MS GENERAL CERTIFICATE OF EDUCATION TYSTYSGRIF ADDYSG GYFFREDINOL MARKING SCHEME. MATHEMATICS - C1-C4 & FP1-FP3 AS/Advanced

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1 MS. GENERAL CERTIFICATE OF EDUCATION TYSTYSGRIF ADDYSG GYFFREDINOL MARKING SCHEME MATHEMATICS - C-C4 & FP-FP AS/Advanced SUMMER 8

2 INTRODUCTION The marking schemes which follow were those used by WJEC for the Summer 8 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. WJEC regrets that it cannot enter into any discussion or correspondence about these marking schemes.

3 Mathematics C May 8 Solutions and Mark Scheme. (a) Gradient of AB increase in y increase in x Gradient of AB / (or equivalent) (b) A correct method for finding the equation of AB using the candidate s value for the gradient of AB. Equation of AB : y 4 / [x ( 7)] (or equivalent) (f.t. the candidate s value for the gradient of AB) Equation of AB : x + y (f.t. one error if both s are awarded) (c) A correct method for finding the length of AB AB 5 (d) A correct method for finding E E(, 5) (e) Either: An attempt to find the gradient of a line perpendicular to AB using the fact that the product of the gradients of perpendicular lines. An attempt to find the gradient of the line passing through C and D ( 5) (Equating expressions for gradient) 6 k k (f.t. candidate s gradient of AB) Or: An attempt to find the gradient of a line perpendicular to AB using the fact that the product of the gradients of perpendicular lines. An attempt to find the equation of line perpendicular to AB passing through C(or D) 5 (k 6) (substituting coordinates of unused point in the equation) k (f.t. candidate s gradient of AB)

4 . (a) Either: ( 6 ) 4 (c.a.o.) Or: ( 6 ) ( 6 ) 4 (c.a.o.) (b) 5 5 (5 5 )(4 5) (4 + 5)( 4 5) Numerator: Denominator: (c.a.o.) Special case If not gained, allow for correctly simplified numerator or denominator following multiplication of top and bottom by dy 6x 8 dx (an attempt to differentiate, at least one non-zero term correct) An attempt to substitute x in candidate s expression for dy m dx Value of dy at P 4 (c.a.o.) dx Gradient of normal m candidate s value for dy dx y-coordinate of P Equation of normal to C at P: y / 4 (x ) (or equivalent) (f.t. one slip in either candidate s value for dy or candidate s value for the dx y-coordinate at P provided and both m s awarded)

5 4. (a) y + δy 5(x + δx) + (x + δx) 4 Subtracting y from above to find δy δy xδx + 5(δx) + δx Dividing by δx and letting δx dy limit δy x + (c.a.o.) dx δx δx (b) dy 8 x + x /, dx Either 4 or 4 / 6 dy (c.a.o) dx 4 5. (a) a b (b) b on its own or least (minimum) value b, with correct explanation or no explanation x a Note: Candidates who use calculus are awarded B, B 6. (5 + x) 5 + 5x + 6x + 8x Two terms correct Three terms correct All 4 terms correct 7. (a) Use of f () + 4p + q Use of f ( ) p + + q 9 Solving simultaneous equations for p and q p 4, q 6 (convincing) Note: Candidates who assume p 4, q 6 and then verify that x is a factor and that dividing the polynomial by x + gives a remainder of 9 may be awarded M A (b) f (x) (x )(4x + ax + b) with one of a, b correct f (x) (x )(4x + 4x ) f (x) (x )(x )(x + ) (f.t. only for f (x) (x )(x + )(x ) from 4x 4x )

6 8. (a) y (, 4) (, ) O (5, ) x Concave down curve with maximum (, 4) or (5, 4) Two of the three points correct All three points correct (b) y (, 7) (, 5) O x Concave down curve with maximum (, 7) or (, ) Maximum point (, 7) Point of intersection with y-axis (, 5) 4

7 9. dy 6x + 6x + dx Putting derived dy dx x, (both correct) (f.t. candidate s dy) dx Stationary points are (, 5) and (, ) (both correct) (c.a.o) A correct method for finding nature of stationary points (, 5) is a maximum point (f.t. candidate s derived values) (, ) is a minimum point (f.t. candidate s derived values). (a) Finding critical points x 5, x A statement (mathematical or otherwise) to the effect that x 5 or x (or equivalent) (f.t. candidate s critical points) B Deduct mark for each of the following errors the use of strict inequalities the use of the word and instead of the word or (b) (i) An expression for b 4ac, with b ( )6 and at least one of a or c correct b 4ac [( )6] 4 m b 4ac < m m > (c.a.o.) (ii) x 4x + 7 x + k No points of intersection x 6x + (7 k) has no real roots (allow one slip in quadratic) m 4 > k (if candidate uses (b)(i), f.t. candidate s inequality for m) 5

