GCE MARKING SCHEME SUMMER 2017 MATHEMATICS - C1 0973/01. WJEC CBAC Ltd.

Transcription

1 GCE MARKING SCHEME SUMMER 7 MATHEMATICS - C 97/

2 INTRODUCTION This marking scheme was used by WJEC for the 7 examination. It was finalised after detailed discussion at examiners' conferences by all the examiners involved in the assessment. The conference was held shortly after the paper was 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 conference was to ensure that the marking scheme was 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' conference, teachers may have different views on certain matters of detail or interpretation. WJEC regrets that it cannot enter into any discussion or correspondence about this marking scheme.

3 Mathematics C May 7 Solutions and Mark Scheme. (a) (i) Gradient of AB = increase in y increase in x Gradient of AB = / (or equivalent) (ii) Use of gradient L gradient AB = (or equivalent) A correct method for finding the equation of L using candidate s gradient for L Equation of L : y 5 = (x 4) (or equivalent) (f.t. candidate s gradient for AB provided that both the rd and 4 th marks (, ) have been awarded) (b) (i) An attempt to solve equations of L and L simultaneously x = 7, y = 4 (convincing) (ii) A correct method for finding the length of AC(BC) AC = BC = 9 Cos BCA = BC = 9 CA (f.t. candidate s derived values for AC and BC) Cos BCA = (c.a.o.) (c) (i) A correct method for finding D D(, 4) (ii) Isosceles E. (a) 55 9 = (55 9)( 5) + 5 ( + 5)( 5) Numerator: Denominator: = 7 5 (c.a.o.) + 5 Special case If not gained, allow B for correctly simplified numerator or denominator following multiplication of top and bottom by + 5 (b) () = 5 B 7 8 = 4 B 599 = 5 B () = 5 (c.a.o.) B

4 . (a) dy = x 4 (an attempt to differentiate, dx at least one non-zero term correct) An attempt to substitute x = 6 in candidate s expression for dy m dx Value of dy at P = 5 (c.a.o.) dx Equation of tangent at P: y ( 7) = 5(x 6) (or equivalent) (f.t. candidate s value for dy provided and m both awarded) dx (b) Use of gradient of tangent = (o.e.) gradient of normal An attempt to put candidate s expression for dy = dx (f.t candidate s derived value for gradient of tangent) m x-coordinate of Q = (c.a.o.) 4. (a) a = B b = 5 B c = 85 B (b) Stationary value = 85 (f.t. candidate s value for c) B This is a maximum B 5. (a) x + 4 = x 4 + 4x + 6x + 4x + 4 x x x x x (4 or 5 terms correct) B (If B not awarded, award B for correct terms) x + 4 = x 4 + 8x (5 terms correct) B x x x 4 (If B not awarded, award B for or 4 correct terms) ( for further incorrect simplification) (b) Coefficient of x = 6 C a 5 (x) B Coefficient of x = 6 C a 4 (x ) B 5 a 4 m = 6 a 5 (m = 4 or ) a = 5 (c.a.o.)

5 6. Finding critical values x = /, x = 4 B A statement (mathematical or otherwise) to the effect that x 4 or / x (or equivalent, f.t. candidate s derived critical values) 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 7. (a) Use of f () = 8k = k = 8 (convincing) Special case Candidates who assume k = 8 and then either show that f () = or that x is a factor by long division are awarded B (b) f (x) = (x )(8x + ax + b) with one of a, b correct f (x) = (x )(8x + 8x 5) f (x) = (x )(4x )(x + 5) (f.t. only 8x 8x 5 in above line) Special case Candidates who find one of the remaining factors, (4x ) or (x + 5), using e.g. factor theorem, are awarded B (c) Attempting to find f ( /) Remainder = If a candidate tries to solve (c) by using the answer to part (b), f.t. for and when candidate s expression is of the form (x ) two linear factors

6 8. (a) y (, ) ( 4, ) O (8, ) x Concave down curve with maximum at (, a) Maximum at (, ) Both points of intersection with x-axis B B B (b) The stationary point will always be a minimum E The y-coordinate of the stationary point will always be 6 E 9. (a) y + y = 5(x + x) 7(x + x) + B Subtracting y from above to find y y = xx 5(x) 7x Dividing by x and letting x dy = limit y = x 7 (c.a.o.) dx x x (b) dy = 6 x /4 + 5 x 4 (completely correct answer) B dx 4 (If B not awarded, award B for at least one correct non-zero term) 4

7 . (a) (i) dy = x 8x + 5 B dx Putting candidate s derived dy = dx x =, 5 (both correct) (f.t. candidate s dy) dx Stationary points are (, 7) and (5, 5) (both correct) (c.a.o) (ii) A correct method for finding nature of stationary points yielding either (, 7) is a maximum point or (5, 5) is a minimum point (f.t. candidate s derived values) Correct conclusion for other point (f.t. candidate s derived values) (b) y (, 7) O x (5, 5) Graph in shape of a positive cubic with two turning points Correct marking of both stationary points (f.t. candidate s derived maximum and minimum points) (c) Use of both k = 5, k = 7 to find the range of values for k (f.t. candidate s y-values at stationary points) k < 5 or 7 < k (f.t. candidate s y-values at stationary points) GCE Maths - C MS Summer 7/ED 5

8 GCE MARKING SCHEME SUMMER 7 MATHEMATICS - C 974/

9 INTRODUCTION This marking scheme was used by WJEC for the 7 examination. It was finalised after detailed discussion at examiners' conferences by all the examiners involved in the assessment. The conference was held shortly after the paper was 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 conference was to ensure that the marking scheme was 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' conference, teachers may have different views on certain matters of detail or interpretation. WJEC regrets that it cannot enter into any discussion or correspondence about this marking scheme.

10 Mathematics C May 7 Solutions and Mark Scheme (5 values correct) B (If B not awarded, award B for either or 4 values correct) Correct formula with h = 5 I 5 { ( )} I I I 4 78 (f.t. one slip) Special case for candidates who put h = (all values correct) (B) Correct formula with h = 4 () I 4 { ( )} I I I 4 76 (f.t. one slip) () Note: Answer only with no working shown earns marks

11 . (a) sin + 6( sin ) + sin =, (correct use of cos = sin ) An attempt to collect terms, form and solve quadratic equation in sin, either by using the quadratic formula or by getting the expression into the form (a sin + b)(c sin + d), with a c = candidate s coefficient of sin and b d = candidate s constant m 5 sin sin 6 = (5 sin + )(sin ) = sin =, (sin = ) (c.a.o.) 5 = 58, 6 4 B B Note: Subtract (from final two marks) mark for each additional root in range from 5 sin + =, ignore roots outside range. sin =, f.t. for marks, sin = +, f.t. for mark (b) A = B B C = B + B + C = 8 (f.t. candidate s value for A) B = 46, C = 4 (f.t. one error). (a) (x + ) = x + (x + 5) x (x + 5) cos 6 (o.e.) (correct use of cos rule) x x 4 = (convincing) An attempt to solve the given quadratic equation in x, either by using the quadratic formula or by getting the expression into the form (ax + b)(cx + d), with a c = and b d = 4 (x + 8)(x ) = x = (b) sin ACB = sin 6 7 (substituting the correct values in the correct places in the sin rule) ACB = 8 (Allow ft for x> obtained in (a) for )

