MATHEMATICS - C1-C4 & FP1-FP3

Transcription

1 GCE MARKING SCHEME MATHEMATICS - C-C4 & FP-FP AS/Advanced SUMMER 5

2 INTRODUCTION The marking schemes which follow were those used by WJEC for the Summer 5 eamination in GCE MATHEMATICS - C-C4 & FP-FP. They were finalised after detailed discussion at eaminers' conferences by all the eaminers 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 eaminers. 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 eaminers' 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. Page C C 6 C C4 8 FP FP 8 FP

3 C. (a) (i) Gradient of AB = increase in y increase in Gradient of AB = (or equivalent) (ii) A correct method for finding the equation of AB using candidate s gradient for AB Equation of AB : y = [ ( 7)] (or equivalent) (f.t. candidate s gradient of AB) Equation of AB : + y = (convincing) (iii) Use of m L m AB = A correct method for finding the equation of L using candidate s gradient for L () (to be awarded only if corresponding is not awarded in part (ii)) Equation of L : y 5 = [ ( )] (or equivalent) (f.t. candidate s gradient of AB) Note: Total mark for part (a) is 7 marks (b) An attempt to solve equations of AB and L simultaneously = 4, y = (convincing) (c.a.o.) (c) A correct method for finding at least one coordinate of the mid-point of AB y-coordinate of the mid-point of AB = 5 (or -coordinate = 5) D is not the mid-point of AB or L is not the perpendicular bisector of AB or the mid-point does not lie on L Alternative Mark Scheme A correct method for finding the lengths of two of AB, AD, BD Two of AB = 9, AD =, BD = 4 D is not the mid-point of AB or L is not the perpendicular bisector of AB or the mid-point does not lie on L

4 (d) A correct method for finding the length of BD(CD) BD = 4 (or equivalent) CD = Substitution of candidate s derived values in tan ABC = CD m BD tan ABC = (c.a.o.) Special Case A candidate who has been awarded M A A m A may be awarded SC for one of AB = 9, AC =, BC = 5. (a) 4 = (4 )( ) + ( + )( ) Numerator: 4 + Denominator: 8 4 = 5 (c.a.o.) + Special case If not gained, allow SC for correctly simplified numerator or denominator following multiplication of top and bottom by + (b) 7 = p4, where p is a fraction equivalent to / = q4, where q is a fraction equivalent to 7 / = 4 (c.a.o.) 4

5 . (a) y-coordinate of P = 4 dy = d (an attempt to differentiate, at least one non-zero term correct) An attempt to substitute = in candidate s epression for dy m d Value of dy at P = 5 (c.a.o.) d Gradient of normal = m candidate s value for dy d Equation of normal to C at P: y ( 4) = / 5 ( ) (or equivalent) (f.t. candidate s value for dy and the candidate s derived y-value at d = provided and both m s awarded) (b) Putting candidate s epression for dy = 8 d An attempt to collect terms, form and solve quadratic equation in a (or ) either by correct use of the quadratic formula or by getting the equation into the form (ma + n)(pa+ q) =, with m p = candidate s coefficient of a and n q = candidate s constant m a a 5 = a = or 5 (both values) (c.a.o.) 4. (a) 4( ) 5 (b) 4( ) = 5 (f.t. candidate s values for a, b, c) ( ) = (±) 5 (f.t. candidate s values for a, b, c) m =, 9 (both values) 5. (a) An epression for b 4ac, with at least two of a, b or c correct b 4ac = (k 5) 4 k (k 6) Putting b 4ac < m k < 5 (or equivalent) 4 (b) k = 5 [f.t. the end point(s) of the candidate s range in (a)] 4

6 6. (a) 8 = ( for further incorrect simplification) (b) First term = n n = n = 5 Second term = n n a a = (f.t. candidate s value for n) 7. (a) y + y = 9( + ) 8( + ) Subtracting y from above to find y y = 8 + 9() 8 Dividing by and letting dy = limit y = 8 8 (c.a.o.) d (b) dy = ( 6) / d 8. (a) Use of f () = 7p = p = 6 (convincing) Special case Candidates who assume p = 6 and show f () = are awarded (b) f () = ( )(6 + a + b) with one of a, b correct f () = ( )( ) f () = ( )( )( + 4) (f.t. only in above line) Roots are =,, 4 (f.t. for factors, 4) Special case Candidates who find one of the remaining factors, ( ) or ( + 4), using e.g. factor theorem, are awarded 4

7 9. (a) y (, ) O (4, ) ( 4, ) Concave up curve and y-coordinate of minimum = -coordinate of minimum = 4 Both points of intersection with -ais (b) Either: Any graph of the form y = af () (with a ) will intersect the -ais at ( 6, ) and (, ) and thus not pass through the origin. Or: f () and since a, af (). Thus any graph of the form y = af () will not pass through the origin. E. (a) L = + y 8 = y (both equations) L = + 6 (convincing) (b) dl = + 6 ( ) d Putting derived dl= d = 4, ( 4) (f.t. candidate s dl) d Stationary value of L at = 4 is 8 (c.a.o) A correct method for finding nature of the stationary point yielding a minimum value (for > ) 5

8 C (5 values correct) B (If B not awarded, award for either or 4 values correct) Correct formula with h = 5 I 5 { ( )} I I I 4689 (f.t. one slip) Special case for candidates who put h = (all values correct) Correct formula with h = 4 I 4 { ( )} I I I 4688 (f.t. one slip) Note: Answer only with no working shown earns marks 6

