MATHEMATICS - C1-C4 & FP1-FP3

Transcription

1 GCE MARKING SCHEME MATHEMATICS - C-C4 & FP-FP3 AS/Advanced SUMMER WJEC CBAC Ltd.

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. WJEC CBAC Ltd.

3 C. (a) Gradient of AB = increase in y increase in x Gradient of AB = 4 / 3 (or equivalent) (b) A correct method for finding C C(, 3) (c) Use of m AB m L = to find gradient of L A correct method for finding the equation of L using candidate s coordinates for C and candidate s gradient for L. Equation of L: y 3 = 3 / 4 [x ( )] (or equivalent) (f.t. candidate s coordinates for C and candidate s gradient for AB) Equation of L: 3x 4y + 5 = (convincing, c.a.o.) (d) (i) Substituting x = 7, y = k in equation of L k = 9 (ii) A correct method for finding the length of CA(DA) CA = 5 (f.t. candidate s coordinates for C) DA = 5 (iii) Sin ADC = CA = 5 DA 5 (f.t. candidate s derived values for CA and DA) Sin ADC = CA = (c.a.o.) DA 5

4 . (a) = (7 ) 7 + (7 + )(7 ) Denominator: = (7 ) = (7 ) = 4 4 (c.a.o.) Special case If not gained, allow for correctly simplified denominator following multiplication of top and bottom by 7 + (b) (43) = = 563 = 5 7 (43) (8 5) 563 = 3 (c.a.o.) 7 3. (a) dy = 4x (an attempt to differentiate, at least dx one non-zero term correct) An attempt to substitute x = in candidate s expression for dy m dx Use of candidate s numerical value for dy as gradient of tangent at P dx m Equation of tangent at P: y ( ) = 3(x ) (or equivalent) (c.a.o.) (b) Gradient of tangent at Q = 9 An attempt to equate candidate s expression for dy and candidate s dx derived value for gradient of tangent at Q 4x = 9 x = 5 (f.t. one error in candidate s expression for dy) dx 4. ( x) 6 = x + 6x 6x ( for further incorrect simplification) 5. (a) a = 3 b = c = 7 (b) Stationary value = 7 (f.t. candidate s value for c) This is a minimum

5 6. (a) An expression for b 4ac, with at least two of a, b, c correct b 4ac = (k ) 4(k k + ) b 4ac = 7 (c.a.o.) candidate s value for b 4ac < ( no real roots) (b) Finding critical values x = 6, x = / 3 A statement (mathematical or otherwise) to the effect that x < 6 or / 3 < x (or equivalent) (f.t critical values 6, / 3 only) 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 + y = 3(x + x) 7(x + x) + 5 Subtracting y from above to find y y = 6xx + 3(x) 7x Dividing by x and letting x dy = limit y = 6x 7 (c.a.o.) dx x x (b) Required derivative = / 3 / 4 x 3/4 + ( 3) x 4, 8. (a) Attempting to find f (r) = for some value of r f () = x is a factor f (x) = (x )(6x + ax + b) with one of a, b correct f (x) = (x )(6x 7x 3) f (x) = (x )(3x + )(x 3) (f.t. only 6x + 7x 3 in above line) x =, / 3, 3 / (f.t. for factors 3x, x 3) Special case Candidates who, after having found x as one factor, then find one of the remaining factors by using e.g. the factor theorem, are awarded for final 4 marks (b) Use of g(a) = a 3 53 = a = 4

6 9. (a) y (, ) O (7, ) x (3, 7) Concave up curve and y-coordinate of minimum = 7 x-coordinate of minimum = 3 Both points of intersection with x-axis (b) y (, ) O (4, ) x ( 4, O 7) Concave up curve and y-coordinate of minimum = 7 x-coordinate of minimum = 4 Both points of intersection with x-axis

7 . (a) dy = 3x + 6x dx Putting derived dy = dx x =, (both correct) (f.t. candidate s dy) dx Stationary points are (, ) and (, 3) (both correct) (c.a.o) A correct method for finding nature of stationary points yielding either (, ) is a minimum point or (, 3) is a maximum point (f.t. candidate s derived values) Correct conclusion for other point (f.t. candidate s derived values) (b) (, 3) y O x (, ) 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) One positive root (f.t. the number of times the candidate s curve crosses the positive x-axis)

8 C (5 values correct) B (3 or 4 values correct) Correct formula with h = 5 I 5 { ( )} I I I 657 (f.t. one slip) Special case for candidates who put h = (all values correct) Correct formula with h = I { ( )} I I I 6495 (f.t. one slip) Note: Answer only with no working earns marks

9 . (a) cos + 3 cos = 4( cos ) (correct use of sin = cos ) An attempt to collect terms, form and solve quadratic equation in cos, either by using the quadratic formula or by getting the expression into the form (a cos + b)(c cos + d), with a c = candidate s coefficient of cos and b d = candidate s constant m 4 cos + 3 cos = ( cos + )(7 cos ) = cos =, cos = (c.a.o.) 7 = 73 4, 86 6 =, 4 Note: Subtract mark for each additional root in range for each branch, ignore roots outside range. cos = +,, f.t. for 3 marks, cos =,, f.t. for marks cos = +, +, f.t. for mark (b) 3x = 54, 34, 36, 594 (one value) x = 85, 9 Note: Subtract (from final two marks) mark for each additional root in range, ignore roots outside range. (c) Use of sin = tan cos tan = = 3, 9 3 (f.t tan = a) 3. (a) = 5 + x 5 x / 5 (correct use of cos rule) An attempt to collect terms, form and solve quadratic equation in x, either by using the quadratic formula or by getting the expression into the form (x + b)(x + d), with b d = candidate s constant m x 4x 96 = x = (c.a.o.) (b) sin XZY = sin (substituting the correct values in the correct places in the sin rule) XZY = 44, 36 (at least one value) Use of angle sum of a triangle = 8 YXZ = 7, 5 (both values) (f.t. candidate s values for XZY provided both M s awarded)

