GCE MARKING SCHEME. MATHEMATICS AS/Advanced

Transcription

1 GCE MARKING SCHEME MATHEMATICS AS/Advanced SUMMER

2 INTRODUCTION The marking schemes which follow were those used by WJEC for the Summer 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. Paper Page C C 7 C C4 7 FP FP 7 FP

3 C. (a) (i) Gradient of AC increase in y increase in x Gradient of AC 4/ (or equivalent) (ii) A correct method for finding the equation of AC using candidate s gradient for AC Equation of AC: y 4 4/[x ( 6)] (or equivalent) (f.t. candidate s gradient for AC) Equation of AC: 4x y + 6 (convincing) (iii) Gradient of BD increase in y increase in x (to be awarded only if corresponding is not awarded in part (i)) Gradient of BD /4 (or equivalent) An attempt to use the fact that the product of perpendicular lines (or equivalent) Gradient AC Gradient BD AC, BD perpendicular (iv) A correct method for finding the equation of BD using the candidate s gradient for BD (to be awarded only if corresponding is not awarded in part (ii)) Equation of BD: y / 4 [x ( 7)] (or equivalent) (f.t. candidate s gradient for BD) Note: Total mark for part (a) is 9 marks (b) (i) An attempt to solve equations of AC and BD simultaneously x, y 8 (convincing) (ii) A correct method for finding the length of BE BE 5

4 . (a) 5 7 (5 7 )( 7 + ) 7 ( 7 )( 7 + ) Numerator: Denominator: (c.a.o.) 7 Special case If not gained, allow B for correctly simplified numerator or denominator following multiplication of top and bottom by 7 (b) 5 B 75 5 B 6 B 5 ( 5 ) 75 6 (c.a.o.) B 5. (a) dy x 8 dx An attempt to differentiate, at least one non-zero term correct) An attempt to substitute x in candidate s expression for dy dx m Value of dy at P (c.a.o.) dx Gradient of normal m candidate s value for dy dx Equation of normal to C at P: y ( 5) / (x ) (or equivalent) (f.t. candidate s value for dy provided and both m s awarded) dx (b) Putting candidate s expression for dy 4 dx x-coordinate of Q 6 y-coordinate of Q c 6 (f.t. candidate s expression for dy and at most one error in the dx enumeration of the coordinates of Q for all three A marks provided both s are awarded)

5 4. (a) ( + x) 6 + 6x + 5x + x +... All terms correct Three terms correct B B (b) An attempt to substitute x (or x ) in the expansion of part (a) (f.t. candidate s coefficients from part (a)) ( 99) (At least three terms correct, f.t. candidate s coefficients from part (a)) ( 99) (correct to four decimal places) (c.a.o.) 5. (a) a B b B c 5 B (b) 6x + 6x 7 [a(x + b) + c] + k (k, candidate s a, b, c) Least value c + 4 (candidate s c) 6. (a) An expression for b 4ac, with at least two of a, b or c correct b 4ac k 4 8 Candidate s expression for b 4ac < m < k < (c.a.o.) (b) Finding critical values x 5, x 6 B A statement (mathematical or otherwise) to the effect that x 5 or 6 x (or equivalent) (f.t. only x ± 5, x ± 6) B Deduct mark for each of the following errors the use of < rather than the use of the word and instead of the word or 7. (a) y + δy (x + δx) + 5(x + δx) 9 B Subtracting y from above to find δy δy xδx (δx) + 5δx Dividing by δx and letting δx dy limit δy x + 5 (c.a.o.) dx δx δx (b) dy x / + ( ) x B, B dx 4 Either 8 / or second term ( ) 4 (or equivalent fraction) B 4 5 dy (or equivalent ) (c.a.o.) B dx 64

6 8. (a) Use of f ( ) k k 9 Special case Candidates who assume k 9 and show f ( ) are awarded B (b) f (x) (x + )(x + ax + b) with one of a, b correct f (x) (x + )(x 5x ) f (x) (x + )(4x )(x + )(f.t. only x + 5x in above line) Special case Candidates who find one of the remaining factors, (4x ) or (x + ), using e.g. factor theorem, are awarded B (c) Attempting to find f (/) Remainder 5 4 If a candidate tries to solve (c) by using the answer to part (b), f.t. when candidate s expression is of the form (x + ) two linear factors 4

7 9. (a) y (, 6) ( 4, ) O (6, ) x Concave down curve with maximum at (, a), a Maximum at (, 6) Both points of intersection with x-axis B B B (b) (, ) y ( 6, ) O (4, ) x Concave down curve with maximum at (b, ), b Maximum at (, ), Both points of intersection with x-axis B B B 5

8 . (a) dy x 6 B dx Putting derived dy dx x, (both correct) (f.t. candidate s dy) dx Stationary points are (, ) and (, 5) (both correct) (c.a.o.) A correct method for finding nature of stationary points yielding either (, ) is a maximum point or (, 5) is a minimum point (f.t. candidate s derived values) Correct conclusion for other point (f.t. candidate s derived values) 6

9 C (5 values correct) B ( or 4 values correct) B Correct formula with h 5 I 5 { ( )} I I 498 I (f.t. one slip) Special case for candidates who put h (all values correct) B Correct formula with h I { ( )} I 959 I 959 I (f.t. one slip) Note: Answer only with no working earns marks. (a) ( sin θ ) 5 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 sin θ + 5 sin θ (4 sin θ )( sin θ + ) sin θ, sin θ (c.a.o.) 4 θ 4 48, 65 5 B θ 8, 8 9 B B Note: Subtract mark for each additional root in range for each branch, ignore roots outside range. sin θ +,, f.t. for marks, sin θ,, f.t. for marks sin θ +, +, f.t. for mark 7