9 . (a) 4( sin ) sin sin + 8 =, (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 epression into the form (a sin + b)(c sin + d), with a c = candidate s coefficient of sin and b d = candidate s constant m 6 sin + sin = ( sin + )( sin 4) = sin =, sin = 4 (c.a.o.) sin no such can eist (f.t. only if candidate has real values for sin, neither of which satisfies sin ) E (b) 75 =,, 9, (one value) =, 4 Note: Subtract (from final two marks) mark for each additional root in range, ignore roots outside range. (c) 4 sin + 7 sin cos = or 4 tan + 7 tan cos = or sin = cos sin = (or tan = ), cos = 4 (both values) 7 =, 8 (both values) = 4 85 (c.a.o.) Note: Subtract a maimum of mark for each additional root in range for each branch, ignore roots outside range. Special Case: No factorisation but division throughout by sin (or tan ) to yield cos = (or equivalent) = (a) sin ACB = sin 5 9 (substituting the correct values in the correct places in the sin rule) ACB = 4, 8 (both values) (b) (i) BÂC = 8 (f.t. either of candidate s values for ACB) BÂC = 7 (f.t. candidate s obtuse value for ACB) (ii) Area of triangle ABC = 9 sin 7 (substituting 9, and candidate s derived value for BÂC in the correct places in the area formula) Area of triangle ABC = cm. (c.a.o.) 7

10 4. (a) (i) nth term = 4 + 6(n ) = 4 + 6n 6 = 6n (convincing) (ii) S n = (6n 8) + (6n ) S n = (6n ) + (6n 8) Reversing and adding Either: S n = (6n + ) + (6n + ) (6n + ) + (6n + ) Or: S n = (6n + ) +... (n times) S n = n(6n + ) S n = n(n + ) (convincing) (b) (i) a + 9d = 4 (a + 4d) a + 7d = 5 (a + 4d) = a + 7d = 4 An attempt to solve the candidate s two derived linear equations simultaneously by eliminating one unknown d =, a = 7 (c.a.o.) (ii) 7 + (k ) = (f.t. candidate s derived values for a and d) k = 7 (c.a.o.) 5. (a) r = 4 = 4 (c.a.o.) 576 t 5 = 576 (f.t. candidate s value for r) 4 t 5 = 9 (c.a.o.) (b) (i) ar = 4 ar + ar + ar = 56 An attempt to solve the candidate s equations simultaneously by eliminating a r = 4 r + r + = (convincing) r + r + r 56 (ii) r = (r = discarded, c.a.o.) a = 6 (f.t. candidate s derived value for r, provided r < ) S = 6 (use of formula for sum to infinity) ( / ) (f.t. candidate s derived values for r and a) S = 6 (f.t. candidate s derived values for r and a) 8

11 6. (a) / 6 7/ + c, / 7/ ( if no constant term present) (b) (i) = 4 An attempt to rewrite and solve quadratic equation in, either by using the quadratic formula or by getting the epression into the form ( + a)( + b), with a b = candidate s constant m ( + )( ) = = (c.a.o.) (ii) Use of integration to find the area under the curve 6 d = 6, 5 d = 5, d = (/), (correct integration) Correct method of substitution of candidate s limits m [6 + (5/) (/) ] = (8 + 45/ 9) ( 6 + 5/ ( /)) = 4/ Use of a correct method to find the area of the triangle (f.t. candidate s coordinates for A) Use of and candidate s value for A as limits and trying to find total area by subtracting area of triangle from area under curve m Shaded area = 4/ 8 = 5/ (c.a.o.) 7. (a) Let p = log a, q = log a y Then = a p, y = a q (the relationship between log and power) = a p = a p q (the laws of indicies) y a q log a /y = p q (the relationship between log and power) log a /y = p q = log a log a y (convincing) (b) log a ( ) log a = log a (subtraction law) 4 log a = log a 4 (power law) = 4 (removing logs) An attempt to solve quadratic equation with three terms in, either by using the quadratic formula or by getting the epression into the form (a + b)(c + d), with a c = candidate s coefficient of and b d = candidate s constant m = ( )( ) = = /, / (both values, c.a.o.) Note: Answer only with no working earns marks 9

12 8. (a) (i) A(, ) (ii) A correct method for finding radius Radius = 9 (convincing) (b) Either: RQ = 8 or RP = 98 (o.e.) Correct substitution of candidate s values in an epression for sin Q, cos Q or tan Q PQR = 66 8 (c.a.o) Or: RQ = 8 or RP = 98 Correct substitution of candidate s values in the cos rule to find cos Q PQR = 66 8 (c.a.o) (c) AT = 65 (f.t. candidate s coordinates for A) Use of ST = AT AS with candidate s derived value for AT ST = 6 (f.t. one slip ) 9. Area of sector AOB = / r 6 Area of triangle AOB = / r sin 6 Area of minor segment = / r 6 / r sin 6 = 4r Use of a valid method for finding the area of the major segment Area of major segment = 99r area of major segment area of minor segment (convincing)

13 C. (a) / / / / (5 values correct) B (If B not awarded, award for either or 4 values correct) Correct formula with h = /9 I /9 { + ( ) + 4[( 6456) + ( 69478)] + ( 66559)} I ( /9) I I 67 (f.t. one slip) Note: Answer only with no working shown earns marks (b) 4/9 ln (sec ) d 67 (f.t. candidate s answer to (a))

14 . (a) 7 cosec 4 (cosec ) = cosec (correct use of cot = cosec ) An attempt to collect terms, form and solve quadratic equation in cosec, either by using the quadratic formula or by getting the epression into the form (a cosec + b)(c cosec + d), with a c = candidate s coefficient of cosec and b d = candidate s constant m cosec 5 cosec = (cosec )( cosec + 4) = cosec =, cosec = 4 sin =, sin = (c.a.o.) 4 = 9 47, 6 5 = 4, 8 59 Note: Subtract mark for each additional root in range for each branch, ignore roots outside range. sin = +,, f.t. for marks, sin =,, f.t. for marks sin = +, +, f.t. for mark (b) sec, cosec and thus 4 sec + cosec 7 E. (a) d( ) = d() = d( /4) = d d d d( cos y) = ( sin y) dy + cos y d d d(y ) = y dy d d dy = (c.a.o.) d (b) d y = d( y) = dy + y d d d Substituting y for dy in candidate s derived epression for d y d d d y = ( y) + y = 4 y + y (o.e.) (c.a.o.) d

15 4. (a) candidate s -derivative = + t candidate s y-derivative = t dy = candidate s y-derivative d candidate s -derivative dy = + t d t (b) d dy = t + (o.e.) dtd Use of d y = d dy candidate s -derivative d dtd d y = ( t + )( + t ) (o.e.) (f.t. one slip) d d y = t = (c.a.o.) d d y = = (f.t. candidate s derived value for t) d 4