10 4. (a) S n = a + [a + d] [a + (n )d] (at least 3 terms, one at each end) 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) a + d + a + 3d + a + 9d = 79 a + 5d + a + 6d = 6 An attempt to solve the candidate s linear equations simultaneously by eliminating one unknown a = 3, d = 5 (both values) (c.a.o.) (c) a = 5, d = S n = n[ 5 + (n )( )] (f.t. candidate s d) S n = n(6 n) (c.a.o.) 5. (a) a + ar = 7 a + ar = An attempt to solve candidate s equations simultaneously by correctly eliminating a 3r 5r = (convincing) (b) An attempt to solve quadratic equation in r, either by using the quadratic formula or by getting the expression into the form (ar + b)(cr + d), with a c = 3 and b d = (3r + )(r ) = r = / 3 a ( / 3 ) = 7 a = 8 (f.t. candidate s derived value for r) S = 8 (correct use of formula for S, f.t. candidate s ( / 3 ) derived values for r and a) S = 8 (c.a.o.)

11 6. (a) 3 x 3/ x /3 + c 3/ /3 ( if no constant term present) (b) (i) 36 x = 5x 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 = 36 m (x 4)(x + 9) = A(4, ) (c.a.o.) B(6, ) (ii) Area of triangle = 4 (f.t. candidate s coordinates for A) 6 Area under curve = (36 x ) dx (use of integration) 4 36 dx = 36x and x dx = x 3 3 Area under curve = [(6 6/3) (44 64/3)] (substitution of candidate s limits) m = 64/3 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 = /3 = 84/3 (c.a.o.)

12 7. (a) Let p = log a x Then x = a p x n = a pn log a x n = pn log a x n = pn = n log a x (relationship between log and power) (the laws of indices) (relationship between log and power) (convincing) (b) Either: (x/ 3) log 9 = log 6 (taking logs on both sides and using the power law) x = (log log 9) log 9 x = 7 63 (f.t. one slip, see below) Or: x/ 3 = log 9 6 (rewriting as a log equation) x = (log ) x = 7 63 (f.t. one slip, see below) Note: an answer of x = from x = (log 6 3 log 9) log 9 earns A an answer of x = 3 85 from x = log log 9 log 9 earns A an answer of x = 98 from x = (log log 9) log 9 earns A an answer of x = 4 63 from x = log log 9 log 9 earns A Note: Answer only with no working earns marks (c) log a (x ) + log a (4x + ) = log a [(x ) (4x + )] (addition law) log a (x 3) = log a (x 3) (power law) (x ) (4x + ) = (x 3) (removing logs) x = (c.a.o.) Note: Answer only with no working earns marks 8. (a) A(, 3) A correct method for finding the radius Radius = (b) AT = 6 (f.t. candidate s coordinates for A) Use of RT = AT AR RT = 7 (f.t. candidate s radius and coordinates for A )

13 9. Area of sector POQ = / r Area of triangle POQ = / r sin ( ) 35 = / r / r sin ( ) (f.t. candidate s expressions for area of sector and area of triangle) r = 35 (o.e.) (c.a.o.) ( 9) r = 9 7 (f.t. one numerical slip)

14 C3. (a) (5 values correct) B (3 or 4 values correct) Correct formula with h = 5 I 5 { ( ) 3 + ( 84547)} I I I 4637 (f.t. one slip) Note: Answer only with no working shown earns marks (b) e x² + 3 dx = e 3 e x² dx e x² + 3 dx = (f.t. candidate s answer to (a)) Note: Answer only with no working shown earns marks

15 . (a) = 36 or = and noting that cos = cos sin sin (including correct evaluations) (b) 3 tan = 5( + tan ) + 6 tan. (correct use of sec = + tan ) An attempt to collect terms, form and solve quadratic equation in tan, either by using the quadratic formula or by getting the expression into the form (a tan + b)(c tan + d), with a c = candidate s coefficient of tan and b d = candidate s constant m 8 tan 6 tan 5 = (4 tan 5)( tan + ) = tan = 5, tan = (c.a.o.) 4 = 5 34, 3 34 = 53 43, Note: Subtract mark for each additional root in range for each branch, ignore roots outside range. tan = +,, f.t. for 3 marks, tan =,, f.t. for marks tan = +, +, f.t. for mark 3. (a) d(x 3 ) = 3x d( 3x ) = 3 dx dx d( 4x y) = 4x dy 8xy dx dx d(y 3 ) = 6y dy dx dx x = 3, y = dy = 6 = (c.a.o.) dx 4 7 (b) (i) Differentiating sin at and cos at with respect to t, at least one correct candidate s x-derivative = a cos at, candidate s y-derivative = a sin at (both values) dy = candidate s y-derivative dx candidate s x-derivative dy = tan at (c.a.o.) dx (ii) d dy = a sec at (f.t. candidate s expression for dy) dtdx dx Use of d y = d dy candidate s x-derivative dx dtdx d y = sec 3 at (c.a.o.) dx

16 4. f (x) = cos x 5x + An attempt to check values or signs of f (x) at x =, x = /4 f () = 3 >, f ( /4) = < Change of sign f (x) = has root in (, /4) x = 6 x = x = x 3 = x 4 = = (x 4 correct to 5 decimal places) An attempt to check values or signs of f (x) at x = , x = f ( ) = >, f ( ) = < Change of sign = correct to five decimal places Note: change of sign must appear at least once 5. (a) dy = a + bx (including a =, b = ) dx 7 + x 3x dy = 6x dx 7 + x 3x (b) dy = e tan x f(x) (f (x), ) dx dy = e tan x sec x dx (c) dy = 5x f (x) + sin x g(x) (f (x), g(x), ) dx dy = 5x f (x) + sin x g(x) dx (either f (x) = or g(x) = x) x dy = 5x + x sin x dx x