8 Mathematics C May 8 Solutions and Mark Scheme ( values correct) (4 values correct) Correct formula with h I { ( )} I I 7 (f.t. one slip) Special case for candidates who put h (all values correct) Correct formula with h 5 I 5 { ( )} I I 78 (f.t. one slip) 6

9 . (a) tan θ 5 (c.a.o.) θ 56 θ 6 Note: Subtract (from final two marks) mark for each additional root in range, ignore roots outside range. (b) x 5 84, 4 6, 85 84, (one value) x 8 6 (f.t candidate s value for x) x 9, 8 6 (f.t candidate s value for x), Note: Subtract (from final two marks) mark for each additional root in range, ignore roots outside range. (c) sin θ 4( sin θ ) 8 sin θ (correct use of cos θ sin θ ) An attempt to collect terms, form and solve quadratic equation in sin θ, either by using the quadratic formula or by getting the expression into the form (a sin θ + b)(c sin θ + d), with a c coefficient of sin θ and b d constant m 5 sin θ 8 sin θ 4 (5sin θ + )(sin θ ) sin θ, (sin θ ) (c.a.o.) 5 θ 58, 6 4 Note: Subtract (from final two marks) mark for each additional root in range from 5sin θ +, ignore roots outside range. sin θ, f.t. for marks, sin θ +, f.t. for mark. (a) 5 x (x + 4) sin 5 (correct use of area formula) Either x(x + 4) 6 or expressing the equation correctly in the form ax + bx + c x 6 (negative value must be rejected) (c.a.o.) (b) BC cos5 (correct substitution of candidate s derived values in cos rule) BC 5 5 cm (f.t. candidate s derived value for x) 7

10 4. (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 + a + (n )d] + [a + a + (n )d] [a + a + (n )d] (reverse and add) S n n[a + (n )d] S n n[a + (n )d] (convincing) (b) [a + 9d] a + 9d 64 [a + ()d] + [a + 5(6)d] 66 [a + d] + [a + 5d] 66 a + 6d 66 An attempt to solve the candidate s two equations simultaneously by eliminating one unknown d 6, a 5 (both values) (c.a.o.) 5. a + ar 7 a r A valid attempt to eliminate a ( r) + ( r)r 7 (a correct quadratic in r) r 8 (r 8) (c.a.o.) r 8 and a 4 (f.t. candidate s positive value for r) 8

11 6. (a) 5 x / 4 x / ( + c), / / (b) (i) 4 x x An attempt to rewrite and solve quadratic equation in x, either by using the quadratic formula or by getting the expression into the form (x + a)(x + b), with a b 4 m (x )(x + 4) x ( 4) (c.a.o.) A(, ) (f.t. candidate s x-value, dependent on only) B(, ) (ii) Area of triangle / (f.t. candidate s coordinates for A) Area under curve (4 x ) dx (use of integration) [4x (/)x ] (correct integration) B [(8 8/) (4 /)] (substitution of candidate s limits) m 5/ Use of candidate s, x A, x B as limits and trying to find total area by adding area of triangle and area under curve m Total area / + 5/ 9/6 (c.a.o.) 9

12 7. (a) Let p log a x Then x a p (relationship between log and power) x n a pn (the laws of indicies) log a x n pn (relationship between log and power) log a x n pn n log a x (convincing) (b) log a (x + 4) log a x log a. log a x + 4 log a (use of subtraction law) x (use of power law) x + 4 (removing logs) (c.a.o) x x 8 (f.t. for 6, 9 only) (c) Either: (y + ) log 4 log 7 (taking logs on both sides and using the power law) y log 7 ± log 4 m log 4 y 99 (c.a.o.) Or: y + log 4 7 (rewriting as a log equation) y log 4 7 ± m y 99 (c.a.o.) 8. (a) (i) A(5, ) (ii) A correct method for finding radius Radius 65 (iii) Equation of C: (x 5) + (y ) ( 65) (f.t. candidate s coordinates of A.) (b) Either: An attempt to substitute the coordinates of R in the equation of C Verification that L.H.S. of equation of C 65 R lies on C Or: An attempt to find AR AR 65 R lies on C (c) Either: RQ 6 (RP 4) sin QPR 6 cos QPR QPR 8 4 Or: RQ 6 or RP 4 ( 6) ( 4) + ( 65) ( 4) ( 65) cosqpr (correct use of cos rule) QPR 8 4

13 9. Let AÔB θ radians (a) 6 θ 5 4 θ 9 Area of sector AOB 6 θ Area of sector AOB 6 cm (convincing ) (b) Area of triangle AOB 6 sin θ Area of triangle AOB 4 cm (f.t. candidate s value for θ ) Shaded area cm (f.t. candidate s value for θ )

