F Further IAL Pure PAPERS: Mathematics FP 04-6 AND SPECIMEN Candidates sitting FP may also require those formulae listed under Further Pure Mathematics FP and Core Mathematics C C4. Area of a sector A = r dθ (polar coordinates) Comple numbers θ e i = cosθ + isinθ n n { r(cosθ + isinθ )} = r (cos nθ + isin nθ ) n k i e π n The roots of z = are given by z =, for k = 0,,, K, n Maclaurin s and Taylor s Series r ( r) f( ) = f(0) + f (0) + f (0) + K + f (0) + K! r! r ( a) ( a) ( r) f( ) = f( a) + ( a) f ( a) + f ( a) + K + f ( a) + K! r! r ( r) f( a + ) = f( a) + f ( a) + f ( a) + K + f ( a) + K! r! e = ep( ) = + +! r + K + +K r! r ln ( + ) = + K + ( ) + r r for all + K 5 r + r sin = + K + ( ) +K! 5! (r + )! 4 r r cos = + K + ( ) +K! 4! (r)! 5 r+ r arctan = + K + ( ) + K 5 r + ( < ) for all for all ( ) Edecel AS/A level Mathematics Formulae List: Further Pure Mathematics FP Issue September 009 9
IAL F SPECIMEN PAPER
JUNE 04 F IAL. (a) Show that (b) Hence, or otherwise, find r r r r r r r () n r r r r giving your answer as a single fraction in its simplest form. (4). Use algebra to find the set of values of for which 6 (7). Solve the equation z 5 = 6 6i giving your answers in the form re iθ where θ is in terms of π and 0 θ < π. (5) 4. A transformation from the z-plane to the w-plane is given by z w, z z Under this transformation, the circle z = in the z-plane is mapped onto a circle C in the w-plane. Determine the centre and the radius of the circle C. (7)
5. d y dy y 0 (a) Show that d y d y 4 4 a b where a and b are constants to be found. (5) Given that y = and d y at = 0, (b) find a series solution for y in ascending powers of up to and including the term in 4. (5) (c) use your series to estimate the value of y at = 0., giving your answer to four decimal places. () 6. dy y, > 0 Find the general solution of the differential equation, giving your answer in the form y = f(). (9) 7. The point P represents a comple number z on an Argand diagram, where z + = z and the point Q represents a comple number w on the Argand diagram, where w = w + i Find the eact coordinates of the points where the locus of P intersects the locus of Q. (7)
8. (a) Show that the substitution = e t transforms the differential equation into the differential equation d y dy 5 0, > 0 (I) y d y dy 4 y 0 dt dt (b) Hence find the general solution of the differential equation (I). (7) (5) 9. Figure Figure shows the curve C with polar equation r asin, 0 θ, and the circle C with polar equation r = a, 0 θ π, where a is a positive constant. (a) Find, in terms of a, the polar coordinates of the points where the curve C meets the circle C. () The regions enclosed by the curve C and the circle C overlap and the common region R is shaded in Figure. (b) Find the area of the shaded region R, giving your answer in the form a p q, where p and q are integers to be found.
JUNE 05 F IAL. (a) Using algebra, find the set of values of for which 5 (7). (a) Epress in partial fractions. ( r6)( r8) (b) Hence show that n r n( an b) ( r 6)( r 8) 56( n 7)( n 8) () where a and b are integers to be found. (4). (a) Show that the substitution z = y transforms the differential equation dy y e y (I) into the differential equation dz 4z e (II) (b) Solve differential equation (II) to find z as a function of. (c) Hence find the general solution of differential equation (I), giving your answer in the form y = f(). () (4) (5)
4. A transformation T from the z-plane to the w-plane is given by z w, z z The line in the z-plane with equation y = is mapped by T onto the curve C in the w- plane. (a) Show that C is a circle and find its centre and radius. (7) The region y < in the z-plane is mapped by T onto the region R in the w-plane. (b) Show that C is a circle and find its centre and radius. () 5. Given that y = cot, (a) Show that d y d cot cot () (b) Hence show that d y d 4 pcot qcot r where p, q and r are integers to be found. (c) Find the Taylor series epansion of cot in ascending powers of up to and including the term in. () 6. (a) Find the general solution of the differential equation () d y dy y sin (I) (8) Given that y = 0 and d y when = 0