16 5. (a) y y = cos y = 5 / O Correct shape for y = cos A straight line with negative y-intercept and positive gradient intersecting once with y = cos in the first quadrant. (b) = 4 = ( correct, at least 4 places after the point) = = = = 46 ( 4 correct to 4 decimal places) Let h() = cos 5 + An attempt to check values or signs of h() at = 465, = 465 h( 465) = >, h( 465) = 86 4 < Change of sign = 46 correct to four decimal places 4

17 6. (a) (i) dy = a + b (including a =, b = ) d 4 5 dy = 8 d 4 5 (ii) dy = e f() (f (), ) d dy = e / d (iii) dy = (a b sin ) m cos (a + b sin ) k cos d (a b sin ) (m = ± b, k = ± b) dy = (a b sin ) b cos (a + b sin ) ( b) cos d (a b sin ) dy = ab cos d (a b sin ) (b) d (cot ) = d (tan ) = ( ) (tan ) f() (f (), ) d d d (tan ) = ( ) (tan ) sec d d (tan ) = cosec (convincing) d 5

18 7. (a) (i) (7 ) d = 7 d d Correctly rewriting as two terms and an attempt to integrate (7 ) d = 7 ln + c (ii) sin( / ) d = k cos ( / ) + c (k =, /, /, / ) sin( / ) d = / cos ( / ) + c Note: The omission of the constant of integration is only penalised once. (b) (5 4) /4 d = k (5 4) /4 (k =, 5, / 5 ) /4 (5 4) /4 d = / 5 (5 4) /4 /4 A correct method for substitution of the correct limits limits in an epression of the form m (5 4) /4 6 (5 4) /4 d = 8 (= 867) 5 (f.t. only for solutions of 8 (= 9 ) and 4 (= ) from k =, k = 5 respectively) Note: Answer only with no working shown earns marks 8. (a) Trying to solve either 5 or / (both inequalities) Required range: 4 / (f.t. one slip) Alternative mark scheme ( 5) (squaring both sides, forming and trying to solve quadratic) Critical values = 4 / and = Required range: 4 / (f.t. one slip in critical values) (b) 4 / /y (f.t. candidate s a b, a >, b > ) / y / 4 (f.t. candidate s a b, a >, b > ) 6

19 9. = 4 y y = f () y = f ( + 4) (, ) O (, ) Correct shape, including the fact that the y-ais is an asymptote for y = f () at y = f () cuts -ais at (, ) Correct shape, including the fact that = 4 is an asymptote for y = f ( + 4) at y = f ( + 4) cuts -ais at (, ) (f.t. candidate s -intercept for f ()) The diagram shows that the graph of y = f () is steeper than the graph of y = f ( + 4) in the first quadrant. (a) Choice of h, k such that h() = k() + c, c Convincing verification of the fact that h () = k () (b) (i) y = ln (4 + 5) An attempt to epress candidate s equation as an eponential equation = (e (y )/ 5) (c.a.o.) 4 f - () = (e ( )/ 5) 4 (f.t. one slip in candidate s epression for ) (ii) D(f - ) = [, 4] (iii) gf () = e ln (4 + 5) + e ln (4 + 5) = (4 + 5) gf () = e (4 + 5) (c.a.o.) 7

20 C4. (a) f () A + B + C (correct form) ( + ) ( + ) ( ) A( ) + B( + )( ) + C( + ) (correct clearing of fractions and genuine attempt to find coefficients) m A = 7, C =, B = (all three coefficients correct) A If A not awarded, award for at least one correct coefficient (b) f () d = 7 + ln ( ) ( + ) (f.t. candidate s values for A, B, C) f () d = 7 + ln ln = 8 6 Note: Answer only with no working earns marks (c.a.o.). (a) 4 + dy + 6y 4ydy = dy + 6y d d d 4 4ydy d dy = 4 + 6y (convincing) d 4y (b) 4y = Either: Substituting for y in the equation of C and an 4 attempt to collect terms m 4 = 6 = () y = (for both values of ) (f.t. 4 = a, a 6, provided both values are checked) Or: Substituting 4y for in the equation of C and an attempt to collect terms m y = 9 y = () y = = (f.t. y = b, b 9) 8

21 . (a) tan + tan 45 = 8 tan tan tan 45 (correct use of formula for tan ( + 45 )) Use of tan 45 = and an attempt to form a quadratic in tan by cross multiplying and collecting terms 8tan 7 tan + = (c.a.o.) Use of a correct method to solve the candidate s derived quadratic in tan m = 4 8, (both values) (f.t. one slip in candidate s derived quadratic in tan provided all three method marks have been awarded) (b) (i) R = 7 Correctly epanding sin ( ), correctly comparing coefficients and using either 7 cos = or 7 sin = 6 or tan = 6 to find (f.t. candidate s value for R) = 59 (c.a.o) (ii) sin ( ) = 4 7 (f.t. candidate s values for R, ) 59 = 4 85, 4 85, 5 5, (at least one value, f.t. candidate s values for R, ) = 4 5, 7 85 (c.a.o.) 4. (a) a V = (m) d (m) d = m V = m a (c.a.o.) (b) (i) Substituting b for m in candidate s derived epression for V a V = b a (c.a.o.) (ii) This is the volume of a cone of (vertical) height a and (base) radius b E 9

22 5. + / = < 8 or 8 < < 8 + (f.t. candidate s coefficients) 6 5 Either: 449 (c.a.o.) 4 Or: 48 (c.a.o.) (a) (i) candidate s -derivative = at candidate s y-derivative = a (at least one term correct) and use of dy = candidate s y-derivative d candidate s -derivative dy = a = d at t Gradient of tangent at P = (c.a.o.) p (ii) Equation of tangent at P: y ap = ( ap ) p (f.t. candidate s epression for dy) m d Equation of tangent at P: py = + ap (b) (i) Gradient PQ = ap aq ap aq Use of ap aq = a(p + q)(p q) Gradient PQ = (c.a.o.) p + q (ii) As the point Q approaches P, PQ becomes a tangent Limit (gradient PQ) = =. E q p p p