10 (b) x 58,, (at least one value) B x 6, 5, (both values) B Note: Subtract a maximum of mark for additional roots in range, ignore roots outside range. (c) sin φ + sin φ cos φ or tan φ + tan φ cos φ or sin φ + cos φ sin φ (or tan φ ), cos φ (both values) φ, 8 (both values) φ Note: Subtract a maximum of mark for each additional root in range for each branch, ignore roots outside range. Special Case: No factorisation but division throughout by sin φ (or tan φ) to yield + cos φ (or equivalent) φ. (a) AC sin (substituting the correct values in the correct places in the area formula) AC (AC 6) BC cos (substituting the correct values in the correct places in the cos rule, (f.t. candidate s value for AC) BC 4 (f.t. candidate s value for AC) (b) sin XZY sin 6 (substituting the correct values in the correct places in the sine formula) sin XZY sin 6 m sin XZY x / 6 x x + c B, B, B / ( if no constant term present) 8

11 5. (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 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) n[ 4 + (n ) ] 46 Either: Rewriting above equation in a form ready to be solved n + 6n 9 or n + n 46 or n(n + ) 46 or: n, n n (c.a.o.) (c) a + 4d 9 B (a + 5d) + (a + 9d) 4 B An attempt to solve the candidate s two linear equations Simultaneously by eliminating one unknown d 4 (c.a.o.) a 7 (f.t. candidate s value for d) 6. (a) r 6 B S 4 ( 6) S 5 (c.a.o.) (b) (i) ar 8 B ar + ar + ar 4 8 B An attempt to solve these equations simultaneously by eliminating a r 8 r 5r + (convincing) r + r + r 4 8 (ii) r 5 (r discarded, c.a.o.) B a 64 (f.t. candidate s value for r, provided r < ) B 7. Area x + x dx (use of integration) 5 x + x 4 (correct integration) B, B 4 5 Area (7/ + 8/) (/ + /) (an attempt to substitute limits) m Area 6 (c.a.o.) 9

12 8. (a) Let p log a x Then x a p (relationship between log and power) B x n a pn (the laws of indices) B log a x n pn (relationship between log and power) log a x n pn n log a x (convincing) B (b) Either: (y ) log 6 log 4 (taking logs on both sides and using the power law) y log 4 + log 6 log 6 y 887 (f.t. one slip, see below) Or: y log 6 4 (rewriting as a log equation) y log y 887 (f.t. one slip, see below) Note: an answer of y from y log 4 log 6 log 6 earns A an answer of y 774 from y log 4 + log 6 log 6 earns A (c) log a 4 4 a / (rewriting log equation as power equation) a 6

13 9. (a) A(4, ) B A correct method for finding radius Radius (b) (i) Either: An attempt to substitute the coordinates of P in the equation of C Verification that x 7, y satisfy equation of C and hence P lies on C Or: An attempt to find AP AP P lies on C (ii) A correct method for finding Q Q(, ) (f.t. candidate s coordinates for A) (c) An attempt to substitute (x 4) for y in the equation of the circle x 4x + (or 5x x + 5 ) x, x (correctly solving candidate s quadratic, both values) Points of intersection are (, ), (, ) (c.a.o.). (a) (i) L Rθ rθ B (ii) K R θ r θ B (b) An attempt to eliminate θ Use of R r (R + r)(r r) m r K R L

14 C ( values correct) B (5 values correct) B Correct formula with h I { ( ) + ( 558)} I I 5469 I 546 (f.t. one slip) Note: Answer only with no working earns marks. (a) e.g. θ π cos θ + cos 4θ (choice of θ and one correct evaluation) B cos θ + cos θ (both evaluations correct but different) B (b) (sec θ ) sec θ + 8 (correct use of tan θ sec θ ) An attempt to collect terms, form and solve quadratic equation in sec θ, either by using the quadratic formula or by getting the expression into the form (a sec θ + b)(c sec θ + d), with a c coefficient of sec θ and b d constant m sec θ sec θ ( sec θ 5)( sec θ + ) sec θ 5, sec θ cos θ, cos θ (c.a.o.) 5 θ 66 4, 9 58 B θ, 4 B B Note: Subtract mark for each additional root in range for each branch, ignore roots outside range. cos θ +,, f.t. for marks, cos θ,, f.t. for marks cos θ +, +, f.t. for mark. (a) d(y 4 ) 4y dy B dx dx d(4x y) 4x dy + 8xy B dx dx d(x 5x) 9x 5 B dx dy 9x 5 8xy (c.a.o.) B dx 4y + 4x

15 (b) dx 4 sin t, B dt dy cos t B dt Use of dy dy dx dx dt dt Substituting π for t in expression for dy m dx dy dx 4. f (x) 4x x 5 An attempt to check values or signs of f (x) at x, x f () <, f () > Change of sign f (x) has root in (, ) x x 766 (x correct, at least 5 places after the point) B x x x (x 4 correct to 5 decimal places) B An attempt to check values or signs of f (x) at x 5, x 5 f ( 5) 97 4 <, f ( 5) > Change of sign α correct to five decimal places Note: Change of sign must appear at least once. 5. (a) (i) dy (7 + x) f (x), (f (x) ) dx dy 6 (7 + x) dx (ii) dy 5 or or 5 dx (5x) (5x) 5x dy 5 dx 5x (iii) dy x f (x) + e 4x g(x) dx dy x f (x) + e 4x g(x) (either f (x) 4e 4x or g(x) x ) dx dy x 4e 4x + e 4x x (all correct) dx