(b) find the particular solution of differential equation (I). (5)
7. Figure shows the curves given by the polar equations r sin, 0 θ π r = + cos θ 0 θ π (a) Verify that the curves intersect at the point P with polar coordinates The region R, bounded by the two curves, is shown shaded in Figure., () (b) Use calculus to find the eact area of R giving your answer in the form a(π ), where a is a constant to be found. (6)
8. (a) Show that 6 z z z k z 6 z z z z where k is a constant to be found. () Given that z = cos θ + i sin θ, where θ is real, (b) show that (i) z n cos n n z (ii) z n isin n n z (c) Hence show that (d) Find the eact value of cos sin sin sin 6 () (4) 8 cos sin d 0 (4) END TOTAL FOR PAPER: 75 MARKS
F IAL JUNE 06. (a) Epress 4r in partial fractions. () (b) Hence prove that n n 4r n r (c) Find the eact value of () 5 5 4r r9 () (Total 6 marks). Use algebra to find the set of values of for which 9 (Total 6 marks). Find, in terms of k, where k is a positive integer, the general solution of the differential equation dy giving your answer in the form y = f(). ( ) k ky ( ), > 0 (Total 6 marks)
4. f ( ) sin (a) Find the Taylor series epansion for f() about in ascending powers of up to and including the term in 4 (6) (b) Hence obtain an estimate of sin, giving your answer to 4 decimal places. () (Total 8 marks) 5. The transformation T from the z-plane to the w-plane is given by z w, z z The circle in the z-plane with equation + y =, where z = + iy, is mapped by T onto the circle C in the w-plane. Find the centre and the radius of C. (Total 7 marks) 6. (a) Find the general solution of the differential equation d y dy y (b) Find the particular solution of this differential equation for which y = 0 and d y = 0 when = 0 (5) (Total 4 marks) (9)
7. Figure shows a sketch of the curves C and C with polar equations C : r = cos, 0 C : r = 9 cos, 0 The curves intersect at the point P. (a) Find the polar coordinates of P. () The region R, shown shaded in Figure, is enclosed by the curves C and C and the initial line. (b) Find the eact area of R, giving your answer in the form pπ +q are rational numbers to be found. where p and q (Total marks) (8)
8. (a) Use de Moivre s theorem to show that cos 5 θ p cos 5θ + q cos θ + r cos θ where p, q and r are rational numbers to be found. (b) Hence, showing all your working, find the eact value of (6) 5 cos d 6 (Total 0 marks) 9. The comple number z is represented by the point P in an Argand diagram. (4) Given that arg z 5. z 4 (a) sketch the locus of P as z varies, (b) find the eact maimum value of z. (Total 7 marks) () (4) TOTAL FOR PAPER: 75 MARKS
WFM0/0: Further Pure Mathematics F Question Scheme Marks (a) r r+ () (b) n = + + +... r (r )(r ) 5 5 8 8 n n+ ft = + n = = n+ (n+ ) * () (c) Sum = f(000) f(99) 000 97 = 0.000 or.0 0 6004 598 () 7 Pearson Edecel International Pearson Education Limited 0 Sample Assessment Materials Advanced Level in Mathematics 59
WFM0/0: Further Pure Mathematics F Question Scheme Marks f ( t) = cos, f (0) = - B f ( t) = ( + sin ), dt f (0) = 0.5 t t f ( t) = f (0) + t f (0) + f (0) + f (0) +...! = 0.5t 0.5 t t +... 5 60 Pearson Edecel International Pearson Education Limited 0 Sample Assessment Materials Advanced Level in Mathematics
WFM0/0: Further Pure Mathematics F Question Scheme Marks (a) ( + 4)( + ) ( + ) = 0, ( + )( + 7+ 0) = 0 so ( + )( + )( + 5) = 0 or alternative method including calculator Finds critical values and -5 Establishes > - ft Finds and uses critical value to give 5 < < - (6) (b) > - Bft () 7 Pearson Edecel International Pearson Education Limited 0 Sample Assessment Materials Advanced Level in Mathematics 6