17 6. (a) (i) 3e x/4 dx = k 3e x/4 + c (k =, / 4, 4, 4) x/4 x/4 3e dx = 4 3e + c (ii) 9 dx = k 9 (x 3) 5 + c (k =,, / ) (x 3) dx = 9 (x 3) 5 + c (x 3) 6 5 (iii) 7 dx = k 7 ln 3x + + c (k =, 3, / 3 ) 3x + 7 dx = 7 / 3 ln 3x + + c 3x + Note: The omission of the constant of integration is only penalised once. (b) sin x dx = k cos x (k =,, /, / ) sin x dx = cos x k (cos a cos ) = / 4 (f.t. candidate s value for k) cos a = / (c.a.o.) a = /6 (f.t. cos a = b provided both M s are awarded) 7. (a) 9 x 3 = 6 x 3 = ± / 3 (f.t. candidate s a x 3 = b, with at least one of a, b correct) x = / 3, 7 / 3 (f.t. candidate s a x 3 = b, with at least one of a, b correct) (b) Trying to solve either 5x 3 or 5x 3 5x 3 x 5x 3 x / 5 (both inequalities) Required range: / 5 x (f.t. one slip) Alternative mark scheme (5x ) 9 (forming and trying to solve quadratic) Critical points x = / 5 and x = Required range: / 5 x (f.t. one slip)

18 8. y ( 5, O x Concave down curve passing through the origin with maximum point in the second quadrant x-coordinate of stationary point = 5 y-coordinate of stationary point = 8 9. (a) (i) f (x) = (x + 5) f (x) (x + 3) g(x) (f (x), g(x) ) (x + 5) f (x) = (x + 5) x (x + 3) x (x + 5) f (x) = 4x (c.a.o.) (x + 5) f (x) < since numerator is negative and denominator is positive (ii) R(f) = ( 3 / 5, ) (b) (i) x = 3 5y (o.e.) (condone any incorrect signs) y x = ( ± ) 3 5y / (f.t. at most one incorrect sign) y x = 3 5y / (f.t. at most one incorrect sign) y f - (x) = 3 5x / (c.a.o.) x (ii) R(f - ) = (, ), D(f - ) = ( 3 / 5, ), (both intervals, f.t. candidate s R(f ))

19 . gg(x) = (3(g(x)) + 7) / or gg(x) = g((3x + 7) / ) gg(x) = (3(3x + 7) + 7) / An attempt to solve the equation by squaring both sides gg(x) = 8 9x = 36 (o.e.) (c.a.o.) x = ± (c.a.o.)

20 C4. (a) f (x) A + B + C (correct form) (x + ) (x ) (x ) + x x A(x ) + B(x + )(x ) + C(x + ) (correct clearing of fractions and genuine attempt to find coefficients) m A =, C = 3, B = ( correct coefficients) (third coefficient, f.t. one slip in enumeration of other coefficients) (b) f (x) = + 6. (o.e.) (x + ) (x ) (x ) 3 (f.t. candidate s values for A, B, C) (at least one of the first two terms) (third term) f () = / 4 (c.a.o.). 3y dy 8x 3x dy 3y = 3y dy 8x dx dx dx 3x dy 3y dx Either dy = 3y + 8x or dy = (o.e.) (c.a.o.) dx 3y 3x dx 3 Equation of tangent: y ( 3) = (x ) 3 f.t. candidate s value for dy dx

21 3. (a) 4( 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 8 sin sin 3 = (4 sin 3)( sin + ) = sin = 3, sin = (c.a.o.) 4 = 48 59, 3 4 =, 33 Note: Subtract mark for each additional root in range for each branch, ignore roots outside range. sin = +,, f.t. for 3 marks, sin =,, f.t. for marks sin = +, +, f.t. for mark (b) (i) R = 7 Correctly expanding sin (x + ) and using either 7 cos = 8 or 7 sin = 5 or tan = 5 to find 8 = 6 93 (f.t. candidate s value for R) (c.a.o) (ii) sin (x + ) = (f.t. candidate s value for R) 7 x = 4 3, 39 68, 4 3, (at least one value on R.H.S., f.t. candidate s values for and R) x = 77 75, (c.a.o.) (iii) Greatest possible value for k is 7 since greatest possible value for sin is (f.t. candidate s value for R) E 4. 4 Volume = x + 5 dx x 3 x + 5 = x x x ax + b + cdx = ax + bx + c ln x, where c and at least one of a, b x Correct substitution of correct limits in candidate s integrated expression of form ax + bx + c ln x, where c and at least one of a, b Volume = 65( ) (c.a.o.)

22 5. + x / = x + x x x 4 x < 3 or 3 < x < 3 6 / + (f.t. candidate s coefficients) (c.a.o.) 5 6. (a) candidate s x-derivative = t candidate s y-derivative = (at least one term correct) and use of dy = candidate s y-derivative dx candidate s x-derivative dy = (o.e.) (c.a.o.) dx t Use of grad normal grad tangent = m Equation of normal at P: y p = p(x p ) (f.t. candidate s expression for dy) m dx y + px = p 3 + p (convincing) (c.a.o.) (b) (i) Substituting x = 9, y = 6 in equation of normal p 3 7p 6 = (convincing) (ii) A correct method for solving p 3 7p 6 = p = p = P is either (, ) or (4, 4) (c.a.o.)

