PhysicsAndMathsTutor.com

Size: px
Start display at page:

Download "PhysicsAndMathsTutor.com"

Transcription

1 . The cartesian equation of the circle C is (a) Find the coordinates of the centre of C and the radius of C. (b) Sketch C. (4) () (c) Find parametric equations for C. () (d) Find, in cartesian form, an equation for each tangent to C which passes through the origin O. (5) (Total 4 marks). The diagram above shows a sketch of the curve C with parametric equations 5t 4, t(9 t ) The curve C cuts the -ais at the points A and B. (a) Find the -coordinate at the point A and the -coordinate at the point B. () Edecel Internal Review

2 The region R, as shown shaded in the diagram above, is enclosed b the loop of the curve. (b) Use integration to find the area of R. (6) (Total 9 marks). The diagram above shows a sketch of the curve with parametric equations cos t, 6sint, 0 t (a) Find the gradient of the curve at the point where t. (4) (b) Find a cartesian equation of the curve in the form f(), k k, stating the value of the constant k. (4) Edecel Internal Review

3 (c) Write down the range of f (). () (Total 0 marks) 4. (a) Using the identit cosθ sin θ, find sin θ dθ. () The diagram above shows part of the curve C with parametric equations tanθ, sinθ, 0 θ < The finite shaded region S shown in the diagram is bounded b C, the line and the - ais. This shaded region is rotated through radians about the -ais to form a solid of revolution. (b) Show that the volume of the solid of revolution formed is given b the integral k sin 6 0 θ dθ where k is a constant. (5) Edecel Internal Review

4 (c) Hence find the eact value for this volume, giving our answer in the form p + q, where p and q are constants. () (Total 0 marks) 5. The curve C shown above has parametric equations t 8t, t where t is a parameter. Given that the point A has parameter t, (a) find the coordinates of A. () The line l is the tangent to C at A. (b) Show that an equation for l is (5) The line l also intersects the curve at the point B. (c) Find the coordinates of B. (6) (Total marks) Edecel Internal Review 4

5 6. l P C R O 4 The diagram above shows the curve C with parametric equations 8cost, 4sin t, 0 t. The point P lies on C and has coordinates (4, ). (a) Find the value of t at the point P. () The line l is a normal to C at P. (b) Show that an equation for l is + 6. (6) The finite region R is enclosed b the curve C, the -ais and the line 4, as shown shaded in the diagram above. (c) Show that the area of R is given b the integral 64 sin t cos t. (4) (d) Use this integral to find the area of R, giving our answer in the form a + b, where a and b are constants to be determined. (4) (Total 6 marks) Edecel Internal Review 5

6 7. C R O ln ln 4 The curve C has parametric equations ln( t + ),, t > ( t + ) The finite region R between the curve C and the -ais, bounded b the lines with equations n and n 4, is shown shaded in the diagram above. (a) Show that the area of R is given b the integral. ( t + )( t + ) 0 (4) (b) Hence find an eact value for this area. (6) (c) Find a cartesian equation of the curve C, in the form f(). (4) (d) State the domain of values for for this curve. () (Total 5 marks) Edecel Internal Review 6

7 8. A curve has parametric equations tan t, sin t, 0 < t <. (a) Find an epression for in terms of t. You need not simplif our answer. () (b) Find an equation of the tangent to the curve at the point where t. 4 Give our answer in the form a + b, where a and b are constants to be determined. (5) (c) Find a cartesian equation of the curve in the form f(). (4) (Total marks) O 0.5 The curve shown in the figure above has parametric equations sin t, sin ( t + ), < t <. 6 (a) Find an equation of the tangent to the curve at the point where t. 6 (6) (b) Show that a cartesian equation of the curve is + ( ), < < () (Total 9 marks) Edecel Internal Review 7

8 0. a A O B a The curve shown in the figure above has parametric equations a cost, a sin t, 0 t The curve meets the aes at points A and B as shown.. 6 The straight line shown is part of the tangent to the curve at the point A. Find, in terms of a, (a) an equation of the tangent at A, (6) (b) an eact value for the area of the finite region between the curve, the tangent at A and the -ais, shown shaded in the figure above. (9) (Total 5 marks). R O The curve shown in the figure above has parametric equations t sin t, cos t, 0 t Edecel Internal Review 8

9 5 (a) Show that the curve crosses the -ais where t and t. () The finite region R is enclosed b the curve and the -ais, as shown shaded in the figure above. (b) Show that the area of R is given b the integral (c) 5 ( cost ). Use this integral to find the eact value of the shaded area. () (7) (Total marks). C O The curve C has parametric equations,, t <. + t t (a) Find an equation for the tangent to C at the point where t. (7) (b) Show that C satisfies the cartesian equation. () Edecel Internal Review 9

10 The finite region between the curve C and the -ais, bounded b the lines with equations and, is shown shaded in the figure above. (c) Calculate the eact value of the area of this region, giving our answer in the form a + b ln c, where a, b and c are constants. (6) (Total 6 marks). A curve has parametric equations cot t, sin t, 0 < t. (a) Find an epression for in terms of the parameter t. (4) (b) Find an equation of the tangent to the curve at the point where t 4. (4) (c) Find a cartesian equation of the curve in the form f(). State the domain on which the curve is defined. (4) (Total marks) 4. C P R O a Edecel Internal Review 0

11 The diagram above shows a sketch of the curve C with parametric equations t sin t, sec t, 0 t <. The point P(a, 4) lies on C. (a) Find the eact value of a. () The region R is enclosed b C, the aes and the line a as shown in the diagram above. (b) Show that the area of R is given b 6 0 (tan t + t). (4) (c) Find the eact value of the area of R. (4) (Total marks) 5. (metres) C R A B (metres) The diagram above shows a cross-section R of a dam. The line AC is the vertical face of the dam, AB is the horizontal base and the curve BC is the profile. Taking and to be the horizontal and vertical aes, then A, B and C have coordinates (0, 0), (, 0) and (0, 0) respectivel. The area of the cross-section is to be calculated. Initiall the profile BC is approimated b a straight line. Edecel Internal Review

12 (a) Find an estimate for the area of the cross-section R using this approimation. () The profile BC is actuall described b the parametric equations. 6t, 0 sin t, t. 4 (b) Find the eact area of the cross-section R. (7) (c) Calculate the percentage error in the estimate of the area of the cross-section R that ou found in part (a). () (Total 0 marks) 6. The curve C is described b the parametric equations cos t, cos t, 0 t. (a) Find a cartesian equation of the curve C. (b) Draw a sketch of the curve C. () () (Total 4 marks) 7. Edecel Internal Review

13 The curve shown in the diagram above has parametric equations cos t, sin t, 0 t <. (a) Find an epression for in terms of the parameter t. () (b) Find the values of the parameter t at the points where d 0. () (c) (d) Hence give the eact values of the coordinates of the points on the curve where the tangents are parallel to the -ais. Show that a cartesian equation for the part of the curve where 0 t < is ( ). () () (e) Write down a cartesian equation for the part of the curve where t <. () (Total marks) Edecel Internal Review

14 . (a) ( 4) 6 + ( ) ( 4) ( ) 9 M A Centre (4, ), radius A A 4 (b) (4, ) O 4 B, B (c) 4 + cos t + sin t (0 t < ) M A A t (4, ) (d) Line through origin m -coordinate of points where this line cuts C satisfies ( + m ) 8 6m M A As line is tangent this equation has repeated roots m 0, m (8 + 6m) 4( + m ) m + 4m 6 + 6m 4m 7m 4 7 M A Equations of tangents are 0, 4 A 5 7 [4] Edecel Internal Review 4

15 . (a) 0 ( 9 ) ( )( ) 0 t t t t + t t 0,, An one correct value B At t 0, 5 (0) 4 4 At t, 5 () 4 4 ( At t, 5( ) 4 4) Method for finding one value of At A, 4; at B, 4 Both A M (b) d 0t Seen or implied B t( 9 t ) 0t M A ( 90t 0t ) 90 5 t 0t ( 4) M A 648 (units ) A 6 [9]. (a) d 4sin t, 6 cos t B, B d 6cost M 4sin t 4sin t At t, m 4 accept equivalents, awrt 0.87 A 4 Alternatives to (a) where the parameter is eliminated ( 8 9) d (8 9) ( 9) At t, cos B B 9 (7) M A 4 Edecel Internal Review 5

16 8 9 9 B At t, 6sin B 9 M A 4 (b) Use of cost sin t M cos t,sin t 6 6 M Leading to ( 8 9)( ( ) ) cao A k B 4 (c) 0 f() 6 either 0 f() or f() 6 B Full correct. Accept 0 6, [0, 6] B [0] sin θ θ θ θ θ θ C) M A 4 4. (a) d ( cos ) d sin ( + (b) tan θ sec dθ θ dθ (sin θ ) sec θ dθ M A dθ ( sinθ cosθ ) dθ M cos θ 6 sin θ dθ k 6 A 0 tanθ 0 θ 0, tanθ θ B V sin θ dθ 0 Edecel Internal Review 6