WFM0/0: Further Pure Mathematics F Question Scheme Marks 4(a) Modulus = 6 B Argument = π arctan( ) = () (b) π π z = 6 (cos( ) + isin( )) = 6 (cos π + isin π) =4096 or 6 () (c) π π π π 4 4 w = 6 (cos( ) + isin( )) = (cos + isin ) 6 6 OR + i OR ior i ( = + i) ft A (,0) (5) 0 6 Pearson Edecel International Pearson Education Limited 0 Sample Assessment Materials Advanced Level in Mathematics
WFM0/0: Further Pure Mathematics F Question Scheme Marks 5(a).5 + sin θ = sin θ = 0.5 π 5π θ = or 6 6,, π 5π and θ = or 8 8 () (b) Area = = θ, - π 9, 5π 8 (.5 + sin θ + ( cos 6 θ))dθ- π 9 π 8 5π 8 (.5 + sin θ) d π 8 = (.5θ cos θ + ( θ sin 6 θ)) 6 5π 8 π 8-9 π 5π = 4 6 (7) 0 Pearson Edecel International Pearson Education Limited 0 Sample Assessment Materials Advanced Level in Mathematics 6
WFM0/0: Further Pure Mathematics F Question Scheme Marks 6(a) Imaginary Ais Re(z) = 0 6 Real ais Vertical Straight line Through on real ais B B () (b) These are points where line = meets the circle centre (, 4) with radius 5. The comple numbers are + 9i and i. () (c) 0 0 z 6 z w 6 w = = 0 6w = 0 5 w = 5 This is a circle with Cartesian equation ( u 5) v 5 + = (5) 0 64 Pearson Edecel International Pearson Education Limited 0 Sample Assessment Materials Advanced Level in Mathematics
Question WFM0/0: Further Pure Mathematics F Scheme Marks 7(a) d y d d = y z. and d y z dz dz = so d y. d z = z Substituting to get dz z z z = and thus d z ztan = *. 4 tan (5) (b) tan I.F. = e = ln cos e = cos d ( cos ) = z cos = z cos cos zcos = = + + z tan sec csec (cos + ) sin = + +c 4 (6) (c) y ( tan sec csec ) = + + Bft () Pearson Edecel International Pearson Education Limited 0 Sample Assessment Materials Advanced Level in Mathematics 65
WFM0/0: Further Pure Mathematics F Question Scheme Marks 8(a) Differentiate twice and obtaining dy d y = λsin 5 + 5λcos5 and = 0λ cos 5 5λ sin 5 Substitute to give λ = 0 (4) (b) Complementary function is y = Acos5+ Bsin 5 or Pe 5i 5i + Qe So general solution is y = Acos5+ Bsin 5+ sin 5 0 or in eponential form ft () (c) y= 0 when = 0 means A = 0 B dy 5B cos5 + + cos5 and at = 0 d y 5 0 So B = So y = sin 5 + sin 5 0 (5) (d) "Sinusoidal" through O amplitude becoming larger Crosses ais at π π π 4π,,, 5 5 5 5 B B () 4 66 Pearson Edecel International Pearson Education Limited 0 Sample Assessment Materials Advanced Level in Mathematics
Question F IAL JUNE 04 MARK SCHEME Marks.(a) (a) Way ( r )( r ) ( r )( r ) r ( r) ( r )( r )( r ) ( r )( r )( r ) ( r )( r )( r ) Attempt common denominator Correct proof ( r )( r )( r ) r r r r r r ( r )( r ) ( r )( r ) : Factor of and attempt partial fractions r : Correct proof () (b) Other methods: Complete method scores All work correct inc final answer reached n ( )( )( ) r r r r... 4 ( n )( n ) ( n )( n ) ( n)( n) n 5n 6 6 ( n)( n) n 5 n ( 5) or n n * ( n )( n ) ( n )( n ) Attempt at least the first pair and the last pair of terms as shown. Must start at and end at n Identifies that the first and last terms do not cancel. Correctly combined fractions Allow either form isw attempts to multiply out the denominatot (4) Total 6
Question. Scheme Marks 6 Way 6 6 0 6 4 Attempt to combine fractions and 0 0 factorise the numerator, 4 Correct critical values, 4 Follow through their 4 ft Identifies as a critical value B : Attempt inside region : Correct inequality (7) Way 6 4 0 Multiplies both sides by and attempt to factorise, 4 Correct critical values, 4 Follow through their 4 ft Identifies as a critical value B : Attempt inside region : Correct inequality (7) Way 0 6 Multiplies both sides by and attempt to factorise 4 0 Must state 0 4 Correct critical values 4 Follow through their 4 ft Identifies as a critical value B 0 6 Correct critical value 6 4 0 : Attempt inside region : Correct inequality (7)
Question. r 5 Scheme Marks 5 6 (6 ) r Correct value for r B z 5 5 arg(6 6i ) Allow or 7 9 5,,, 5 B n, n,,, 4 At least values which must be positive. May be implied by correct final answers. i i 7 i i 9 i 5 5 5 5 5 e,e,e,e,e 5 5 i i or, e or e B (all 4 values) Total 5 (5)