23 7. (a) u = x du = dx (o.e.) dv = e x dx v = e x (o.e.) x e x dx = x e x e x dx x e x dx = x e x e x + c (c.a.o.) 4 (b) dx = k du (k = / 3 or 3) x( + 3 ln x) u a du = a ln u u e 4 e dx = k [ ln u ] or k [ln ( + 3 ln x)] x( + 3 ln x) e dx = 46 (c.a.o.) x( + 3 ln x) 8. (a) dv = kv 3 (where k > ) dt (b) dv = k dt (o.e.) V 3 V = kt + c c = (c.a.o.) 7 V = 36 = 36 (convincing) 7kt + at + where a = 7k (c) Substituting t = and V = 5 in expression for V a = Substituting V = 7 in expression for V with candidate s value for a t = 7 9 (c.a.o)

24 9. (a) An attempt to evaluate a.b Correct evaluation of a.b and a.b a and b not perpendicular (b) (i) AB = i + j + k (ii) Use of a + AB, a + (b a), b + AB or b + (b a) to find vector equation of AB r = 4i + j 6k + (i + j + k) (o.e.) (f.t. if candidate uses his/her expression for AB) (c) 4 + = + = = p + 3 (o.e.) (comparing coefficients, at least one equation correct) (at least two equations correct) Solving the first two of the equations simultaneously m (f.t. for all 3 marks if candidate uses his/her expression for AB) =, = 3 (o.e.) (c.a.o.) p = 7 from third equation (f.t. candidates derived values for and ). a = 5b (5k) = 5b b = 5k 5 is a factor of b and hence 5 is a factor of b a and b have a common factor, which is a contradiction to the original assumption

25 FP Ques Solution Mark Notes n n 3 S r r (a) (b) 3(a) (b) n r r n ( n ) n( n ) = 4 n( n ) = n n 4 n( n )( n )( n ) = 4 ( i) 4i 4i = 3 + 4i ( 3 4i)( i) z = ( i)( i) 6 8i 3i 4i = 5 i = (.4 +.i) cao 5 r 5 (.4) tan ( 5.5).39 ( 79.6º) or tan (5.5).39 (79.6º) =.75 (.3º), 3 3 = 3 ( ) 3( ) 3 ( / ) 3 ( / ) = = 8 Consider The required equation is x x (8x x 8 ) cao 8 Award for 3 reasonable terms. numerator, denominator FT arithmetic slip from line FT on line above for r. FT on line above for this FT only if in nd or 3 rd quad

26 4(a)(i) (ii) (b) 3 9 Cofactor matrix = Adjugate matrix = 9 3 Determinant = 3(7 ) +4(6 7) +(5 4) = Inverse matrix = 9 3 x y z = Award if at least 5 cofactors are correct. No FT on cofactor matrix. FT the adjugate FT their inverse matrix. 5(a) (b) By reduction to echelon form, 3 x 5 y z k 6 It follows now that k 6 = k = 4 Put z = α. Then y = 5α And x = 7α 6 Putting n =, the expression gives 3 which is divisible by 3 so the result is true for n = Assume that the formula is true for n = k. ( k 3 k is divisible by 3 or 3 k k 3N ) ). Consider (for n = k + ) 3 ( k ) ( k ) 3 k 3k 3k k = 3N k 3k 3k k = 3( N k k ) (This is divisible by 3), therefore true for n = k true for n = k + and since true for n =, the result is proved by induction. Award this only if it is clearly stated that this is an assumption Do not award the second if this is stated as an assumption but the three s may be awarded if either of the s is awarded

27 7(a) (b) Ref matrix in y = x = Translation matrix = Ref matrix in x-axis = T = = or = y x y x y = x, x = y, y x cao

28 8(a) Taking logs, ln f ( x) xln x Differentiating, f ( x) ln x f ( x) x f ( x) x (ln x ) for LHS, for RHS (b) At a stationary point, f ( x) ln x / e x ; y (.368,.69) e e (c) Differentiating the expression in (a), x f ( x) x (ln x )(ln x ) x x x x = x x ( ln x). f ( / e).88 Since this is positive it is a minimum. x each term Accept Since the first term is positive and the second term zero, it is a minimum FT the final on the line above 9(a) (b)(i) (ii) x iy u iv u iv = u v u x u v v y u v We are given that v mu u v u v v mu u v u v mu v (This is the equation of a circle). Completing the square or quoting the standard results, FT on their circle equation (iii) Radius = m Centre m, v Accept y

29 FP Ques Solution Mark Notes Putting x =, 4a 8 = 8 b The two derivatives are ax and 3x b Putting x =, 4a = b Solving, a =, b = 4 cao x x u e du e dx, [,] [, e] e du / u I u 4 u / e du = u 4 u = tan ( ) =.36 e 3 Put t tan(x/ ) 3 t t t t ( t 5) t = giving x/ = n x = n (36nº) t 5 giving x/.56.. n x =.3 + n (36nº + 3º) t 5 giving x/.56.. n x =.3 + n (36nº 3º)

30 4(a) Let 3x 4x ( x )( x ) A Bx C x x (b) A( x ) ( Bx C)( x ) ( x )( x ) x = gives A = Coeff of x gives A + B = 3, B = Const term gives A C =, C = 4 3 f ( x)dx 4 3 dx x 4 3 x dx x 4 = ln( x ) 4 3 ln( x ) 3 = ln ln + ln7 ln 34 7 = ln or ln 5 5(a) (b) Consider f ( x) ( x) sin( x) = x sin x f ( x) f is therefore odd. n sin x is odd and x is even if n is even and odd if n is odd. si So g is even if n is odd and g is odd when n is even. Accept a specific value for x. For a valid attempt. Accept a specific value for x.

31 6(a) Putting x = gives (, /3) Putting y =, 6 x x 3 ( x 3)(6 x) x 9x giving (4,) ; (5,) cao (b) Differentiating, ( x 3) ( x 3) (c) x 3 (4.4), y x 3 (.59), y 3 The asymptotes are x = 3 y = x 6 3(.7) )( 5.83) Award A for the x values only. y (d) x G General shape of both branches. G Correct shape including asymptotic behaviour.