17 (c) 6 sin θ V 6 θ 4 M 0 6 sin (0 0) Use of correct limits *M p, q A 8 [0] 5. (a) At A, & ( ) A(7,) A(7, ) B (b) t 8t, t, t d 8, t t Their d t 8 t d divided b their M Correct A ( ) At A, m(t) ( ) Substitutes for t to give an of the four underlined oe: T: (their ) m r ( (their 7)) Finding an equation of a tangent with their point and their tangent gradient or 9 ( 7) + c c or finds c and uses dm (their gradient) + c. 9 Hence T: 5 5 gives T: AG A cso 5 (c) (t 8t) 5t 9 0 Substitution of both t 8t and t into T t 5t 6t 9 0 (t + ){(t 7t 9) 0} (t + ){(t + )(t 9) 0} A realisation that (t + )is a factor. dm 9 {t (at A) t at B} t 9 A M Edecel Internal Review 7

18 8 8 ( 9 ) 8( 9 44 ) or awrt 55. Candidate uses their value of t to find either the or coordinate ddm 9 8 ( ) 0.5 or awrt 0. One of either or correct. A 44 8 Hence (, ) Both and correct. A 6 B awrt [] 6. (a) At P(4, ) either 4 8cost or 4 sin t onl solution is t where 0 t 4 8cost or 4 sin t M t or awrt.05 (radians) onl stated in the range 0 t A (b) 8 cos t, 4 sin t 8sin t, 8 cos t 8 cos( ) At P, 8sin ( ) ( ) 8 awrt 0.58 ( ) ( 8) Hence m(n) or N: ( 4) N: + 6 AG or (4) + c c so N: [ ] Edecel Internal Review 8

19 Attempt to differentiate both and wrt t to give ±p sin t and ±q cos t respectivel M Correct and A Divides in correct wa round and attempts to substitute their value of t (in degrees or radians) into their epression. M* You ma need to check candidate s substitutions for M* Note the net two method marks are dependent on M* Uses m(n) dm* their m( T) Uses (their m N )( 4) or finds c using 4 and and uses (their m N ) + c. dm* + 6 A cso AG 6 (c) A A A A sin t.( 8sin t) sin t.sin t 64.sin t cos t 64.sin t cos t ( sin t cos t).sin t attempt at A M correct epression (ignore limits and ) A Seeing sin t sin t cos t anwhere in this part. M Correct proof. Appreciation of how the negative sign affects the limits. A AG 4 Note that the answer is given in the question. Edecel Internal Review 9

20 (d) {Using substitution u sin t {change limits: du cost} when t, u & when t, u } sin A 64 A 64 A 64 t 8 (Note that a. u or A , b 8) k sin t or ku with u sin t Correct integration ignoring limits. M A Substitutes limits of either (t and t ) or (u and u ) and subtracts the correct wa round. dm 64 8 A aef isw 4 Aef in the form a + b, with awrt. and anthing that cancels to a 64 and b 8. [6] 7 (a) ln( t + ),, t + t + ln 4 Area (R) ; ln t + 0 t + t + Changing limits, when: ln ln ln(t + ) t + t 0 ln4 ln4 ln(t + ) 4 t + t Hence, Area (R) 0 ( t + )( t + ) Edecel Internal Review 0

21 Must state d t t + B Area. Ignore limits. t + M;. Ignore limits. t + t + A AG changes limits t so that ln 0 and ln4 B 4 (b) A B + ( )( ) t + t + ( t + ) ( t + ) A(t + ) + B(t + ) Let t, A() A Let t, B( ) B 0 ( t + )( t + ) 0 ( t + ) ( t + ) [ ln( t + ) ln( t + ) ] 0 (ln ln 4) (ln ln ) Takes out brackets: ln ln 4 + ln ln ln ln A B + with A and B found ( t + ) ( t + ) Finds both A and B correctl. Can be implied. Writing down + ( t + )( t + ) ( t + ) ( t + ) Writing down ( t + )( t + ) ( t + ) ( t + ) means first MA0. means first MA. M A Either ± a ln(t + ) or ± b ln(t + ) dm Both ln terms correctl ft. Aft Substitutes both limits of and 0 and subtracts the correct wa round. ddm ln ln 4 + ln or ln ln or ln ln or ln 4 (must deal with ln ) A aef isw 6 Edecel Internal Review

22 (c) ln(t + ), t + e t + t e e + e Attempt to make t the subject giving t e Eliminates t b substituting in giving e M A dm A 4 Aliter Wa t + t or t (t + ) t + t t ln + or ln + ln + e + e e Attempt to make t the subject Giving either t or t Eliminates t b substituting in giving e M A dm A 4 Edecel Internal Review

23 Aliter Wa e t + t + e t + e Attempt to make t + the subject giving t + e Eliminates t b substituting in giving e M A dm A 4 Aliter Wa 4 + t + t + + or t + + ln + or ln ln + e + e e Attempt to make t + the subject + Either t + + or t + Eliminates t b substituting in giving e M A dm A 4 (d) Domain: > 0 > 0 or just > 0 B [5] Edecel Internal Review

24 8. (a) tan t, sin t (tan t) sec t, cos t (*) 4 cos t cos t tan t sec t sin t B Correct and ± cos t M their + cos t Aft (*) their (b) When t,, (need values) 4 cos When t, m( T ) 4 4 d tan sec 4 4.()().() 4.() B, B The point (, ) or (, awrt 0.7) These coordinates can be implied. ( sin is not sufficient for B) 4 B aef an of the five underlined epressions or awrt Edecel Internal Review 4

25 T: ( ) (Note: The and coordinates must be 4 the right wa round.) T: + or + (*) or () + c c (*) A candidate who incorrectl differentiates tan d t to give sec t d or sec 4 t is then able to fluke the correct answer in part (b). Such candidates can potentiall get: (a) B0MAft (b) BBBMA0 cso. Note: cso means correct solution onl. Note: part (a) not full correct implies candidate can achieve a maimum of 4 out of 5 marks in part (b). Hence T: + or Mft aef Finding an equation of a tangent with their point and their tangent gradient or finds c using (their gradient) + c. A aef cso Correct simplified EXACT equation of tangent (c) Wa tan sin t t sin t cos t sin t sin t ( ) + ( + ) + M Uses cos t sin t M Eliminates t to write an equation involving and. ddm Rearranging and factorising with an attempt to make the subject. A + 4 Edecel Internal Review 5

26 Aliter Wa + cot t cosec t sin t Hence, + Hence, or ( + ) + M Uses + cot t cosec t M implied Uses cosec t sin t ddm Eliminates t to write an equation involving and. A ( or + ) + Aliter Wa tan t sin t + tan t sec t cos t sin t Hence, + Hence, M ( or + ) + Uses + tan t sec t M Uses sec t cos t ddm Eliminates t to write an equation involving and. A ( or + ) + Edecel Internal Review 6

27 Aliter Wa 4 sin t cos t sec t ( + tan t) Hence, ( + ) M or + Uses sin t cos t M Uses cos t sec ddm then uses sec t + tan t A ( or + ) + t + is an acceptable response for the final accurac A mark. Edecel Internal Review 7

28 Aliter Wa 5 tan t tan t tan t sin t ( + ) t Hence, sin t + M Draws a right-angled triangle and places both and on the triangle M Uses Pthagoras to deduce the hpotenuse ddm Eliminates t to write an equation involving and A + is an acceptable response for the final accurac A mark. + There are so man was that a candidate can proceed with part (c). If a candidate produces a correct solution then please award all four marks. if the use a method commensurate with the five was as detailed on the mark scheme then award the marks appropriatel. If ou are unsure of how to appl the scheme please escalate our response up to our team leader. [] 9. (a) sin t sin(t + 6 ) M Attempt to differentiate both and wrt t to give two terms in cos cost, cos ( t + ) 6 d Correct and A Edecel Internal Review 8

29 When t, 6 cos ( + ) 6 6 cos ( ) 6 awrt 0.58 Divides in correct wa and substitutes for t to give an of the four underlined oe: Ignore the double negative if candidate has differentiated sin cos A when t,, 6 The point ( ), or (, 0.87 ) awrt B T: ) ( Finding an equation of a tangent with their point and their tangent gradient or finds c and uses (their gradient) + c. dm Correct EXACT equation of tangent oe. A oe + c c or ( ) 6 or T: [ + ] 6 (b) sin (t + 6 ) sin t cos 6 + cos t sin 6 M Use of compound angle formula for sine. Nb: sin t + cos t cos t sin t sin t gives cos t ( ) M Use of trig identit to find cos t in terms of or cos t in terms of. Edecel Internal Review 9