14 A LEVEL MATHEMATICS PAPER C Summer 8 Marking Scheme. h 5 (h 5, correct formula) Integral.5 [ ( correct values) 4 ( ) ( correct values) + ( 67489)] 64 (F.T. one slip). (a) π θ, for example (choice of θ and one correct 6 evalution) 4 tan θ tan π 7. tan θ 866 ( correct evaluations) + tan θ + (Statement is false) (b) ( + tan θ) 8 tan θ (use of correct formula sec θ + tan θ) tan θ + tan θ 6 (attempt to solve quadratic) ( tan θ )(tan θ + ) tan θ, θ 56 º, 6 º, 6 6º, 96 6º (56 º, 6 º) (allow to nearest degree) (6 6º) (96 6º) 8

15 . x + sin y + x cos y dy dx dy + y (sin y + x cos y dy dx (y dy dx ) dx ) + dy dx + π dy dx (correct differentiation of x, π, ) dy 699 dx π (allow 7()) (F.T. d (sin y) cos y) dy 4 4. (a) dy e t dt (ket, k,, ) (k ) dy dt t dy dx tet (o.e.) (C.A.O) (b) d dy t t te + 4te dt dx (f (t) e t + t g (t)) (f(t), g(t) e t ) d y t te ( + t) dx (o.e.) (use of d y d dy dx ) dx dt dx dt (F. T candidates d dy dt dx ) 8

16 5. (a) dy dx x x, dy dx x x 9x ( x ) 9x 9x + (move one term to other side, and attempt to square both sides) (b) x 9x 9x + 78 Change of sign indicates α is between and (attempt to check values or signs) (correct values or signs and conclusion) x, x, x 5, x 557 (x ) x 54 (x ) Check 55, 545 x 9 x 9x + (attempt to check values 55 or signs) (correct values or signs, F.T. x ) Change of sign indicates α is 54 correct to 5 dec places 4

17 6. (a) ( correct x values) (y for st pt) (all correct, correct shape) (b) 4 x (a x b, a 4, b ) x ± 4 (both values, F.T. a, b) (c) x 9 > x > 6 or x 9 < x > x > 6 or x > (o.e.) e.g. (, ) (6, ) (answer must contain 'or') (o.e.) alternatively (x 9) > 9 4x 6x + 7 > x 9 x + 8 > (for x > 6 ) (x ) ( x 6) > for x < x > 6 or x < for union of intervals 9 5

18 7. (a) (i) cos x ( + C ) (k cos x, k ±,, ) (k ) (ii) ln x + 5 (+ C) (k ln x + 5 ) (iii) e x+ 4 ( + C) (k ) (ke x+4 ) (k ) (b) 6(x + ) (a (x + ), a ± 6,, 8 ) (a ) (C.A.O, either answer) 8 6

19 8. (a) cosec x (k cosec x) (k ) (b) x x + x ln x x + x ln x (x f(x) + lnxg (x)) (f(x), g (x) x x and simplification) (c) ( x )( x) ( x + )( x) ( x ) ( x ) f ( x) ( x + ) g( x) ( x ) (f(x) x, g(x) x) 6x ( x ) (simplified form, C.A.O.) 7 9. (a) Range is [, ), (b) Let y (x + ) (y ± (x + ) ) y + (x + ) and attempt to isolate x x ± y + A ( for + y + Take ve square root since domain is x (correct choice of sign) x x + f (x) x + (F.T expression for x) Domain is [, ), Range is (, ] (both F.T. from (a)) (or x ) (or f (x) ) 7 7

20 . (a) f( ) e 76 gf cannot be formed since domain of g is x (b) fg(x) e lnx e lnx x Domain is [, ), range is [, ) (o.e.), 7 8

21 MATHEMATICS C4. (a) Let x (x ) A x + B x C + x (Correct form) Ax (x ) + B(x ) + Cx x B( ) B x C 4 x C 4 A + C A (no need for display) (correct clearing and attempt to substitute) ( constants C.A.O.) (third constant, F.T. one slip) (b) 4 + dx x x x ln x + + ln x,. x ( + C) 7 dy dy dy. x + x + y + 4y ( x + y) dx dx dx dy (4y ) dx dy dx 5 Gradient of normal (C.A.O.) ( 5 dy candidate' s dx numerical value), Equation of normal is y ( x + ) 5 (F.T. candidate s value) 5. (a) R sin α, R cos α ( R ) R, α tan ( ).7 or4 (correct method forα ) ( α 4 ) (b) cos( x.7 ) x.7 7.9,86. (one value) x 7.6, 9. 8, 6 9

22 4. Volume π x + dx x π 4 4 x x + dx x (attempt to square, at least correct terms) (all correct) 4 x π + 4x + 9ln x A (integration of terms, F.T. similar work c Ax + B x + ) x π [ ln x] ~9. or 6.48π (C.A.O.) 7 5. (a) dy dx sin t dy y ( ) 4cost dx x ( 4 cost ) ( k sin t, k, ±, ) ( k ) 4sin t cost sin t (correct use of formula) 4cost (C.A.O.) (b) Equation of tangent is y cos p sin p( x 4sin p) ( y y m( x ) ) x xsin p + y cos p + 4sin p sin p + 4sin p (attempt to use correct formula) + sin p 9

23 x e x x 6. (a) ( x + ) e dx ( x + ) e dx ( f ( x)( x ) f ( x) x e x e 4 ( x + ) ( + C) + dx ), x ( f ( x) ke k, ( k ) (F.T. one slip), ) (b) π π 6 9 9sin θ. cosθdθ π π 6 π π 6 9cos θdθ (limits) + cos θ dθ π π 6 (for first line, unsimplified) (simplified without limits) ( cos θ a + bcos θ ) ( a b ) 9θ 9sin θ b + 4 ( k sin θ, k ±,b, b ) b ( k ) 9 π 8 or.764 (C.A.O.)