32 7(a)(i) (ii) (iii) Completing the square, ( y ) 8x 4 The vertex is therefore (3,) In the usual notation, a = si The focus is (5,) The equation of the directrix is x = FT on arithmetic slip (b)(i) (ii) Substituting y = mx, m x mx 8x 5 For coincident roots, ( m 8) m 3m m Solving using a valid method, m =, /3 8(a) The result is true for n = since it gives (cos isin ) cos isin Let the result be true for n = k, ie k (cos isin ) cos k isin k Consider k k (cos isin ) (cos isin ) (cos isin ) = (cos k isin k )(cos isin ) cos k cos sin k sin i(sin k cos cos k sin ) = cos( k ) isin( k ) True for n = k true for n = k + and since true for n = the result is proved by induction. (b)(i) Consider w(cos /3 isin /3 3 w (cos isin ) = z = z Showing that w (cos /3 isin /3) is a cube root of z. 3 (ii) The real cube root of 8 is. The other cube roots are (cos /3 isin/3) 3i (cos 4 /3 isin4/3) 3i

33 FP3 Ques Solution Mark Notes x sinh xdx xcosh x cosh xdx = cosh sinh x = cosh sinh = = e e e e e Do not accept an argument which evaluates this as and shows that this is also the numerical value of /e. (a) (b) The equation can be rewritten as sinh x sinh x k The condition for no real roots is 4( k ) 4k 3 3 k 4 sinh x sinh x (sinh x )(sinh x ) sinh x x sinh ln( 5) m 3 Let f ( x) tan x p f () 4 f ( x) ; q f () x f x f () f ( x) ; r ( x ) 4 ( x ) ( x ).4x x) ( x ) ( 4 ; s f () 6

34 4(a) Consider y rsin = sin cos sin d y cos sin sin cos d = cos sin The tangent is parallel to the initial line where cos sin tan.554, r.8 Accept 3.7 (b) The curves intersect where cos sin sin cos sin EITHER Putting t tan( / ) OR ( t ) 4t t t 3t 4t 4 8 tan( / ) (.55..) 6.44, r.4 Putting cos sin r cos( ) = /4 r = cos / 4.44, r.4 Accept 4.3 Accept 4.3

35 5 dt Putting t tan( x / ) gives dx t (, / ) (,) = = dt /( t ) I = 4( t )/( t ) 3 dt = 7 t 7 t t ln or tanh ln 7 or 7 7 t tanh tanh 7 7 =.3 7 ln() 7 7 () 7 6(a) I n / n / n sin n sind n / n = n n sind / n = n cos n( n ) In n = n( n ) In. (b)(i) / / sin I cosd I4 I 4 4 I =.479 (b)(ii) / 5 sind / 5 / cos 5 = 5I cosd FT their answer from (b)(i)

36 7(a) (b) The Newton-Raphson iteration is ( xn tanh xn) xn xn ( sech x ) x x x xn xnsech xn xn tanh xn = sech xn xn sinh xn cosh xn = cosh xn sinh xn xn = cosh x n n Rounding to three decimal places gives.95 Let f(x) = x tanhx 4 f(.955) = 4. 4 f(.945) = 4. The change of sign shows =.95 correct 3dp The values are required 8(a) The curve cuts the x-axis where x cosh = dy sinh x dx y d sinh x cosh dx x Seen or implied Arc length = dy dx = cosh xdx = sinh x dx = 3 (3.46) cao (b) Curved surface area = y dy dx dx = ( cosh x)cosh xdx = 4 cosh x dx cosh x = 4 sinh x sinh x x = cosh =3.5 A Minus each error

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38 GCE MARKING SCHEME MATHEMATICS - -M3 & S-S3 AS/Advanced SUMMER WJEC CBAC Ltd.

39 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. WJEC CBAC Ltd.

40 Q Solution Mark Notes (a). T 8 5g NL dim correct equation attempted T, 5g opposing T 5g = 5 a Any form correct equ. T = 5( ) T = 9 (N) cao (b) R 8 Mg NL attempted R, Mg opposing, no extra forces R Mg = Ma Any form correct equ. 696 = M( ) M = 6 (kg) cao

41 Q Solution Mark Notes (a). R F 3g Resolve vertically R = 3g May be implied 6 F = R = F = 3.6 (N) NL F = ma used ±3.6 = 3a a = -. (ms - ) needs to see - (b) Using v = u + as with u=9, v=,a=(-). allow sign errors, oe = 9 + (-.) s s = (m) allow

42 Q Solution Mark Notes A B 6 kg kg v A v B 3(a) Conservation of momentum dim correct equation (-3) = 6v A + v B v B = v A m used 4 6 = 6v A + v A 36 = v A v A = 3.6 v B = 7. (ms - ) 3(b) Restitution equation attempted, ft c s vs, e on correct side. No more than one sign error = -e(-3 7) 3.6 = e e =.36 cao 3(c) I = 7. (-3) allow 6(7-3.6) I = I =.4 (Ns) cao

43 Q Solution Mark Notes 4. 4g T T 3g Mg Apply NL to B dim correct equation Mg T = Ma Apply NL to A dim correct equation T 3g = 3a Adding m correct method. dep on both M s Mg 3g =.4g(M + 3) M 3 =.4M +..6M = 4. M = 7 cao T = T = 4.6 (N) cao Alternative solution Apply NL to A dim. correct equation T 3g = 3a T = 3( ) T = 4.6 (N) cao Apply NL to B dim correct equation Mg T = Ma 9.8M.4 9.8M = 4.6 m 5.88M = 4.6 M = 7 cao

44 Q Solution Mark Notes 5. R F 39g 5(a) Resolve perp to plane allow sin or cos R = 39gcos R = = 35.8 N 3 F = R F = m F = 5.84 N si NL down slope dim correct equation, -F 39gsin - F = 39a a a =.554 a =.6 (ms - ) 5(b) NL up slope dim correct equation, all forces, sin/cos, -F T 39gsin - F = 39a T = T = (N) cao