12 4. (a) S n = a + [a + d] [a + (n )d] (at least terms, one at each end) B S n = [a + (n )d] + [a + (n )d] a In order to make further progress, the two expressions for S n must contain at least three pairs of terms, including the first pair, the last pair and one other pair of terms Either: S n = [a + a + (n )d] + [a + a + (n )d] [a + a + (n )d] Or: S n = [a + a + (n )d] n times S n = n[a + (n )d] S n = n[a + (n )d] (convincing) (b) 8 (a + 7d) = 56 B a + 7d = 9 6 (a + 5d) = 76 B a + 5d = 95 An attempt to solve the candidate s two derived linear equations simultaneously by eliminating one unknown d = 7, a = 5 (c.a.o.) (c) d = 9 B A correct method for finding (p + 8) th term (p + 8) th term = 9 (c.a.o.) 5. (a) a =, r = Value of donation in th year = Value of donation in th year = 74 (b) ( n ) = 5474 n = ( ) m n = 948 n = log 948 m log n = 9 cao

13 6. (a) x 4 6 x 7/4 + c B, B 4 7/4 ( if no constant term present) (b) (i) 6 a = 4 B (ii) dy = x dx Gradient of tangent = 8 (f.t. candidate s value for a) b = (convincing) (iii) Use of integration to find the area under the curve (6 x ) dx = 6x (/)x (correct integration) Correct method of substitution of candidate s limits m [6x (/)x ] = [ 64 ( 64/)] = 8/ 4 Area of the triangle = 64 (f.t. candidate s value for a) B Use of candidate s value for a and as limits and trying to find total area by subtracting area under curve from area of triangle m Shaded area = 64 8/ = 64/ (c.a.o.) 7. (a) Let p = log a x, q = log a y Then x = a p, y = a q (the relationship between log and power) B x = a p = a p q (the laws of indicies) B y a q log a x/y = p q (the relationship between log and power) log a x/y = p q = log a x log a y (convincing) B (b) log b x 5 = log b x 5, 4 log b /x = log b 4 /x 4 (one correct use of power law) B log b x 5 log b 7x + 4 log b /x = log b x 5 4 (addition law) B 7x x 4 (subtraction law) B log b x 5 log b 7x + 4 log b /x = log b (c.a.o.) B (c) log d 5 = 5 = d / (rewriting log equation as power equation) d = 5 4

14 8. (a) (i) A( 5, 4) B A correct method for finding radius Radius = (ii) Either: A correct method for finding AP AP = 5 (> ) P is outside C (f.t. candidate s coordinates for A) Or: An attempt to substitute x =, y = in the equation of C () ( ) + + ( ) 8 + = 5 (> ) P is outside C () (b) An attempt to substitute (x + 4) for y in the equation of the circle 5x + x + 5 = Either: Use of b 4ac m Discriminant =, y = x + 4 is a tangent to the circle x =, y = Or: An attempt to factorise candidate s quadratic (m) Repeated (single) root, y = x + 4 is a tangent to the circle () x =, y = () 9. (a) (i) L = R + r B (ii) K = R r B (b) K = (R + r)(r r) L = (R + r), R r = x (both expressions) m K = Lx Alternative solution K = θ(r r ) K = θ((r+x) r ) () K = θ(rx + x ) K = xθ(r + x) (m) K = xθ(r + r) K = Lx () 5

15 . (a) t = 67 B t = 7 (f.t. candidate s value for t ) B (b) is of the form k (not k + ) (o.e.) OR The number does not end in a or a 7 E GCE Maths - C MS Summer 7/ED 6

16 GCE MARKING SCHEME SUMMER 7 MATHEMATICS - C 975/

17 INTRODUCTION This marking scheme was used by WJEC for the 7 examination. It was finalised after detailed discussion at examiners' conferences by all the examiners involved in the assessment. The conference was held shortly after the paper was 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 conference was to ensure that the marking scheme was 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' conference, teachers may have different views on certain matters of detail or interpretation. WJEC regrets that it cannot enter into any discussion or correspondence about this marking scheme.

18 Mathematics C June 7 Solutions and Mark Scheme. (a) (5 values correct) B (If B not awarded, award B for either or 4 values correct) Correct formula with h = 5 I 5 { ( ) + ( 6979)} I I I 7 (f.t. one slip) Note: Answer only with no working earns marks (b) ln dx = ln dx ln ( + x ) dx ( + x ) ln ( + x ) dx 6 (f.t. candidate s answer to (a)) B 5 7 ln dx 6 = 4 ( + x ) (f.t. candidate s answer to (a)) 5

19 . (a) 6(sec ) 6 = 4 sec + 5 sec. (correct use of tan = sec ) An attempt to collect terms, form and solve quadratic equation in sec, either by using the quadratic formula or by getting the expression into the form (a sec + b)(c sec + d), with a c = candidate s coefficient of sec and b d = candidate s constant m sec 5 sec = ( sec + )(sec 4) = sec =, sec = 4 cos =, cos = (c.a.o.) 4 = 8, 8 9 B B = 75 5, B Note: Subtract mark for each additional root in range for each branch, ignore roots outside range. cos = +,, f.t. for marks, cos =,, f.t. for marks cos = +, +, f.t. for mark (b) Correct use of sec = and tan = sin (o.e.) cos cos sin = 5 =, 6 87 (f.t. for sin = a)

20 . (a) d(y ) = 6y dy B dx dx d( x y) = x dy 6xy B dx dx d(x 4 ) = 4x, d( 4x) = 4, d(7) = B dx dx dx dy = 4 4x + 6xy (o.e.) (c.a.o.) B dx 6y x (b) (i) candidate s x-derivative = 7 + 4t B candidate s y-derivative = (7 + 4t )r (4 + t )m, (7 + 4t ) where r, m are integers candidate s y-derivative = (7 + 4t ) (4 + t )4 (7 + 4t ) dy = candidate s y-derivative dx candidate s x-derivative dy = 5 (c.a.o.) dx (7 + 4t ) (ii) d dy = 5 4 (o.e.) dtdx (7 + 4t ) 4 (f.t. candidate s expression of correct given form for dy) B dx Use of d y = d dy dx dx dtdx dt (f.t. candidate s expression for d dy) dtdx d y = 6 (c.a.o.) dx (7 + 4t ) 5