24 7. (a) kw dt (b) dw ( k > ) dw kdt W (attempt to separate variables) lnw kt + C (allow absence of C) t, W., C ln. (value of C) W ln kt. (use of logs or exponentials) kt W e. k.7 (value of k) W t.e 7 8. (a)(i) AB i + j k (AB) (ii) Equation of AB is r 4i j +k + λ (i + j k) (reasonable attempt to write equations) (must contain r, F.T. candidate s AB) (b) (Point of intersection is on both lines) Equate coeffs of i and j (any two of i,j,k) (attempt to write equations using candidate s equation) + µ 4 + λ ( correct equations, F.T. candidate s equations) λ µ (attempt to solve) (C.A.O.) i j kposition vector is + (F.T. value of λ or µ ) ijkijk(c) angle between + and + is required (coeffs of λ and µ ) ijkjik+, + (for one modulus) ijkijk+. cosθ (use of correct formula) ( )( + ) cosθ (l.h.s. unsimplified) θ 5. (C.A.O.)

25 + x 9. ( )( ) ( +L) x ( + x) + ( )( x) + ( )( ) ( x) ( + x)( + x + x +L)!, (unsimplified) 9x + 4x + + L ( + 4x ) 9x ( ) Exparsim is valid for x <. ( x + 49 < 4x) x 4x + 49 < ( x 7) < x 7 is not real contradiction (accept impossibility) 49 x + 4 for all real and positive x x 5 4

26 A/AS Level Mathematics - FP June 8 Markscheme Draft 4 (Post Stand) S n r + r n r n r n ( n + ) n( n + )(n + ) n( n + )(n + n + 4n + ) n( n + )(n + 7n + ) n( n + )( n + )(n + ) (a) deta ( ) + 4( ) +( ) Cofactor matrix 4 [Accept matrix of minors] 4 Adjugate matrix [Award AA if rule of signs not used] Inverse matrix 4 (b) x 4 8 y 8 z 5 4 [FT from their inverse] (a) ( i) 4 4i + i 4i 7 4i (7 4i)( i) i + i ( + i)( i) z 4i + i 8 7i (cao) (b) Mod(z) 58 (7.6) tan (7 / ).7 or 67 Arg(z) 4. or 47 [FT from (a) but only for the nd in arg if their z is in the nd or rd quadrant] 4

27 4 (a) x + y + z 5 5y + z 7 5y z k [Or equivalent row operations] x + y + z 5 5y + z 7 k [Award this for -7 k ] k (cao) (b) Let z α. α 7 y 5 6 8α x 5 [FT from (a)] Other possible solutions are Let x α 8 α 6 5 y ; z α 8 8 Let y α z 5α + 7; x 8α 8 5 The statement is true for n since is divisible by 6. k Let the statement be true for n k, ie is divisible by 6 (so that k N). Consider (6N 5) + 5 4N This is divisible by 6 so true for n k true for n k +, hence proved by induction. 5

28 6 (a) Let the roots be α,β,γ. Then b α + β + γ a c βγ + γα + αβ a d αβγ a [Award this if candidates go straight to what follows] Now use the fact that the roots are α, α and 4α. Then b ( + + 4)α a c ( ) α a d ( 4) α a [Award - for each error] It follows that 7α 4α b c a 8α a a d 98 bc 8 ad 4bc 49ad 4 (b) 7 α so α 8 4 The roots are,, 4 A 6

29 7 7 (a) Rotation matrix Translation matrix T [Award this for writing their matrices in the right order] (b) Fixed points satisfy y x y x giving x y + y x + The solution is, ), ( y x. (c) Under T, + + x y y x or + x y y x So y x becomes ) ( + y x 6 + y x [FT their x,y] Alternatively, they might show that the image of (x, x ) is ( x, x + ) for which award, then for eliminating x.

