Mark (Results) Summer 00 GCE Core Mathematics C3 (6665) Edexcel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London WCV 7BH
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June 00 6665 Core Mathematics C3 Mark sinθ cosθ. (a) + cos θ (b) sinθ cosθ cosθ cosθ = tanθ (as required) AG A cso tanθ = tanθ = M θ = awrt 6.6 A θ = awrt 53.4 M A () [5] (a) M: Uses both a correct identity for sin θ and a correct identity for cos θ. Also allow a candidate writing + cosθ = cos θ on the denominator. Also note that angles must be consistent in when candidates apply these identities. A: Correct proof. No errors seen. (b) st M for either tanθ = A: awrt 6.6 A : awrt 53.4 or θ = 80 + θ or tanθ =, seen or implied. Special Case: For candidate solving, tan θ = k, where k, to give θ and θ = 80 + θ, then award M0A0B in part (b). Special Case: Note that those candidates who writes tanθ =, and gives ONLY two answers of 45 and 35 that are inside the range will be awarded SC M0A0B. GCE Core Mathematics C3 (6665) Summer 00
. At P, y = 3 B d y 3( ) ( 5 3 ) 3 8 = x ( 3) or 3 dx (5 3 x) MA dy 8 3 d x = (5 3()) { 8} M m(n) = or 8 8 M N: y 3 = ( x ) 8 M N: x 8 y+ 5 = 0 A [7] st 3 M: ± k(5 3 x) can be implied. See appendix for application of the quotient rule. nd M: Substituting x = into an equation involving their d y d 3 rd M: Uses m(n) = their m( T ). 4 th M: y y = m( x ) with their NORMAL gradient or a changed tangent gradient and their y. Or uses a complete method to express the equation of the tangent in the form y= mx+ c with their NORMAL ( changed numerical) gradient, their y and x =. Note: To gain the final A mark all the previous 6 marks in this question need to be earned. Also there must be a completely correct solution given. x ; GCE Core Mathematics C3 (6665) Summer 00
3. (a) f (.) = 0.496655..., f (.3) = 0.0487987... Sign change (and as f ( ) x is continuous) therefore a root α is such that [.,.3] α MA (b) 4cosecx 4x + = 0 4x = 4cosecx + M x = cosecx + x = + 4 sinx 4 A (c) x = + M sin(.5) 4 x =.303757858..., x =.86745793... A x 3 =.974463... A (d) f (.905) = 0.00044566695..., f (.95) = 0.004750778... Sign change (and as f ( x) is continuous) therefore a root α is such that α α =.9 (3 dp) (.905,.95) (a) M: Attempts to evaluate both f (.) and f (.3) and evaluates at least one of them correctly to awrt (or truncated) sf. A: both values correct to awrt (or truncated) sf, sign change and conclusion. (b) M: Attempt to make 4x or x the subject of the equation. A: Candidate must then rearrange the equation to give the required result. It must be clear that candidate has made their initial f ( x ) = 0. (c) M: An attempt to substitute x 0 =.5 into the iterative formula Eg = +. sin(.5) 4 Can be implied by x = awrt.3 or x = awrt 46. A: Both x = awrt.3038 and x = awrt.867 A: x 3 = awrt.97 (d) M: Choose suitable interval for x, e.g. [.905,.95] or tighter and at least one attempt to evaluate f(x). A: both values correct to awrt (or truncated) sf, sign change and conclusion. M A () () () [9] GCE Core Mathematics C3 (6665) Summer 00