21 4. (a) (i) V(x) = 5 x (x + 4) (x ) = 5 x + x 8x 5 = (convincing) (ii) Let f (x) = x + x 8x 5 Use of a correct method to find f (x) when x = 5 and x = 6 f (5) = 5 (< ), f (6) = 9 (> ) Change of sign 5 < x < 6 (b) x = 6 x = (x correct, at least places after the point) B x = x = x 4 = = 5 7 (x 4 correct to decimal places) B An attempt to check values or signs of f (x) at x = 5 65 and x = 5 75 f (5 65) = 78 (< ), f (5 75) = 75 (> ) Change of sign x = 5 7 correct to two decimal places 5. (a) (i) dy = (x + 5x) / f (x) (f (x) ) dx dy = (x + 5x) / (6x + 5) dx (ii) dy = or or dx ( (x) ) ( (x) ) ( x ) dy = dx ( 9x ) (b) x = cot y dx = cosec y B dy dx = ( + cot y) B dy dx = ( + x ) B dy dy = (c.a.o.) B dx + x 4

22 6. (a) (i) 8e 5x dx = k 8 e 5x + c (k =, 5, / 5, / 5 ) 8e 5x dx = 8 e 5x + c 5 (ii) 6(4x 7) / = 6 k (4x 7) / + c (k =, 4, / 4 ) / 6(4x 7) / = 6 /4 (4x 7) / + c / 6(4x 7) / = 9 (4x 7) / + c 4 (iii) cos7x 9 dx = k sin 7x 9+ c (k =, 7 /, / 7,, / 7 / 7 ) cos7x 9dx = sin 7x 9+ c 7 Note: The omission of the constant of integration is only penalised once. (b) (i) dy = a + bx (including a =, b = ) dx x 8 dy = 6x dx x 8 (ii) 6 6 x dx = r [ln (x 8)] x 8 where r is a constant 6 6 x dx = [ln (x 8)] x 8 6 x dx = r {ln (8 8)) ln( 8)} m x 8 6 x dx = ln (5) (c.a.o.) x 8 5

23 7. (a) Choice of negative x Correct verification that L.H.S. of inequality > 5 and a statement to the effect that this is in fact the case (b) a = B b = 6 B 8. (a) y = B (5x 4) An attempt to isolate 5x 4 by crossmultiplying and squaring x = (c.a.o.) 5 (y ) f - (x) = (x ) (f.t. one slip in candidate s expression for x) (b) D(f - ) = (, 5] B B 9. (a) R(f) = [8 + k, ) B (b) 8 + k k ( least value of k is ) (f.t. candidate s R(f) provided it is of form [a, ) ) (c) (i) gf(x) = (4x + k) 9 B (ii) (4 + k) 9 = 7 (substituting for x in candidate s expression for gf(x) and putting equal to 7) Either k + 6k + 48 = or (8 + k) = 6 (c.a.o.) k = 4, (f.t. candidate s quadratic in k) k = 4 (c.a.o.) GCE Maths C MS Summer 7/ED 6

24 GCE MARKING SCHEME SUMMER 7 MATHEMATICS - C4 976/

25 INTRODUCTION This marking scheme was used by WJEC for the 7 examination. It was finalised after detailed discussion at examiners' conferences by all the examiners involved in the assessment. The conference was held shortly after the paper was 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 conference was to ensure that the marking scheme was 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' conference, teachers may have different views on certain matters of detail or interpretation. WJEC regrets that it cannot enter into any discussion or correspondence about this marking scheme.

26 Mathematics C4 June 7 Solutions and Mark Scheme. (a) f (x) A + B + C (correct form) (x ) (x ) (x + 4) 8x + 7x 5 A(x + 4) + B(x )(x + 4) + C(x ) (correct clearing of fractions and genuine attempt to find coefficients) m A =, C =, B = 5 (all three coefficients correct) A (If A not awarded, award for either or correct coefficients) (b) 9x + 5x 4 = 8x + 7x 5 + x x + (x ) (x + 4) (x ) (x + 4) (x ) (x + 4) x x + = (x ) (x + 4) x + 4 9x + 5x 4 = (x ) (x + 4) (x ) (x ) (x + 4) (f.t. candidate s values for A, B, C). (a) 6y 5 dy x 9x dy 8xy = 6y 5 dy x B dx dx dx 9x dy 8xy B dx dy = 6xy + 4x dx y 5 x (convincing i.e intermediary line required) B (b) y = x = or x = B At (, ), dy = 8 B dx At (, ), dy = 8 B dx

27 . (a) 5 cos + 7 sin cos = sin (correct use of sin = sin cos ) An attempt to form a quadratic equation in tan by dividing throughout by cos and then using tan = sin m cos tan 4 tan 5 = (c.a.o.) tan =, tan = 5 (c.a.o.) = 6 57 B = B Note: F.t. candidate s derived quadratic equation in tan. Do not award the corresponding B if the candidate gives more than one root in that particular branch. Ignore roots outside range. (b) (i) R = 4 B Correctly expanding cos ( ) and using either 4 cos = 5 or 4 sin = or tan = to find 5 (f.t. candidate s value for R) = 56 (c.a.o) (ii) Least value of =. 5 cos + sin k + 6 (k = or ) (f.t. candidate s value for R) Least value = (f.t. candidate s value for R) Corresponding value for = 56 (o.e.) (f.t. candidate s value for )

28 4. / Volume = (cos x + sec x) dx /6 Correct use of cos x = ( + cos x) Integrand = ( + cos x) + + sec x (c.a.o.) a cos x dx = a sin x (a ) B b dx = bx and sec x dx = tan x (b ) B Correct substitution of correct limits in candidate s integrated expression of the form px +qsin x + tan x ( p,q ) Volume = ( ) = 7 74 (c.a.o.) Note: Answer only with no working earns marks B 5. (a) ( + 4x) / = x + 6x +... ( x) B (6x ) B x < / 4 or / 4 < x < / 4 B (b) + 4y + 8y = + 4(y + y ) ( + 4y + 8y ) / = (y + y ) + 6(y + y ) +... (f.t. candidate s expression from part (a)) m ( + 4y + 8y ) / = y + y +... (f.t. candidate s expression from part (a))

29 6. (a) candidate s x-derivative = at candidate s y-derivative = bt (at least one term correct) B dy = candidate s y-derivative dx candidate s x-derivative dy = bt (o.e.) (c.a.o.) dx a Equation of tangent at P: y bp = bp (x ap ) a (f.t. candidate s expression for dy) m dx ay = bpx abp (convincing) (b) Substituting 4a for x and 8b for y in equation of tangent 6ab = abp abp p p + 6 = (convincing) (p )(p + p 8) = (p )(p )(p + 4) = p = corresponds to (4a, 8b) p = 4 (c.a.o.) 7. (a) u = ln x du = dx B x dv = x 4 dx v = x (o.e.) B ln x dx = x ln x x dx (o.e.) x 4 x ln x dx = x ln x x + c (c.a.o.) x 4 9 (b) x (x + ) 4 dx = f (u) u 4 du (f (u) = pu + q, p, q ) x (x + ) 4 dx = (u ) u 4 du (pu 5 + qu 4 ) du = pu 6 + qu 5 B 6 5 Either: Correctly inserting limits of, in candidate s pu 6 + qu 5 or: Correctly inserting limits of, in candidate s p(x + ) 6 + q(x + ) x (x + ) 4 dx = 4 = 5 (c.a.o.) m 4