Question 4 z w w z z w w z w w w 9 u v 4 u 4v Scheme z w z : Attempt to make z the subject : Correct epression for z : Uses z to obtain an equation in u and v Pythagoras must be used correctly. No i seen : Any correct equation in u and v Isw attempts to simplify Marks 5u 5v 8u 4 0 u 4 6 4 v 5 5 5 4 Centre,0 5 0 Rearrange to a suitable form for a circle and attempt centre and/or radius. -6/5 (from completing the square) may be omitted. May be implied by centre and radius correct for their previous equation oe Radius 6 5 oe (7) Total 7
Question 5 d y d Scheme dy 0 y (a) y y y y( 0) y y y y y y y( 0) y y y y y 4 y y, y 0 y 4 v (b) 0 0 0 (c) y y 4 6 ( 0.) ( 0.) ( 0.) 6 4 : Attempt to differentiate including use of the product rule on dy Equation may have been re-written as y'' =... before differentiating : Correct differentiation : Second use of product rule. Dependent on first. : Correct differentiation NB A simpler form is obtained if y ''' y '' 0 is used. Cao and cso B: y 0 : Attempts y 0 and y v 0 : All correct and obtained from correct epressions : Correct use of Maclaurin series : Fully correct epansion. Use of the correct Maclaurin series and substitution of 0. (y =) 0.597 Allow awrt Marks d B, Total (5) (5) ()
Question 6. d y ( ) y ( ) Integrating factor I Scheme dy ( ) y e d Divides by - may be implied by subsequent working Correct attempt at I, including an attempt at the integration. ln must be seen if not fully correct. ln e Correct epression d Marks e No errors in working allowed e y e e e e e c 4 ce y 4 Multiply through by their IF and integrate LHS. Can be given if y their IF is seen : Correct application by parts ie differentiate and attempt to e integrate : Correct epression : Complete the integration to a correct result. Constant not required. Oe Must have y =... Must include a constant. ft their previous line ft (9) Total 9
Question Scheme 7 P: z z z iy iy iy y y Q : w w i w iy iy iy i y y y : Correct use of Pythagoras : Any correct equation : Correct use of Pythagoras. Allow with u and v instead of and y : Any correct equation Must have and y now. Alternative : identifies perpendicular bisector of (0, 0) and (, -) : y = Marks or y y Attempt to solve simultaneously tie obtain an equation in one variable and get to =... or y =..., or y, Both (oe), and, Both (oe) Pairs must be clearly identifiable but coordinate brackets not needed. (7) Total 7
Question Scheme 8. e t d d d (a) y y t dt dy dy dy t dt e dt d y d d d y, y t dt dt d d d or y y, y dt dt d y dy 5 y 0 d y dy dy. 5 y 0 dt dt dt d y dy 4 y 0* dt dt Attempt to use an appropriate version of the chain rule Oe : Use of the product rule (penalise chain rule errors by loss of A mark or marks) Note dt t ln : Substitutes their first and second derivatives into the given differential equation Depends on both M marks above : Correct completion to printed answer Marks, dd (b) m 4m 0 m 4 6 5 m t y e Acost B sin t i Attempt to solve the auiliary equation Correct roots May be implied by a correct GS i i or e t y A Be Correct GS t ln B Acosln Bsin ln y or iln iln e y A Be (7) (5) Total
Question June 05 WFM0 Further Pure Mathematics F Mark Scheme Scheme Notes. 5 Seen anywhere in solution Critical Values and 5 Both correct BB; one correct BB0 0 5 4 0 5 4 5 0 Critical values 4 and 5 4, 5, 4, Attempt single fraction and factorise numerator or use quad formula Correct critical values May be seen on a graph or number line. d: Attempt an interval inequality using one of or 5 with another cv, : Correct intervals Can be in set notation One correct scores A0 Award on basis of the inequalities seen - ignore any and/or between them Set notation answers do not need the union sign. Marks B, B d, ALT Critical Values and 5 Seen anywhere in solution B, B 5 5 5 5 5 0 5, 4, Multiply by 5 5 4 0 and attempt to factorise a quartic or use quad formula Critical values 4 and Correct critical values d: Attempt an interval 5 4, inequality using one of or 5 with another cv, : Correct intervals Can be in set notation One correct scores A0 (7) d, (7) Any solutions with no algebra (eg sketch graph followed by critical values with no working) scores ma BB