23 7. (a) d = k du (k = /, /, or ) ( ) u a du = a u - u Either: Correctly inserting limits of, 4 in candidate s bu or: Correctly inserting limits of, in candidate s b( ) d = (c.a.o.) ( ) 8 Note: Answer only with no working earns marks (b) (i) u = du = d (o.e.) dv = cos d v = sin (o.e.) cos d = sin sin d cos d = sin + cos + c (c.a.o.) 4 (ii) sin d = k m cos d (o.e.) (k =,, m =, ) sin d = d cos d sin d = sin cos + c (f.t. only candidate s answer to (b)(i)) 8. (a) (i) AB = i j + 7k (ii) Use of a + AB, a + (b a), b + AB or b + (b a) to find vector equation of AB r = 5i j k + ( i j + 7k) (o.e.) (f.t. if candidate uses his/her epression for AB) (b) 5 = + = = 4 (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 equation of AB) =, = 4 (o.e.) (c.a.o.) Correct verification that values of and satisfy third equation Position vector of point of intersection is 6i + j 8k

24 (f.t. one slip) 9. (a) dp = kp dt (b) dp = k dt P = kt + c (o.e.) P c = (c.a.o.) A = kt kt = P A = t (convincing) P A A P k PA (c) 8 A =, 9 A = 4 (both equations) k 8A k 9A An attempt to solve these equations simultaneously by eliminating k A = 6 (c.a.o.). Assume that 4 is a factor of a + b. Then there eists an integer c such that a + b = 4c. Similarly, there eists an integer d such that a b = 4d. Adding, we have a = 4c + 4d. Therefore a = c + d, an even number, which contradicts the fact that a is odd.

25 FP Ques Solution Mark Notes h h f h f ) ( ) ( ) ( ) ( = ) )]( ( ) [( )] ( ) [( h h h h = ) )]( ( ) [( )] [ h h h h h = ) )]( ( ) [( h h h h h h f h f h f ) ( ) ( lim ) ( = ) )]( ( ) [( lim h h h h = ) ( oe (a) (b) The reflection matri for y = is The reflection matri for y = is It follows that T = = cao T therefore corresponds to a rotation through 8 about the origin. cao Allow the use of matrices Special case for matrices the wrong way round Do not award this for a matri (a) i) i)( ( i) i)( ( i i = i i i i i = i Let z = + iy so that y z i ( + iy) i( iy) = i ; y y i so 6 7 ; 6 5 z y FT their above result

26 (b) Mod = ( ) ( ) 9 tan.8 or 46.4 si Arg =.8 + =.95 or = 6.4 Accept. or.6 4(a) (b)(i) (ii) det(m) = ( ) ( 4) ( 5) = M is therefore singular. Using row operations, y z It follows that = so = 4 Let z =. Then y =. and = 6. 5 Let the roots be a, ar, ar. Then, a ar ar 4 a r a r a r 8 Dividing, ar = k a r 8 Allow a/r, a, ar 6(a) (b) (c) It follows that A B 4 y 6 8 z 6 B Award if error, B more than error A for B FT their A 4

27 7(a) (b) Let S n A B n( n ) n n A ; B A( n ) Bn n( n ) n( n ) n ( n ) ( n ) n ( n ) ( n ) ( n ) ( n ) ( n )( n ) ( n ) ( n ) ( n )( n ) n 5n ( n )( n ) 8(a) A 4 A I = Hence equal. 4 (b) METHOD Let the result be true for n = k, ie A k = ka (k )I Consider, for n = k +, A k+ = ka (k )A = k(a I) (k )A = (k + )A ki Hence true for n = k true for n = k + and since trivially true for n = (A = A), the result is proved by induction. Award this for a correct concluding statement and correct presentation of proof byinduction 5

28 METHOD Let the result be true for n = k, ie A k = ka (k )I = k Consider, for n =k +, k A k = ( k ) Hence true for n = k true for n = k + and since trivially true for n = (A = A), the result is proved by induction. Award this for a correct concluding statement and correct presentation of proof byinduction METHOD Let the result be true for n = k, ie k Consider, for n =k +, k k ( k ) k ( k ) = k ( k ) But the assumed result for n = k can be written as k k Hence true for n = k true for n = k + and since trivially true for n = (A = A), the result is proved by induction. Award this for a correct concluding statement and correct presentation of proof byinduction 9(a) Taking logs correctly, lnf() = ln + lnsin Differentiating, f ( ) ln cot f ( ) f ( ) sin (ln cot ) (b) Stationary value where f ( ) cot ( ln ) cao =.8 A Condone ignoring sin = Award for.96 6

29 (a)(i) 6 9 y z + = k z i Putting z = + iy, ( ) y k k ( y ) k k y k y k (ii) (b)(i) ( k ) ( k ) y 6 k y k 9 (which is the equation of the circle.) Rewriting the equation in he form 6 k 9 k y y ( k ) ( k ) ( k ) Completing the square, k k y k k Centre =, k k 6 + y + 8 = terms involving k m Award full credit for the use of the standard result for the coordinates of the centre (ii) It is the perpendicular bisector of the line joining the points (,) and (,) 7

30 Ques Solution Mark Notes (a) Let 5 A B C ( )( ) (b) ( A B)( ) C( ( )( ) A ; B ; C 5 ( )( ) ( ) u = tan du = sec d [,/4] [, ] 5 du I ( u) ( u ( u )du u u = ln( u ) tan =.6 cao ) FP ) u ln( u) Award M if no working (a) Denoting the two functional epressions by f, f f ( ) 4, f( ) a b Therefore a + b = 4 f f a ( ), ( ) b f ( ), f () a b Therefore a + b = 9 Solving, a, b (b) Solving 9 ; (a) Modulus of cube roots = R (cos π / 4 isin π / 4) = i R (cosπ / isinπ /).7.6i R (cos9π / isin9π /).6.7i FT one slip in equations FT if possible Award for attempting to solve this equation Use of de Moivre s Theorem FT their modulus Addition of / to argument Penalise accuracy only once 8

