MATHEMATICS (New Specification)

Transcription

1 MS WELSH JOINT EDUCATION COMMITTEE. CYD-BWYLLGOR ADDYSG CYMRU General Certificate of Education Advanced Subsidiary/Advanced Tystysgrif Addysg Gyffredinol Uwch Gyfrannol/Uwch MARKING SCHEMES JANUARY 6 MATHEMATICS (New Specification)

2 INTRODUCTION The marking schemes which follow were those used by the WJEC for the 6 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. The WJEC regrets that it cannot enter into any discussion or correspondence about these marking schemes.

3 MATHEMATICS C. (a) Gradient of AB = (b) 6 M (correct method) A (grad = ) Gradient of k 6 k 6 M (use of m m =, seen or implied) Then k 6 M (method of finding equation in k) k 6 = k = 5 A (C.A.O.) (convincing) (c) Gradient of L = B (F.T. for same gradient as BC or ) gradab Equation of L is y + = (x + ) B (give mark here) (d) Coords of D : x =, y = 7 B(F.T. unsimplified equation of L) CD (5 ) ( 7) 5 M (correct formula) A ( single no, F.T. derived coords of D)

4 . (a) 4 B, B, B = 5 B (F.T. one slip, answer of form k ) (b) ( ( 7)( 7)( 7) 7) M (correct rationalising) 7 A numerator with ( 7) 7, allow A (denominator with no surds) A (F.T. one slip) 8. dy 8x 7 dx B (correct differentiation) = 9 at (, 4) B (numerical result, F.T. one slip) Gradient of normal = 9 gradient of tgt Equation is y 4 = (x ) A (F.T. candidate's gradient 9 of tangent) 4 4. (a) M (reflection in x axis) A (correct points) (b) M (x translation to right) B (st.pt) (on diagram or stated) A (points of intersection) 5

5 5. For no real roots, M (b 4ac, correct b, a or c correct) 4 4 (k + )(k + 5) < A (correct) 4 k 7k < M (b 4ac < ) k + 7 k + 6 > A (convincing) (k + 6)(k + ) > B (fixed pts,, 6) k < 6 (and/or) k > or B (F.T. fixed points) k < 6, k > < k < 6 or k > or k < 6 All gain B 7 6. (a) f() = 4 M (remainder theorem or division with remainder equated to 4) 8a = 4 8a = 6 a = A (b) f () = = M (correct use of factor theorem with appropriate factor mentioned) x is a factor A (correct factor) x x 7x + 6 = (x ) (x + x 6) m (x + ax + b, a or b correct, any method) A (correct quad. factor) = (x )(x ) (x + ) A ( factors and roots, F.T. one slip) Roots are,, 7 7. (a) (x + ) = (x) + (x) () + (x)() + = 7x + 54x + 6 x + 8 B ( for each error, any method) (b) n( n ) () = (n) M ( n C p = k n C, k=, n( n ) A ( =(n)) n = A (C.A.O.), p =,) 7 8. (a) Let y = x 5x + y + y = (x + x) 5(x + x) + B y = 4xx + (x) 5 M (attempt to find y) A y 4x + x 5 x M (divide by x and let x o) Let x o dy Lt y dx x x 4x 5 A (award only if some mention of limit, clear presentation and no abuse of notation)

6 dy a (b) x (o.e.) B, B dx x a 4 4 a = 6 7 M ( f (4)=7, reasonable diffn) A (C.A.O.) 9 9. (a) + 6x x = (x ) B ( (x ) B () Greatest value = B (F.T. candidate's b and a) when x = B (b) 6x x = 7 6x x A Least value = 9 A (F.T. b, a) 6. (a) dy dy B dx dx x 6x = dy M dx x =, A (either root, F.T. one slip) When x =, y = ; when x =, y = A (both C.A.O.) d y x d x M (any method) d y x =, dx = > min. A (F.T. x values) d y x =, dx = - < max. A (b) B (shape, negative cubic) B, B (derived stat pts) (c) One real root as M (reason) graph crosses x axis once A (F.T. graph) 4