Question (a) Scheme Notes Marks 6 8 r r oe r6r8 Correct partial fractions, any equivalent form (b)... 7 9 8 0 9 n 5 n 7 n 6 n 8 Epands at least terms at start and at end (may be implied) The partial fractions obtained in (a) can be used without multiplying by. Fractions may be etc These comments apply to both and 7 9 B () 7 8 n7 n8 n n n 56n7n8 n5n 56n7n8 5 7 8 56 5 Identifies the terms that do not cancel Attempt common denominator Must have multiplied the fractions from (a) by now cso (4) Total 5
Question Scheme Notes Marks dy (a) z y y z (b) dy z dz z dz y e y : dy kz dz : Correct differentiation z e z Substitutes for dy/ dz Correct completion to printed answer 4z e * with no errors seen cso (a) Alternative dz dz : ky y oe dy dy : Correct differentiation dz y y e y Substitutes for dy/ dz Correct completion to printed answer 4z e * with no errors seen dz y dz dy (a) Alternative dz dy : ky inc chain rule : Correct differentiation y y e y Substitutes for dy/ dz 4z e * 4 I e e ze e e Correct completion to printed answer with no errors seen : : I e e 4 e d d z I I c e q q pe c z ce e Or equivalent (c) ce e y y ce e y e () b ke Bft (4) (5) () Total 0
Question Scheme Notes Marks z w z 4(a) z w wz w z z... z Attempt to make z the subject w z w Correct epression in terms of w u iv u iv Introduces u + iv and multiplies top and bottom by the comple u iv u iv conjugate of the bottom u v, y v...... y v u v Uses real and imaginary parts and y = to obtain an equation connecting u and v Can have the on the wrong side. Processes their equation to a form that is recognisable as a circle u v 4 ie coefficients of u and v are the same and no uv terms 5 : Correct centre (allow -½i) Centre (0, ), radius : Correct radius, Special Case: iy i i w : rationalise the denominator, i y i i may have or y 4 i 4 : Correct result in terms of only. Must have rational denominator shown, but no other simplification needed (7) (b) R Bft: Their circle correctly positioned provided their equation does give a circle B: Completely correct sketch and shading Bft B () Total 9
Question Scheme Notes Marks 5 y cot (a) dy cosec d y d cosec cosec cot cosec cot cot cot * : Differentiates using the chain rule or product/quotient rule : Correct derivative : Correct completion to printed answer cot cosec or cos sin must be used Full working must be shown cso* Alternative: cos dy sin cos y sin sin sin d y sin cos... : Correct completion to printed answer see above () (b) d y cosec 6cot cosec Correct third derivative B cot 6cot cot Uses cot cosec 4 6cot 8cot cso (c) 4 8 6 f,f,f,f : Attempts all 4 values at No working need be shown () 4 4 8 y 9 : Correct application of Taylor using their values. Must be up to and including : Correct epression Must start y =... or cot f cot or y f() allowed provided defined here or above as Decimal equivalents allowed (min sf apart from 0.77), 0.578,., 0.770, (0.7698.., so accept 0.77) 0.889 Total 9 ()
Question Scheme Notes Marks 6(a) d y dy y sin AE: m m 0 m m m 0..., Forms Auiliary Equation and attempts to solve (usual rules) e y A Be Cao PI: y psin qcos Correct form for PI B y p cos qsin y psin qcos psin qcos pcos qsin psin q cos sin Differentiates twice and substitutes q 4p, 4q p 0 Correct equations p, q 5 5 both correct A0 one correct y cos sin 5 5 Follow through their p and q and their y Ae Be cos sin 5 5 CF Bft (8) (b) y Ae Be sin cos 5 5 Differentiates their GS : Uses the given conditions to give 0 A B, A B two equations in A and B 5 5 : Correct equations A, B 0 Solves for A and B Both correct y e e cos sin 0 5 5 Sub their values of A and B in their GS ft (5) Total