30 8 (a) ln f ( x) cos x ln x f ( x) cos x sin x ln x + f ( x) x cos x cos x f ( x) x sin x ln x + x (b) At a stationary point, cos x sin x ln x x tanxlnx x leading to the printed result. (c) When α.7, αlnαtanα or αlnαtanα When α.8, αlnαtanα.55.. or αlnαtanα Since these values are either side of (or ), the root lies between.7 and.8. [The actual values and an appropriate comment are required for ] 8

31 9 9 (a) i i i v u v u v u y x Equating real and imaginary parts, v u v y v u u x ; (b) Substituting v u v v u u 4 Simplifying, 4 + v u

32 A/AS Mathematics -FP June 8 Markscheme Final Draft x (a) Odd because f ( x) f ( x) ( x) + x (b) Neither because f ( x) e + which is neither f(x) or f(-x). For x <, f ( x) ax For x, f ( x) bx + 8a 4b, a 4b The solution is a, b cao (a) u x du xdx and [, ] [, ] du I 9 + u u arctan 6 π 4 (b) dx 5 9x dx 5/ 9 x sin x 5 sin [FT on minor arithmetic error, do not FT the omission of the factor /] 4 (a) Substituting, t t + + t + t 4t + t + t t t AG (b) ± t 4 (-.66,.66) Either θ θ or General solution is θ θ n π or nπ θ.7 / nπ, θ.88 + nπ [FT from their particular solution]

33 5 (a) dy dy / dp a dx dx / dp ap p Gradient of normal p Equation of normal is y ap p( x ap ) whence the printed result. (b)(i) Putting y, x a ( + p ) Coords of R are (x,y) ( ap + a{ + p }), ( + ap) ( a ( + p ), ap) (ii) Eliminating p, y x a a or y a( x a) a a 5a Coordinates of focus (a +,), ie (,) 4 4 z n 6 (a) cos( nθ) + isin( nθ) cos nθ isin nθ n z n z cos nθ + isinnθ (cos nθ isin nθ) isinnθ (b) z z z. + z. z z z z z z z (isinθ) isin θ 6isinθ sin θ sin θ sin θ 4 4 z m

34 7 (a) Let 5 x A B A( x ) + B( x ) + ( x )( x ) x x ( x )( x ) x gives A - ; x gives B - (b) f ( x) + ( x ) ( x ) This is the sum of positive quantities which can never be zero so no stationary points. [Accept a solution which obtains the derivative from the original form of f(x) and then shows that the numerator has no real zeroes] (c) y G O x [- each region omitted] (d) (i) (5/, ); (,5/) (ii) x, x, y Consider 5 x ( x )( x ) f 5 x x 4x + x x x -, ( A) (, ) (5 /, ) 8 (a) Modulus 8, Argument 9 First cube root 8, 9 / + i Second cube root 8, (9 / + ) + i Third cube root 8, (9 / + 4) i o

35 A/AS Level Mathematics FP June 8 Markscheme Final Draft (a) y G x (b) [Award G for a graph of the difference] The two graphs intersect at the two points shown, one being x and the other x >. The Newton-Raphson iteration is (cosh x sin x) x x (sinh x cos x) Starting with x.5, we obtain So root.948 (correct to 4 dps) dx coshθdθ ; [,] [,sinh ] I sinh sinh (.88) + sinhθ + sinh θ sinhθ + coshdθ cosh θ dθ sinh ( + cosh θ) dθ sinh sinh θ + θ (FT on minor integration error] sinh + sinh(sinh ).5 (cao)

36 (a) f() / f ( x) x, f () 5 / f ( x) x, f ( ) 4 4 ( x ) + ( x )... x 8 (b) 8 f (a) Consider (b) (or ) or ( 9 ) cao 6 (e (e e + e x x tanh x x x ) ) x x x x (e + e ) (e e ) x x (e + e ) 4 x x (e + e ) sec h x AG 5 5 tanh x tanh x 5 tanh x tanh x + 6 tanh x, 5 tanhx has no solution. + / 5 x ln / 5 ln Alternative solution: Substitute for sech and tanh in terms of exponential functions, 4 e x e x e x 4 or e x has no solution e x 4 gives x ln4 x ln 4

37 n 5 (a) I n (ln x) d x I n n x (ln x) x n ln x). dx. ( x n n (ln ) I n (ln I (ln ) ln I ) x (ln ) ln + (ln ) ln (b) 6 (a) dx dθ dy + ( cos θsin θ) + (sin θcos θ) dθ 4 4 9cos θsin θ + 9sin θcos θ 9sin θcos θ(cos θ + sin θ) π / sinθcosθ sin θ AG (b)(i) Length dx dy + dθ dθ dθ π / sin θ dθ π / cos θ 4 (ii) π / dx dy CSA y + dθ dθ dθ π / π sin θ.sin θcosθ dθ answer given π 5 [ sin θ] π / 5 6π 5 (.77) 5

38 7 (a)(i) dy d [( θ )sinθ ] dθ dθ ( θ )cosθ sinθ when ( - θ) tanθ leading to the printed answer. (ii) Area ( θ) dθ (b)(i) ( θ) 6 At the point of intersection, θ θ θ + θ ( θ )( θ + ) θ r / (c) Area (θ ) dθ [ θ 5 ] 5 8 [Award for evaluating this integral but then mis-using it subsequently].mathematics C C4 & FP FP MS (Summer 8)/LG August 8 6

39 WJEC 45 Western Avenue Cardiff CF5 YX Tel No Fax exams@wjec.co.uk website:

40

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42 MS. GENERAL CERTIFICATE OF EDUCATION TYSTYSGRIF ADDYSG GYFFREDINOL MARKING SCHEME MATHEMATICS - -M & S-S AS/Advanced SUMMER 8

43 INTRODUCTION The marking schemes which follow were those used by WJEC for the Summer 8 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. WJEC regrets that it cannot enter into any discussion or correspondence about these marking schemes.