30 sin t + cost gives + ( ) AG A cso Substitutes for sin t, cos 6, cost and sin 6 to give in terms of. Aliter Wa (a) sin t sin (t + ) sin t + cos cos t sin 6 M (Do not give this for part (b)) d Attempt to differentiate and wrt t to give in terms of cos and in the form ± a cost ± b sin t [9] d cos t; cost cos sin t sin 6 6 A d Correct and When t 6, cos cos sin sin cos 6 Divides in correct wa and substitutes for t to give an of the four underlined oe A When t,, 6 B The point, or,awrt 0.87 T: Finding an equation of a tangent with their point and their tangent gradient or finds c and uses (their gradient) + c. dm Correct EXACT equation of tangent oe. A oe or ( ) + c c 6 or T: [ + ] Edecel Internal Review 0

31 Aliter Wa (a) + ( ) d + ( ) ( ) Attempt to differentiate two terms using the chain rule for the second term. Correct M A + ( (0.5) ) d ( (0.5)) Correct substitution of into a correct A When t, 6,, or (, 0.87 ) awrt The point ( ) B T: ) ( Finding an equation of a tangent with their point and their tangent gradient or finds c and uses dm (their gradient) + c Correct EXACT equation of tangent A oe oe. + c c or ( ) 6 or T: [ + ] 6 Aliter Wa (b) sin t gives sin t + ( sin t) M Substitutes sin t into the equation give in. Nb: sin t + cos t cos t sin t t Cost ( sin ) M Use of trig identit to deduce that cos t ( sin ) t gives sin t + cost Edecel Internal Review

32 Hence sin t cos + cost sin sin ( t + ) A cso Using the compound angle formula to prove sin ( t + ) 6 0. (a) a sin t, a cost therefore cost sin t M A When 0, t 6 B Gradient is M 6 Line equation is ( a) ( 0) M A 6 6 (b) Area beneath curve is a sin t ( a sin t) M a (cos t cos 4t) M a [ sin t sin 4t] M A 4 Uses limits 0 and a to give 6 6 A Area of triangle beneath tangent is a a a M A 4 Thus required area is a 4 a a A N.B. The integration of the product of two sines is worth marks (lines and of to part (b)) If the use parts sin t sin t cos t sin t + cost cost M cos t sin t + cos t sin t + 9 sin t sin t 8I cost sint cost sint M A [5]. (a) Solves 0 cost to obtain t or 5 (need both for A) M A Or substitutes both values of t and shows that 0 Edecel Internal Review

33 (b) cost M A 5 Area ( 5 cos t ) ( cos t) ( cost) AG B (c) Area 4 cost + 4 cos t terms M 4 cost + (cos t + ) (use of correct double angle formula) M 4 cost + cos t M A [t 4 sin t + sin t] M A Substitutes the two correct limits t 5 and and subtracts. M 4 + AA 7 []. (a) and B, B ( + t) ( t) ( + t) and at t ½, gradient is 9 M Acao ( t) M requires their / / their / and substitution of t. At the point of contact and B Equation is -9 ( ) M A 7 Edecel Internal Review

34 (b) Either obtain t in terms of and i,e, t or t (or both) M Then substitute into other epression f() or g() and rearrange (or put and rearrange) M To obtain (*) a ( + t) Or Substitute into M + t ( + t) t A (*) M (c) Area u + du du u 4 u putting into a form to integrate + B M u + ln u 4 4 M A + ln 4 4 M + ln or an correct equivalent. 6 4 A 6 [6] Or Area + putting into a form to integrate B M + ln( ) 4 M A Edecel Internal Review 4

35 ln ln dm A Or Area t ( + t) B A B C + + ( t) ( + t) ( + t) M putting into a form to integrate ln( t ) ln( + t) + ( + t) 4 4 M aft Using limits 0 and ½ and subtracting (either wa round) dm + ln or an correct equivalent. 6 4 A 6 Or Area then use parts ln( ) ( ) M ln( ) ( ) ln( ) 4 MA ln ln + DM ln 6 4 A 6 B. (a) d cosec d t, 4 sin t cos t both M A sin t cost ( sin t cos t) M A 4 cosec t (b) At t 4,, B both and Substitutes t into an attempt at to obtain gradient 4 M Equation of tangent is ( ) M A 4 Accept + 4 or an correct equivalent Edecel Internal Review 5

36 (c) Uses + cot t cosec t, or equivalent, to eliminate t M + A correctl eliminates t cao A The domain ir 0 B 4 Alternative for (c): sin t ; cos t sin t sin t + cos t + 4 M A 8 Leading to 4 + A [] 4. (a) 4 sec t cos t, t M, A a sin B (b) a A 0 d M Change of variable sect [sint + t cost] M d Attempt (6 tan t, + 6t) (*) A, Acso 4 0 Final A requires limit stated (c) A [6 ln sec t + t ] 0 Some integration (M) both correct (A) ignore lim. M, A (6 ln + 9 ) (0) Use of M 6 ln + A 4 [] Edecel Internal Review 6

37 5. (a) Area of triangle 0 ( 444.) B Accept 440 or 450 (b) Either Area shaded 0sin t.t M A 4 + [ 480t cos t 480 cos t] M A 4 [ 480t cos t + 40sin t] 4 A ft 40( ) M A 7 or 4 60 cos t.(6t ) M A [(0 sin t ( 6t ) 480t cos t + 480cos t ] M A [ 480t cos t + 40sin t] 4 A ft 40( ) M A 7 4 (c) 40( ) estimate Percentage error 00.6% 40( ) M A (Accept answers in the range.4% to 4.4%) [0] 6. (a) Attempt to use correctl stated double angle M formula cos t cos t, or complete method using other double angle formula for cost with cos t + sin t to eliminate t and obtain or an correct equivalent.(even cos(cos ) A (b) Edecel Internal Review 7

38 4 4 shape B position including restricted domain < < B [4] 7. (a) d d d cos sin t, cos t d sin tt M A A 5 7 (b) cos t 0 t,,, 5 7 So t,,, M A A (c),), ),), ) M A (d) sin t cos t M cos t cost M A (e) B [] Edecel Internal Review 8

39 . No Report available for this question.. Part (a) was well done. The majorit of candidates correctl found the -coordinates of A and B, b putting 0, solving for t and then substituting in 5t 4. Full marks were common. Part (b) proved difficult. A substantial minorit of candidates failed to substitute for the when d substituting into d or used rather than. A surprising feature of the solutions seen d t 9 t 0t, were unable was the number of candidates who, having obtained the correct ( ) 4 to remove the brackets correctl to obtain ( 90t 0t ). Weaknesses in elementar algebra flawed man otherwise correct solutions. Another source of error was using the - coordinates for the limits when the variable in the integral was t. At the end of the question, 4 man failed to realise that ( 90t 0t ) gives onl half of the required area. 0 Some candidates made either the whole of the question, or just part (b), more difficult b eliminating parameters and using the cartesian equation. This is a possible method but the indices involved are ver complicated and there were ver few successful solutions using this method.. Nearl all candidates knew the method for solving part (a), although there were man errors in d differentiating trig functions. In particular (cost) was often incorrect. It was clear from both this question and question that, for man, the calculus of trig functions was an area of weakness. Nearl all candidates were able to obtain an eact answer in surd form. In part (b), the majorit of candidates were able to eliminate t but, in manipulating trigonometric identities, man errors, particularl with signs, were seen. The answer was given in a variet of forms and all eact equivalent answers to that printed in the mark scheme were accepted. The value of k was often omitted and it is possible that some simpl overlooked this. Domain and range remains an unpopular topic and man did not attempt part (c). In this case, inspection of the printed figure gives the lower limit and was intended to give candidates a lead to identifing the upper limit. 4. The responses to this question were ver variable and man lost marks through errors in manipulation or notation, possibl through mental tiredness. For eamples, man made errors in manipulation and could not proceed correctl from the printed cos θ sin θ to sin θ θ cos θ and the answer sin θ was often seen, instead of sin θ. In part 4 4 (b), man never found or realised that the appropriate form for the volume was dθ dθ. dθ However the majorit did find a correct integral in terms of θ although some were unable to use the identit sin θ sinθ cosθ to simplif their integral. The incorrect value k 8 was ver common, resulting from a failure to square the factor in sin θ sinθ cosθ. Candidates were epected to demonstrate the correct change of limits. Minimall a reference to the result tan Edecel Internal Review 9