45 Q Solution Mark Notes 6. T N 3 N Resolve vertically Tsin = 4g Resolve horizontally Tcos = 3 4g Dividing tan = 3 m dep on both M s = 5.5(7) cao T = (4 9.8) + (3) m T = (N) cao

46 Q Solution Mark Notes 7(a) Using v = u + at with u=, a=(±)9.8, t=5 v = v = 49 (ms - ) accept -49 7(b) v ms 49 4 O 5 5 t s units, labels and correct shape starting (,) (, ) to (5, v) (5, v) to (5, 4) (5, 4) to (, 4) 7(c) Distance = Area under graph oe Distance = (4 + 49) any one correct area, ft graph Distance = Distance = 87.5 (m) ft graph

47 Q Solution Mark Notes 8. R A x 4 x B C 5g g 8(a) Resolve vertically R = 5g + g R = 7g (N) 8(b) Moments about C dim correct equation, no extra forces 5gx = g(.4 x) rhs correct lhs correct 5x =.8 x 7x =.8 x =.4 cao AC =.4 (m) Alternative solution Moments about A dim correct equation 7gx = g.4 rhs correct lhs correct x =.4 (m) cao SC No marks at all, one correct moment, sc.

48 Q Solution Mark Notes 9. G F D (i) (iii) 6 A E (ii) 8 C B 9(a) Area from AG from AB (i) 4 6 correct distances (ii) 5 correct distances (iii) correct distances Lamina 54 x y areas all correct Moments about AG 54x = ft table if or more B marks for distances gained. 9 x = = 3. 9 cao Moments about AB 54y = ft table 38 y = = 4. 9 cao 9(b) x = tan y correct triangle 9 = tan correct equation, ft x, y =.5 ft x and y

49 M Q Solution Mark Notes. s = 6 4cos t dt limits not required s = [sint] correct integration s = sin - 3 s = 3 =.73 cao (a) NL T = 7.5g 45x Hooke s Law T = 5 / 3 (= 47x) = 47x x =.5 cao (b) x Elastic Energy = l used 45 5 EE = 5/ 3 EE = (J) ft c s x value dr 3(a). v = dt v = ( + 4t)i + (3t )j used we required v.(-i + j) = -( + 4t) + (3t ) = m - -4t + 6t -4 = t = 5 t =.5 cao 3(b) a = dv dt used a = 4i + 3j ft c s v provided constant a is independent of t and constant. a = 4 3 = 5 ft constant a=xi+yj

50 Q Solution Mark Notes 4. R 6 N g P 75 4(a) T = = v 5 T = 3 N NL up plane dim correct, all forces T gsin - 6 = a A - each error a = a =. (ms - ) cao 4(b) 9 T = v NL up plane dim correct, all forces T gsin - 6 = a a = m si 9 = 776 v v = 5.7 (ms - ) cao 5. KE at A =.5. v PE at A = PE at B = both or difference WD against resistance = 6. Work-energy principle all terms included.5 v = correct equation v = 6.64 v =.7 (ms - ) cao

51 Q Solution Mark Notes 6(a). u H = Vcos (=.8V) attempt to resolve u V = Vsin (=.6V ) both answers correct 6(b) Consider horizontal motion.8v T = VT = 5 correctly obtained 6(c) Consider vertical motion s = ut +.5at with s=(±)5.4, u=.6v, t=t a=(±) =.6VT 4.9T -5.4 = T 4.9T = 4.4 T = 7 cao V = 5 7 V = 8.75 cao 6(d) Using v = u + at with u=5.5, a=(±)9.8, t= 7 v = v = -.55 si, cao u H = = 7 Speed =.55 7 Speed = 3.5 (ms - )

52 Q Solution Mark Notes 7. T 3v r 3g 7(a) Resolve vertically Tcos = mg 39 8 = cos - 88 = 7.5 cao 7(b) NL towards centre attempted Tsin = ma a = r m used T sin r = m length of string = l lsin = r m r l = sin T 88 l = = m 3 8 l = 3.75 (m) cao Alternative Solution Nl towards centre attempted Tsin = ma a = r m used 88.sin = 3 r.8 r = m r AP = sin m AP = 3.75 (m) cao

53 Q Solution Mark Notes 8(a) v = 3 [(4i 5j) (8i + 7j)] v = 3 (6i - j) v = (i - 4j) 8(b) r S = (8i + 7j) + (i - 4j)t r S = (8 + t)i + (7 4t)j 8(c) r B = (xi + yj)(t ) r B = x(t )i + y(t-)j At t = 5 r S = r B 8 + t = x(t ) m 4x = 8 x =.7 cao = 4y y = cao Alternative solution At t = 5 r S = 8i 93j r B = 4xi + 4yj r S = r B si 4x = 8 m x =.7 cao 4y = -93 y = cao

54 Q Solution Mark Notes 9(a). Conservation of energy mu = mv + mgl( - cos) At max height, v=, cos= 3, l=. u = 9.8.( - ) 3 u = m u =.8 (ms - ) cao v v = u gl( - cos) = (- cos) v = 3.5cos cao 9(b) NL towards centre T - mgcos = mv /l 3 T = 3 9.8cos + (3.5cos ) m T = 9.4cos cos T = 88.cos cao 9(c) Greatest value of T when cos = T = T = 49 (N) Least value of T when cos = 3 T = T = 9.6 (N)

55 M3 Q Solution Mark Notes (a) NL 45 t 3 7 t 3 = dv dt = 6a +/-, no additional terms m use of dv/dt 45 v = - (+ C) t 3 k/(t+3) completely correct When t =, v = m use of initial conditions C = 5 45 v = 5 - t 3 As t, v 5 ft similar expression dx 45 (b) v = = 5 - dt t 3 x = 5t 45 ln(t + 3) (+ C) ft similar expressions t =, x = C = 45 ln3 ft 3 x = 5t + 45 ln t 3 When t = 6 3 x = ln 9 m x = 9 45 ln(3) x = 4.56 (m) cao