7 MATHEMATICS C. (a) h =. M (correct formula h=.). Integral [ ( B (4 values) )] B ( values).4497 A (F.T. one slip) S. Case h = 6 M (correct formula, h = 6 ) Integral [ ( B (all values) )] A (F.T. one slip) 4. (a) 4 cos cos = ( cos ) M (correct use of sin + cos =) 6 cos cos = M (attempt to solve quadratic in cos, correct formula) ( cos )( cos + ) = cos =, A (C.A.O.) = 48.º,.8º, º, 4º B (48.,.8º) B (º) B (4º) (b) tan = = º, º B, B (c) = º, 5º, 9º, 5º B (one value) = 5º, 75º, 95º, 55º B ( values) B ( values) 5

8 . (a) Sine rule sin 45 sin ACB ˆ M (correct use of sine rule) sin ˆ sin 45 ACB ACB ˆ ABC ˆ = 58.5º or.95º A (one value) = 8º (45º º) = 76.95º A (F.T. one slip) or 8 (45º +.95º) =.5º A (b) Area = sin 76.95º 58.4 cm ( ) M (correct formula) or sin.5º.4 cm A (both) (F.T. candidate values) 6 4. (a) nth term = ar n B Let S n = a + ar + ar ar n + ar n () rs n =ar + ar ar n + ar n + ar n () B (at least terms, one at each end) M (multiplication by r and subtract) () () S n rs n = a ar n S n ( r) = a ( r n ) n a( r ) S n = r A (convincing) 6

9 (b) (i) ar =, ar 6 = 54 B (both) r = 7 M (eliminate a) r = A (ii) a S 7 ( ) B (F.T. one slip) M (correct formula, F.T. candidate values) A (iii) n 7 = 5 B (F.T. candidate values) n = 6875 (n ) ln = ln 6875 M (attempt to take logs) n = +.5 = 4.5 A (C.A.O.) Least value is 5 A (F.T. candidate's n) 4 5. (a) a + d = () B (a + a + d = ) a + 7d = 47 () B Solve (), () d = 7, a = M (attempt to solve) A (C.A.O.) (b) S = [ + 9 7] M (correct formula with candidate values) = 9 A (F.T. candidate values) x 4 4 x (+ C) B, B 5 4 x 4 x 7. (a) B y =, 4 x = M (setting y = ) x = B (, ) A A 4 x = x M (equating ys) x + x 4 = M (correct attempt to solve quad) x = 4, A(, ) A 7

10 (b) Area = xdx (4 x ) dx M (use of integration to find area) M (addition of areas) = x x 4x B (integration) = = M (use of candidate's limits, any order) A (C.A.O.) 8. (a) Centre (4, ) B Radius = 4 ( ) M (correct method of finding radius) A (b) Centre is (, ), radius = a B (both) Distance between centres = 4 B Circles touch if = a + M (F.T. distance) a.47 A (F.T. distance) 7 9. (a) 4 ( + ) = 5. M (use of correct formula) 5. + =. 9 8 () A (convincing) (b) 4 4 =. M (use of correct formula) =.8 () A Solve (), () =.5, =.55 M (attempt to solve) A (F.T. one slip) 6 8

11 . (a) Let x = a p, y = a q B (properties of log a x = p, log a y = q log a x and x = a p ) xy = a p + q = a p + q B (laws of indices) (b) log a (xy) = p + q = log a x + log a y log xdx logdx dx B (convincing) B laws of logs) B =.568 +, x B (integration) = B =

12 MATHEMATICS C. (a) h =.5. 5 Integral [ M (h=.5 use of correct formula) 4( ) B ( values) + ( ] B ( further values).768 A (F.T. one slip) 4. (a) = º, for example M (substitution of in both, tan =.7 one correct value) tan.5 A (both correct and clearly unequal) ( tan tan ) (b) 4 (cosec ) = 4 cosex M (correct use of cosec = + cos ) 4 cosec + 4 cosec 5 = M (grouping terms and attempting to solve quad. in cosec or sin ( cosec )( cosec + 5) = correct formula or (a cosec + b)(cosec + d) 5 cosec =, where ac = coefft of cosec bd = constant term) sin =, A (CAO) = 4.8º, 8.º,.6º, 6.4º B (4.5º 4º) B (5º 4º) B (6º 6.5º). (a) 4 y dy dy x x y x 4 dx dx B (4y dy ) dx B (x dy +x y) dx B (x + 4) dy dy dx dx 8 dy 4 dx (o.e.) B (C.A.O.)