40 , 6 or an equivalent, was required. Those who had complete solutions usuall gained the two method marks in part (c) but earlier errors often led to incorrect answers. 5. Part (a) was answered correctl b almost all candidates. In part (b), man candidates correctl applied the method of finding a tangent b using parametric differentiation to give the answer in the correct form. Few candidates tried to eliminate t to find a Cartesian equation for C, but these candidates were usuall not able to find the correct gradient at A. In part (c), full correct solutions were much less frequentl seen. A significant number of candidates were able to obtain an equation in one variable to score the first method mark, but were then unsure about how to proceed. Successful candidates mostl formed an equation in t, used the fact that t + was a factor and applied the factor theorem in order for them to find t at the point B. The then substituted this t into the parametric equations to find the coordinates of B. Those candidates who initiall formed an equation in onl went no further. A common misconception in part (c), was for candidates to believe that the gradient at the point B would be the same as the gradient at the point A and a significant minorit of candidates attempted to t solve to find t at the point B. t In part (a), man candidates were able to give t. Some candidates gave their answer onl in degrees instead of radians. Other candidates substituted the -value of P into 4 sin t and found two values for t, namel t,. The majorit of these candidates did not go on to 6 reject t. In part (b), man candidates were able to appl the correct formula for finding 6 in terms of t, although some candidates erroneousl believed that differentiation of 4 sin t gave either 8 cos t, 8cos t or cos t. Some candidates who had differentiated incorrectl, substituted their value of t into and tried to fudge their answer for as, after realising from the given answer that the gradient of the normal was. The majorit of candidates understood the relationship between the gradient of the tangent and its normal and man were able to produce a full correct solution to this part. A few candidates, however, did not realise that parametric differentiation was required in part (b) and some of these candidates tried to convert the parametric equations into a Cartesian equation. Although some candidates then went on to attempt to differentiate their Cartesian equation, this method was rarel eecuted correctl. Few convincing proofs were seen in part (c) with a significant number of candidates who were not aware with the procedure of reversing the limits and the sign of the integral. Therefore, some candidates convenientl differentiated 8 cos t to give positive 8sin t, even though the had previousl differentiated 8 cos t correctl in part (a). After completing this part, some candidates had a crisis of confidence with their differentiation rules and then went on to amend their correct solution to part (a) to produce an incorrect solution. Other candidates differentiated 8 cos t correctl but wrote their limits the wrong wa round giving Edecel Internal Review 40

41 A 4 sin t.( 8 sin t) and after stating sin t cos t sin t (as man candidates were able to do in this part) wrote A 64 sin answer b arguing that all areas should be positive. t cos t. These candidates then wrote down the given Part (d) was unstructured in the sense that the question did not tell the candidates how to integrate the given epression. Onl a minorit of candidates spotted that a substitution was required, although some candidates were able to integrate 64 sin t cos t b inspection. Man candidates replaced sin t with ( cos t) but then multiplied this out with cos t to give cos t cos t cos t. Ver few candidates correctl applied the sum-product formula on this epression, but most candidates usuall gave up at this point or went on to produce some incorrect integration. Other candidates replaced sin t with cos t, but did not make much progress with this. A significant number of candidates used integration b parts with a surprising number of them persevering with this technique more than once before deciding the could make no progress. It is possible, however, to use a loop method but this was ver rarel seen. It was clear to eaminers that a significant number of stronger candidates spent much time tring to unsuccessfull answer this part with a few candidates producing at least two pages of working. 7. The majorit of candidates were able to show the integral required in part (a). Some candidates, however, did not show evidence of converting the given limits, whilst for other candidates this was the onl thing the were able to do. In part (b), it was disappointing to see that some strong candidates were unable to gain an marks b failing to recognise the need to use partial fractions. Those candidates who split the integral up as partial fractions usuall gained all si marks, while those candidates who failed to use partial fractions usuall gave answers such as ln(t + t + ) or ln(t + ) ln(t + ) after integration. A few candidates onl substituted the limit of, assuming that the result of substituting a limit of 0 would be zero Few candidates gave a decimal answer instead of the eact value required b the question. Part (c) was well answered b candidates of all abilities with candidates using a variet of methods as identified in the mark scheme. Occasionall some candidates were able to eliminate t but then failed to make the subject. The domain was not so well understood in part (d), with a significant number of candidates failing to correctl identif it. 8. In part (a), a significant number of candidates struggled with appling the chain rule in order to differentiate tan t with respect to t. Some candidates replaced tan sin t t with and proceeded cos t to differentiate this epression using both the chain rule and the quotient rule. Ver few Edecel Internal Review 4

42 candidates incorrectl differentiated sin t to give cos t. A majorit of candidates were then d able to find b dividing their b their. In part (b), the majorit of candidates were able to write down the point,, and find the equation of the tangent using this point and their tangent gradient. Some candidates found the incorrect value of the tangent gradient using t even though the had correctl found in 4 part (a). There were a significant number of candidates, who having correctl written down the equation of the tangent as ( ) were unable to correctl rearrange this equation 8 into the form. a + b. In part (c), there were man was that a candidate could tackle this question and there were man good solutions seen. Errors usuall arose when candidates wrote down and used incorrect trigonometric identities. It was disappointing to see a number of candidates who used trigonometric identities correctl and reached ( ) but were then unable to rearrange this or, worse still, thought that this was the answer to the question. 9. Part (a) was surprisingl well done b candidates with part (b) providing more of a challenge even for some candidates who had produced a perfect solution in part (a). In part (a), man candidates were able to appl the correct formula for finding in terms of t, although some candidates erroneousl believed that differentiation of a sine function produced a negative cosine function. Other mistakes included a few candidates who either cancelled out t + 6 cos in their gradient epression to give or substituted t t 6 into their and epressions before proceeding to differentiate each with respect to t. Other candidates made life more difficult for themselves b epanding the epression using the compound angle formula, giving them more work, but for the same credit. Man candidates were able to substitute t 6 into their gradient epression to give simplified, but it was not uncommon to see some candidates who incorrectl to give The majorit of candidates wrote down the point (, ) and understood how to find the equation of the tangent using this point and their tangent gradient. Whilst some candidates omitted part (b) altogether, most realised the needed to use the compound angle formula, though it was common to see that some candidates who believed that sin (t + 6 )could be rewritten as sint + sin 6. Man candidates do not appreciate that a proof requires evidence, as was required in establishing that cos t, and so lost the final two marks. There were, however, a significant number of candidates who successfull obtained the required Cartesian equation. Edecel Internal Review 4

43 0. This question proved a significant test for man candidates with full correct solutions being d rare. Man candidates were able to find and, although confusing differentiation with integration often led to inaccuracies. Some candidates attempted to find the equation of the tangent but man were unsuccessful because the failed to use t in order to find the 6 gradient as. 6 Those candidates who attempted part (b) rarel progressed beond stating an epression for the area under the curve. Some attempts were made at integration b parts, although ver few candidates went further than the first line. It was obvious that most candidates were not familiar with integrating epressions of the kind sin at sin bt. Even those who were often spent time deriving results rather than using the relevant formula in the formulae book. Those candidates who were successful in part (a) frequentl went on to find the area of a triangle and so were able to gain at least two marks in part (b).. The majorit of candidates gained the marks in part (a) and a good proportion managed to produce the given result in part (b). Some candidates suggested that the area of R was, which made the question rather trivial; although that happened to be true here as working was needed to produce that statement. 5, The integration in part (c), although well done b good candidates, proved a challenge for ( ) man; weaker candidates integrating (-cost) cost as or something similar. It ma have been that some candidates were pressed for time at this point but even those who knew that a cosine double-angle formula was needed often made a sign error, forgot to multipl their epression for cos t b 4, or even forgot to integrate that epression. It has to be mentioned again that the limits were sometimes used as though 80, 5 t 4sin t + sin t became [ 900.] [ 80.]. so that [ ]. (a) A number of candidates lost marks in this question. Some confused differentiation with integration and obtained a logarithm, others made sign slips differentiating, and a number who obtained the correct gradient failed to continue to find the equation of the tangent using equations of a straight line. Edecel Internal Review 4

44 (b) There was a lack of understanding of proof with a number of candidates merel substituting in values. Better candidates were able to begin correctl but some did not realise that if the answer is given it is necessar to show more working. (c) Ver few got this correct. There was a tendenc to use parts and to be unable to deal with the integral of ln( ). The most successful methods involved dividing out, or substituting for (-). Those who tried a parametric approach rarel recognised the need for partial fractions.. This proved a testing question and few could find both d and d correctl. A common error was to integrate, giving ln ( sin t). Most knew, however, how to obtain d from and d and were able to pick up marks here and in part (b). In part (b), the method for finding the equation of the tangent was well understood. Part (c) proved ver demanding and onl a minorit of candidates were able to use one of the trigonometric forms of Pthagoras to eliminate t and manipulate the resulting equation to obtain an answer in the required form. Few even attempted the domain and the full correct answer, 0, was ver rarel seen 4. Part (a) was often answered well but some candidates who worked in degrees gave the final answer as 90. Part (b) proved more challenging for man; some did not know how to change the variable and others failed to realize that d required the chain rule. Most candidates made some progress in part (c) although a surprising number thought that tan t d t was sec t. The eaminers were encouraged to see most candidates tring to give an eact answer (as required) rather than reaching for their calculators. 5. This was found to be the most difficult question on the paper. Some ecellent candidates did not appear to have learned how to find the area using parametric coordinates and could not even write down the first integral. A few candidates used the formula from the formula book. The integration b parts was tackled successfull, b those who got to that stage and there were few errors seen. The percentage at the end of the question was usuall answered well b the few who completed the question. Edecel Internal Review 44