4. (a) y ( 0, 5 ) MA O 5 (,0 ) x (b) x = 0 B 0 x 5 = (5 + x) ; x = M;A oe. 3 (c) fg() = f( 3) = ( 3) 5; = = M;A (d) g( x) = x 4x+ = ( x ) 4 + = ( x ) 3. Hence gmin = 3 M Either gmin = 3 or g( x) 3 or g(5) = 5 0 + = 6 B 3 g( x) 6 or 3 y 6 A (a) M: V or or graph with vertex on the x-axis. 5 A: (, { 0 }) and ({ } ) 0, 5 seen and the graph appears in both the first and second quadrants. (b) M: Either x 5 = (5 + x) or (x 5) = 5+ x (c) M: Full method of inserting g() into f( x) = x 5 or for inserting x = into ( x 4x + ) 5. There must be evidence of the modulus being applied. (d) M: Full method to establish the minimum of g. Eg: ( x ± α ) + β leading to g min = β. Or for candidate to differentiate the quadratic, set the result equal to zero, find x and insert this value of x back into f ( x ) in order to find the minimum. B: For either finding the correct minimum value of g (can be implied by g( x) 3 or g( x) > 3 ) or for stating that g(5) = 6. A: 3 g( x) 6 or 3 y 6 or 3 g 6. Note that: 3 x 6 is A0. Note that: 3 f( x) 6 is A0. Note that: 3 g( x) 6 is A0. Note that: g( x) 3 or g( x) > 3 or x 3 or x > 3with no working gains MBA0. Note that for the final Accuracy Mark: If a candidate writes down 3 < g( x) < 6 or 3 < y < 6, then award MBA0. If, however, a candidate writes down g( x) 3, g( x) 6, then award A0. If a candidate writes down g( x) 3 or g( x) 6, then award A0. () () [0] GCE Core Mathematics C3 (6665) Summer 00
5. (a) Either (b) When (c) y = or ( ) x 0, B =, y = (8 0 + )e = 0e = 0 B (x 5x ) 0 ( x )(x ) 0 + = = M Either x = (for possibly B above) or x = A dy dx. x x = (4x 5)e (x 5x+ )e MAA (d) (4 5)e x x x (x 5x+ )e = 0 M x 9x 7 0 + = (x 7)( x ) = 0 M x = 7, A When x 7 7 y =, = 9e, when x, y e = = ddma x (b) If the candidate believes that e = 0 solves to x = 0 or gives an extra solution of x = 0, then withhold the final accuracy mark. x (c) M: (their u )e + (x 5x+ )(their v') A: Any one term correct. A: Both terms correct. (d) st M: For setting their d y dx found in part (c) equal to 0. nd M: Factorise or eliminate out e x correctly and an attempt to factorise a 3-term quadratic or apply the formula to candidate s ax + bx + c. See rules for solving a three term quadratic equation on page of this Appendix. 3 rd ddm: An attempt to use at least one x-coordinate on y = (x 5x+ )e x. Note that this method mark is dependent on the award of the two previous method marks in this part. Some candidates write down corresponding y-coordinates without any working. It may be necessary on some occasions to use your calculator to check that at least one of the two y-coordinates found is correct to awrt sf. 7 7 Final A: Both{ x = }, y = e and { x = }, y = 9e. cao Note that both exact values of y are required. () (5) [] GCE Core Mathematics C3 (6665) Summer 00
6. (a) (i) ( 3, 4 ) B B (ii) ( 6, 8) B B (b) y (4) 5 B B B O x ( 3, 4) (3, 4) (c) f( x) ( x 3) 4 = or f( x) x 6x 5 (d) Either: The function f is a many-one {mapping}. Or: The function f is not a one-one {mapping}. = + MA (b) B: Correct shape for x 0, with the curve meeting the positive y-axis and the turning point is found below the x-axis. (providing candidate does not copy the whole of the original curve and adds nothing else to their sketch.). B: Curve is symmetrical about the y-axis or correct shape of curve for x < 0. Note: The first two BB can only be awarded if the curve has the correct shape, with a cusp on the positive y-axis and with both turning points located in the correct quadrants. Otherwise award BB0. B: Correct