16 (b) d (tan x) cos x m cos x sin x k sin x (m ±, k ±) dx cos x d (tan x) cos x cos x sin x ( sin x ) dx cos x d (tan x) cos x + sin x dx cos x d (tan x) sec x (convincing) dx cos x 6. (a) (i) (7x 9) / dx k (7x 9) / + c (k, 7, / 7 ) / (7x 9) / dx / 7 (7x 9) / + c / (ii) e x/6 dx k e x/6 + c (k, 6, / 6 ) e x/6 dx 6 e x/6 + c (iii) 4 dx 4 k ln 5x + c (k, 5, / 5 ) 5x 4 dx 4 / 5 ln 5x + c 5x (b) (x 4) dx k (x 4) (k,, / ) (x 4) dx 8 / (x 4) Correct method for substitution of limits 4 8 (x 4) dx 5 5 (f.t. for k, only) 6 4

17 7. (a) Trying to solve either x + 5 or x + 5 x + 5 x 4 / x + 5 x (both inequalities) Required range: x 4 / (f.t. one slip) Alternative mark scheme (x + ) 5 (forming and trying to solve quadratic) Critical points x and x 4 / Required range: x 4 / (f.t. one slip in critical points) (b) (i) y y f (x) O x Correct graph B (ii) y f (x ) + y (, 5) O (, ) x Translation of graph of f (x) x with vertex at (±, ±) Coordinates of vertex (, ) Crosses y-axis at (, 5) 5

18 8. (a) g (x) f (x) + f (x) 4x + 9 g (x) 8x + 4x + 9 g (x) 4x + 8x + 8 (x + ) (convincing) 4x + 9 4x + 9 (b) (i) At stationary point, (x + ) 4x + 9 or 8x + 4x + 9 (x + ) only when x / 4x + 9 (ii) g (x) > either side of x / (or at all other points) Stationary point is a point of inflection 9. (a) y 5 ln (x ) B An attempt to express candidate s equation as an exponential equation x (e y 5 + ) (f.t. one slip) f - (x) (e x 5 + ) (f.t. one slip) (b) D(f - ) [5, ) B. (a) R(f) [, ) B R(g) [, ) B (b) gf(x) (x + 4). gf(x) x + 5 (c) fg(x) (x + 4) (correct composition) B [fg(x)] 7 (candidate s fg(x)) x 44 (f.t. one numerical slip) x ± (c.a.o.) 6

19 C4. (a) f (x) A + B + C (correct form) x (x ) (x ) 8 x x A(x ) + Bx(x ) + Cx (correct clearing of fractions and genuine attempt to find coefficients) m A, C, B ( coefficients, c.a.o.) (third coefficient, f.t. one slip in enumeration of other coefficients) (b) f (x) + (at least one of first two terms) B x (x ) (x ) (third term) B (f.t. candidates values for A, B, C) f () (c.a.o.) B. x + 4x dy + 4y y dy 4x dy + 4y B dx dx dx y dy B dx dy (o.e.) (c.a.o.) B dx 4 Use of grad normal grad tangent Equation of normal: y ( ) 4(x ) f.t. candidate s value for dy dx. (a) ( cos θ ) 9 cos θ + 7 (correct use of cos θ cos θ ) An attempt to collect terms, form and solve quadratic equation in cos θ, either by using the quadratic formula or by getting the expression into the form (a cos θ + b)(c cos θ + d), with a c coefficient of cos θ and b d constant m 4 cos θ 9 cos θ 9 (4 cos θ + )(cos θ ) cos θ, (cos θ ) (c.a.o.) 4 θ 8 59, 4 B B Note: Subtract (from final two marks) mark for each additional root in range from 4 cos θ +, ignore roots outside range. cos θ, f.t. for marks, cos θ +, f.t. for mark 7

20 (b) (i) R B Correctly expanding sin (x α) and using either cos α 5 or sin α or tan α to find α 5 (f.t. candidate s value for R) α 67 8 (c.a.o) (ii) Least value of. 5 sin x cos x + (±) + (f.t. candidate s value for R) Least value (f.t. candidate s value for R) Corresponding value for x 57 8 (o.e.) (f.t. candidate s value for α ) 4. π/ Volume π sin x dx π/6 Use of sin x (± ± cos x) Correct integration of candidate s (± ± cos x) Correct substitution of correct limits in candidate s integrated expression Volume π 8(467...) (c.a.o.) B 5. x / x x x B x B 8 x < 4 or 4 < x < 4 B (f.t. candidate s coefficients) B 8 8. (convincing) B 64 8