31 (b)(i) (ii) z n is real when n = 4 and imaginary when n =. B Award for n = 8 4 METHOD Combining the first and third terms, π π cos cos cos 6 6 π cos (cos ) 6 Either cos =, π n π or (n ) π Or cos 6 π n or n π 6 nπ π n or 4 6 METHOD cos cos sin sin cos cos sin sin cos cos sin sin Combining appropriate terms, cos cos cos cos 6 sin sin cos sin 6 cos cos sin cos Either cos = Or n π, π tan n n 6 or (n ) for combining two terms Accept equivalent answers Accept equivalent answers Accept equivalent answers Accept equivalent answers 9

32 5(a) (b) (c) d d d e d Put e u du e dy y ; d y d du e dy = e u u dy du d Do not accept integration followed by differentiation Do not accept integration followed by differentiation e u d d du e u du e u du u cao e du e e Award this for the difference of integrals 6(a) We are given that ( y ) ( y ) y 6y 9 y y 6y 9 Do not accept solutions which assume the equation given the focus and directri (b)(i) (ii) (iii) (iv) 6t ;y 6t showing that the point (6t,t ) lies on C. dy dy d d dt dt 6t = t 6 The equation of the tangent is y t t( 6t) y t t Substituting (, ) into the equation, = t t = Since the positive gradient of the tangent is equal to, the angle between the tangent and the y-ais is equal to tan (/ ). The angle between the tangents is therefore equal to tan (/ ) = 5. or.97 rad Award for any valid method Accept 6.9 or. rad

33 7(a) =, = (b) f() = giving the point (,) f() = = / giving the point (/,) (c) 4 f ( ) ( ) ( ) At a stationary point, 4 ( ) ( ) ( ) ( ) giving (,) and (4/,9) 8 f ( ) ( ) ( ) f ( ) so that (,) is a maimum f ( 4/) so that (4/,9) is a minimum Award A if only values given Accept any valid method including looking at appropriate values of f() or f () (d) G G Award G for correct branches (e)(i) (ii) f( ) = 5/6 ; f() = f(s) = [5/6,] Solve = 5 f ( S) [/, 5] [ 5, )

34 Ques Solution Mark Notes (a) (b) Epanding the right hand side, 5cosh sinh rcosh cosh rsinh sinh Therefore r cosh 5 and rsinh Squaring and subtracting, r (cosh sinh ) 5 so that r 4 Dividing, sinh tanh cosh 5 tanh.69 5 Substituting, 4cosh(.69) (.69) cosh 4.69 cosh 4 =.6,.874 FP Condone the absence of here EITHER I / e d(sin ) / / = e sin e sin d / = e e d( cos ) = e e cos 4I = e 4I e I 5 /

35 (a)(i) (ii) (b)(i) (ii) OR I / e cosd / / e = cos e sin d / e = sin d / e sin 4 4 = e I 4 4 e / 4 / e π I 5/ 4 5 = I f ( ) 6 6 f (.4) f (.6).8... The change in sign shows that lies between.4 and.6. Since satisfies f ( ), it follows that 6 6 so that Let ( ) F 4 F ( ) F (.5) The sequence converges because F (.5) < Using the iterative formula, successive values are etc =.57 (to dps)

36 4(a) sinh f ( ) cosh cosh ( cosh ) sinh f ( ) ( cosh ) cosh = ( cosh ) = cosh (b) sinh ( ) ( cosh ) f cosh ( cosh ) term inc sinh f ( ) 4 ( cosh ) f () ln, f (), f () f (), f () 4 The Maclaurin series for f() is 4 ln FT their derivatives 5(a) d dy cos t; dt dt sin t (b)(i) d dt dy cos t cos dt = (+cost) = 4cos t Arc length = cos t dt = 4sin t = 4 t sin t Convincing 4

37 (ii) CSA = y d dt dy dt dt = ( cos t) cos t dt = 4 (cos t dt cos t cos t ) dt = 4 sin t 6 = sin t Or = 8 sin 6 sin t cos tdt t 6(a) (b) (c)(i) d (4 d I n ) (4 ) (4 ) ( ) n / d ((4 ) d n d n / n / = (4 ) (4 ) d I n Evaluate n n = (4 ) 4 d n = 4I n I n 4( n ) In n I 4 d Put sin, d cos d,[,] [, / ] Convincing (ii) / I 4 cos d // cos d = sin = π I4 I I = = / 5

38 7(a) (b) Consider r cos = tan cos d sec cos tan sin d The tangent is perpendicular to the initial line where sec cos tan sin sin tan tan cos tan tan tan Putting t tan, t t t t t 4 4t t 5 t 5.95 (5.8) r = t =.486 Area = r / d = tan d / = (sec ) d / = tan = (.5) 4 6

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40 GCE MARKING SCHEME MATHEMATICS - -M & S-S AS/Advanced SUMMER 5

41 INTRODUCTION The marking schemes which follow were those used by WJEC for the Summer 5 eamination in GCE MATHEMATICS -M & S-S. They were finalised after detailed discussion at eaminers' conferences by all the eaminers 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 eaminers. 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 eaminers' 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. Page M M 8 S 5 S 8 S

42 Q Solution Mark Notes. R a Mg NL applied to man R and Mg opposing. dim correct R Mg = Ma 68 = M(9.8 +.) M = 68 cao NL applied to Lift and Man T and weight opposing. dim correct. T 868g = 868a ft M T = 868 (N) ft M

43 Q Solution Mark Notes. Apply NL to B dim correct, all forces. allow 5a RHS 5g T = 5g and T opposing. Resolve perpendicular to plane for A allow sin R = 4gcos Apply NL to A Friction opposes motion. Allow 4a RHS and/or cos T 4gsin - F = At limiting equilibrium F = R used F 45g 5 = convincing R 48g 6 T = 5g = 49 45g 44 F = T 4gsin = = =.9 48g 5 R = 4g = =

44 Q Solution Mark Notes (a) Conservation of momentum attempted, equation, dim correct = v A + 5 v B v A + 5 v B = 4 Restitution v B v A = - ( 8) v B v A = v A + 5 v B = 4 -v A + v B = 6 Adding m dep on both M s 8v B = 4 v B = 5 (ms - ) cao v A = (ms - ) cao (b) Impulse = change of momentum used I = = 5 (Ns) ft v A or v B