13 (b) (i) dy t (o.e.) M A dx 6t (ii) d dy d y dt dx (o.e.) M (use of correct formula) dx dx 6t dt A (F.T. one slip) 8 x 4. (a) 8x x x B x kx M x A (k = ) x 8x 8x = A (correct unsimplified equation, F.T. one slip) 4x + x = A (convincing) (b) x 4x x Change of sign indicates presence of root M (attempt to find signs or values) A (correct signs, values and conclusion) x =., x =.996, x = B (x ) x = Check.99645, B (x, rounded or unrounded) M (attempt to find signs or values) A (correct) x 4x + x Root lies between and and is therefore to 6 decimal places A (correct conclusion) 5. (a) e x sin x + e x cos x M (f(x)e x + g(x) cos x) A (f(x) = sin x, g(x) = ke x ) A (k =, correct unambiguous answer)

14 (x )4x (x (b) (x ) x (x ) )6 x (x ) f ( x) (x M (x ) A (f(x) = 4x, g(x) = 6x) A (C.A.O.) ) g( x) (c) x sec (5x + ) M (sec (5x + ), allow g(x) = ) A (correct unambiguous answer) (d) k M (, allow k =, ) x x x A (simplified answer) (e) (x) 9x M ( A (k = ) k (x) (o.e.), allow k = ) 6. (a) x 8 5 x B x 8 5 M x x (or x and x ) A (must indicate both conditions apply) (b) Graphs M (for x, V shape through origin) A (translation in +ve y direction, cusp at (, )) A (cusp at (, ) A (correct relative positions) 7 7. (a) (i) 4 5 ln 7x 7 6(x ) M (kln (7x + )) 4 (+C) A k 7 k M ( x ) 5 A k (o.e) 6

15 (ii) sin x (+C) M (k sin x, k =,,,) A (k = ) (b) 4 x x e M ( ke, k,,) A (k = ) = e e M (ke ke, allowable ks) ~.8 A (C.A.O. at least sig. figs) 8. (a) Let y = x + 4 M (x = f(y)) x = y 4 x = y 4 A Choose + domain of f is x x = y 4 f (x) = x 4 A (F.t. one slip) A (F.T. one slip) Domain is x 4, range is f (x) (o.e.) Domain is x 4, range is f (x) > B, B (b) Graphs M (full or half parabola passing through (,b)) A (r.h. branch and minimum at (,4)) A F.T. b, full or half parabola)) 9 9. (a) Domain is (, ) B (b) (fg(x) = 5) x 4) e 5 = 5 M (correct order) ln( x 4 = 5 or ln (x 4) = ln 5 x = 9 x = ( not in domain) A (either) A A (with reason) 5

16 MATHEMATICS FP. Mod = B Arg = tan 6 () B i (cos isin ) 6 6 B Arg of ( n n i) 6 MA Power is real when n is an integer multiple of 6 A Smallest n = 6. A. Det =. + + ( 4) MmA = A EITHER = ( / ) 5 / 9 M > for all A OR ' b 4ac' So determinant not equal to zero for any real. M A. f(x + h) f(x) = MA ( x h) x x [ ( x h) ] = [ ( x h) ]( x ) m h(x h) = [ ( x h) ]( x ) A lim (x h) ( x) h [ ( x h) ]( x ) M x = ( x ) A f 4. 7i ( 7i)(- i) MA i ( i)(- i) 8 4i 9 i A (x + iy) + (x iy) = 9 i M Equating real and imaginary parts m x = and y = A 4