21 6. (a) Use of dy dy dx and at least one of dx, dy 4 correct dx dt dt dt t dt dy t (o.e.) dx Equation of tangent at P: y 4p p x p (f.t. candidate s expression for dy) m dx y p x + 8p (convincing) (b) Substituting x, y in equation of tangent 4p 8p + p, (both values, c.a.o.) Points are (4, ), ( 4 /, 6) (f.t. candidate s values for p) 7. (a) x ln x dx f (x) ln x f (x) g(x) dx f (x) x 4, g(x), 4 x x ln x dx x 4 ln x x 4 + c (c.a.o.) 4 6 (b) x(x ) 4 dx f (u) u 4 k du (f (u) pu + q, p, q and k / or ) x(x ) 4 dx (u + ) u 4 du (au 5 + bu 4 ) du au 6 + bu 5 (a, b ) B 6 5 Either: Correctly inserting limits of, in candidate s au 6 + bu or: Correctly inserting limits of, in candidate s a(x ) 6 + b(x ) 5 m 6 5 x(x ) 4 dx (c.a.o.) 9

22 8. (a) dv kv B dt (b) dv k dt (o.e.) V kt + c V c (c.a.o.) V (convincing) kt + at + (c) Substituting t and V 9 in expression for V a 6 Substituting t 4 in expression for V with candidate s value for a V 7 (c.a.o) 9. (a) (i j + k).(i 4j + 8k) 8 B i j + k 9, i 4j + 8k 8 (one correct) B Correctly substituting in the formula a.b a b cosθ θ 48 (c.a.o.) (b) (i) AB i j + 7k B (ii) Use of a + λab, a + λ(b a), b + λab or b + λ(b a) to find vector equation of AB r i j + k + λ ( i j + 7k) (o.e.) (f.t. if candidate uses his/her expression for AB) (c) λ + μ λ 4 + μ + 7λ μ (o.e.) (comparing coefficients, at least one equation correct) (at least two equations correct) Solving two of the equations simultaneously m (f.t. for all marks if candidate uses his/her expression for AB) λ, μ 4 (o.e.) (c.a.o.) Correct verification that values of λ and μ satisfy third equation Position vector of point of intersection is i 6k (f.t. one slip). Assume that positive real numbers a, b exist such that a + b < ab. Squaring both sides we have:(a + b) < 4ab a + b + ab < 4ab a + b ab < (a b) < This contradicts the fact that a, b are real and thus a + b ab B B B

23 FP. f ( x + h) f ( x) + ( x + h) + x f + x x xh h [ + ( x + h) ]( + x ) xh h [ + ( x + h) ]( + x ) lim xh h x) h h[ + ( x + h) ]( + x ) ( x ( + x ). 5z z z 5( + i) i i 5( + i) i ( + i)( i) 5(4 + 4i ) i 5 [Award for either a correct numerator or denominator] B 5i Modulus 6 (5.) B tan ( y / x).7 (78.7 ) B Argument 4.5 ( 58.7 ) B

24 . (a) Determinant 8) (5 5) (4 ) 5 ( + + λ λ λ λ λ Singular when 5 4 λ + λ λ, /4 (b) When λ, A (i) Determinant B Cofactor matrix 7 5 Adjugate 7 5 A 7 5 (ii) z y x

25 4. α + β, αβ B Consider α + β β α α + β + α β [( α + β ) αβ ] α + β α β [4 6] α β αβ + β α α β αβ α + β + α β αβ The required equation is x + α β x B 5. The statement is true for n since 4 5 B Let the statement be true for n k, ie 4 k k is divisible by 5 or 4 5N Consider ( 4 ) 6 4 6( + 5N) 5N 5 This is divisible by 5 so true for n k true for n k +, hence proved by induction.

26 A B A( r + ) + Br + 6. (a) Let r ( r + ) r r + r( r + ) A /, B / (b) S n n n n + n + 4 ( n + ) ( n + ) n + + n + 4 ( n + )( n + ) n + 4 ( n + )( n + ) 7. (a) ln f ( x) xln x f ( x) ln x x f ( x) x f (x) (ln x + ) x x (b) At a stationary point, f ( x) so lnx x /e (.68) e y (e ) /.9 It is a maximum because f(.).6 and f(.4).8 [Accept f (.).84 and f (.4).] 4

27 5 8. (a) Rotation matrix B Translation matrix B T (b) Fixed points satisfy y x y x y x x + y (x,y) ), ( (c) Let ), ( ), ( y x y x under T. + x y y x We are given that + x y so that ) ( + + x y y x 4 [Special case: x + y giving y x + award /4]

28 u + iv 9. (a) x iy x + iy ( x) + y x u ( x) + y v y ( x) + y (b) (c) Dividing, v y u x The equation of the locus is v u. Fixed points satisfy z z or z z + Solutions are ± ± i z corresponding to points,± Alternative solution: Fixed points satisfy x y x ; y ( x) + y ( x) + y x x giving x ½ + y ± 4 giving y 6

29 FP. u x x d u xdx B and [,] [, ] B du I 9 u u sin ( ).8. (a) r 5 B θ tan (4/).97 (5. ) B (b) First root ( 5,. 9 ).6 +.5i / Second root (5,.9 + π / ).6 +.5i / Third root (5,.9 + 4π / ).64.67i. Substituting for sin and cos, 5 t 5( t ) + t + t t 5 + 5t + t t + 5t Solving, t ½,. x x tan nπ ( n ) x.97 + nπ (5. + 6n ) x x tan.5 + nπ ( n ) x.5 + nπ ( 4 + 6n ) 7