30 8. (a) dn = kn B dt (b) dn = k dt N N / = kt + c / Substituting 56 for N and 5 for t and 4 for N and 7 for t in candidate s derived equation m = 5k + c, 4 = 7k + c (both equations) (c.a.o.) Attempting to solve candidate s derived simultaneous linear equations in k and c (k = 4, c = ) m N = (t + 6) (o.e.) (c.a.o.) 9. (a) AD = AO + OD = a +b B Use of a + AD (o.e.) to find vector equation of AD Vector equation of AD: r = a + ( a +b) r = ( )a + b (convincing) (b) BC = BO + OC = 5a b B Vector equation of BC: r = b + (5a b) r = 5a +( )b (o.e.) B (c) = 5 = (comparing candidate s coefficients of a and b and an attempt to solve) = 4 or = (f.t. candidate s derived vector equation of BC) 9 9 OE = 5a + 8b (f.t. candidate s derived vector equation of BC) 9 9. a = 7b (7k) = 7b b = 7k B 7 is a factor of b and hence 7 is a factor of b B a and b have a common factor, which is a contradiction to the original assumption B GCE Maths - C4 MS Summer 7/ED 5

31 GCE MARKING SCHEME SUMMER 7 MATHEMATICS - FP 977-

32 INTRODUCTION This marking scheme was used by WJEC for the 7 examination. It was finalised after detailed discussion at examiners' conferences by all the examiners involved in the assessment. The conference was held shortly after the paper was 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 conference was to ensure that the marking scheme was 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' conference, teachers may have different views on certain matters of detail or interpretation. WJEC regrets that it cannot enter into any discussion or correspondence about this marking scheme.

33 FP June 7 Markscheme Ques Solution Mark Notes (a) det(m) = ( 4) +(8 9) = 8 7 (b)(i) adj(m) = Award if at least 5 correct 7 5 elements 8 7 (ii) M = 7 5 B FT if at least one awarded (c) x 8 7 FT inverse in (b)(ii) y 7 5 z 7 = n S n ( r ) S n r 9 n r r n r r 4 n r 9n( n )(n ) n( n ) = 4n 6 n9( n )(n ) 6( n ) 4 = 6 n8n 7n 9 6n 6 4 = 6 = n n n EITHER i 5 ; i arg( i).7;arg( i).8; arg( i).49 ; i B B For both moduli and arguments, B for correct values Accept 6.4 o, 6.56 o,7.56 o 5 z 5 cao arg( z ) cao Accept 5 o

34 Ques Solution Mark Notes OR i) ( 5i) 5 ( i) ( i) i)( ( = i) i)( ( i) 5i)( 5 ( = ) ( i = +i 5 ) arg( 5; z z or.68 rad () () () () () (BB) FT from line above provided both M marks awarded and arg is not in the st quadrant 4(a) (b) Reflection matrix = Translation matrix = Rotation matrix = T = = or = Fixed points satisfy y x y x x = y y = x These equations have no solution because, for example, x = y = y + therefore no fixed points or algebra leading to = or equivalent B B B Convincing, answer given both equations Convincing FT from line above provided it leads to no fixed point

35 Ques Solution Mark Notes 5(a) (b) Using row operations, x + y z = 7y 4z = 4y 8z = λ It follows that λ = λ = 5 Let z = α FT from (a) 4 y 7 5 x 7 6 Putting n = states that 8 is divisible by 8 which is correct so true for n =. Let the result be true for n = k, ie k 9 is divisible by 8or 9 k = 8N + B Consider (for n = k + ) k 9 k 9 9 7(a) (b) 9(8N ) 7N 8 Both terms are divisible by 8 Hence true for n = k true for n = k + and since true for n =, the result is proved by induction. Taking logs, lnf(x) = tanxlntanx Differentiating, f ( x) tan xsec x sec xln tan x f ( x) tan x tan x f ( x) (tan x) sec x( ln tan x) Stationary points satisfy + lntanx = tan x e x =.5 Only award if all previous marks awarded for LHS, for RHS

36 Ques Solution Mark Notes 8(a) u iv x iy u iv u iv u iv = u v u v x ; y u v u v (b)(i) (ii) (c) Putting x + y = gives u v u v u v u v This is the equation of a circle Completing the square, u v The centre is, The radius is Putting w = z, z giving z The two possible positions are (,) and (,) m FT from (a) Allow working in terms of x,y,u,v 4

37 Ques Solution Mark Notes 9(a)(i) B (ii) (b) 4 ( ) ( ) = ( 4) ( ) ( 4) = 7 = 6 There are two complex roots and one real root Let the roots be a,b,c. a b c = ( ) ( ) = ( ) = ( 4) = 7 bc + ca + ab = 6 abc = 4 The required equation is 7 x x x (or equivalent) 6 4 B B B Allow a less specific correct comment, eg not all the roots are real Can be implied by final answer FT their previous values Award for correct numbers irrespective of signs GCE Maths (FP) MS Summer 7 5

38 GCE MARKING SCHEME SUMMER 7 MATHEMATICS - FP 978-

39 INTRODUCTION This marking scheme was used by WJEC for the 7 examination. It was finalised after detailed discussion at examiners' conferences by all the examiners involved in the assessment. The conference was held shortly after the paper was 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 conference was to ensure that the marking scheme was 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' conference, teachers may have different views on certain matters of detail or interpretation. WJEC regrets that it cannot enter into any discussion or correspondence about this marking scheme.

40 FP June 7 - Mark Scheme Ques Solution Mark Notes Consider f ( x) sec( x) ( x) tan( x) M if particular value used sec x x tan x ( f ( x)) This line must be seen Therefore f is even. x 5 dx x 4 x 8 = dx x 4 x = x = 4 8 tan x dx 4 B Award the B for a correct k integration of x 4 8i 8(cos 7 isin7 ) Root = (cos9 isin9 ) = i Root = (cos isin ) = i Root = (cos isin ) = i BB B modulus, B argument for mod, for arg/ Special case B for spotting i 4(a) (b) Using demoivre s Theorem, n n z z cos n isinn cos( n ) isin( n ) cos n isinn cos( n) isin( n ) = cosnθ n n z z cos n isinn cos( n) isin( n) = isinnθ 5 5 ( z z ) z 5z +z +z + 5z + z 5 oe 5 5 = ( z z ) 5( z z ) ( z z ) cos5 cos cos 5 cos cos5 cos cos 5 5 cos 5 cos5 cos cos 6 6 8