45 6. Part (a) was generall answered quite well and most of the candidates used the correct double angle formula, as on the mark scheme. Quite a few used sin + cos to obtain /9 + ( )/. A sizeable minorit also eliminated t to obtain cos(cos - (/)). Some of these candidates arrived at a straight line equation, e.g.. If the double angle formula was mis-quoted it tended to be b stating cos cos or cos sin. Quite a few candidates, however found and and then stopped. Presumabl the differentiated before the read the question! These candidates seemed to have no concept of tring to eliminate t. In Part (b) ver few correct graphs were seen, and a large number of answers were on graph paper. Man candidates achieved a B for the shape, but ver few realised the importance of the restricted domain.. Some candidates sketched their derived Cartesian equation, and others worked successfull from the original parametric equations. 7. No Report available for this question. Edecel Internal Review 45

PhysicsAndMathsTutor.com

PhysicsAndMathsTutor.com . A curve C has parametric equations x sin t, y tan t, 0 t < (a) Find in terms of t. (4) The tangent to C at the point where t cuts the x-axis at the point P. (b) Find the x-coordinate of P. () (Total

More information

PhysicsAndMathsTutor.com

PhysicsAndMathsTutor.com . A curve C has parametric equations x = sin t, y = tan t, 0 t < (a) Find in terms of t. (4) The tangent to C at the point where t = cuts the x-axis at the point P. (b) Find the x-coordinate of P. () (Total

More information

Time: 1 hour 30 minutes

Time: 1 hour 30 minutes Paper Reference(s) 6665/01 Edecel GCE Core Mathematics C Silver Level S Time: 1 hour 0 minutes Materials required for eamination papers Mathematical Formulae (Green) Items included with question Nil Candidates

More information

PhysicsAndMathsTutor.com

PhysicsAndMathsTutor.com . (a) Simplify fully + 9 5 + 5 (3) Given that ln( + 9 5) = + ln( + 5), 5, (b) find in terms of e. (Total 7 marks). (i) Find the eact solutions to the equations (a) ln (3 7) = 5 (3) (b) 3 e 7 + = 5 (5)

More information

Time: 1 hour 30 minutes

Time: 1 hour 30 minutes Paper Reference(s) 6666/0 Edecel GCE Core Mathematics C4 Gold Level (Harder) G Time: hour 0 minutes Materials required for eamination Mathematical Formulae (Green) Items included with question papers Nil

More information

Time: 1 hour 30 minutes

Time: 1 hour 30 minutes Paper Reference(s) 6666/0 Edexcel GCE Core Mathematics C4 Silver Level S5 Time: hour 0 minutes Materials required for examination papers Mathematical Formulae (Green) Items included with question Nil Candidates

More information

PhysicsAndMathsTutor.com

PhysicsAndMathsTutor.com . f() = 4 cosec 4 +, where is in radians. (a) Show that there is a root α of f () = 0 in the interval [.,.3]. Show that the equation f() = 0 can be written in the form = + sin 4 (c) Use the iterative formula

More information

A booklet Mathematical Formulae and Statistical Tables might be needed for some questions.

A booklet Mathematical Formulae and Statistical Tables might be needed for some questions. Paper Reference(s) 6663/01 Edexcel GCE Core Mathematics C1 Advanced Subsidiary Quadratics Calculators may NOT be used for these questions. Information for Candidates A booklet Mathematical Formulae and

More information

6665/01 Edexcel GCE Core Mathematics C3 Bronze Level B3

6665/01 Edexcel GCE Core Mathematics C3 Bronze Level B3 Paper Reference(s) 6665/0 Edecel GCE Core Mathematics C3 Bronze Level B3 Time: hour 30 minutes Materials required for eamination papers Mathematical Formulae (Green) Items included with question Nil Candidates

More information

Core Mathematics C3 Advanced Level

Core Mathematics C3 Advanced Level Paper Reference(s) 666/0 Edecel GCE Core Mathematics C Advanced Level Wednesda 0 Januar 00 Afternoon Time: hour 0 minutes Materials required for eamination Mathematical Formulae (Pink or Green) Items included

More information

Time: 1 hour 30 minutes

Time: 1 hour 30 minutes Paper Reference(s) 6665/01 Edecel GCE Core Mathematics C3 Gold Level (Harder) G3 Time: 1 hour 30 minutes Materials required for eamination Mathematical Formulae (Green) Items included with question papers

More information

Time: 1 hour 30 minutes

Time: 1 hour 30 minutes Paper Reference(s) 6666/ Edecel GCE Core Mathematics C4 Gold Level (Hardest) G4 Time: hour minutes Materials required for eamination Mathematical Formulae (Green) Items included with question papers Nil

More information

Time: 1 hour 30 minutes

Time: 1 hour 30 minutes Paper Reference(s) 666/0 Edecel GCE Core Mathematics C Bronze Level B4 Time: hour 0 minutes Materials required for eamination papers Mathematical Formulae (Green) Items included with question Nil Candidates

More information

C4 QUESTIONS FROM PAST PAPERS DIFFERENTIATION

C4 QUESTIONS FROM PAST PAPERS DIFFERENTIATION C4 QUESTIONS FROM PAST PAPERS DIFFERENTIATION. A curve C has equation + y = y Find the eact value of at the point on C with coordinates (, ). (Total 7 marks). The diagram above shows a cylindrical water

More information

Time: 1 hour 30 minutes

Time: 1 hour 30 minutes Paper Reference(s) 6666/0 Edexcel GCE Core Mathematics C4 Silver Level S Time: hour 0 minutes Materials required for examination papers Mathematical Formulae (Green) Items included with question Nil Candidates

More information

6664/01 Edexcel GCE Core Mathematics C2 Bronze Level B2

6664/01 Edexcel GCE Core Mathematics C2 Bronze Level B2 Paper Reference(s) 6664/01 Edexcel GCE Core Mathematics C Bronze Level B Time: 1 hour 30 minutes Materials required for examination papers Mathematical Formulae (Green) Items included with question Nil

More information

C4 mark schemes - International A level (150 minute papers). First mark scheme is June 2014, second mark scheme is Specimen paper

C4 mark schemes - International A level (150 minute papers). First mark scheme is June 2014, second mark scheme is Specimen paper C4 mark schemes - International A level (0 minute papers). First mark scheme is June 04, second mark scheme is Specimen paper. (a) f (.).7, f () M Sign change (and f ( ) is continuous) therefore there

More information

Time: 1 hour 30 minutes

Time: 1 hour 30 minutes Paper Reference(s) 6665/0 Edecel GCE Core Mathematics C3 Bronze Level B Time: hour 30 minutes Materials required for eamination papers Mathematical Formulae (Green) Items included with question Nil Candidates

More information

6664/01 Edexcel GCE Core Mathematics C2 Bronze Level B3

6664/01 Edexcel GCE Core Mathematics C2 Bronze Level B3 Paper Reference(s) 666/01 Edecel GCE Core Mathematics C Bronze Level B Time: 1 hour 0 minutes Materials required for eamination papers Mathematical Formulae (Green) Items included with question Nil Candidates

More information

Time: 1 hour 30 minutes

Time: 1 hour 30 minutes Paper Reference(s) 666/01 Edexcel GCE Core Mathematics C1 Silver Level S4 Time: 1 hour 0 minutes Materials required for examination Mathematical Formulae (Green) Items included with question papers Nil

More information

PhysicsAndMathsTutor.com

PhysicsAndMathsTutor.com C Trigonometry: Trigonometric Equations. (a) Given that 5sinθ = cosθ, find the value of tan θ. () (b) Solve, for 0 x < 360, 5sin x = cos x, giving your answers to decimal place. (5) (Total 6 marks). (a)

More information

GCE Edexcel GCE. Core Mathematics C4 (6666) June Edexcel GCE. Mark Scheme (Final) Core Mathematics C4 (6666)

GCE Edexcel GCE. Core Mathematics C4 (6666) June Edexcel GCE. Mark Scheme (Final) Core Mathematics C4 (6666) GCE Edecel GCE Core Mathematics C4 (6666) June 006 Mark (Final) Edecel GCE Core Mathematics C4 (6666) June 006 6666 Pure Mathematics C4 Mark Question. dy dy dy = 6 4y + = 0 d d d Differentiates implicitly

More information

Time: 1 hour 30 minutes

Time: 1 hour 30 minutes Paper Reference(s) 6664/01 Edexcel GCE Core Mathematics C Silver Level S3 Time: 1 hour 30 minutes Materials required for examination Mathematical Formulae (Green) Items included with question papers Nil