Question Scheme Notes Marks 7(a) r sin r cos Alternative: Equate rs: sin cos and verify (by substitution) that Attempt to verify coordinates in at least one of the polar equations Coordinates verified in both curves (Coordinate brackets not needed) is a solution or solve by using t tan or writing sin cos sin 6 Squaring the original equation allowed as is known to be between 0 and Use in either equation to obtain r () (b) ( sin ) d, ( cos ) d sin d, ( cos cos )d Correct formula used on at least one curve (/ may appear later) Integrals may be separate or added or subtracted. cos d, ( cos cos )d Attempt to use sin or cos cos on either integral Not dependent / may be missing sin, sin sin 4 4 0 Correct integration (ignore limits) or A0 0 R 4 4 8 Correct use of limits for both integrals Integrals must be added. Dep on both previous M marks 4 Cao No equivalents allowed, dd (6) Total 8
Question Scheme Notes Marks 8(a) z z z z z z 6 6 z z z z 6 z z 6 z z (a) ALT : Attempt to epand : Correct epansion Correct answer with no errors seen, z z z z z z z z z z z z : Attempt to epand both cubic brackets : Correct epansions 6 z z 6 z z Correct answer with no errors (b)(i)(ii) n z cos n isin n Correct application of de Moivre B n z cos n isin n cos n sin n Attempt z -n but must be different from their z n z n cos n *, z n i sin n n * n n z cos n isin n must be seen * z z (c) (d) z z cos i sin z z 6 z z i sin 6 6i sin 6 Follow through their k in place of z z 64i sin cos i sin 6 6i sin Equating right hand sides and simplifying i (B mark needed for each side to gain M mark) cos sin sin sin 6 * cso 8 8 4 5 6 6 6 cos sin d sin sin 6 d 0 0 : pcos q cos 6 8 cos cos6 : Correct integration 6 0 Differentiation scores M0A0 d: Correct use of limits lower limit to have non-zero result. Dep on previous M mark : Cao (oe) but must be eact B Bft d () () () (4) (4) Total 4
Question 9. Scheme Marks (a) a asin sin... C = C and attempt to solve for 5 sin, 6 6 5 a,, a, 5 or or both Decimals 6 6 allowed (min sf). Both points 5 Can be written r a,, Decimals allowed (min sf). () (b) d sin d r a cos 4 sin a oe Correct epression for the sector B sin cos 4 cos 4 d sin 4 4 I a sin 4 4 sin 4. 0 4 a a a R I a 6 8 6 R a 4 0 Use of correct formula Limits not needed (ignore any shown) cos 4 Uses sin Correct integration Limits not needed (ignore any shown) An attempt to find one or both of the regions either side of the sector. 5 ie uses limits 0, and/or,, limits to be substituted and subtracted (if non-zero after substitution). Limits to be used the correct way round. If two integrals seen award mark if either correct. Both previous method marks must have been scored. Correct strategy for the complete area (sector + I). All areas must be positive. If decimals seen anywhere (either in rt or the limits) this mark is lost. dd Total 0 (7)
JUNE 06 F MARK SCHEME Question (a) (b) (c) Scheme Notes Marks or or equivalent or A B A, B 4r r r r r r r 4r Correct partial fractions or correct values of A and B. Isw if possible so if correct values of A and B are found, award when seen even if followed by incorrect partial fractions. n... r 4r n n Attempt at least first and last terms using their partial fractions. May be implied by e.g. n or or n Correct epression n 4n n n * Correct completion with no errors * Allow a different variable to be used in (a) and (b) but final answer in (b) must be as printed i.e. in terms of n. 5 5f (5) f (8) f (5) f (8) where f n r9 5 4r 5 8 5 5 5 7 5 cao Correct answer with no working in (c) scores both marks. n n B () () () Total 6
Question Scheme 9 Notes (ignore use of < instead of = when finding cv s) 9 0 0... Attempts to solve or 9 OR 9 9 8 0... to obtain two non-zero values of One correct pair of values. Allow the 44 irrational roots to be at least as given OR, 4 here or or awrt., -4. or truncated., -4. Attempts to solve 9 8 0... 9 AND 9 to obtain four non-zero values of Both pairs of values correct. Allow 44 the irrational roots to be at least as AND, 4 given here or or awrt., -4. or truncated., -4. For allow For allow 4 or 44 Allow alternative notation e.g., for allow One correct inequality. 44 but must be eact here., 4,, 4, and 4,, and 4 and 44 Allow alternative notation e.g., for allow Both inequalities correct. 44 but must be eact here., 4,, 4, and 4,, and B B Marks (6) Total 6