30 4. (a) Let x A Bx + C + ( x + )( x + ) x + x + A( x + ) + ( Bx + C)( x + ) ( x + )( x + ) A, B, C (b) dx xdx dx I x x x + [ ln( + ) ] + [ ln( x + ) ] x tan x ln 4 ln + ( ln 6 ln) tan tan.44 cao 5. cos5θ isin 5θ (cosθ isinθ ) B cos θ.isinθ + cos θ.i sin θ + i sin θ + real terms Considering imaginary terms, 4 5 sin 5θ 5cos θ sinθ cos θ sin θ + sin θ sin5θ 4 5cos θ cos θ ( cos θ ) + ( cos θ ) sinθ cos θ cos θ + cos θ + cos θ + cos θ 4 6cos θ cos θ + As θ,cosθ so the required limiting value is 5. [FT their previous expression] 8

31 6. (a) ( x ) x( x ) f ( x) 4 ( x ) Stationary points occur when ( x ) x( x ) m x, y / 4 (b) The asymptotes are x and y. BB (c) y GG x (d) Consider x ( x ) x 5x + x ½, f ( A) [,/ ] [, ) 9

32 7. (a) g ( x) f ( x) + f ( x) g( x), therefore even B h( x) f ( x) f ( x) h( x) therefore odd B The result follows from the fact that f ( x) g( x) + h( x) B (b) (i) g( x) ln( + sin x) + ln( sin x) ln cos x lncosx (ii) h( x) ln( + sin x) ln( sin x) + sin x ln sin x ( + sin x) ln cos x + sin x ln cos x ln(sec x + tan x) 8. (a) Writing the equation in the form x 8y we note that, in the usual notation, a and x,y are interchanged with the negative sign indicating that the graph is below the x-axis. The focus is (, ) and the directrix y. (b) (i) The result follows since (4 p ) + 8( p ) B (ii) (iii) dy 4 p p dx 4 The equation of the tangent is y + p p( x 4 p) y px + p This passes through (λ,) if + p λ p or p pλ B The roots p, p satisfy p p And since the gradient of the tangent, m, satisfies m p, it follows that m m which is the condition for perpendicularity.

33 FP. (a) f ( x) cosh x 4 cosh x + 8 ( cosh x ) 4 cosh x + 8 (cosh x 7cosh x + ) (b) We solve cosh x 7 cosh x + ( cosh x )(cosh x ) coshx ½, coshx ½ has no solution (so only one stationary point given by coshx ) x cosh. 76 (c) f ( x) (4 cosh x sinh x 7sinh x) B f (.76) 8 (exact value ) B Positive therefore minimum. B. dx coshudu ; [,] [, θ ] whereθ sinh BB I θ θ sinh u cosh udu sinh u + sinh u du θ (coshu -) du θ sinh u u.8

34 . (a) x Putting f ( x) x and taking logs, ln f ( x) x ln x f ( x) + ln x f ( x) x f ( x) x ( + ln x) (b) (i) The Newton-Raphson formula is xn ( xn ) xn+ xn xn xn ( + ln xn) Starting with.5, one application of the formula gives.56 (ii) f (.555).6...; f (.565) The change of sign indicates that α.56 correct to dps. (c) (i) ln ln d ln x x e e dx x Putting x.5 gives.489 (.56 gives.444) B B (iii) The iteration converges because this is less than in modulus. B Successive values are The required value is [Award only if calculations done to reasonable accuracy] 4. dy dy x y x ( dx dx y x) B 9x Arc length + d x 4 / 4 9x /

35 5. (a) (i) f() B cosh x f ( x) ; f () + sinh x BB sinh x( + sinh x) cosh x f ( x) ;f ( ) ( + sinh x) BB cosh x( + sinh x) ( + sinh x) cosh x(sinh x ) f ( x) ; f () 4 ( + sinh x) BB (ii) The Maclaurin series is x x x x + x x Both even and odd powers of x are present or equivalent argument based on series. B (b) x x x + x x x + x.96 [FT from (a)] 6. (a) Consider x rcosθ cos θ + sinθ cosθ dx cosθ sinθ + cos θ sin θ dθ d x At the required point,, giving dθ sin θ cos θ tan θ θ tan. 55 r.9 [Treat consideration of y rsinθ as a special case and award BB for (.8,.4)]

36 π / (b) Area (cosθ sinθ ) dθ + / π (cos θ + 4sinθ cosθ + 4sin θ )d θ π / n 7. (a) I cos xd(sin x) I n ( cos θ sin θ cos θ )dθ π / 5θ cos θ sin θ π + π / π / + π / n n [ cos x sin x] sin x.( n ) cos x sin xdx π / n ( n ) ( cos x) cos xdx ( n ) I n ( n ) In n n n I n (b) (i) I 4 I I 4 4 π / dx 8 π (.589) 6 π / 5 (ii) Integral cos x ( cos x)dx I5 I π / cos x dx GCE Mathematics C-C4 & FP-FP Mark Scheme - Summer / JSM 4

37 INTRODUCTION The marking schemes which follow were those used by WJEC for the Summer 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. Paper Page M 5 M 9 S 5 S 9 S

38 Mathematics Notes: cao correct answer only, oe or equivalent, si seen or implied, ft follow through (c) candidate's value acceptable. (a) Use of v u + as with u (±)., a (±)9.8, s (±)5.4 v v 7.5 (ms - ) cao (b) Use of v u + at with v 7.5(c), a (±)9.8, u (±). oe t t 7 cao. (a) v ms t s attempt at v-t graph with one correct section and axes second correct section completely correct graph with labels (b) Distance.5(9 + ).7 attempt to calculate total area any correct value for an area B 8.5 (m) cao (c) R a 75g.7 a ( 8) B 5 Apply NL to woman R 75g 75a all forces, dim correct correct equation R 75( ) (N) ft a