45 Q Solution Mark Notes 4 Moments about -ais =5(-) + () y = 6 si y = cao Moments about y-ais = (-) + 6(-) 6 = si = cao 4

46 Q Solution Mark Notes 5(a) R A R B A.9 C B 5g 8g Moments about A terms, dim correct Equation required.8r B = 8g.9 + 5g.4 correct equation any correct moment R B = 5.5 (N) cao Vertical forces in equilibrium all forces, no etra R A + R B = 8g + 5g R A = 65.5 (N) cao 5(b).5R R A.9 C B 5g 8g Resolve vertically.5r + R =95g R = 8g Moments about A terms, dim correct.8r = 8g.9 + 5g oe 7 = =. (m) 75 cao 5

47 Q Solution Mark Notes 6(a) v ms - O +T t s labels, units and shape (, ) to (, ) (, ) to (+T, ) 6(b) v = u + at, v=, u=, t= = + a a = (ms - ) 6(c) Total distance = area under graph attempted D =.5 + T one correct area D = + T (m) cao 6(d) s = ut +.5at, u=, t=5+t, a= s =.5 (5+T) D = 5 + T + T 5 + T + T = + T Ft ep for D in (d) and (c) T T 75 = (T + 5)(T 5) = T = 5 cao D = 4 (m) cao 6

48 Q Solution Mark Notes 7 Resolve in 8 N direction Equation required 8 = Pcos6º + Qcos45º Resolve in 5 N direction Equation required 5 = Psin6º - Qsin45º 6 = P + Q 5 = P- Q Adding m dep on both M s ( + )P = P = 76.9 cao Q = 58.8 cao penalise once if not d.p. 7

49 Q Solution Mark Notes 8(a) Use of v =u +as with u=(±).,a=(±)9.8, s=(±)4. v = v = 9. allow - 4 speed of rebound = 9. 7 m = 5. (ms - ) convincing 8(b) n 4 We require smallest n st 7 9. < oe, si trial & error 4 bounces 8

50 Q Solution Mark Notes 9 BCD 45 9 (5) for 9 ABDE 6 8 (5) Circle 9 7 (5) both 8 and 7 required Lamina 5-9 (y) epressions for areas, oe Moments about AE (5-9) = signs correct. Ft table if at least one for c of m gained. =.96 cao y = 5 9

51 M Q Solution Mark Notes..y = used (sin i + cos j).(i j) (= ) correct method for dot product, no i, j s sin -cos (= ) sin - ( sin ) = m cos = sin depends on both M s sin + sin - = (sin - )(sin + ) = sin =.5 5 =, 6 6 both values sin = - =

52 Q Solution Mark Notes (a) Apply NL to object 6 R = 5a 6 kt = 5a R = kt When t =, a= -4 m used 6 k = 5 (-4) k = 9 dv 6 9t = 5 dt d v = 8t dt convincing (b) dv 8t dt increase in power at least once v = t 9t (+ C) When t =, v = 4 m used C = C = cao v = 9t + t + When v = 8, 8 = 9t + t + m substitution of v=8 in c s epression for v(t). 9t t + 5 = (9t 5)(t ) = t = 9 5, cao

53 Q Solution Mark Notes. NL dim correct, all forces T mgsin - R = ma correct equation P T = v used si 5P R = 6 correct equation in P & R 5P - R = 9 6 NL with a = dim correct, all forces T mgsin - R = correct equation P R = correct equation in P & R P - R = 7 6 Solving simultaneously m eliminating one variable, depends on both M s P 6 = P = 96; R = 8 both answers cao

54 Q Solution Mark Notes 4(a) NL (4t - )i + (t 5t)j =.5a use of F = ma a = (8t - 6)i + (6t t)j cao v = a dt attempted, i, j retained, power of t increased once v = (4t 6t)i + (t 5t )j + (c) ft a of same diff, not F When t =, v = 8i 7j c = 8i 7j v = (4t 6t)i + (t 5t )j + 8i 7j v = (4t 6t + 8)i + (t 5t -7)j 4(b) Impulse = change in momentum attempted, vector form required When t =, v = 6i + j si ft c s v.5(i + yj).5(6i + j) = i 9j (i + yj) = i 6j cao ft c s, y Speed = 4 ms - cao Speed = 6

55 Q Solution Mark Notes 5(a) T = 5g si Hooke s Law 47 T l =.4 (m) cao 5(b) Let PE be zero at the natural length level. PE = mgh used Initial PE = (-.6) Initial PE = -.5 J Initial EE = used l 47 6 Initial EE = 4 Initial EE = 47.4 J Final KE =.5mv Final KE = 7.5v Final PE = Final PE = -7.5 J 47 5 Final EE = 4 Final EE = J Conservation of energy equation, all types 7.5v = all correct, any form v =.55 v =.878 =.87 (ms - )(to d.p.) 4

56 Q Solution Mark Notes 6(a) Initial u H = 5cos = (5.6 = ) (ms - ) si Initial u V = 5sin = (5.8 = 8) (ms - ) si use of s= ut+.5at with s=,u=8(c),a=(±)9.8 complete method = 8t+.5(-9.8)t ft u V t(8 4.9t) = 4 t = (), 7 Total distance travelled by ball = 7 4 Ball will not fall into lake. = (m) (b) time to tree = 6 Use v=u+at with u=8(c),a=(±)9.8,t=5/6(c) 5 v = oe complete method v = 9 (= 9.8) 6 9 speed = m 6 speed = 8.89 (ms - ) cao = tan m = 4.6º cao 5

57 Q Solution Mark Notes 7 a R g Resolve vertically equation, dim correct No etra force Rcosº = g R =.7 (N) NL towards the centre of motion dim correct, no etra force v Rsinº = 8 v =.9 cao 6

58 Q Solution Mark Notes 8(a)(i) Conservation of energy KE and PE.5 5 = 9.8.8( cos )+.5 v 5 = v cos v = cos cao 8(a)(ii) NL towards centre of motion dim correct, terms T, gcos opposing v T gcos = 8 T = gcos +.75( cos ) m ft v of form a±bsin/cos T = cos cao 8(b) Greatest value of occurs when T= cos = ft Tof form a±bsin/cos 4 95 cos = - 88 =.4º ft a+bcos Motion stops being circular when =.4º as string cannot support negative tension. P moves under the action of gravity only. E ft >9º 7