41 Ques Solution Mark Notes (c) / / cos d cos5 cos cos d FT from (b) 5 = sin = = Rewrite the equation in the form sinθsinθ = sinθ sinθ(sinθ ) = 5 sin sin 8 / No A marks if no working Award FT mark only if answer less that Accept answers in degrees Either sinθ = nπ θ = nπ giving Or sin n π giving n π 4 6 Accept equivalent forms 6(a) Let (b)(i) (ii) 4x x 9 A B C ( x )(x )(x ) x x x A(x )(x ) B( x )(x ) C( x )(x ) ( x )(x )(x ) x = gives A = x = ½ gives B = x = / gives C = 6 f ( x) dx dx x ln( x ) ln( x ) ln( x ) ( ln ln 5 ln 7) dx x 6 dx x ln 75 cao The integral cannot be evaluated because the interval of integration contains points at which the integrand is not defined. A B FT their A,B,C if possible Their answer should be ln( A 5 B/ 7 C/ ) but only FT if this gives lnn Award for correct integrals

42 7(a) ( x a) y x a ( x a) y ( x a x ax a y x ax a y 4ax ) Convincing (b) EITHER dx dy at, a dt dt dy a dx at t OR dy y 4a dx dy a dx y t () () (c) Gradient of normal = t The equation of the normal is y at t( x at ) EITHER The normal intersects the parabola again where as at t( as at ) at( s t)( s t) Cancelling a(s t) both sides because s t, = t( s t) s t t OR The normal intersects the parabola again where as = ats + at +at ts + s t t = Solving, s = t, t t (Rejecting t), s = t t 4 8t t 4t 4 () () () () () y = tx + at + at

43 8(a)(i) (ii) (b) x y x f ( x) ( x ) B B B Stationary points occur where f ( x) ( x ) Giving (,4) and (,) cao (c)(i) (ii) f ( x) ( x ) f ( ) therefore (,4) is a minimum f ( ) therefore (,) is a maximum B B B k B FT for deriv = ( x ) (d) G G G G each branch, G asymptotes correctly positioned cao (e) Consider x 5 x x x x =.68,.68 f (S) = [.68,.68] Accept 5, 5 GCE Maths (FP) MS Summer 7 4

44 GCE MARKING SCHEME SUMMER 7 MATHEMATICS - FP 979-

45 INTRODUCTION This marking scheme was used by WJEC for the 7 examination. It was finalised after detailed discussion at examiners' conferences by all the examiners involved in the assessment. The conference was held shortly after the paper was 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 conference was to ensure that the marking scheme was 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' conference, teachers may have different views on certain matters of detail or interpretation. WJEC regrets that it cannot enter into any discussion or correspondence about this marking scheme.

46 FP June 7 - Mark Scheme Ques Solution Mark Notes EITHER Rewrite the equation in the form e e e e e 4 e e 4e 4 6 e 6 = ,(.5...) θ =.47 OR Let sinhθ + coshθ = rsinh(θ + ) = rsinhθcosh + rcoshθsinh Equating coefficients, rcosh = ; rsinh = Solving, - r ; tanh (.5) (.549..) Consider sinh(θ + ) = θ + = sinh (/ ) (= ) θ = =.47 () () () () () () () Putting t tan x [,/] becomes [,] dt dx t I dt /( t ) t /( t ) ( t dt = 4 t t dt = 4 4 ( t ) = ln t t = ln )/( t ) B B m M no working Accept = dt t t ln( t) ln( t) = = ln

47 Ques Solution Mark Notes dy y x, x B dx CSA = π y dy dx 4 = π x 9x dx dx 4 Put u 9x du 6x dx,[,] [,] CSA = = π u d 6 / u / π u 54 = π / 7 =.56

48 Ques Solution Mark Notes 4(a) EITHER f ( x) cos ln( x) f ( x) sin ln( x) x B ( x) f ( x) sin ln( x) B ( x) f ( x) f ( x) cos ln( x) x ( x ) f ( x) ( x) f ( x) f ( x) Convincing OR f ( x) cos ln( x) f ( x) sin ln( x) x (B) f ( x) cos ln( x) sin ln( x) (B) ( x) ( x) (b) ( x) f ( x) ( x) f ( x) f ( x) cos ln( x) sin ln( x) sin ln( x) cos ln( x) Using the above results, f ( ), f (), f () Differentiating again, ( x) f ( x) ( x) f ( x) f ( x) ( x) f ( x) f ( x) Therefore f ( ) The Maclaurin series is x x.. giving 6 x x... () () B Convincing Award B for two correct values convincing (c) Differentiating, sin ln( x ) x x... x sin ln( x ) ( x)( x x...)) = x x x... = x x...

49 Ques Solution Mark Notes 5(a) tan(.9)tanh(.9) = B tan(.)tanh(.) = B The change of sign indicates a root between.9 and. B (b)(i) (ii) (c)(i) (ii) d tan sech d tanh tanh tanh tanh = tanh EITHER For θ >, tanhθ lies between and. Therefore tanh tanh so that the modulus of the above derivative is less than therefore convergent. OR For θ =, tanh.66 tanh This is less than therefore convergent. Successive iterations give etc The value of α is.98 correct to decimal places. B B (B) (B) Do not award the second if the required result is not derived 4

50 Ques Solution Mark Notes 6(a) / 4 n In tan x tan xdx I n / 4 tan n x(sec n / 4 tan x = In n n = I n x ) dx convincing (b) / 4 tan x π I 4 dx / 4 I I = 9I 9dx / 4 6I π 4 π 6tan xdx I 4 / 4 tan I 4 I 4 Substituting above, / 4 π π π ( tan x ) d 9 6( ) = π 4 xdx B B B 5

51 Ques Solution Mark Notes 7(a) For C consider x r cos sin cos sin π It follows that x is maximised at P when. 4 For C consider y r sin sin cos (b)(i) (ii) sin It follows that y is maximised at Q when Therefore O, P and Q lie on the line The graphs intersect where sin cos tan π, r 6 Area of region = / 6 π sin or cos 6 6 / 6 sin d / π 4 π. oe 4 / / 6 cos d = ( cos )d ( cos )d oe 4 4 = 4 =. sin / / 6 sin / / 6 Convincing the integrals, for addition Limits si Award for one correct integration, for fully correct line GCE Maths (FP) MS Summer 7 6

52 GCE MARKING SCHEME SUMMER 7 MATHEMATICS - 98-

53 INTRODUCTION This marking scheme was used by WJEC for the 7 examination. It was finalised after detailed discussion at examiners' conferences by all the examiners involved in the assessment. The conference was held shortly after the paper was 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 conference was to ensure that the marking scheme was 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' conference, teachers may have different views on certain matters of detail or interpretation. WJEC regrets that it cannot enter into any discussion or correspondence about this marking scheme.