39 . a R F sinα 5 5g cosα Resolve perpendicular to plane R 5gcosα Use of F μr m si 94.8 (N) Apply NL to object down slope 5gsinα - F 5a a a.96 (ms - ) Dim correct, all forces cao 4. ms - 7 ms - A kg B kg v A v B (a) attempt at conservation of momentum equation + 7 v A + v B v A + v B 45 attempt at restitution equation v B - v A -.6(7 ) -v A + v B 9 attempt to solve simultaneously dep. 5 v A 6 v A 7. (ms - ) v B. (ms - ) Both M s m cao cao (b) Use of Impulse change in momentum I (. 7) 9.6 (Ns) ft sensible results only

40 5. R a F A T a 6g T B (a) Apply NL to B/A 4g T 4a 4g Apply Nl to other particle T F 6a Resolve vertically, particle A R 6g F μr.4 6g.4g attempt to solve equations simultaneously 4g.4g a a.6g.568 (ms - ) T.98 (N) si B B m cao cao (b) Light strings enable the assumption that tension is constant throughout the string to be used. B 6. Attempt to resolve in direction of N force Y - 5 sin 6 - sin 45 Y.5 Attempt to resolve in perpendicular direction X 5 cos 6 - cos 45 X. Resultant R ( ).5 +..(48) (N) ft θ tan Dir of R is 4.6 to the right with the N force ft

41 7. T A T B A C 4 g 6g B Moments about A dim. correct equation, all forces any correct moment B.4 T B.7 6g +. g T B 5.4 (N) cao Resolve vertically dim correct, all forces oe T A + T B 6g T A 6.4 (N) ft T B 8. Use of s ut +.5at with s 95, t u +.5 a 5 Use of v u + at with t 7. v u + 7a attempt to solve simultaneously.8 4.5a a.4 u m cao cao 9. (a) Lamina Area from AD from AB ABCD XYZ 9 Decoration 89 x y one correct pair of distances B all four correct B correct areas B Moments about AD 89 x ft x.9 (cm) cao Moments about AB 89y y 4.8 (cm) ft cao (b) x θ tan - y correct triangle.9 tan ft 6.9 4

42 Mathematics M. (a) Attempt to integrate a v 4t dt When t, v - Therefore C - v t t (+C) use of initial conditions m v -t + t (b) When P is at rest, v -t + t - t t + (t )(t ) t.5, ft C (c) Attempt to integrate v Distance required t + t dt limits oe m 4 t + t t ft v cao. (a) dr v dt v 6t i + ( 4t) j used Speed ( 6t) + ( 4t) When t, speed ft (b) velocity is perpendicular to (i j) when v.(i j) t + 4t t 6 method for dot product cao dv (c) Acceleration of P a used dt a 6i 4j independent of t ft v Magnitude ft (d) Let θ be the required angle. Use of a.b a b cosθ with b v when t (6i 4j).(i + 5j) 5 cos θ ft 7 5 cos θ θ 56. cao 5

43 . (a) Use of Hooke's Law 94x 9.8 x. (m) T λx l cao (b) Use of loss in potential energy mgh Loss in PE 9.8 (.8 +.) si ft x 9.4 (J) Gain in KE.5 v.5 v (J) B Use of gain in elastic energy λ l energies 94. Gain in EE ft x.94 (J) Use of conservation of energy v +.94 ft v 4. (ms - ) cao x 4. (a) N a T α R 5g P Use of T v T ( 75 N) 8 si NL up slope all forces, dim correct equation T R 5gsinα 5a - each error A a a.9 (ms - ) cao (b) At maximum attainable speed, a used Apply NL to particle up the slope T R + mgsinα v v.5 (ms - ) 6 49 cao 6

44 5. (a) Initial vertical speed Vsin (.5V) B Use of v u + as with u Vsin, v, a (±)9.8, s (±)4.9.5V + (-9.8)(4.9) V 9.6 (b) Use of s ut +.5at with s (±)9., a (±)9.8, u (±).5 9.6(c) t 4.9t t t 8 attempt to solve quadratic m t 4s (c) initial horizontal velocity 9.6cos (6.97) B Use of v u + at with u 9.8(c), a (±)9.8, t v v -9.6 Speed (ms - ) 6. v 4 F R 6g Let the maximum speed be v ms -. On the point of slipping, friction is limiting and F μr.5r Resolving vertically Rcos mg + Fsin R R 6g 8 R(4 ) 48g R (N) Apply NL horizontally Rsin + Fcos ma v a 4 dim correct equation all forces, dim correct R + 4 R v.5v v 9.5 (ms - ) cao 7

45 7. (a) Use of conservation of energy.5m.5mv + mg(.5)( - cosθ) KE PE v ( - cosθ) v + 49cosθ v ( + 49cosθ) cao When cosθ.5 v v (.) ft (b) Apply NL towards centre all forces, dim correct mv T - mg cosθ r ( + 49cosθ ) T 9.8 cosθ + substitution m.5 T cosθ cao (c) Minimum value of cosθ is -. Therefore T > for all values of θ. Therefore P describes complete circles. 8