59 M Q Solution Mark Notes (a). Use of NL F = 4a dv dv 5 4v use of a= v v d d dv 5 4vv d (b)(i) 5d 4v vdv 5 v 4 sep variables v C When =, v =, hence C = 8 v = 5 v m any correct form (b)(ii)when v =.5 = 4(9 + 9) m = m = 5.7 m 5 cao a = a = 5 m substitution of and v= =.4 (ms - ) cao 5 8

60 Q Solution Mark Notes (a)(i). NL.5a = -6.5 v dimensionally correct d d d d a= dt dt dt, v=. d t d d = dt d t (a)(ii) Ailliary equation m + 4m + = m = - ± i C. F. is = e -t (Asint + Bcost) ft m if comple d When t=, =6, = dt m used B = 6 d = -e -t (Asint + Bcost) dt + e -t (Acost - Bsint) ft e kt (Asinpt + Bcospt) -B + A = A = 5 Solution is = e -t (5sint + 6cost) cao When t is large, (b) Try PI = at + b 4a + (at + b) = 9t + 5 a = 9 m equating coefficients a = 7 4a + b = 5 b = - cao both G.S. is = e -t (Asint + Bcost) +7t 9

61 Q Solution Mark Notes (a) NL 5a = 5g 5v dimensionally correct d v 5 d t = 5g v convincing (b) 5dv 5g v -5ln dt separation of variables 5 g v = t (+C) correct integration When t =, v = m used -5ln 5 g = C 5 t t 5g v ln 5g 5 5ge 5g v m correct inversion t v 5g e 5 cao When t = 5, v = 5g( e - ) =.974 (ms - ) cao numerical answer. (c) t d = 5g - 5 5ge v = dt t 5 = 5gt + 5ge (+C) correct integration ft similar epression d dt When t =, = m used C = -5g t 5 = 5gt + 5ge -5g When t = 5, = 5ge - = 9. (m) cao

62 Q Solution Mark Notes 4(a) A y C.4-y B Tension of spring at A = Tension of spring at B = 5( y ). ( 4 y 6).6 When in equilibrium T A = T B 5( y ) ( 4 y 6) =..6 all correct y 9 = 6 y 5y = 5 y =.5 (m) convincing

63 Q Solution Mark Notes 4(b)(i) A.5 C.9- B 5( ) T A =. ( ) T B =.6 either ( ) 5( ) Force to right = -.6. allow =/- 5 = - d 5 Apply NL to P, 7.5 = - dt d = dt 9. si or =/9 Period = Therefore motion is SHM with = = 5 convincing 4(b)(ii) Amplitude =.5 (m) 4(b)(iii)Use v = (a - ), =,a=.5,=. ft a and. oe v =( ) (.5 -. ) v =.5 (ms - ) cao 4(b)(iv) = acos(t) oe allow sin/cos, c s a,.. =.5cos( t) t = cos - ( ) 5 t =.9 (s) cao

64 Q Solution Mark Notes 5. A v J l l 6º J u B 8 ms - v 5(a) Sine rule sin sin l l sin =.5= º = 6º º = º 5(b) Impulse = change in momentum used. Allow sin/cos. Apply to B J = 5 8cosº - 5v Apply to A J = v Solving simultaneously m 4-5v = v Speed of A = v = u = 8sinº = 4 (ms - ) 5 = 4. (ms - ) cao 5 Speed of B = 4 m = 5.9 (ms - ) cao J = v =.99 (Ns) ft c s v

65 Q Solution Mark Notes 6 A S 8g g R.6R B Resolve vertically equation, no missing and no etra force. R = 8g + g (= g) Resolve horizontally equation, no missing and no etra force. S =.6R = 6g = 588 (N) Moments about B equation, no missing and no etra force. Dimensionally correct. 8g 5cos + g cos = S 6sin A - each error 6sin = 46cos 46 = tan - 6 = 5.95º cao The ladder is modelled as a rigid rod. 4

66 Ques Solution Mark Notes (a) (b) E(X) =,Var(X) =. si E(Y) = E(X) + = 7 Var(Y) = 4Var(X) = 8.4 P(Y = 7) = P(X = ) (a) S 7 =..7 =.67 P(AB) = P(A) + P(B) P(AB) oe P(AB) = P(AB) P(AB) =. Award just for this line Award MA for no working Accept or Award for a valid verification (b) P( A B) P( A B) P( B). =. 6.5 Accept the use of a Venn diagram in (b) and (c) (c) P( B A) P( B A ) ( = P( A).5. =.4 = (.) P( B) P( B A) ) P( A) (a) P(A chooses G) =. (b) (c) 8 P(B chooses Y) = =. P(Diff colours) = = Accept. without working Accept 5 5 C C C C C C C 4(a)(i) (ii) e P(X = 9) = 9! =.5 P(X < ) = Accept or Award M if no working seen Award A if in adjacent row or column (b) Looking at the appropriate section of the table, n = 9 Award A for 8 or 5

67 5(a)(i) (ii) P(male and bike) = =. 45 P(owns a bike) = =.57 (b). P(female bike) =.57 =. (4/9) cao num, denom FT denominator from (a) 6(a) (i) (ii) (b) Let X = no. of defective cups so X is B(5,.5) 5 48 P ( X ).5.95 =.6 P( X 8) = or =.4587 Let Y = no. of defective plates so Y is B(5,.5) Po(.75) si.75 4 e.75 P( Y = 4) = 4! =.94 si Accept or MA if no working Award no marks if no working seen MA if no working 7(a) (b) k k 4 k E (X ) =. Or equivalent. Accept verification (c) The possible pairs are (,4), (4,), (,6),(6,) Prob = =. (6/75) for (,4),(,6) AA if factor missing 6