44 Mathematics (June 8) Markscheme.(a) Using v u + at with v, u, a.5 +.5t t.5 s cao.(b) attempt at v-t graph Straight line joining (, 8) to (9, ) Straight line joining c's (9, ) to ((c's 9, ) Straight line joining (c's 9, ) to (c's 9+, ) + units and labels.(a) Using v u + at with v, a (-)9.8, t.5 u + (-9.8).5 u 4.5 ms -.(b) Using s ut +.5at with u 4.5, a (-)9.8, t 4 s (-9.8) m cao.(c) Using v u + as with u 4.5, a (-)9.8, s (-)7 v (-9.8) (-7) 97.5 v 44.4 ms - cao

45 . (a) NLapplied to lift and man -R + 5g 5a 5a a. ms - cao (b) NL applied to man 7g R 7a R 67 N ft a 4.(a) Apply NL to B 9g T 9a Apply Nl to A T 5g 5a Adding 4g 4a m a 4.8 ms - cao 4.(b) T 5(g + a) 5( ) 6 N cao 5.(a) Impulse change in momentum.7(- 5) -. Ns Direction of impulse is opposite to motion. oe 5.(b) Speed after impact.6. ms -

46 6.(a) Resolve perpendicular to slope R 6.5 g cos α F ft.76 N Nl down slope terms, dimensionally correct 6.5g sin α - F 6.5 a a ms - cao 6.(b) NL up slope 4 terms, dimensionally correct T F 6.5 g sin α 6.5 a Constant speed, therefore a s.i. 5 T N cao

47 7. Momentum (-) 7v A + 5v B Restitution v B v A -. (- 5) 5v B 5v A 8 Subtract v A m v A ms - cao v B.6 ms - cao Sphere A is moving in the same direction as before impact. Sphere B has reversed its direction of motion. 8. Moments about P to obtain equation 5(x ) 6 6 R Q 5x R Q 6 R Q 5x For limiting equilibrium R Q si 5x 5x x 6.6 cao 4

48 9. Resolve perpendicular to N X 9 8 cos Resolve parallel to N Y 8 sin Magnitude of resultant N ft X, Y 8 θ tan, cao.(a) Area from AB from AC triangle ABC rectangle DEFG 5.5 lamina 88 x y Moments about AB to obtain equation x 8 6 ft x 6. cao Moments about AC to obtain equation y 8 4 ft y 4.4 cao.(b) 4.4 tan θ 6. θ 4.9 ft x, y 5

49 Mathematics M (June 8) Markscheme λx.(a) Using T with T, x.55., l. l (.55.) λ.. λ N cao (b) Using EE ft λ..5 J ft λ. T 6 N si NL up plane T F 9 g sin α 8 Resistive force F N cao 6

50 .(a) Using F ma 5a 5t 6t a t t When t a 4 - Therefore magnitude of acceleration - ms -.(b) v t t dt t 6t (+ C) When t, v 5 m C 5 v t 6t + 5.(c) Least value of v when a t(t 4) t ( or) 4 ft v Therefore least value of v ms - ft v 8.(d) Required distance t 6t + 5 dt attempt to integrate v 8 4 t t + 5t 4 correct integration ( ) ( ) m 8 58 m cao 4.(a) Difference in PE 9 g g + PE ( ) 978 J Difference in KE KE J Work done against resistance 678 J 4.(b) Work done 5 R 678 R 7.55 N ft WD if M's gained in (a) 7

51 5.(a) Using s ut +.5at with u 4, a (-)9.8, s t 4.9t 7t t + attempt to solve m (7t 6)(t ) t 6/7, As particle is on the way down, t Therefore horizontal distance of wall 4 m cao 5.(b) Using v u + at with u 4, a (-)9.8, t v ft t ms - ft t Therefore speed of motion ms - ft v 5.6 θ tan 5 to the horizontal. ft v 6.(a) AB (i j + k) ( i + j + k) i j + k cao 6.(b) Work done by F F.AB (i 4j + k).(i j + k) ft m J ft dv 7.(a) a used dt cos t i sin 5t j + 9t k A 7.(b) r A (-8t -)i + (t + )j r B (-6t + )i + (9t 8)j r A r B (8t )i + (-6t + )j ra r B (- + 8t) + ( 6t) 69 8t + 64t + t + 6t 9 4t + t Minimum when t 4 m t.7 cao Minimum distance 9 4(.7) + (.7) cao 8