54 MATHEMATICS (June 7) Markscheme Q Solution Mark Notes (a) 5 a g NL applied to lift,upwards +ve dimensionally correct 5, g opposing No extra forces. 5 g = a = a a =.7 (b) R. 5g NL applied to crate, down +ve dimensionally correct R and 5g opposing. No extra forces. 5g R = 5a R = 5(9.8.) R = 48 (N)

55 Q Solution Mark Notes (a) Impulse on Q = (7.5 (-)) I = (Ns) magnitude required. (b) Conservation of momentum equation required. Allow sign error (-) = 6v v =.5 (ms - ) cao speed required (c) Restitution equation allow one sign error Ft v = -e(- 5) Ft v e =.75 cao (d) speed after rebound = = 4.5 (ms - ) cao allow -4.5

56 Q Solution Mark Notes. R C R D A C.5m.7m D.5m B g 4g (a) Moments about D dimen correct equation All forces, no extra 4g. + g.7 = R C.4 B any correct moment correct equation R C = 6(N) cao Resolve vertically dimen correct equation All forces, no extra R C + R D = 4g + g R D = 46(N) cao Alternative method Two simultaneous equations award B cao cao (b) R C R D A C.5m.7m.7m x D.5m B 4g g Moments about C dimen correct equation All forces, no extra oe 4g(x.9) + R D.4 = g.7 Equilibrium on point of collapse when R D =. or if moments about point not C R C =6g, (and R D = implied). 4g(x.9) = g.7 x =.5(m) cao

57 Q Solution Mark Notes 4(a) using v=u+at, u=, v=5, t=5 5 = + 5a a =. (ms - ) cao 4(b) NL T R = ma dim correct equation R = 8. Ft a R = - 4 R = 6 (N) cao 4(c) using s=ut+.5at, u=, a=.(c), t=5 oe s =.5. 5 FT a s = 75 Distance used in braking = 5 75 = 5 Using v =u +as, u=5, v=, s=5(c) oe = 5 + a 5 5 a = 5 a = (-)(.9) = (-)7 B ft a NL B R = ma dim correct equation B= 66 (N) cao Alternative (-)F = 8 (-)(.9) (B) F = 7 Force exerted by brakes = 7 6 () = 66 (N) () cao 4

58 Q Solution Mark Notes 5 T Q T P 4g 6g 5(a) sin = 5 4g T = 4a B NL applied to second particle Dim correct equation. T and weight opposing sin/cos required. T 6gsin = 6a Adding 4g 6g 5 = a m a =.4g =.9(ms - ) cao mag req. accept.4 T =.84g = 7.6(N) cao accept 7.6/7 5(b) Using v =u +as, u=, a=.9(c), s=.5 oe v =.4g.5 Ft a v = g =.844(ms - ) cao 5 5(c) Using v=u+at, v=, u= 5 g (c), a=(±).6g oe g = -.6gt 5 Ft v from (b) t =.844 cao Required time =.7(s) Ft t, dp required. 5

59 Q Solution Mark Notes 6. A B.g g.8g Take moments about B dimensionally correct 4 terms equation, condone no g throughout. (.g + g +.8g)x =.g.5 + g.5 +.8g.6 B any correct moment correct equation x =.6 (m) 6

60 Q Solution Mark Notes 7 R T R T F F 45g 45g Resolve perpendicular to plane accept sin R = 45g cos = (6g = 5.8) F =.5 R = (8g = 76.4) m NL parallel to plane or NL with a= Dimensionally correct All forces, T and wt opp. For greatest T T = 45g sin + F a= T = 7g + 8g T = 45g = 44(N) cao NL parallel to plane or NL with a= Dimensionally correct All forces, T and wt opp. F in opposite direction to previous NL. For least T 45g sin = T + F a= T = 45g sin - F T = 7g - 8g T = 9g = 88.(N) cao Condone absence of greatest/least but if present must be correct for. 7

61 Q Solution Mark Notes 8(a). Area from AF(x) from AB(y) ABEF B BCD B Lamina 7 x y B areas correct, allow areas in proportion ::. Moments about AF 7x = x = 5 5 x = = 8. cao Moments about AB 7y = y = 6 y = 8 cao 8(b) Identification of correct triangle 5/ tan = = tan - Ft x, y = 9.5 ( º ) or =.65 (c) FT x, y units not required but if present must be correct. GCE Maths () MS Summer 7 8

62 GCE MARKING SCHEME SUMMER 7 MATHEMATICS - M 98-

63 INTRODUCTION This marking scheme was used by WJEC for the 7 examination. It was finalised after detailed discussion at examiners' conferences by all the examiners involved in the assessment. The conference was held shortly after the paper was 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 conference was to ensure that the marking scheme was 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' conference, teachers may have different views on certain matters of detail or interpretation. WJEC regrets that it cannot enter into any discussion or correspondence about this marking scheme.

64 Mathematics M (June7) Markscheme Q Solution Mark Notes d (a)(i) v = r dt differentiation attempted Vector required v = (sint + t cost)i + (cost t sint)j (mod v) = (sint + t cost) + (cost t sint) = sin t + t sint cost + t cos t + cos t - t sint cost + t sin t Ft similar expressions = + t Speed of P = t cao (a)(ii) Momentum vector = mv = [(sint + t cost)i + (cost t sint)j] B ft v(c) = (sint + t cost)i + (cost t sint)j (b) At t = 6, r = sin i + cos j r = i + j B If perpendicular, r.( b i + j) = ( i + j).(b i + j) = b + method correct, no i, j b + = b+ = b = - cao

65 Q Solution Mark Notes (a) x = 4 t 6t 7 dt at least one power increased. x = t 4 t + 7t + (C) correct integration When t =, x = 5 m initial conditions used C = 5 x = t 4 t + 7t + 5 When t = m used x = x = x = (m) cao dv (b) a = dt a = t 6 at least one power decreased. F = ma =.8(t 6) Ft a When t = F =.8( 6) F = 8.6 (N) cao

66 Q Solution Mark Notes (a). T = T = P v = (4) B NL dimensionally correct 4 terms, allow sin/cos T mg sin - R = ma = a a =. (ms - ) cao (b) NL dimensionally correct 4 terms, allow sin/cos a = T v - mg sin - R = - v = v - v - 4 = v v 4v = v + 4v - = v = m dep on both M v =.49 cao answer rounding to.5.