46 Mathematics M. N a R T (a) Use of 7g P T v 8 T v si Apply NL to car dim correct equation T R ma 8 dv 9v 7 v dt Divide by 9 and multiply by v throughout dv 9 v 8v dt (b) Attempt to separate variables 8v 9 v dv dt Integrating -4 ln 9 v t + C correct ln term all correct t -4 ln 9 v - C Required time [ 4 ln 9 v ] subtraction of t values 5 correct limits oe ln ln 5 4 ln(.75).4 (s) cao 9

47 . (a) λ 5 At equilibrium g 75 λ 764 (N) use of Hook's Law (b) Consider a displacement x from the equilibrium position. Apply NL g T x g - λ ( 5 + x) 75 x ft λ x -(4) x Therefore is SHM (with ω 4). Amplitude.5 (m) π π Period s ω 7 B B (c) Maximum speed aω used (ms - ) ft a (d) Use of v ω (a x ) with ω 4, a.5(c), x. v 4 (.5. ) ft a 4.4 v.56 (ms - ) cao (e) Displacement from Origin x x (-).5cos(4t) When t.6 x (-).5 cos(4.6) x (-).46 (m) ft a cao

48 . Auxiliary equation 4m m + 9 B (m ) m.5 (twice) B Complementary function x (A + Bt)e.5t ft B d x For PI, try x at + b, a dt -a + 9(at + b) 8t 87 9a 8 comparing coefficients m a b -87 b -7 both General solution x (A + Bt)e.5t + t -7 ft B Use of initial conditions t, x 5, A 7 5 A d x dt in general solution cao d x (A + Bt)(.5)e.5t + Be.5t + correct diff. ft B dt.5a + B + B - cao x ( t)e.5t + t 7

49 4. When the string jerks tight, each particle begins to move in direction PQ with equal speeds v. P P v C 6 6 Q J 6 6 J u 8 ms C Q v Just before jerk Just after jerk cos CPQ sin CPQ si B Use of impulse change in momentum Applied to P J v B Applied to Q J 5 8sin 6-5v Attempt to solve simultaneously m v 4-5v 5 v 4. (ms - ) cao Speed of particle P is 4. ms -. Magnitude of impulsive tension J v 5.99 (Ns) cao units B Perpendicular to PQ, there is no impulse Speed of particle Q perpendicular to PQ 8cos6 4 ms - 5 Speed of particle Q ms - cao B

50 5. (a) Use of NL 5g v 5a 5g v dv 5v d s (b) Attempt to separate variables 5v dv v 5g ds 5 ln v 5g s ( + C) correct ln all correct Use of boundary conditions s, v m 5 ln 95g C 5 75 s ln cao v 5g (c) v 4 used 5 75 s ln s.49 cao (d) Removing ln 75 exp s 5 v 5g ft v 5g 75exp s 5 v 5g + 75exp s 5 cao v exp s 5

51 6. Q A θ S 8g g R B F (a) Use of Friction μ Normal reaction si Q. S Attempt at taking mom. about B 4 terms, g.5 sinθ + 8g sinθ 4S + Q S +.9S 4.9S 75. S 48 (N) dim correct equation - each error A cao (b) Resolve vertically 4 terms, dim correct Q + R 8g + g R g. 48 R N Resolve horizontally F S ( 48) B Use of F μr 48 μ μ.97 cao 4

52 Mathematics S. (a) P(A B).6. B P(A B) (b) EITHER P (B ) P(B).7 B P( A B ) P(A) + P( B ) P(A)P( B ) OR B. A.4.8 Correct Venn diagram Required prob..88 B. (a) E(Y) 4 Var(Y) 9 8 (b) E ( Y ) Var( Y ) + { E( Y )} [FT arithmetic slip in (a)]. (a) Mean 6 si B 6 (i) 6 Prob e. 89! (ii) Prob [FT on mean] e (b) Prob of no arrivals.t B Attempting to solve e t. 5.t loge log.5 log.5 t.86.log e [Award marks for 4 using tables] 5

53 4. (a) Prob wins on st throw.8..4 (b) Prob wins on nd throw [FT from (a) if awarded in (a)] (c) Prob wins / (.55).56 [For candidates who solve for Bill first, award MA for (a), for.68 in (b), for and for./(.56) 5/ (.68) in (c)] 5. (a) P(correct ans) [Award if and.5 reversed] (b).6 Reqd prob [FT denominator from (a) if answer < ].7 BB 6 7 cao B 6. (a) p x 6 k so k /6 E 6 (b) (i) ( X ) ( ) 5.5 [Accept 84k] (ii) E X [Accept 4k] (c) (i) Possibilities are,5 and, si B Prob ( ) 56 [Award for or terms] 9 56 (.74) [Accept 9/ k ] (ii) Possibilities are, ;, ; 5,5 ; 7,7 si B Prob ( ) 56 [Award for or 4 terms].8 (/64) [Accept 84/ k ] 6

54 7. (a) Number of 6s obtained, X, is B(5,.) B 5 8 (i) Prob..8. or or (ii) P(at least ).556 or (b) Prob of 6s..4 B X is now B(,.4) which is approx P(8) B [FT p from previous line] P( 5 X ) or BB.76 cao B 8. (a) kx ( x )dx 4 Integral k x x B 4 [Limits must appear somewhere for ] k 4 so k 4 (b) E(X) x.4x( x ) dx x x 5 [Limits not required until nd line] 8 5 7