68 8(a) P( st hit with rd throw) =.7.7. =.47 (b)(i) (ii) (iii) P(F wins st throw) = P(G misses) P(F hits) =.8.=.4 P(F wins with nd throw) = P(G miss) P(F miss) P(G miss) P(F hits) = =.44 9(a) P(F wins) = E = (4 )d X = 4 9 =.74 (/7) Award this for realising that the probability is the sum of an infinite geometric series for the integral of f ( ) for completely correct although limits may be left until nd line Award M if no working (b)(i) 4 F( ) u 8 9 (4u u 4 u )du 7 9 Allow as dummy variable Limits may be left until net line but must then be applied (ii) (iii) P(.5 X.75) = F(.75) F(.5) =.565 (9/6) The median m satisfies 4 8m m m 8m.5 FT from (b)(i) if possible FT from (b)(i) if possible 8 m m Condone the absence of 7

69 Ques Solution Mark Notes (a) H : ; H : (b) = Test statistic =. / =. Value from tables =.74 p-value =.486 Strong evidence that the mean speed has changed. Award M if omitted (a) (b) (c) S 95 th percentile = = 86. Let X=weight of a man, Y=weight of a woman z z.5 P( Y.5).9 or P( Y.5).695 P( Y.5).85 or P( Y.5).668 P(64 Y 68).647 Let U = X 4 i i i E(U) = = 56 Var(U) = = z = Prob =.846 Y i FT from line above Accept mean speed has decreased FT the p-value if less than.5 M if no working (a) Let X, Y = measured sugar contents of A,B ( 6; y 584 ).5; y 98 (b).5.5 SE of diff of means= (.75) % confidence limits for the difference are [.57, 5.4] z.75 z =.75 Confidence level = 9% m M if 8 omitted or only one term Award this for z if m given FT from (a) 8

70 4(a) (b) Under H, X is B(,.4) si P( X ). P( X 4).65 X 4 has significance level closest to % Let Y = number of hits Under H, Y is B(,.4) N(48, 8.8) si Test statistic = 8.8 =. p-value =. Insufficient evidence to conclude that his shooting has improved Award for valid attempt at using tables Award A for or 5 Award A for incorrect or no continuity correction but FT for following marks No cc gives z =., p =.968 Wrong cc z =.4, p =.88 FT the p-value 5 Let X = score on a single die. Then E(X) =.5 and Var(X) = Let Y = mean of scores on dice. Then by the Central Limit Theorem, Y N(.5, 5/) z 5/ = ().46 Prob =.7 6(a)(i) H :.; H :. (ii) Under H, X is Po() si P ( X 9) =.44 Insufficient evidence to conclude that the (mean) number of breakdowns has decreased. (b) Under H, Y is Po() N(,).5 z.69 p-value =.455 Strong evidence to conclude that the (mean) number of breakdowns has decreased. m FT their mean and variance Use of continuity correction gives z =.4, p =.764 Accept in place of. FT the p-value Award A for incorrect or no continuity correction but FT for following marks No cc gives z =.7, p=.48 Wrong cc, z =.78, p =.75 FT the p-value if less than.5 9

71 7(a)(i) (ii) (b) P( Y y) P( X y) P( X y y a b a Attempting to differentiate, y giving b a y f ( y) for a y b b a = otherwise ) We are given that a b 5.5 and Solving, a =.5, b = 8.5 ( b a)

72 Ques Solution Mark Notes The sample space is as follows. EITHER S OR Samples R M,,,,4,,6 5,,6 5,,4,,6 5,,6 5,4,6 5 4,4,6 5 4,6,6 5 6,,4,,6 4,,6 4,4,6 4 4,4,6 4 4,6,6 4 6,4,6 4 4,4,6 4 4,6, ,6,6 6 Samples R M No. of ways,,,,4,,6 5 4,4,6 5 4,6,6 5 6,,4,,6 4,4, ,6, ,6,6 6 for the samples column for the R column for the M column Minus if or rows omitted Minus A if or 4 rows omitted for columns and 4 for the R column for the M column Minus if or rows omitted Minus A if or 4 rows omitted

73 The probability distribution of R is therefore r 4 5 P(R=r) / / / 8/ 7/ The probability distribution of M is therefore FT for both tables from (a) if sum of probabilities is (a) (b) m 4 6 P(M=m) / 6/ 4/ 9.9; 8.9 UE of = UE of =.64 DF = si Crit value =.6 99% confidence limits are giving [4.9,7.] Must be seen No working need be seen M division by Answer only no marks FT their s and mean M use of z-values M if omitted Answer only no marks (a) (b) H : a b; H : SE = (=.8 ) Test stat =.8 =.68 Tabular value =.4648 p-value =.996 Insufficient evidence to conclude that there is a difference in mean weight. a b Treat taking the variances as SDs as a misread, giving SE =.9, Test stat =.6, p-value =.55 M if omitted FT the p-value 4(a) 54 p ˆ.6 si ESE = =.56.. si 9 9% confidence limits are giving [.55,.685]

74 (b)(i) p ˆ.5965 Only award if used to find SE in (b)(i) (ii) n.96 n n = 4 cao Number of red squirrels = = 4 m Award this even if.6 used instead of.5965 FT the n from (b)(i) 5(a) (b)(i), y 76.6, 5, y 4485 S / 5 5 si y S 5 /5 5 si 5 b = a = = SE of a = % confidence limits are [.,.9] (.54..) B Minus each error FT from (a) (ii).5 SE of b = (.58 ) Test stat =.58.4 Critical value =.96 or p-value =.64 Reject H FT from (a)

75 6(a)(i) (ii) (b)(i) (ii) (c) E( X ) 4( 6 ) 4 Therefore E( X ) 4 si E( U ) a(4 ) b for all 4 a ; b 4 U X Var( X ) 4 9 6( 6 ) (4 ) ( 5 ) Var( X ) Var( U ) a n ( 5 ) 5n Y is B(n, 6) so E(Y) = n( 6) Therefore E(V) = cn( 6) + d = (for all ) c ; d 6n 6 V Y 6 6n Var( V ) c Var( Y) c npq ( 6 ) 6n Var( U) ( 5 ) 6n 6 Var( V ) 5n ( 6 ) 5 This ratio is greater than so that V is the better estimator. Convincing 4

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