52 8. (a) NL towards centre T sin θ mv r 4 7 Resolve vertically T cos θ Dividing tan θ 9. m AOP θ tan cao 9. (b) T o cos( 4.6 ) 54. N ft angle (c) l o 7sin( 4.6 ).6 m ft angle 9

53 9. (a) Energy considerations - g.5 cos θ +.5 mv - g.5 cos v g cos θ -.5 g + 6 v 6 + g (cos θ -.5) g cos θ +. (b) NL towards centre mv T - mg cos θ r T g cos θ g cosθ - g m.5 g cos θ g cos θ - g 6g cos θ - g + 64 T 58.8 cos θ

54 Mathematics M (June 8) Markscheme,(a) R 5g Friction F µ R 5 g 7 5g 49 N NL dim. correct -F -.7 v 5 a dv dv v a m dt d t dv Divide by v convincing dt (b) 5 dv + dt 7 v - 5 ln (7 + v) t (+ C) Either limits or initial conditions used m t, v 8 C -5 ln 98 t 5 ln 98-5 ln (7 + v) Body come to rest when v used m t 5 ln (98/7) 5 ln(7/5) 6.8 s cao

55 .(a) (i) NL.8 x.v a dim correct d x dx a, v dt dt m Multiply by.5 d x dx 5 + x dt dt (ii) Auxiliary equation 5m + m - (5m )(m + ) m -,.4 General solution is x Ae.4t + Be -t ft m values When t, x, v 7 used A + B dx.4t t v.4ae Be dt ft x.4 A B 7 ft x, v both Solving simultaneously.4 A 7 A 5 B -5 both cao x 5e.4t - 5e -t (iii) e.4t increases with t; e -t decreases with t, therefore e.4t - e -t increases with t. Hence x 5e.4t - 5e -t increases with t..(b) For PI try x at + b a (at + b) t 7 Compare coefficients m -a a - a b -7 b 4 b both cao Therefore general solution is x Ae.4t + Be -t t + ft a, b

56 .(a) v ω (a x ) used At A 5 ω (a 9) At B 4.65 ω (a 6) Therefore 5(a 6) 4.65(a 9) m.975 a a 5 a 5 convincing ω ω.5 π Period ω 8π 5 convincing (b) Let 5 x 5sin t 4 When t, x 5 sin (.5 ).99 m cao (c) Minimum speed ωa used ms - si ft ω We require t such that cos(.5t) cos (.5t).4 t.9 s cao 4.(a) dv + v dv v + a v dx 9 dx d v v ( v 9) dx 9 v 9 dv v + 9 dx 45 ln(v + 9) -x (+ C) When t, x, v 5 m C 45 ln(75 + 9) 45 ln(5) 5 x 45ln oe v + 9 (b) Greatest height when v 5 x 45ln 9 45 ln(.5).4 m

57 5. (a) Moments about A dim correct S.5 g.5cos θ cos θ S N cao (b) Resolve horizontally S sin θ F F N ft S Resolve vertically R g + S cos θ N ft S s.i. Limiting friction F µr used µ cao 6. Using v u + as with u, a (-)9.8, s v v 5.6 ms - Impulse change in momentum applied to both particles For A J 7 v' For B J v' ft v Solving 7v' 6.8 v' m v' 6.8 v'.68 ft v J.76 Ns ft v 4

58 A/AS Mathematics - S June 8 Mark Scheme (a) P( red) or 9 (/6) 9 8 P( grn) or (/) 4 4 P( yel) or (/6) 5 P(same) (b) P(diff) P(same ) 8 (a) Using P ( A B) P( A) + P( B) P( A B).4. + P(B).P(B).8P(B). P(B).5 (b) EITHER P(A only). 5 P(B only). Reqd prob.5 OR P(Exactly one) P(A B) P(A B) (c)..75 Reqd prob.5 /7 [FT for P(B) in (b) and (c)] 5

59 (a)(i) Total number of poss 6 [Award if any evidence is seen that there are 6 possibilities] Number giving equality 6 Reqd prob 6/6 /6 (ii) Number giving smaller score Reqd prob 5/6 5/ (b) EITHER Reduce the sample space to (5,), (4,), (,), (,4), (,5) Reqd prob /5 OR Prob P(,) P(Sum 6) / 6 5 / 6 /5 4 (a) Prob or (cao) (b)(i) Prob e. 7 5! (ii).5.5 P(X ) e [Special case : Award for.7 with no working] 5 (a) P(B).5, P(C).5, P(D).5 [Award A for ¼, ¼, ¼ with no working] P(Win) (b).5.6 P(D win) [FT their initial probs in (b) but not in (a)] 6 (a) Σp 9k k /9 AG (b) 4 E(X) (/9) (c) 4 E + + X (9/54) 6

