Mark Scheme 2605 June 2005

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Ronald Floyd
5 years ago
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4 Mark Scheme 6 June

5 (i α, αβ, αβγ (ii α ( α αβ ( 7 ( (iii Roots cannot all be real One real, two comple (conjugate roots (iv α β ( α ( αβ (v ( ( ( 6 αβγ βγ ( γα γβ α αβ βγ ( α βγ α( αβγ βγ α βγ ( γ α( α β αβ α ( ( 94 (vi Sum αβ ( 6 Sum in pairs 94 Product αβγ ( α β αβγ ( 4 Equation is (ag (ag 4 4 For correct formula Or there are comple roots Accept imaginar for comple When α >, cannot be awarded, but give B for one real and two comple or α β ( α ( α and α α α Dependent on previous Forming cubic equation (with numerical coefficients A if omitted α

6 OR Let, 9 4 ( 64 6 ( ( A Give if just one slip made

7 (a(i e ( e e ce e c c ± 4c 4 or M for a complete alternative method c ± c ln( c ± c ( c c ( c c c ( c Hence ln( c c ln( c c ln( c c (ii cosh cosh 9 cosh cosh (cosh (cosh cosh ± ln( OR Writing in eponential form and obtaining quadratic factors (e 4e (e e M e ± ln( ± 6 Using cosh sinh (using wrong identit is M Solving quadratic Or ln( ± (A if an other solutions given for ( e 4e Obtaining a value for in log form f ( (b(i ( ( ( ( f (

8 (ii f( arcsin 7 f (, f ( 4 64 f( f ( f (...! 7 arcsin 4... ft Evaluating f ( or f ( For 7 p and q 4 (ft requires non-zero values (iii arcsin ft ft requires three non-zero terms Evaluating three non-zero terms (i (A (cosθ jsinθ (cosθ jsinθ (ii (iii cosθ (B ( e e 9 C js e e e e e 6 cosθ ( [e e n ( ( e cosθ C sinθ S e j π 6, e j π 6 e 9e n ] jnθ ( e ( e e e 6cosθ n, n n jnθ 6 cosθ... cos nθ sin nθ 6 cosθ e j π n n n j( n θ cos( n θ sin( n θ (ag For or θ e j cosθ jsinθ ( cosθ 9sin Obtaining a geometric series Summing a geometric series Using conjugate of denominator Epression with real denominator and numerator multiplied out Equating real or imaginar parts Correctl obtained Summing to infinit can earn all the M marks but no A marks If B, give B for arguments correct for arguments correct θ

9 (iv w e j θ π π π where θ,, 6 6 ( w( w* ( e 6cosθ ( e,, 4 Implied if net B earned Accept 4.,. 4(a(i r sinθ a(sinθ 4sin d a(cosθ sinθ cosθ dθ θ when cosθ or sinθ When sin θ, maimum a( 4 64 a 6 (ag Differentiating d Solving dθ For sin θ r sinθ OR r sinθ a(sinθ 4sin θ a ( sinθ 6 4 a 6 Completing the square (ii Correct shape in st or nd quadrant Correct shape in rd or 4th quadrant Full correct, with a, a, 9a shown, and zero gradient when crossing the -ais

10 (iii Area 4 π π a a ( 4sinθ dθ θ 4cosθ 4sin θ a (π a ( 4sinθ cos θ dθ π ft 6 Integral of r Correct integral epression For sin θ ( cos θ Integrating Accept.9a a bsinθ and c cos θ (b(i For an ellipse Ellipse with O as RH focus (ii BS OA k, π S is ( (iii OP PS length of major ais 7 k

11 6 - Pure Mathematics General Comments There was a wide range of performance on this paper, with about a quarter of the candidates scoring marks or more (out of 6, and about a quarter scoring less than marks. Almost ever candidate answered questions and ; then about % chose question and onl % chose question 4. Comments on Individual Questions Roots of a cubic equation This was b far the best answered question, with half the attempts scoring 7 marks or more (out of. For man candidates this question provided a high proportion of their total mark. Parts (i and (ii were almost alwas answered correctl. In part (iii, most candidates mentioned the eistence of comple roots; but relativel few earned both marks b stating that one root is real and two are comple. Parts (iv and (v were ver often answered efficientl and correctl, although some candidates set off on the wrong algebraic track and wasted a lot of time in fruitless effort. Finding the new cubic equation in part (vi was well understood, and the product of the new roots was ver often found correctl. Man candidates did not realise that the had alread found the sum of products in pairs, and calculated this again, often obtaining a value different from their answer to part (v. The average mark on this question was about. (a Hperbolic functions In part (i, most candidates were able to show that ln( c ± c, but onl a few then showed correctl that this is equivalent to the desired result ± ln( c c. In part (ii, those who used sinh cosh were usuall able to obtain cosh and hence write in logarithmic form, but the other solution cosh was sometimes not rejected. Those who wrote the original equation in eponential form ver rarel made an progress.

12 (b Inverse circular functions and Maclaurin series Comple numbers In part (i, the double differentiation of arcsin( caused a surprising number of problems, notabl sign errors. In part (ii, the Maclaurin series usuall followed correctl from the results in part (i, although man forgot to divide f ( b when finding q. Most candidates knew what to do in part (iii, but.arcsin(.6 was often evaluated as arcsin(.6, and degrees were sometimes used instead of radians. The average mark on this question was about. Part (i was generall answered well, but the responses to part (ii ranged quite uniforml from ver poor to full correct. Most candidates began b considering C js, but some made no progress beond this. A common stumbling block, when the geometric series had been summed, was the failure to make the n jnθ denominator real. Careless errors such as ( e e, and sign errors, spoilt some otherwise good attempts, and the epression for S often included in the numerator. In part (iii, the three cube roots were ver often given correctl, but a surprising number of candidates had all three arguments wrong. Part (iv was also correctl answered b man candidates, although the connection with part (i(b was not alwas seen. Some confused w * with w. 4 Polar coordinates This was the worst answered question, with an average mark of about. In part (a(i, most candidates did not even make the first step of epressing r sinθ in terms of θ. In part (a(ii, there were some good attempts to sketch the curve, although few earned full marks; the most common error was to draw a cusp at π. θ In part (a(iii, there was a lot of good work, and the area was often found correctl. Man made slips in the integration, and the overall factor of was sometimes missing. In part (b(i, the ellipse was often drawn correctl, but onl a few candidates could answer parts (b(ii and (b(iii.

The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 72.

The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 72. ADVANCED GCE UNIT 7/ MATHEMATICS (MEI Further Methods for Advanced Mathematics (FP THURSDAY 7 JUNE 7 Additional materials: Answer booklet (8 pages Graph paper MEI Examination Formulae and Tables (MF Morning