56 Q Solution Mark Notes (a). Using v = (a x ) used.9 3 = (a.6 ).4 5 = (a.8 ).7 =.8 m =.5. =.5(a.64) a =. Period = Period = 4 used (b) x = - x used x =.5.6 x =.5 (ms - ) (c) x =.sin(.5t) used, accept cos At A,.6 =.sin(.5t) t = sin - (.5) =.47 or.944 At B,.8 =.sin(.5t) t = sin - (.667) =.4595 or.68 Required t = Required t =.4 (s) cao (d) x = asin(t) x =.sin(.5t) x =.sin(.5 /3) x =.39 (m) (e) v = acos(t) oe v =..5cos(.5t) v =.6cos(.5t) When t = /3, v =.6cos(.5/3) v =.6cos(/3) v =.3 (ms - ) cao

57 Q Solution Mark Notes 3. Auxiliary equation m + 5m + = cao (m + )(m + ) = m = -.5, - cao CF is x = Ae -.5t + Be -t ft solutions for m For PI, try x = at + b dx dt = a 5a + (at + b) = 6t + 5 Comparing coefficients m a = 6 a = b = 5 b = -5 both answers cao General solution is x = Ae -.5t + Be -t + 3t -5 ft CF and PI When t =, x = 3 use of conditions in GS 3 = A + B -5 A + B = 8 dx = -.5Ae -.5t Be -t + 3 ft similar expressions dt dx When t =, = dt = -.5A B +3.5A + B = A + 4B = A + B = 8 3B = -6 B = - cao A = cao

58 Q Solution Mark Notes 4(a) NL F = ma used, no extra term 4 dv 5v x dx m use of vdv/dx 8 dx x v dv separating variables 4 ln x v C kln(x+) all correct v = 8ln x + C When x = 3, v = 4 m 6 = 8ln7 + C C = 6 8 ln7 ft kln(x+) + C x v 8ln 6 7 When x = v 8ln 6 7 v = 8ln3 + 6 v = 4.98 (ms - ) cao 4(b) x v = 6, 6 8ln 6 7 allow similar expressions x ln 7 8 x + = 7e 5/ m correct inversion x =.5[7e 5/ - ] x = 4. (m) cao

59 Q Solution Mark Notes 5. Using v = u + at with u=, a=(±)9.8, t=.5 v = v = 4.9 ms - Impulse = Change in momentum used For A J = 5v For B J = 4.9 v Solving 5v = 9.8 v m 7v = 9.8 v =.4 (ms - ) cao J = 5.4 J = 7 (Ns) cao

60 Q Solution Mark Notes 6.(a) T B R F A N F = R = 4 3 R Moments about B dim correct equation, 3 terms, perp distance R cos + F sin = cos A3 - each error 3 5 R + R = 3 5 4R + R = 5 R = 8 (N) cao 6(b) Resolve vertically Tsin = R Tsin = 3 Resolve horizontally Tcos = F Tcos = Tcos = 6 T = 3 6 m oe T = 43 (N) cao 3 = tan - 6 m oe = 65. cao

61 S Ques Solution Mark Notes (a)(i) (ii) (b) P(AB) = P(A) + P(B) =.8 P(AB) = P(A)P(B) =.5.3 P(AB) = P(A) + P(B) P(AB) = =.65 P(AB) = P(A) + P(B) P(AB) =. Award for using formula Award for using formula P( A B) P(B A) = P( A) Award for using formula =. (a) E( X ) Var( X ) [ E( X )] Award for using formula (b) = 66 E(Y) 3E(X) + 4 = 8 Award for using formula 3(a) Var(Y) = 3 Var( X ) = P(no white) = or Award for using formula (b) = P( white) = 3 or = M if 3 omitted. (c) EITHER P( blue) = 3 or = 4 3 P( the same) = 4 9 = cao 4 M if 3 omitted OR P( the same) = = cao 4 M if 3 omitted Accept

62 4(a)(i) (ii) (b) 5(a) 4 6 P(X = 4) = =.6 Let Y denote the number of games won by Dave so that Y is B(,.5). si We require P(Y 4) =.99 The number of games lasting less than hr, G, is B(45,.8) Poi(3.6). si P(G > 6) =.733 P(CB) = 6 = m Accept or Award A for use of adjacent row or column. FT their mean Use of Law of Total Prob (Accept tree diagram) (b) P(F CB) = /.6 =. cao FT denominator from (a) num, denom 6(a) (b) (c) (d) ' and / 6 Prob = 5/36 6 = Award if only 3 rd term given. FT their answer to (a) 7(a)(i) (ii) e P(X = ) =! =.5 P(X >) =.347 =.658 Working must be shown. Accept or Award for adjacent row/col (b) Using tables, we see that P(X 8) =.966 He needs to take 8 jars. Award A for 7 or 9

63 8(a) (b) (c)(i) (ii).3 E ( X ) (.3 ) 3 4(.7 ) = 3.4 E(X) is therefore independent of E( X ) 4(.3 ) 9 6(.7 ). =.4 Var(X) = = =.8 =. cao Possibilities are 3,3; 4, si P(Sum = 6) = =.8 Accept use of <. Use of xp x with. Need not be seen Must include FT their E(X) if possible Award A if is missing in st term or present in nd term FT their value of if sensible 9(a)(i) (ii) (b)(i) (ii) (iii) E(X) = x(x 3x E( X = x 3 = x 4 ) dx ) x (x 3x ) dx = x x 4 5 =.6 Var(X) =.6.59 =.8 x F( x) (t 3t )dt t t 3 = x 3 = ( x x ) cao P(X.4) = F(.4) =.7 The lower quartile is less than.4 since F(.4) is more than.5. for the integral of xf(x), for completely correct although limits may be left until nd line. For evaluating the integral Integral and limits Correct evaluation of integral FT their E(X) Limits may be left until nd line FT their F(x) if possible FT their answer to (a)(ii)