60 7 (a) The number of sales, X, is B(5,.) si 5 8 (i)p(x )..8 or or or. (ii) P( X 4) or (iii) Prob (b) The week s pay Y is given by Y + 5X si E(Y) SD [FT on their sensible Y] 8 (a) P(.5 X.75) F(.75) F(.5) (b) F(.6).475 Since F(med).5, it follows that the median is greater than.6. [FT on their F(.6)] (c) f(x) F ( x) ( x x ) (d) ( X ) E x( x x ) dx 4 5 x

61 A/AS level Mathematics - S June 8 Mark Scheme x, y 5 6 SE of difference of means The 95% confidence limits are ± m giving [.9,.687] (a) E(X) 4, Var(X).4 Using Var(X) E( X ) [ E( X )] E ( X ) E(Y) 9, Var(Y) 6. Similarly, E ( Y ) (b) E(U)E(X)E(Y) 6 E( X Y ) E( X ) E( Y ) 66. Var(U) E( X Y ) [ E( XY )] (a) (i) 49 5 z. 5 Prob.85 (ii) Number weighing less than 49 g is B(6,.85) si 6 Required prob [Award A if combinatorial term omitted] (b) X-Y is N(5-8, + 9. ), ie N(- 4, 6.96) z 4 ( ± ) Prob.66 8

62 4 (a) 48. x 8 ( 6.4).5 SE of X 8 (.767 ) [Award only if the 8 appears in the denominator] 99% conf limits are 6.4 ± m [ correct form,.576] giving [5.99,6.9] (b) We solve n n giving n 4 [FT on their n] 5 (a) H : p.6 versus H : p <. 6 (b) Under H, X is B(,.6) si and Y (No not germ) is B(,.4) (si) p-value P(X 9 H ) P(Y H ).75 (c) X is now B(,.6) which is approx N(,48).5 z 48 [A if no continuity correction] -.67 p-value.79 Strong evidence to support Malcolm (or germination prob less than.6). [No c/c gives z -.74, p.7, wrong c/c gives z -.8, p.48] 9

63 6 (a)(i) AC X + X X (si) We require P( X > 8) P( X > 8/ ) 6 8/ ( - v) (ii) Area X X We require P( < ) P(X < ) 4. 6 (v5 ) 6 4 (b) EITHER f ( x).5 for 4 x 6 Expected area 6 4.5x x.5 8.5dx 6 4 OR E ( ) X 8 Expected area E ( X ) Var( X ) ( E( X )) 7 (a) H : µ.5 versus H : µ. 5 (b)(i) Under H S is Po(5) P(S 8).74 P(S ).7 Significance level.7 (ii) If µ, S is Po() We require P(9 X ) (c) Under H, S is now Po(5) N(5,5) z 5. [Award A if the c/c is incorrect or omitted] Prob from tables.9 p-value.86 Accept H (context not required but accept if correct)). [No c/c gives z.6, prob.8, p-value.76; incorrect c/c gives z., prob.968, p-value 96]

64 AS/A Mathematics - S June 8 Markscheme. (a) P( )) P( ) P(, ) P(, ) 9 8 [Do not accept sampling with replacement] The sampling distribution is Total Prob / / / /6 [- each error] A [FT on their probabilities] (b) E( Total) Mean value +.6 as (a)(i) 6 pˆ. 55 (ii) ESE (b) 95% confidence limits are.55 ± giving [.497,.55] (c) No, because.6 is not in the confidence interval. (a) Σx 58; Σx 884 UE of µ UE of σ 46.8 (b) H µ versus H : µ : < Test stat / -.7 DF (i) At the 5% significance level, critical value.796 Accept farmer s claim with correct reasoning. (ii) At the % significance level, critical value.6 Reject farmer s claim with correct reasoning. [FT in (b) from (a) ; FT conclusions from their critical values but not from p-values]

65 4 (a) UE of µ 85/75 ( 4.) UE of Var(X) (.9) [Accept division by 75 giving.87] ESE (.87 gives.7) Approx 9% confidence limits for µ are 4. ± giving [.8, 4.6] (cao) (b) No since the Central Limit Theorem ensures the approximate normality of X. 5 (a) H : µ x µ y versus H : µ x µ y (b) x 56 / 6( 5.); y 5 / 6( 5.5) s x (/6 gives 5.9) s y (/6 gives 5.) [Accept division by 6] Test stat (±).94 (.95) Prob from tables.76 (.7) p-value.47 (.4) Insufficient evidence to doubt that the mean weights are equal. / 6 (a) P(X > ) ( + λx)dx / λ λ x + x + 8 AG (b)(i) Y is B(n,/+λ/8) λ E(Y) n + 8 λ E( U ) 8 n + n 4 λ 8 (ii) λ λ λ Var(Y) n + n Var( U ) n SE n 6 λ n λ 4 64

66 7 (a) S / 6 7 xy Sxx 9 7 / 6 75 b a (b) Estimated length ( 88.5) SE (.85) 6 75 The 99% confidence interval is 88.5 ± giving [88., 88.7] [FT from (a)]

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