67 Q Solution Mark Notes 4(a) initial vertical vel of P = 5sin6º 5 = =.99 initial vertical vel of Q = vsinº B either correct expression use of s = ut+.5gt 5 height of P at time t = t.5gt height of Q at time t =.5vt.5gt either For collision 5 t.5gt =.5vt.5gt m v = 5 = 5.98 accept 6 4(b) initial horiz vel of P = 5cos6º = 7.5 initial horiz vel of Q = 5cosº =.5 B either For collision, 7.5t +.5t = 8 t =.6 (s) convincing 5 4(c) use of v=u+at, u= (c),a=±9.8, t=.6 5 v = Ft u v = 7. speed = accept candidate s values =.(ms - ) cao 4

68 Q Solution Mark Notes 5 5 ms - m sin ms - KE at t= =.5 4 v= or 5 KE at t= = 8 (J) PE at t = = KE at t=8 = KE at t=8 = 5 (J) PE at t =8 = h PE at t =8 = sin PE at t =8 = 588 (J) WD by engine = 4 8 WD by engine = 44 (J) B Work-energy principle KE, PE and WD( terms) = WD correct equation WD = 4 (J) cao 5

69 Q Solution Mark Notes 6 O T P 5g 6(a) conservation of energy KE and PE in equation.5mu mglcos6º =.5mv - mglcos v = u.8g +.6gcos cao v = u cos 6(b) NL towards centre dim correct equation T and 5gcos opposing 5v T - 5gcos =.8 5 T = 5gcos + ( u.8g +.6gcos).8 m subt v equivalent expressions T = 6.5u 5g + 5gcos cao, any correct expression T = 6.5u cos 6(c) For complete circles, T when =8º, (cos=-). 6.5u u.6 u 5.6 cao 6(d) For complete circles, v when =8º, (cos=-). u u.5 u 4.85 cao 6

70 Q Solution Mark Notes 7. O l T r g 7(a) Resolve vertically Tcos = g allow m NL towards centre of motion Tsin = r Tsin = l sin use of r=l sin T = l l cos = g g cos = l convincing 7(b)(i) Tcos = g, T = g cos =. B 7(b)(ii) cos =. and g = g, cos = l g. = l g l = or g=lxg convincing 7(b)(iii)Hooke s Law x T = natural length used, condone natural length=/, but x not/or ( ) g = one of /- or correct = 8g = 764 cao 7

71 x 7(b)(iv)EE = ( nat len) used 764 EE = 98 EE = =.67 (J) cao GCE Maths (M) MS Summer 7 8

72 GCE MARKING SCHEME SUMMER 7 MATHEMATICS - M 98-

73 INTRODUCTION This marking scheme was used by WJEC for the 7 examination. It was finalised after detailed discussion at examiners' conferences by all the examiners involved in the assessment. The conference was held shortly after the paper was 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 conference was to ensure that the marking scheme was 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' conference, teachers may have different views on certain matters of detail or interpretation. WJEC regrets that it cannot enter into any discussion or correspondence about this marking scheme.

74 Mathematics M (June7) Markscheme Q Solution Mark Notes (a) dx dt = x dx dt x sep variables, (-x) required -ln x = t + (C) correct integration ft x- When t =, x = m use of initial conditions C = -ln ft if ln present. t = ln x When x = t = ln = (.69) cao e -t x = m correct method inversion x = ( - e -t ) any correct exp. cao d dx (b) = - dtx dt d x = - ( x) = x dt d x = ( - e -t ) dt m substitute for x d x = -e -t dt Alternative x = ( - e -t ) dx dt = e -t ()()ft similar expressions d x = -e -t dt () d x ft =-e -t only. dt

75 Q Solution Mark Notes P(kg) J J 8 ms - Q(7kg) v ms - v ms - Impulse = change in momentum Applied to Q allow +/-J J = 7 8 7v Applied to P J = v B v = 56 7v m v = 5.6 (ms - ) cao J = 6.8 (Ns) cao

76 Q Solution Mark Notes (a) d x d x - 6 dt d t + 5x = Auxilliary equation m 6m + 5 = (m )(m 5) =, m =, 5 G.S. is x = Ae t + Be 5t ft real roots d x When t =, x = 8 and =6 dt m used both A + B = 8 d x = Ae t + 5Be 5t dt B ft similar expressions A + 5B = 6 Solving, A = 6, B = both values cao x = 6e t + e 5t (b) d x d x - 6 dt d t + x = Auxilliary equation m 6m + = m = ± i C.F. is x = e t (Asint + Bcost) ft complex roots Using initial conditions m used both B = 8 d x =e t (Asint + Bcost)+e t (Acost - Bsint) dt B ft similar expression 6 = 4 + A, A = -8 both values cao x = 8e t (-sint + cost) (c) d x dx - 6 dt d t = (t - 6), Auxilliary equation m 6m = m =, 6 C.F. is x = A + Be 6t ft, another real root For P.I. try x = at + bt allow at+b a 6(at + b) = t 6 correct LHS a = - m comparing coefficients a 6b = -6, b = 4 both values cao x = A + Be 6t t + 4t 8 = A + B d x = 6Be 6t t + 4 dt B ft similar CF+PI 6 = 6B + 4 B =, A = 6 both values cao x = e 6t t + 4t + 6

77 Q Solution Mark Notes 4(a) NL applied to P Dimensionally correct All forces -v dv =.5 dt d v = -6v dt convincing 4(b) dv - 6 dt v separating variables = 6t + (C) v correct integration When t=, v= m use of initial conditions C = = 6t + v v = t cao, any correct exp. 4(c) d v v = -6v d x d v dx = -6v dv 6 dx v m separating variables ln v = -6x + (C) correct integration when x =, v = m use of initial conditions C = ln -6x = ln v ln v = e -6x cao, any correct exp. 4(d) Rate of work = F.v used Rate of work = v v Rate of work = (e -6x ) Rate of work = 4e -8x cao, any correct exp. 4

78 Q Solution Mark Notes 5(a) v = -4x + 8x + attempt to differentiate d v v = -8x + 8 d x or dv/dx= d v v = -4(x ) d x d x = -4(x ) dt d y d x d y d x Let y = x, =, dt d t =, dt dt d y = -4y = - y dt Hence motion is simple harmonic convincing Centre of motion is x = B 5(b) = B Period = = B convincing Amplitude is given by x- when v = v= -4x + 8x + = -4(x ) + 5 = (x ) = ±.5 Amplitude = a =.5 cao Alternative solution v = [a y ] v = [.5 -(x ) ] Hence = (B) Period = = (B) Amplitude = a =.5 () cao () attempt to write equation in correct form Alternative solution Amplitude is given when v = () used -4x + 8x + = (x + )(x 7) = x = -.5,.5 amplitude =.5 =.5 () cao 5

79 5(c) (x ) =.5 sin(t) x =.5 sin(t) + - =.5 sin(t) m use of -centre t = sin -.5 m inversion ft a,,centre t =.9795 t =.466 (s) cao 6

80 6(a) Q Solution Mark Notes A S 4 W 4 R F α B Resolve vertically R = W Resolve horizontally S = F = µr = µw B B Moments about B W 4cosα = S 8sinα dim correct, all forces no extra except friction A 6W = µw 8 µ = cao 7

81 Q Solution Mark Notes 6(b) A G S x W 8-x R F α B F =.6R G =.6S B both Resolve vertically dimensionally correct All forces, no extra G + R = W.6S + R = W Resolve horizontally dimensionally correct All forces, no extra S = F S =.6R.6.6R + R = W.6R = W Moments about A dimensionally correct All forces, no extra Wxcosα+.6R 8sinα = R 8cosα A - each error 4 4.6Rx + 4.8R = 8R m substitute to obtain one common factor force 5.44x = 5.44x = x = =.5 (m) 7 cao GCE MATHEMATICS - M Mark Scheme Summer 7 8

82 GCE MARKING SCHEME SUMMER 7 MATHEMATICS - S 98-

83 INTRODUCTION This marking scheme was used by WJEC for the 7 examination. It was finalised after detailed discussion at examiners' conferences by all the examiners involved in the assessment. The conference was held shortly after the paper was 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 conference was to ensure that the marking scheme was 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' conference, teachers may have different views on certain matters of detail or interpretation. WJEC regrets that it cannot enter into any discussion or correspondence about this marking scheme.