64 S Ques Solution Mark Notes (a) E( X ) Var( X ) [ E( X )] Award for using formula = 7 Similarly, E ( Y ) 39 (b) E(U) = E(X)E(Y) = 3 E ( X Y ) E( X ) E( Y ) 739 Var(U) = E( X Y ) [ E( XY )] = = 53 FT their E ( X ), E( Y ) but not their E(X),E(Y) Award for using formula (a)(i) (ii) (b)(i) (ii) (iii) z.5. P(X > 4.5) = th percentile =. 645 = 4.73 E(Y X) =.8 Var(Y X) = 4Var(Y) + Var(X) =.3.8 z =. (Accept ).3 We require P(Y X < ) Prob =.3 Let total weight = S E(S) = = 6.6 Var(S) = = z = Prob =.946 m Award only for z FT their values from (b)(i) 3(a) (b) 69.9 x (=.93) 75. SE of X (=.547 ) 75 9% conf limits are giving [.93,.95] If the method for finding the confidence interval is repeated a large number of times, then 9% of the intervals obtained will contain μ (or equivalent) correct form, correct z. SE must have 75 in denom for. Award B for any solution which suggests that the calculated interval contains μ with a probability of.9

65 4(a) (b)(i) (ii) 5(a) (b) The total number of errors, X, is Poi(8) P(X < 5) =.94 =.996 H :.8; H :.8 Under H, number of errors is Poi(64) N(64,64) z = 8 =.4375 p-value =.33 Insufficient evidence to reject H /Accept H H ; H : : D F x D ( 48.4); xf ( 46.5) si SE of difference of means= (.866..) Test statistic = =.9 Prob from tables =.46 p-value =.85 Strong evidence that there is a difference in mean distances for the two players. OR Strong evidence that David s mean is larger than Frank s mean. D F Award A for use of adjacent row/column Award A for incorrect or no continuity correction No c/c gives z =.5, p =.3 Incorrect c/c gives z =.565, p =.9 FT their p-value FT arithmetic slip in evaluating means FT from previous line FT on their p-value

66 6(a) A B D C (b)(i) Drop a perpendicular from A to BC. X = BD = ABcos = 4cos The probability density function of is f ( ) (for / ) E(X) = / 4cosd 8 / sin = = 8/ cao si Accept any valid method Must be convincing Limits not required, award for K 4cos d, K Limits required here (ii) 7(a) (b) P(X 3) = P(cos.75) = P(.73) /.73 = / =.54 Let X denote the number of white flowers produced. If bag is Type B, X is B(,.7) N(84,5.) P(label A) = P(X < 7) z = 5. =.89 (Accept ) Prob =.93 If bag is of Type A, X is B(,.5) N(6,3) P(label B) = P(X 7) z = 3 =.73 (Accept ) Prob =.48 An answer of.46 is given AA Award A for incorrect or no c/c. 7.5 z =.69, p = z =.79, p = z =.99, p = z = 3.9, p =. Award A for incorrect or no c/c. 7.5 z =.9, p = z =.83, p = z =.64, p = z =.55, p =.657

67 S3 Ques Solution Mark Notes (a) The possibilities are Numbers drawn Mean Median 3 4 7/3 5 8/ / / / / The sampling distribution of the mean is x 7/3 8/3 3 /3 /3 4 Prob / / / / / / / each column Special case B if one combination is missing. No FT from earlier work. (b) The sampling distribution of the median is Median 3 4 Prob 3/ 4/ 3/ (a) UE of = 99.3 x UE of 9 9 =.8 (.778 ) No working need be seen Answer only no marks (b)(i) H : ; H : (ii) 99.3 t.778../ =.3 DF = 9 si Critical value =.6 M if treated as z Insufficient evidence to reject the manager s claim or Accept the manager s claim Because.3 <.6 or equivalent using the term acceptance region or by means of a diagram FT their critical value but not their p-value obtained from using the normal distribution

68 3(a)(i) p ˆ.45 (ii) ESE = = si 95% confidence limits are giving [.36,.539] (b)(i) This time, p ˆ =.58 (ii) Width of CI =.645 n = =.6 Solving, 3.9 n.6 = Attempting to solve for n (iii) x = 7.58 = 88 FT their n 4(a) H : A B : H : A B (b) x 5.3; y s x s y [Accept division by 5 giving.49 and.4] SE = 5 5 =.48.. (.44..) Test stat =.48.. =.6 (.8) p-value =.39 (.38) Strong evidence for believing there is a difference in mean distances travelled (or that the Model A mean is less than the Model B mean). FT their p-value

69 5(a) x 5, x 55, y 345.5, xy 3. B Minus each error. S xy / FT I error in sums. S xx 55 5 /6 7.5 a = b = = 9.4 (accept 9.3) (b).75 SE of b = (.79 ) % confidence limits for β are giving [4.8,5.7] FT their values from (a) 6(a) E(X) = a x x dx a 3 x = 3a a = 3 a Limits not required in this line x E( X ) x dx a = x 4a a = 4 a Limits not required in this line a 4a Var(X) = 9 a = 8

70 (b)(i) E(U) = c E( X ) (or ce(x)) = E ( U) a c 3 9 Var(U) = Var( X ) 4 9 a = 4 8n a = 8n a c 3 Penalise the omission of E once in the question (ii) na E(V) = d E( Y) = d n n E( V ) a d n n Var(V) = Var( Y) n n na = n ( n )(n ) a = 4n( n ) (iii) Var( U) a a Var( V ) 8n 4n( n ) n = V is the better estimator Because (for n > ) it has the smaller variance

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