Q Alternative by squaring (ignore use of < instead of = when finding cv s) 4 9 8 8 4 4 4 4 80 0... 44 or, 4 44 and, 4 4 or 4 and Squares and attempts to solve a quartic equation to obtain at least two values of that are non-zero. One pair of values correct as defined above : Obtains four non-zero values of. : Both pairs of values correct as defined above See notes above See notes above In an otherwise fully correct solution, if any etra incorrect regions are given, deduct the final B mark. B B
Scheme Notes Marks dy ky k dy ky k k I e k y k Divides by ( + ) including the ky term d: Attempt integrating factor. k I e is sufficient for this mark but must include the k. Condone omission of. : k Reaches k y their I their I 5 c 5 or by parts Correct integration 5 4 c 5 5 c y 5 k y or e.g. 4 5 c 5 k or e.g. 4 5 y c 5 or e.g. k k k 5 0 4 c y k 5 or e.g. k 5 k k y c 5 Correct answer with the constant correctly placed. Allow any equivalent correct answer. d (6) Total 6
Question 4(a) (b) f f f f f Scheme Notes Marks sin : Attempt first 4 derivatives. Should be ' sin cos sin cos sin. I.e. ignore cos signs and coefficients. '' 9 4sin ' '' 9 :f cos and f 4sin ''' 7 8 cos ''' 7 '''' 8 :f 8 cos and f 6 sin '''' 8 6 sin Allow un-simplified e.g. f ''. sin ' '' 9 ''' '''' 8 y, y 0, y, y 0, y 4 6 Attempts at least derivative at d: Correct use of Taylor series. I.e. f f f f... 4 9 7 f 8 8 f 0.485 Evidence of at least one term of the correct n f n structure i.e. and not a n! Maclaurin series. Dependent on the previous method mark. : Correct epansion. Allow equivalent single fractions for 9 7 and/or and allow 8 8 decimal equivalents i.e..5 and 0.0975. Ignore any etra terms.! : Attempts f or states : 0.485 cao A d (6) () Total 8
Question 5 z Scheme Notes Marks w z w w uiv w u iv u v u v 0 u v u 0 8 8 u v 5 5 8 64 8 v 5 49 5 7 u,0, 8 64 8 8 : Attempt to make z the subject as far as z = : Correct equation Uses z and introduces w = u + iv : Correct use of Pythagoras. Condone missing brackets provided the intention is clear and allow e.g. (v) = v but there should be no i s. Attempt to complete the square on the equation of a circle. I.e. an equation where the coefficients of u and v are the same and the other terms are in u, v or are constant. (Allow slips in completing the square). Dependent on all previous M marks. 5 : Centre,0 8 : Radius 7 8 dddd Alternative for the first marks w : Attempt to make z the subject z w : Correct equation 5u u v u iv u iv u iv 7v iy u iv u iv u iv u v u v 5u u v 7v y u v u v Introduces w = u + iv, multiplies numerator and denominator by the comple conjugate of the denominator and uses y correctly to obtain an equation in u and v. i (7) Total 7
Question Scheme Notes Marks d y dy y 6(a) Correct roots (may be implied by their m m 0 m, CF) B y Ae Be : CF of the correct form : Correct CF y a b c Correct form for PI B dy d y a b, a a a b a b c : Differentiates twice and substitutes into the lhs of the given differential equation and puts equal to or substitutes into the lhs of the given differential equation and compares coefficients with. For the substitution, at least one of y, y or y must be correctly placed. a 7 7 : Solves to obtain one of b or c 6a b b c 4 : Correct b and c 7 7 Correct ft (their CF + their PI) but y Ae Be 4 must be y = Bft (9) (b) 7 Substitutes = 0 and y = 0 into their 0 A B 4 GS dy 7 7 Ae Be 0 A B Attempts to differentiate and substitutes = 0 and y = 0 7 7 Solves simultaneously to obtain 0 A B, 0 A B A.., B.. 4 values for A and B A, 4 B 5 Correct values 7 7 Correct ft (their CF + their PI) y e 5e 4 4 but must be y = Bft (5) Total 4
Question Scheme Notes Marks 7. 9 C: r cos, C: r cos (a) 9 cos cos... Puts C = C and attempt to solve for or or r Eliminates cos θ and solves for r cos r r r... Correct θ or correct r. or r 6 4 Allow θ = awrt 0.54, r = awrt. Correct r and θ (isw e.g., ) r and 6 4 4 6 Allow θ = awrt 0.54, r = awrt. ()