turning points of ( 3, 4) and ( 3, 4 ). Also, ({ 0,5 } ) is marked where the graph cuts through the y-axis. Allow (5, 0) rather than (0, 5) if marked in the correct place on the y-axis. (c) M: Either states f ( x ) in the form ( x ± α) ± β ; α, β 0 Or uses a complete method on f( x) = x + ax + b, with f (0) = 5 and f = 4 to find both a and b. A: Either ( x 3) 4 or x 6x + 5 (d) B: Or: The inverse is a one-many {mapping and not a function}. Or: Because f (0) = 5 and also f (6) = 5. Or: One y-coordinate has corresponding x-coordinates {and therefore cannot have an inverse}. B () () [0] GCE Core Mathematics C3 (6665) Summer 00
7. (a) R = 6.5 or.5 B tanα = = α = awrt 0.6435 MA.5 3 4 (b) (i) Max Value =.5 B π (ii) sin ( θ 0.6435) = or θ theirα = ; θ = awrt. M;A (c) H Max = 8.5 (m) B (d) 4πt sin 0.6435 = or 5 4π t = their (b) answer ; t = awrt 4.4 M;A 5 4π t 4π t 6 +.5sin 0.6435 = 7 ; sin 0.6435 = = 0.4 M;M 5 5.5 4π t 0.6435 = sin (0.4) or awrt 0.4 A 5 Either t = awrt. or awrt 6.7 A c So, 0.6435 = { π 0.457... or.730076... } 4π t 5 ddm Times = { 4 : 06, 8: 43} A (6) [5] (a) B: R =.5 or R = 6.5. For R = ±.5, award B0. M:.5 tanα =± or tanα =± A: α = awrt 0.6435 (b) B.5 :.5 or follow through the value of R in part (a). sin θ theirα = M: For ( ) π A : awrt. or + their α rounding correctly to 3 sf. (c) B : 8.5 or 6 + their R found in part (a) as long as the answer is greater than 6. 4π t M: sin ± their α = or 4 π t = their (b) answer 5 5 A: For sin (0.4) This can be implied by awrt 4.4 or awrt 4.40. 4π t 6 + their sin ± their α = 7, M: 5 4π t sin ± their α = 5 their R (d) M: ( R) A: For sin (0.4). This can be implied by awrt 0.4 or awrt.73 or other values for different α ' s. Note this mark can be implied by seeing.055. A: Either t = awrt. or t = awrt 6.7 ddm: either π their PV c. Note that this mark is dependent upon the two M marks. This mark will usually be awarded for seeing either.730 or 3.373 A: Both t = 4 : 06 and t = 8: 43 or both 6 (min) and 403 (min) or both hr 6 min and 6 hr 43 min. GCE Core Mathematics C3 (6665) Summer 00
8. (a) (b) ( x+ 5)(x ) = ( x+ 5)( x 3) x + 9x 5 ln = x + x 5 x + 9x 5 = e x + x 5 (x ) ( x 3) x = e 3e = x(e ) x 3 x = 3e e (a) M: An attempt to factorise the numerator. B: Correct factorisation of denominator to give ( x+ 5)( x 3). Can be seen anywhere. (b) M: Uses a correct law of logarithms to combine at least two terms. This usually is achieved by the subtraction law of logarithms to give x + 9x 5 ln. = x + x 5 The product law of logarithms can be used to achieve ( x + x ) = ( ( x + x )) ln 9 5 ln e 5. The product and quotient law could also be used to achieve x + 9x 5 ln 0. = e( x + x 5) dm: Removing ln s correctly by the realisation that the anti-ln of is e. Note that this mark is dependent on the previous method mark being awarded. M: Collect x terms together and factorise. Note that this is not a dependent method mark. A: 3e 3e or or 3e. aef e e -e Note that the answer needs to be in terms of e. The decimal answer is 9.960559 Note that the solution must be correct in order for you to award this final accuracy mark. Note: See Appendix for an alternative method of long division. M B A aef M dm M A aef cso (4) [7] GCE Core Mathematics C3 (6665) Summer 00
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