More information

Time: 1 hour 30 minutes

Time: 1 hour 30 minutes Paper Reference(s) 6664/01 Edexcel GCE Core Mathematics C Silver Level S4 Time: 1 hour 0 minutes Materials required for examination Mathematical Formulae (Green) Items included with question papers Nil

More information

Time: 1 hour 30 minutes

Time: 1 hour 30 minutes Paper Reference(s) 6664/01 Edecel GCE Core Mathematics C Gold Level G Time: 1 hour 0 minutes Materials required for eamination papers Mathematical Formulae (Green) Items included with question Nil Candidates

More information

6664/01 Edexcel GCE Core Mathematics C2 Silver Level S1

6664/01 Edexcel GCE Core Mathematics C2 Silver Level S1 Paper Reference(s) 6664/01 Edexcel GCE Core Mathematics C Silver Level S1 Time: 1 hour 0 minutes Materials required for examination papers Mathematical Formulae (Green) Items included with question Nil

More information

Time: 1 hour 30 minutes

Time: 1 hour 30 minutes Paper Reference(s) 6666/ Edexcel GCE Core Mathematics C4 Gold Level (Harder) G3 Time: hour 3 minutes Materials required for examination Mathematical Formulae (Green) Items included with question papers

More information

Time: 1 hour 30 minutes

Time: 1 hour 30 minutes www.londonnews47.com Paper Reference(s) 6665/0 Edexcel GCE Core Mathematics C Bronze Level B4 Time: hour 0 minutes Materials required for examination papers Mathematical Formulae (Green) Items included

More information

6664/01 Edexcel GCE Core Mathematics C2 Gold Level G2

6664/01 Edexcel GCE Core Mathematics C2 Gold Level G2 Paper Reference(s) 6664/01 Edexcel GCE Core Mathematics C Gold Level G Time: 1 hour 30 minutes Materials required for examination papers Mathematical Formulae (Green) Items included with question Nil Candidates

More information

PhysicsAndMathsTutor.com

PhysicsAndMathsTutor.com . A curve C has equation x + y = xy Find the exact value of at the point on C with coordinates (, ). (Total marks). The curve C has the equation cosx + cosy =, π π x, 4 4 0 π y 6 (a) Find in terms of x

More information

Report on the Examination

Report on the Examination Version 1.0 General Certificate of Education (A-level) January 01 Mathematics MPC4 (Specification 660) Pure Core 4 Report on the Examination Further copies of this Report on the Examination are available

More information

Examiners Report/ Principal Examiner Feedback. June GCE Core Mathematics C2 (6664) Paper 1

Examiners Report/ Principal Examiner Feedback. June GCE Core Mathematics C2 (6664) Paper 1 Examiners Report/ Principal Examiner Feedback June 011 GCE Core Mathematics C (6664) Paper 1 Edexcel is one of the leading examining and awarding bodies in the UK and throughout the world. We provide a

More information

Time: 1 hour 30 minutes

Time: 1 hour 30 minutes Paper Reference(s) 6663/0 Edecel GCE Core Mathematics C Silver Level S Time: hour 30 minutes Materials required for eamination Mathematical Formulae (Green) Items included with question papers Nil Candidates

More information

PhysicsAndMathsTutor.com

PhysicsAndMathsTutor.com C Integration: Basic Integration. Use calculus to find the value of ( + )d. (Total 5 marks). Evaluate 8 d, giving your answer in the form a + b, where a and b are integers. (Total marks). f() = + + 5.

More information

Core Mathematics C3 Advanced Subsidiary

Core Mathematics C3 Advanced Subsidiary Paper Reference(s) 6665/0 Edecel GCE Core Mathematics C Advanced Subsidiary Thursday June 0 Morning Time: hour 0 minutes Materials required for eamination Mathematical Formulae (Pink) Items included with

More information

FP1 PAST EXAM QUESTIONS ON NUMERICAL METHODS: NEWTON-RAPHSON ONLY

FP1 PAST EXAM QUESTIONS ON NUMERICAL METHODS: NEWTON-RAPHSON ONLY FP PAST EXAM QUESTIONS ON NUMERICAL METHODS: NEWTON-RAPHSON ONLY A number of questions demand that you know derivatives of functions now not included in FP. Just look up the derivatives in the mark scheme,

More information

Time: 1 hour 30 minutes

Time: 1 hour 30 minutes Paper Reference(s) 6663/0 Edexcel GCE Core Mathematics C Gold Level G5 Time: hour 30 minutes Materials required for examination Mathematical Formulae (Green) Items included with question papers Nil Candidates

More information

abc Mathematics Pure Core General Certificate of Education SPECIMEN UNITS AND MARK SCHEMES

abc Mathematics Pure Core General Certificate of Education SPECIMEN UNITS AND MARK SCHEMES abc General Certificate of Education Mathematics Pure Core SPECIMEN UNITS AND MARK SCHEMES ADVANCED SUBSIDIARY MATHEMATICS (56) ADVANCED SUBSIDIARY PURE MATHEMATICS (566) ADVANCED SUBSIDIARY FURTHER MATHEMATICS

More information

ADDITIONAL MATHEMATICS

ADDITIONAL MATHEMATICS ADDITIONAL MATHEMATICS Paper 0606/ Paper Key messages Candidates should be reminded of the importance of reading the rubric on the eamination paper. Accuracy is of vital importance with final answers to

More information

Mark Scheme (Results) Summer 2009

Mark Scheme (Results) Summer 2009 Mark (Results) Summer 009 GCE GCE Mathematics (6668/0) June 009 6668 Further Pure Mathematics FP (new) Mark Q (a) = rr ( + ) r ( r+ ) r ( r+ ) B aef () (b) n n r = r = = rr ( + ) r r+ = + +...... + + n

More information

A booklet Mathematical Formulae and Statistical Tables might be needed for some questions.

A booklet Mathematical Formulae and Statistical Tables might be needed for some questions. Paper Reference(s) 6663/0 Edexcel GCE Core Mathematics C Advanced Subsidiary Inequalities Calculators may NOT be used for these questions. Information for Candidates A booklet Mathematical Formulae and

More information

Version : abc. General Certificate of Education. Mathematics MPC4 Pure Core 4. Report on the Examination

Version : abc. General Certificate of Education. Mathematics MPC4 Pure Core 4. Report on the Examination Version :.0 0608 abc General Certificate of Education Mathematics 6360 MPC4 Pure Core 4 Report on the Examination 008 examination - June series Further copies of this Report are available to download from

More information

Examiner's Report Q1.

Examiner's Report Q1. Examiner's Report Q1. For students who were comfortable with the pair of inequality signs, part (a) proved to be straightforward. Most solved the inequalities by operating simultaneously on both sets and

More information

Mark Scheme (Results) Summer 2008

Mark Scheme (Results) Summer 2008 Mark (Results) Summer 008 GCE GCE Mathematics (6666/0) Edecel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London WCV 7BH June 008 6666 Core Mathematics C4

More information

Coordinate goemetry in the (x, y) plane

Coordinate goemetry in the (x, y) plane Coordinate goemetr in the (x, ) plane In this chapter ou will learn how to solve problems involving parametric equations.. You can define the coordinates of a point on a curve using parametric equations.

More information

Examiners Report Summer 2008

Examiners Report Summer 2008 Examiners Report Summer 008 GCE GCE Mathematics (8371/8374,9371/9374) 8371/8374, 9371/9374 GCE Mathematics Summer 008 Edexcel Limited. Registered in England and Wales No. 4496750 Registered Office: One90

More information

PhysicsAndMathsTutor.com

PhysicsAndMathsTutor.com Edecel GCE Core Mathematics C4 (6666) PhysicsAndMathsTutor.com GCSE Edecel GCE Core Mathematics C4 (6666) Summer 005 Mark (Results) Final Version June 005 6666 Core C4 Mark. ( ) 4 9 9 = 4 B 9 ( ) 9 ( )(

More information

6664/01 Edexcel GCE Core Mathematics C2 Bronze Level B4

6664/01 Edexcel GCE Core Mathematics C2 Bronze Level B4 Paper Reference(s) 6664/01 Edexcel GCE Core Mathematics C Bronze Level B4 Time: 1 hour 30 minutes Materials required for examination papers Mathematical Formulae (Green) Items included with question Nil

More information

Trigonometric. equations. Topic: Periodic functions and applications. Simple trigonometric. equations. Equations using radians Further trigonometric

Trigonometric. equations. Topic: Periodic functions and applications. Simple trigonometric. equations. Equations using radians Further trigonometric Trigonometric equations 6 sllabusref eferenceence Topic: Periodic functions and applications In this cha 6A 6B 6C 6D 6E chapter Simple trigonometric equations Equations using radians Further trigonometric

More information

Topic 6 Part 4 [317 marks]