17 5. (a) = B T = B T T = M = A Note: Multiplying the wrong way gives Award MA (b) Fixed points satisfy y x y x M giving y = x and x + = y A Fixed point is ( /, /). MA FT from wrong answer to (a) the incorrect matrix above leads to ( /,/). 6. (a) Assume the proposition is true for n = k, that is ) ( ) ( k r k r B Consider MA ) ( ) ( k k r k r A ( ) k So, if the proposition is true for n = k, it is also true for n = k +. A (b) Since 4, P is false for n = is therefore false. B 5

18 7. (a) Using reduction to echelon form, 5 x y 6 z 4 5 x y 6 z The solution will not be unique if =. MAA A A (b) The equations are consistent if =. B Put z = M y = 6 A x = 6 4 MA 8. (a) Let the roots be,, +. M Then ( ) + ( + ) + ( )( + ) = 47 m 6 A = 4 A The roots are, 4 and 5. A (b) p =, q = 6BB 9. (a) lnf(x) = ln( x) x M f ( x) ln( x) f ( x) x x ma = ln( x) x A At the stationary point, ln(x) = M e so x = e (.7) and y = e (.44) AA (b) We now need to determine its nature. We see from above that For x < e, f ( x) and for x > e, f ( x) M Showing it to be a maximum. A 6

19 . z w z wz + w = z + z(w ) = w w z = w Since z =, it follows that w = w ( v u) v ( u ) A 9 6u u v u u v leading to u =. Straight line parallel to the v-axis (passing through (,)). ALTERNATIVE METHOD x iy x iy u iv. M x iy x iy ( x )( x ) y iy( x x ) = ( x ) y Equating real and imaginary parts, u = x y 4x x y x v = y x y x Putting x y, u = which is a straight line parallel to the v-axis (passing through (,)). M A A M A A B A M A A M A B 7

20 MATHEMATICS M 8

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26 M ASSESSMENT OBJECTIVES Q.No. AO AO AO AO4 AO5 Total TOTAL

27 MATHEMATICS S. (a) P(Score at least on one die) = 6 4 P(Score at least on both dice) = B MA [Award B for the 6 6 array of possible scores, M for counting the number of these satisfying the required condition and A for the correct answer] (b) Possibilities are 6,;5,;4, and reverse, ie 6 possibilities MA Prob = 6 6 (cao) 6 A. (a) P(A B) = P(A) + P(B) gives P(B) =. MA (b).5p(b) =.5 + P(B).7 MA.5P(B) =. gives P(B) =.4 MA (c).5. =.5 + P(B).7 MA P(B) =.5 A. (a). B 4 4 (b) (i) Prob = e.. 95 A! [Accept or.45.8 from tables, M for at least correct value] (ii) Prob = or =.7977 (cao) BBB (c) E(C) = = MA SD = (6 Var(X)) = 8 [Award M for 64] MA 4. (a) Prob = or MA

28 (b) P( blue) = or ( ) B P( blue) = or ( ) MA Reqd prob = ( ) 56 7 B 5. np = ; np( p) = 6 MAA Valid attempt to solve equations (FT) M p =. B n = B 6. (a) P(4) = 6 4 MA 5 = 4 A (b) P(cube4) = / = (cao) BBB 5/ (a) Number of defective glasses X is B(4,.5) (si) B 4 P(X = ) = MA (b) P( X 5) = or =.4 (cao) BBB (c) Number of defective glasses is now B(,.5) Poi(6) B P(Y < 8) =.744 (or.56) MA 8. (a) [,.7] [Accept (..7)] BB [Award B for {,.,.7} or similar] (b) (i) E(X) = (.7 ) =. MA. = gives =. MA (ii) E( X ) = MA = 5.4 A 6

29 k 4 kx d MA Int = gives k A 9. (a) (i) x x k 4 4 (ii) E(X) = x. x dx (Limits required, here or later) M 4 =. x 4 4 B 85 = (.4) 8 A x (b) (i) F(x) = y dy M = y x 6 A = x 6 A (ii) Prob = F() F() M 9 = (.) 6 A (iii) The median m satisfies. 5 m M 6 m.5 m =.9 (cao) A A GCE M/S Mathematics Spec (New) (Jan 6)/JD 7

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