7(b) 9 cos d or cos d Attempts to use correct formula on either curve. The ½ may be implied by later work. 9 8 8 cos cos 7 7 cos cos 7 7 cos 4 4 Epands to obtain an epression of the form a bcos ccos and attempts to use cos cos 9 97 8 cos d 7 sin sin 8 6 : Attempts to integrate to obtain at least two terms from, sin, sin 97 8 : Correct integration with or without the ½ (NB 7 ) 8 8 6 97 8 97 8 7 sin sin. 7.sin sin. 0 8 6 0 8 6 6 6 6 : Uses the limits 0 and their 6 If the substitution for = 0 evaluates to 0 then the substitution for = 0 does not need to be seen but if it does not evaluate to 0, the substitution for = 0 needs to be seen. 9 9 9 6 6 6 4 cos d cos d sin 6 cos : Uses cos, integrates to obtain at least k sin and uses the limits of their 6 and to find the other area NB can be done as a segment : Allow sin 4 4 their sin their 4 6 4 6 97 5 9 05 45 96 64 6 4 8 : Adds their two areas both of which are of the form a b : Correct answer (allow equivalent fractions for 05 45 and/or 8 ) (8) Total
Question 8(a) WAY Scheme 5 5 z z 5z 0z Notes : Attempt to epand z 5 z 0 5 5 z z z z : Correct epansion with correct powers of z. z cos isin z cos May be implied B 5 z 5 z 5 z 0 z cos5 0cos 0cos 5 z z z Uses at least one of z 5 cos5 or z cos 5 z z 5 5 z cos B z 5 5 5 cos cos5 cos cos Correct epression 6 6 8 WAY (Using e e e 5e 0e 0e 5e e i i 5i i i i i 5i i e ) : Attempt to epand i i e e 5 : Correct epansion i i cos e e May be implied B 5i 5i i i i i e e 5 e e 0 e e cos5 0cos 0cos Uses one of 5 5i 5i i i e e cos5 or e e cos i i 5 e e cos B 5 5 5 cos cos5 cos cos Correct epression 6 6 8 Marks (6) WAY (Using De Moivre on cos 5θ and identity for cos θ) 5 5 4 4 4 5 5 cos isin c 5ic s 0c i s 0c i s 5ci s i s : Attempts to epand. NB may only consider real parts here. : Correct real terms (may include i s) (Ignore imaginary parts for this mark) 5 4 cos5 cos 0cos sin 5cos sin Correct real terms with no i s B cos 5 0cos cos 5cos cos 5 6cos cos5 0cos 5cos cos 4cos cos Uses sin cos to eliminate sin θ Correct identity for cos θ B 5 6cos cos5 5cos 0cos 5 5 5 cos cos5 cos cos Correct epression 6 6 8 (6) (b) 5 5 5 5 cos5 cos cos d sin 5 sin sin ft 6 6 8 80 48 8
: Attempt to integrate Evidence of cos n sin n where n 5 or n ft: Correct integration (ft their p, q, r) 5 5 5 5 5 5 5 5 sin 5 sin sin sin sin sin sin sin sin 80 48 8 80 48 8 80 6 48 8 6 6 Substitutes the given limits into a changed function and subtracts the right way round. There should be evidence of the substitution of and 6 into their changed function for at least of their terms and subtraction. If there is no evidence of substitution and the answer is incorrect, score M0 here. Allow eact equivalents e.g. 49 0 0 60 480 4.9 6 0 If they use the letters p, q and r or their values of p, q and r, even from no working, the M marks are available in (b) but not the A marks. (4) Total 0
Question + - 9(a) (b) Scheme Notes Marks Centre C, y is at.5,.5 c c r.5.5 Ma z OC r.5.5 r z 5 arg z 4 A circle or an arc of a circle anywhere. Allow dotted or dashed. A circle or an arc of a circle (allow dotted or dashed) passing through or touching at and 5 on the positive real ais. (Imaginary ais may be missing) Fully correct diagram with and 5 marked correctly with no part of the circle below the real ais. It must be a major arc and not a semi-circle. The imaginary ais must be present and the arc must not cross or touch the imaginary ais. () May be implied and may appear on the diagram. Can score anywhere e.g. from finding the equation of the B circle in part (a) or as part of the calculation for OC. r yc or equivalent work e.g..5.5 r, r or cos 45 sin 45 seen 58 Oe e.g Special Case correct work with arc below the real ais: Centre C, y is at.5,.5 c c r.5.5 Ma 5 z OC r.5.5 r 58 4.5 4.5, B0 r yc or equivalent work e.g..5.5 r, r or cos 45 sin 45 seen 58 Oe e.g 58 4.5 4.5, Total 7 (4)