Topic 6 Part 4 [317 marks] Topic 6 Part [7 marks] a. ( + tan ) sec tan (+c) M [ marks] [ marks] Some correct answers but too many candidates had a poor approach and did not use the trig identity. b. sin sin (+c) cos M [ marks] Allow

More information

Time: 1 hour 30 minutes

Time: 1 hour 30 minutes Paper Reference(s) 6664/0 Edexcel GCE Core Mathematics C Bronze Level B Time: hour 0 minutes Materials required for examination Mathematical Formulae (Green) Items included with question papers Nil Candidates

More information

Time: 1 hour 30 minutes

Time: 1 hour 30 minutes Paper Reference(s) 6666/0 Edexcel GCE Core Mathematics C4 Silver Level S Time: hour 30 minutes Materials required for examination papers Mathematical Formulae (Green) Items included with question Nil Candidates

More information

PhysicsAndMathsTutor.com

PhysicsAndMathsTutor.com . The diagram above shows a cylindrical water tank. The diameter of a circular cross-section of the tank is 6 m. Water is flowing into the tank at a constant rate of 0.48π m min. At time t minutes, the

More information

GCE EXAMINERS' REPORTS

GCE EXAMINERS' REPORTS GCE EXAMINERS' REPORTS GCE MATHEMATICS C1-C4 & FP1-FP3 AS/Advanced SUMMER 017 Grade boundary information for this subject is available on the WJEC public website at: https://www.wjecservices.co.uk/marktoums/default.aspx?l=en

More information

PhysicsAndMathsTutor.com

PhysicsAndMathsTutor.com C Integration - By sbstittion PhysicsAndMathsTtor.com. Using the sbstittion cos +, or otherwise, show that e cos + sin d e(e ) (Total marks). (a) Using the sbstittion cos, or otherwise, find the eact vale

More information

Mark Scheme (Results) January 2011

Mark Scheme (Results) January 2011 Mark (Results) January 0 GCE GCE Core Mathematics C (6664) Paper Edecel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London WCV 7BH Edecel is one of the leading

More information

Time: 1 hour 30 minutes

Time: 1 hour 30 minutes Paper Reference(s) 666/0 Edexcel GCE Core Mathematics C Gold Level G Time: hour 0 minutes Materials required for examination Mathematical Formulae (Green) Items included with question papers Nil Candidates

More information

PMT. GCE Edexcel GCE Mathematics Core Mathematics C1 (6663) June Mark Scheme (Results) Mathematics. Edexcel GCE

PMT. GCE Edexcel GCE Mathematics Core Mathematics C1 (6663) June Mark Scheme (Results) Mathematics. Edexcel GCE GCE Edecel GCE Mathematics Core Mathematics C (666) June 006 Mark Scheme (Results) Edecel GCE Mathematics June 006 Mark Scheme. 6 + + (+c) A = + + A +c B 4 for some attempt to integrate n n + st A for

More information

4751 Mark Scheme June Mark Scheme 4751 June 2005

4751 Mark Scheme June Mark Scheme 4751 June 2005 475 Mark Scheme June 2005 Mark Scheme 475 June 2005 475 Mark Scheme June 2005 Section A 40 2 M subst of for x or attempt at long divn with x x 2 seen in working; 0 for attempt at factors by inspection

More information

PMT. Mark Scheme (Results) Summer Pearson Edexcel GCE in Further Pure Mathematics FP2R (6668/01R)

PMT. Mark Scheme (Results) Summer Pearson Edexcel GCE in Further Pure Mathematics FP2R (6668/01R) Mark Scheme (Results) Summer 04 Pearson Edecel GCE in Further Pure Mathematics FPR (6668/0R) Edecel and BTEC Qualifications Edecel and BTEC qualifications come from Pearson, the world s leading learning

More information

Time: 1 hour 30 minutes

Time: 1 hour 30 minutes Paper Reference(s) 6665/0 Edexcel GCE Core Mathematics C3 Gold Level (Hardest) G4 Time: hour 30 minutes Materials required for examination Mathematical Formulae (Green) Items included with question papers

More information

abc Mathematics Further Pure General Certificate of Education SPECIMEN UNITS AND MARK SCHEMES

abc Mathematics Further Pure General Certificate of Education SPECIMEN UNITS AND MARK SCHEMES abc General Certificate of Education Mathematics Further Pure SPECIMEN UNITS AND MARK SCHEMES ADVANCED SUBSIDIARY MATHEMATICS (56) ADVANCED SUBSIDIARY PURE MATHEMATICS (566) ADVANCED SUBSIDIARY FURTHER

More information

Level 3, Calculus

Level 3, Calculus Level, 4 Calculus Differentiate and use derivatives to solve problems (965) Integrate functions and solve problems by integration, differential equations or numerical methods (966) Manipulate real and

More information

Edexcel GCE Core Mathematics C2 Advanced Subsidiary

Edexcel GCE Core Mathematics C2 Advanced Subsidiary Centre No. Candidate No. Paper Reference 6 6 6 4 0 1 Paper Reference(s) 6664/01 Edecel GCE Core Mathematics C Advanced Subsidiary Thursday 4 May 01 Morning Time: 1 hour 0 minutes Materials required for

More information

Mark Scheme (Results) January 2007

Mark Scheme (Results) January 2007 Mark Scheme (Results) January 007 GCE GCE Mathematics Core Mathematics C (666) Edecel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London WCV 7BH January

More information

Candidates sitting FP2 may also require those formulae listed under Further Pure Mathematics FP1 and Core Mathematics C1 C4. e π.

Candidates sitting FP2 may also require those formulae listed under Further Pure Mathematics FP1 and Core Mathematics C1 C4. e π. 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

More information

PhysicsAndMathsTutor.com

PhysicsAndMathsTutor.com GCE Edecel GCE Core Mathematics C(666) Summer 005 Mark Scheme (Results) Edecel GCE Core Mathematics C (666) June 005 666 Core Mathematics C Mark Scheme Question Number. (a) Scheme Penalise ± B Marks ()

More information

Cambridge International Advanced Subsidiary Level and Advanced Level 9709 Mathematics November 2011 Principal Examiner Report for Teachers

Cambridge International Advanced Subsidiary Level and Advanced Level 9709 Mathematics November 2011 Principal Examiner Report for Teachers MATHEMATICS Paper 9709/11 Paper 11 Key messages In order to gain as many marks as possible all working should be either exact or correct to at least 4 significant figures. It is important for candidates

More information

MATHEMATICS Paper 980/11 Paper 11 General comments It is pleasing to record improvement in some of the areas mentioned in last year s report. For example, although there were still some candidates who

More information

PhysicsAndMathsTutor.com

PhysicsAndMathsTutor.com M Collisions - Direct impact. A small ball A of mass m is moving with speed u in a straight line on a smooth horizontal table. The ball collides directly with another small ball B of mass m moving with

More information

11.4 Polar Coordinates

11.4 Polar Coordinates 11. Polar Coordinates 917 11. Polar Coordinates In Section 1.1, we introduced the Cartesian coordinates of a point in the plane as a means of assigning ordered pairs of numbers to points in the plane.

More information

ADDITIONAL MATHEMATICS 4037/12 Paper 1 October/November 2016 MARK SCHEME Maximum Mark: 80. Published

ADDITIONAL MATHEMATICS 4037/12 Paper 1 October/November 2016 MARK SCHEME Maximum Mark: 80. Published Cambridge International Eaminations Cambridge Ordinary Level ADDITIONAL MATHEMATICS 07/ Paper October/November 06 MARK SCHEME Maimum Mark: 80 Published This mark scheme is published as an aid to teachers

More information

Mark Scheme (Results) Summer 2007

Mark Scheme (Results) Summer 2007 Mark Scheme (Results) Summer 007 GCE GCE Mathematics Core Mathematics C (666) Edecel Limited. Registered in England and Wales No. 96750 Registered Office: One90 High Holborn, London WCV 7BH June 007 666

More information

Examiners Report/ Principal Examiner Feedback. Summer GCE Core Mathematics C3 (6665) Paper 01

Examiners Report/ Principal Examiner Feedback. Summer GCE Core Mathematics C3 (6665) Paper 01 Examiners Report/ Principal Examiner Feedback Summer 2013 GCE Core Mathematics C3 (6665) Paper 01 Edexcel and BTEC Qualifications Edexcel and BTEC qualifications come from Pearson, the UK s largest awarding

More information

Cambridge International Advanced and Advanced Subsidiary Level 9709 Mathematics June 2010 Principal Examiner Report for Teachers

Cambridge International Advanced and Advanced Subsidiary Level 9709 Mathematics June 2010 Principal Examiner Report for Teachers MATHEMATICS Paper 9709/11 Paper 11 General comments Whilst there were some very pleasing scripts scoring high marks Examiners were disappointed to see quite a significant proportion of scripts scoring