84 S June 7 Markscheme Ques Solution Mark Notes (a) (If A,B are independent,)p(ab)=.. =.6 B (Using P(AB) = P(A) + P(B) P(AB)) EITHER P(AB) = =. OR P(AB) = =.44 B (A and B are not independent because) EITHER..6 OR.4.44 B (b)(i) P( A B) P( A B) P( B) = FT from (a) M if independence assumed So P ( A B) (ii) P( A B) P( A) P( B) P( A B) P( A) P( B) 4 = 5 P( A) P( A B) m M if independence assumed (a) E( E( X X = 4 ) Var( X ) ) (b) (a) (b) E(Y) = E(X) + = Var(Y) = 4Var(X) = 6 4 P( each col) = 6 or P( same col) = = 7 (.86) 4 or 9 = 84 5 (.595) 4 9 Award M for, for 4 A if 6 omitted

85 Ques Solution Mark Notes 4(a)(i).8 P(at least error) = e M exactly, more than =.55 Accept the use of tables (ii) (b)(i) (ii) P( rd page st error) = (.55).55 =. Consider p n = (e.8 ) n = e.8n e.8n <..8nloge < log. giving n > Therefore take n = 9 FT.449 answer to (a) Accept.449 for e.8 can be earned later Allow the use of = Accept solutions using tables or evaluating powers of e.8 5(a)(i) X is B(,.7) B (ii) E(X) = 7 SD(X) =.7. =.45 B Accept., (iii) Let Y = Number of games won by Brian so that Y is B(,.) P(X 6) = P ( Y 4) =.8497 m M no working Accept summing individual probabilities (b) Let G = number of games lasting more than hour G is B(44,.6) which is approx Po(.64) P(G >) = e.64 =.49 B si M no working 6(a) E ( X ) = 4.5 (/7) E(X )= (8..) 54 Var(X) = =.94 (659/79) Allow MR for wrong range. (b) The possible values are 4,5, P(Sum = 4) = 54 =.9 B Accept.9 si

86 Ques Solution Mark Notes 7(a)(i) P(+) = =.67 (ii).5.96 P(disease +) =.67 =.76 cao BB B FT denominator from (i) (b)(i) (ii) 8(a)(i) (ii) (iii) P( nd +) = (.76). =.69 P(disease nd ) =.69 =.99 cao (4/) F() = so k = giving k Use of F(x) =.95 4 x x.4 x.9... x.98 P(X <.5 X <.75) = F(.5) F(.75).5.75 = = 4 FT from (a) Accept Convincing (b)(i) (ii) f ( x) F( x) x x 6 = Use of E X x f ( x) dx = xx x 6 dx for knowing you have to differentiate FT from (b)(i) if answer between and 4 = 6 x 9 =.9 9 / x 5 5 / GCE Maths (S) MS Summer 7

87 GCE MARKING SCHEME SUMMER 7 MATHEMATICS - S 984-

88 INTRODUCTION This marking scheme was used by WJEC for the 7 examination. It was finalised after detailed discussion at examiners' conferences by all the examiners involved in the assessment. The conference was held shortly after the paper was 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 conference was to ensure that the marking scheme was 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' conference, teachers may have different views on certain matters of detail or interpretation. WJEC regrets that it cannot enter into any discussion or correspondence about this marking scheme.

89 (b) S - June 7 - Markscheme Ques Solution Mark Notes (a) E(X) =., E(Y) =.6 B si E(W) = E(X)E(Y) =. Var(X) =.,Var(Y) =.8 B si E( X ) Var( X ) [ E( X )] 5. E( Y ) Var( Y) [ E( Y)].84 Var(W) = E( X ) E( Y ) [ E( X ) E( Y)] = 9.7 P(W = ) = P{(X = )(Y = )} = P(X = ) + P(Y = ) P{(X = )(Y = )} = =. Under H, the number, X, of breakdowns in days is Poi(8) which is approx N(8,8) z 8 =.7 p-value =.48 There is strong evidence to conclude that the mean number of breakdowns per day has been reduced. m BB Allow P(W = ) = P(X )P(Y ) = ( P(X = ))( P(Y = )) = (.6 5 )(.8 8 ) =. Award A for an incorrect or no continuity correction and FT for the following marks 64 z =.79 p-value = z =.84 p-value =.9 FT the p-value (a) (b) (c) 9 th percentile =. 8 = 8 Let X = weight of an apple, Y = weight of a pear Let S denote the sum of the weights of apples Then E(S) = Var(S) = 4 = 96 z = 96 = ( ).6 Prob =.9 Let U = X X X Y Y E(U) = 6 = Var(U) = = We require P(U > ) z = = ( ). Prob =.679 B m m si, condone incorrect notation

90 Ques Solution Mark Notes 4(a) Let x,y denote distance travelled by models A,B respectively. x 66.9; y 6.9 B B (b).5 Standard error = (=.5) 8 95% confidence limits are giving [.55,5.45] The lower end of the interval will be if.5z z.4 Tabular value =.8() cao Smallest confidence level = 98.4% FT their SE and x, y (for the first two marks only) 5(a)(i) Under H, X is B(5,.75) Since p >.5, we consider X which is B(5,.5) P ( X ) P( X 9).87 P ( X 44) P( X 6).94 Significance level =.48 B si (ii) If p =.5, P(Accept H ) = P( X 4) =.9675 =.5 (b)(i) (ii) Let Y now denote the number of heads so that under H, Y is B(,.75) N(5,7.5) z 7.5 =( ).7 Tabular value =.46 p-value =.87 (accept.87) There is insufficient evidence to reject H. B Award A for incorrect or no continuity correction but FT for following marks 9 z =.8 p-value = z =.88 p-value =. Penultimate for doubling line above FT the p-value

91 Ques Solution Mark Notes 6(a)(i) f ( x), a x b b a B Allow < = otherwise (ii) E( X ) b a x dx = = = x b a b a ( b a) ( b a)( b ab a ) ( b a) b a Condone omission of limits (iii) = ( b ab a ) Var(X) = E( X X ) E( ) b ab a a ab b = 4 4b 4ab 4a a 6ab b = ( b a) = Convincing (b)(i) E( Y) dx b a x (ii) = ln x b a b a ln b ln a = b a P( Y y) P y X = P X y b y b a Condone omission of limits