TS = m n = =.5 EITHER p-value = = 0. This is greater than 0.0 so accept that concentrations are equal. OR Critical value =.576 The calculated value is less than this so accept that concentrations are equal. 6. (a) (i) X is B(0,p) (si) Sig level = P(X p = 0.5) = (ii) We require P(X p = 0.7) = P(Y 6p = 0.) = (b) Under H0, X is B(00,0.5) N(00,50) z = 50 m =.76 p-value = Strong evidence to support Dafydd s theory.

64 GCE AS/ MATHEMATICS Specimen Assessment Materials 6 7. (a) (i) H : versus H : 0 (ii) In 5 days, number sold Y is Poi(5) under H0. p-value = P( Y 0 ) = 0.8 We cannot conclude that the mean has increased. (b) Under H0 the number sold in 00 days is Poi(00) N(00,00) z = 00 =.70 p-value = 0.06 Significant at the 5% level because 0.06 < 0.05.

65 GCE AS/A MATHEMATICS Specimen Assessment Materials 65 WELSH JOINT EDUCATION COMMITTEE General Certificate of Education Advanced Level/Advanced Subsidiary CYD-BWYLLGOR ADDYSG CYMRU Tystysgrif Addysg Gyffredinol Safon Uwch/Uwch Gyfrannol MATHEMATICS S Statistics Specimen Mark Scheme 005/006

66 GCE AS/ MATHEMATICS Specimen Assessment Materials 66. (a) The possible combinations are given in the following table. Combination Sum (combs)(sum) The sampling distribution is Sum 5 6 Prob /6 /6 /6 /6 (b) The possibilities are Possibility Sum A(Poss)(sum) ( each error) The distribution is Sum 5 6 Prob /6 /6 5/6 /6 /6

67 GCE AS/ MATHEMATICS Specimen Assessment Materials (a) p (b) SE = The value of p is not known and has to be estimated. (c) (d) 90% confidence limits are giving [0.60,0.6]. No we are not. The confidence interval has a level of confidence associated with it and it is not certain.. (a) H : versus H : 0 y y. 5.5 (b) Test statistic = =.9.. p-value = 0.08 = 0.66 We conclude that there is no difference in mean weight.. (a) (i) = 0. ;. 8 7., = 0.(8..) (ii) 99% confidence limits are giving [.,.]. (b) = 0. 95% confidence limits are giving [.8,.00]

68 GCE AS/A MATHEMATICS Specimen Assessment Materials (a) b = = a = 8 = 7.0 (b) Test statistic = / 8 0. =.08 p-value = 0.00 = 0.00 Very strong evidence that the value of is not 0.. (c) (i) Est length = = 8.65 (ii) SE = 0 ( ) / 8 = % confidence limits are giving [8.0,8.] 6. (a) E E( X ) = E(X) =. Var Var( X ) Var( X) =. 8 = SE( ) (b), N 0 We require P z = / m = 0.6 Prob = 0.7 = RJC/JD/W00(0) -Sep-