More information

MATHEMATICS Unit Further Pure 1

MATHEMATICS Unit Further Pure 1 General Certificate of Education June 008 Advanced Subsidiary Eamination MATHEMATICS Unit Further Pure MFP Monday 6 June 008.30 pm to 3.00 pm For this paper you must have: an 8-page answer book the blue

More information

Cambridge International Advanced Subsidiary Level and Advanced Level 9709 Mathematics June 2015 Principal Examiner Report for Teachers

Cambridge International Advanced Subsidiary Level and Advanced Level 9709 Mathematics June 2015 Principal Examiner Report for Teachers MATHEMATICS Paper 9709/11 Paper 11 Key Messages Candidates who displayed good algebraic skills coped well with paper. The formulae provided were most useful and rarely misquoted. Efficient application

More information

PhysicsAndMathsTutor.com

PhysicsAndMathsTutor.com . The end A of a uniform rod AB, of length a and mass 4m, is smoothly hinged to a fixed point. The end B is attached to one end of a light inextensible string which passes over a small smooth pulley, fixed

More information

PhysicsAndMathsTutor.com

PhysicsAndMathsTutor.com 1. A triangular frame is formed by cutting a uniform rod into 3 pieces which are then joined to form a triangle ABC, where AB = AC = 10 cm and BC = 1 cm, as shown in the diagram above. (a) Find the distance

More information

Mark Scheme (Results) Summer 2007

Mark Scheme (Results) Summer 2007 Mark Scheme (Results) Summer 007 GCE GCE Mathematics Core Mathematics C (4) Edecel Limited. Registered in England and Wales No. 449750 Registered Office: One90 High Holborn, London WCV 7BH June 007 Mark

More information

Name: TT5 remedial work. Date: Time: 0 minute. Total marks available: 60. Total marks achieved: Peter Vermeer. Questions

Name: TT5 remedial work. Date: Time: 0 minute. Total marks available: 60. Total marks achieved: Peter Vermeer. Questions Name: TT5 remedial work Date: Time: 0 minute Total marks available: 60 Total marks achieved: Peter Vermeer Questions Q1. (a) Express 2cosθ sinθ in the form Rcos(θ + a), where R and a are constants, R >

More information

MEI STRUCTURED MATHEMATICS 4753/1

MEI STRUCTURED MATHEMATICS 4753/1 OXFORD CAMBRIDGE AND RSA EXAMINATIONS Advanced Subsidiary General Certificate of Education Advanced General Certificate of Education MEI STRUCTURED MATHEMATICS 4753/1 Methods for Advanced Mathematics (C3)

More information

Time: 1 hour 30 minutes

Time: 1 hour 30 minutes Paper Reference(s) 6665/0 Edecel GCE Core Mathematics C3 Gold Level (Hard) G Time: hour 30 minutes Materials required for eamination Mathematical Formulae (Green) Items included with question papers Nil

More information

Mathematics. Mathematics 2. hsn.uk.net. Higher HSN22000

Mathematics. Mathematics 2. hsn.uk.net. Higher HSN22000 Higher Mathematics UNIT Mathematics HSN000 This document was produced speciall for the HSN.uk.net website, and we require that an copies or derivative works attribute the work to Higher Still Notes. For

More information

Methods for Advanced Mathematics (C3) Coursework Numerical Methods

Methods for Advanced Mathematics (C3) Coursework Numerical Methods Woodhouse College 0 Page Introduction... 3 Terminolog... 3 Activit... 4 Wh use numerical methods?... Change of sign... Activit... 6 Interval Bisection... 7 Decimal Search... 8 Coursework Requirements on

More information

PhysicsAndMathsTutor.com

PhysicsAndMathsTutor.com 1. A particle P of mass 0.5 kg is moving under the action of a single force F newtons. At time t seconds, F = (6t 5) i + (t 2 2t) j. The velocity of P at time t seconds is v m s 1. When t = 0, v = i 4j.

More information

Quadratic Equations. All types, factorising, equation, completing the square. 165 minutes. 151 marks. Page 1 of 53

Quadratic Equations. All types, factorising, equation, completing the square. 165 minutes. 151 marks. Page 1 of 53 Quadratic Equations All types, factorising, equation, completing the square 165 minutes 151 marks Page 1 of 53 Q1. (a) Factorise x 2 + 5x 24 Answer... (2) (b) Solve x 2 + 5x 24 = 0 Answer... (1) (Total

More information

(b) Find det A, giving your answer in terms of a. (1) Given that the area of the image of R is 18, (c) find the value of a. (3) (Total 7 marks)

(b) Find det A, giving your answer in terms of a. (1) Given that the area of the image of R is 18, (c) find the value of a. (3) (Total 7 marks) FP MATRICES PAST EXAM QUESTIONS Matrix questions are pretty much all standard and plain-vanilla. Q. is a bit more difficult just because it is a long question with a lot of working. Q. (d) and Q.3 onwards

More information

Time: 1 hour 30 minutes

Time: 1 hour 30 minutes Paper Reference(s) 6678/01 Edexcel GCE Mechanics M2 Gold Level G3 Time: 1 hour 30 minutes Materials required for examination Mathematical Formulae (Pink) Items included with question papers Nil Candidates

More information

Mark Scheme (Final) January 2009

Mark Scheme (Final) January 2009 Mark (Final) January 009 GCE GCE Core Mathematics C (6666/0) Edecel Limited. Registered in England and Wales No. 96750 Registered Office: One90 High Holborn, London WCV 7BH General Marking Guidance All

More information

Time: 1 hour 30 minutes

Time: 1 hour 30 minutes Paper Reference(s) 6667/0 Edexcel GCE Further Pure Mathematics FP Bronze Level B Time: hour 30 minutes Materials required for examination papers Mathematical Formulae (Green) Items included with question

More information

Mark Scheme (Results) January Pearson Edexcel International Advanced Level. Core Mathematics 3 (6665A) January 2014 (IAL)

Mark Scheme (Results) January Pearson Edexcel International Advanced Level. Core Mathematics 3 (6665A) January 2014 (IAL) January 014 (IAL) Mark (Results) January 014 Pearson Edecel International Advanced Level Core Mathematics (6665A) January 014 (IAL) Edecel and BTEC Qualifications Edecel and BTEC qualifications are awarded

More information

4754 Mark Scheme June 005 Mark Scheme 4754 June 005 4754 Mark Scheme June 005 1. (a) Please mark in red and award part marks on the right side of the script, level with the work that has earned them. (b)

More information

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 4753/0 MATHEMATICS (MEI) Methods for Advanced Mathematics (C3) MONDAY JUNE 007 Additional materials: Answer booklet (8 pages) Graph paper MEI Examination Formulae and Tables (MF) Afternoon

More information

MEI STRUCTURED MATHEMATICS 4751

MEI STRUCTURED MATHEMATICS 4751 OXFORD CAMBRIDGE AND RSA EXAMINATIONS Advanced Subsidiary General Certificate of Education Advanced General Certificate of Education MEI STRUCTURED MATHEMATICS 75 Introduction to Advanced Mathematics (C)

More information

Mark Scheme (Results) Summer GCE Core Mathematics 3 (6665/01R)

Mark Scheme (Results) Summer GCE Core Mathematics 3 (6665/01R) Mark Scheme (Results) Summer GCE Core Mathematics (6665/R) Question Number Scheme Marks. (a) + ( + 4)( ) B Attempt as a single fraction (+ 5)( ) ( + ) ( + )( ) or + 5 ( + 4) M ( + 4)( ) ( + 4)( ), ( +

More information

Mark Scheme (Results) Summer Pearson Edexcel GCE in Core Mathematics 2 (6664_01)

Mark Scheme (Results) Summer Pearson Edexcel GCE in Core Mathematics 2 (6664_01) Mark Scheme (Results) Summer 014 Pearson Edecel GCE in Core Mathematics (6664_01) Edecel and BTEC Qualifications Edecel and BTEC qualifications come from Pearson, the world s leading learning company.

More information

Mark Scheme (Results) January 2008

Mark Scheme (Results) January 2008 Mark Scheme (Results) January 008 GCE GCE Mathematics (6664/01) Edecel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London WC1V 7BH January 008 6664 Core

More information

Unit 3 Notes Mathematical Methods

Unit 3 Notes Mathematical Methods Unit 3 Notes Mathematical Methods Foundational Knowledge Created b Triumph Tutoring Copright info Copright Triumph Tutoring 07 Triumph Tutoring Pt Ltd ABN 60 607 0 507 First published in 07 All rights

More information

Higher Unit 9 topic test

Higher Unit 9 topic test Name: Higher Unit 9 topic test Date: Time: 45 minutes Total marks available: 41 Total marks achieved: Questions Q1. (a) (i) Factorise x 2 12x + 27... (ii) Solve the equation x 2 12x + 27 = 0 (b) Factorise

More information