55 x (c) (i) F(x) 4t( t )dt [Limits not required for ] 4 [ ] x t t [Limits must appear here] 4 x x (ii) (iii) Prob F(.75) F(.5) (.5.5 ).6875 [FT if awarded in (c)(i) and answer sensible] The median m satisfies 4 m m.5 [FT from (c)(i) if awarded there] 4 m 4m m m 4 m.54 cao ± 8

56 Mathematics S. (a) 6 z. 75 (Accept ±) 8 Prob.4 (b) Distribution of total weight T is N(6, 64) 6 z.7 64 P ( T < ).89 [No FT on incorrect variance]. (a) Under H, mean 5 si B P(X 9).699 (Accept.778 from Normal app) B P(X ).5 (Accept.465 from Normal app) B Sig level [FT one slip but treat Normal apps as incorrect here] (b) X is now Po(5) which is approx N(5,5) B [FT their mean] z 5 [Award A for incorrect continuity correction].59 Prob from tables.559 p-value.8 B Insufficient evidence to reject H (Accept Accept H ). B [No c/c gives z.6, prob.56 and p-value. Incorrect c/c gives z.67. prob.475 and pv.95] 9

57 . (a) x (.6) B. SE of X (.5 ) B 95% conf limits are.6 ±.96./ [ correct form,.96] giving [.4,.8] cao (b) H μ ; H : μ B : > y (.) B 4. Test stat. / 4 [Award only if there is division by 4 in the denominator] [FT on slip in calculating y ]. p-value.8 Strong evidence for thinking that μ exceeds. B (c) SE of y x.. + (.5..) 4 9% confidence limits are..6 ± m giving [.5,.85] cao 4. (a) E(X) np B Using Var(X) E( X ) [ E( X )] np( p). 9. Solving, p.7 so p. m n (b) E(Y) 6 B E(XY) 6 8 B E ( Y ) E ( X Y ) Var(XY) m

58 5. (a) A.5 PQ cosθ. PQsinθ 8sinθ cosθ 4sin θ B (b) P ( A ) P(4sin θ ) P (θ sin [/ ]) P ( θ π /) [Accept 5 ] π / π / 4 / (c) 4 f ( θ ) [Only award if quoted or used in (c)] π B π / 4 4 E(A) 4sin θdθ π 8 π cos θ / 4 [ ] π 8 π 6. (a) H p.75 versus H : p. 75 B : < (b) (i) Under H, X (No germ) is B(5,.75) B and Y (No not germ) is B(5,.5) (si) B Sig level P(X < ) H ) P(Y > H ).6 [Award BBA if Normal approx used] (ii) Required prob P(X p.5) P(Y ). [Award A if Normal approx used] (c) X is now B(,.75) which is approx N(5,7.5) BB z [Award A for incorrect continuity correction] p-value.66 [No c/c gives p.56, incorrect c/c gives p.46] Insufficient evidence to doubt the statement on the packet B

59 Mathematics S. (a) 4 p ˆ B (b) ESE (.94..) si 99% confidence limits are.56 ± giving [.48,.64] (c) No because.5 lies within the interval. B. (a) H μ ; H : μ B : < (b) 99.6 x. 996 B s B [Accept division by giving.84] Test stat / [M if square root omitted]. (.4) p-value. Strong evidence to reject H (or accept H ) [FT on p-value] B (c) We assume that the sample mean is normally distributed B. (a) (i) P ( p, p, p) P ( p,p,5p) P ( p,5p,5p) P ( p,p,p) P ( p,p,5p) P ( p,5p,5p) The sampling distribution of T is t p 5p p 5p 4p P(T t) / 6/ / 6/ /

60 4. (a) Σx 8 ; Σx si B UE of μ.5 B 8 UE of σ 5.99 (b) DF si B At the 95% confidence level, critical value. B The 95% confidence limits are ±. [Only award if t-distribution used and square root present] [FT values from (a)] giving [.7,4.7] (c) The claim is justified since all the possible means within the interval are less than 5 in modulus. [FT from confidence interval] B 5. (a) H : μx μ y ; H : μx μ y B (b) x...; y.56.. BB s x.9.. (.96..) B. 86. s y (.86) B [Accept division by 75] SE (.79..,.77) Test stat [M if no square root] (accept.8 or.79) Prob from tables.84 (.75,.67) p-value.768 (.75,.74) [FT from line above] B (c) Insufficient evidence to state there is a difference in mean life-times. B

61 6. (a) E(X) θ + θ 4θ B Var(X) θ + θ ( 4θ ) θ + θ + 8θ 6θ θ ( 8θ ) E( X ) (b) E(U) [A if E omitted] 4 ( 4θ ) 4 θ Var( X ) Var(U) 6 θ ( 8θ ) 6n (c) N is B(n,θ) ; E(N) nθ si B nθ E( V ) θ [B if E omitted] n B Var(N) nθ( θ) si B Var( N) Var(V) 4n θ ( θ ) n (d) θ θ Var(V) Var(U) θ + θ n 8 θ 8n (> ) [FT from previous results] U is better because Var(U) < VarV) B 7. (a), x 9, y 4.9, xy x B [B for or correct] S xy / 6. B S xx 9 / 6 75 B. b a [Working must be seen for marks to be awarded] (b) SE of a. 9 (.86..) 6 75 The 9% confidence interval for α is given by.84 ± (.8,.87) cao GCSE Mathematics -M & S-S Mark Scheme (Summer )/JSM 4

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