June 2009 Markscheme

EDEXCEL CORE MATHEMATICS C3 (6665) – JUNE 2009FINAL MARK SCHEMEQuestionSchemeNumber21. (a) Iterative formula: xn+ 1= + 2 , x20= 2.5( x )nMarks(b) Let2x1= + 22M1(2.5)x1= 2.32, x2= 2.371581451...A1x3= 2.355593575... , x4= 2.360436923...A1 cso (3)3 2f ( x) = − x + 2x+ 2 = 0f (2.3585) = 0.00583577...f (2.3595) = − 0.00142286...Sign change (and f ( x)is continuous) therefore a root α is such that( 2.3585, 2.3595)α ∈ ⇒ α = 2.359 (3 dp) A1 (3)2. (a) cos 2 θ + sin 2 θ = 1 ( ÷ cos2 θ )2 2cos θ sin θ 1+ =2 2 2M1cos θ cos θ cos θ2 21 + tan θ = sec θ2 2tan sec 1M1M1(6 marks)θ = θ − (as required) A1 cso (2)(b) 2tan 2 θ 4secθ sec 2 θ 2, ( eqn *)+ + = 0 ≤ θ

EDEXCEL CORE MATHEMATICS C3 (6665) – JUNE 2009FINAL MARK SCHEMEQuestionNumber53. (a) P = 80e tSchemeMarkst0= 0 ⇒ P = 80e 5 = 80(1) = 80B1 (1)t 1000t5 5(b) P = 1000 ⇒ 1000 = 80e ⇒ = eM180⎛1000⎞∴ t = 5ln ⎜ ⎟⎝ 80 ⎠t = 12.6286...A1 (2)dPt5(c) = 16eM1 A1 (2)dt5(d) 50 = 16e t⎛ 50 ⎞∴ t = 5ln ⎜ ⎟ = 5.69717...⎝ 16 ⎠{ }M11⎛⎛ 50 ⎞⎞⎜5ln⎜⎟⎟5⎝⎝ 16 ⎠⎠P = 80e or1( 5.69717... )P = 80e M1580(50)P = = 250A1 (3)16(8 marks)2

EDEXCEL CORE MATHEMATICS C3 (6665) – JUNE 2009FINAL MARK SCHEMEQuestionNumber4. (i)(a)y = x 2 cos3xSchemeMarksApply product rule:2⎧ u = x v = cos3x⎫⎪⎪⎨dudv⎬⎪ = 2x= −3sin3x⎪⎩ dxdx⎭M1dydx 2= 2xcos3x − 3x sin3xA1 A1 (3)(i)(b)y=2ln( x + 1)x2+ 1udu2x2= ln( x + 1) ⇒ =2dxx + 1M1 A1Apply quotient rule:2 2⎧ u = ln( x + 1) v = x + 1⎫⎪⎪⎨du 2x dv⎬⎪ = = 2x2⎪⎩ dx x + 1 dx⎭M1⎛ 2x⎞ 2 2⎜ ⎟( x + 1)− 2xln( x + 1)x=⎝ ⎠2dx12dy+ 12( x + )A11(ii) y = 4x + 1, x > −4At P, y = 4(2) + 1 = 9 = 3B1dy1 −( 4 1 ) 1 2= x + (4)M1dx2dy2=d x (4x+ 1)12A1At P,dy2dx = M14(2)+1( ) 1 2Hence m(T) = 2 32Either T: y − 3 = ( x − 2) ; M13T: 3 y − 9 = 2( x − 2) ;T: 3 y − 9 = 2x− 4T: 2x− 3 y + 5 = 0A1 (6)(13 marks)3

EDEXCEL CORE MATHEMATICS C3 (6665) – JUNE 2009FINAL MARK SCHEMEQuestionNumber5. (a)ySchemeMarksCurve retains shape1when x > ln k 2B1( 0, k −1)O 1( )ln k , 0x2Curve reflects through the x-axis1when x < ln k 2B1( 0, k 1) and ( 1 ln k , 0)− marked in2the correct positions. B1 (3)(b)yCorrect shape of curve. The curveshould be contained inquadrants 1, 2 and 3(Ignore asymptote)B1( 1 − k , 0)( 0, 1 ln k )2( 1 − k , 0) and ( 0, 1 ln k )2B1 (2)Ox(c) Range of f: f ( x)> − k or y > − k or ( −k, ∞ )B1 (1)2x2(d) y = e − k ⇒ y + k = exM1⇒ ln ( y + k ) = 2x⇒ 1( + ) =Hence−1f ( x)2 ln y k xf ( x) = ln( x + k)A1 cao (3)−1 12: Domain: x > − k or ( −k, ∞ )B1ft (1)M1(10 marks)4

EDEXCEL CORE MATHEMATICS C3 (6665) – JUNE 2009FINAL MARK SCHEMEQuestionNumber6. (a) ( )SchemeMarksA = B ⇒ cos A + A = cos2A = cos Acos A − sin AsinAM12 2cos2A = cos A − sin A and2 2cos A sin A 1+ = gives2 2 2cos2A = 1− sin A − sin A = 1 − 2sin A (as required) A1 (2)(b) C1 = C2⇒23sin 2 4sin 2cos2x = x − xM1⎛1−cos2x⎞3sin 2x= 4⎜⎟ − 2cos 2x⎝ 2 ⎠( )3sin 2x = 2 1− cos2x − 2cos2x3sin 2x = 2 − 2cos2x − 2cos2xM13sin 2x+ 4cos2x= 2A1 (3)(c) 3sin 2x + 4cos2x = Rcos( 2x − α )3sin 2x + 4cos2x = Rcos 2xcosα+ Rsin 2xsinαEquate sin 2 x : 3 = RsinαEquate cos 2 x : 4 = R cosα2 2R = 3 + 4 ; = 25 = 5B1tanα= ⇒ α = 36.86989765...34Hence, 3sin 2x 4cos2x 5cos( 2x36.87)(d) 3sin 2x+ 4cos2x= 25cos( 2x − 36.87)= 2oM1 A1+ = − A1 (3)2cos( 2x − 36.87)= M15( )2x − 36.87 = 66.42182...( )2x − 36.87 = 360 − 66.42182...ooHence, x = 51.64591... o, 165.22409... o A1 A1 (4)A1(12 marks)5

EDEXCEL CORE MATHEMATICS C3 (6665) – JUNE 2009FINAL MARK SCHEMEQuestionNumber7. (a)2 x − 8f ( x) = 1 − +( x + 4) ( x − 2)( x + 4)Schemex ∈ R, x ≠ −4, x ≠ 2.Marksf ( x)=( x − 2)( x + 4) − 2( x − 2) + x − 8( x − 2)( x + 4)M1 A1=2x x x x+ 2 − 8 − 2 + 4 + − 8( x − 2)( x + 4)=x2+ x − 12[( x + 4)( x − 2) ]A1=( x + 4)( x − 3)[( x + 4)( x − 2) ]M1=( x − 3)( x − 2)A1 cso (5)(b)g( x)=xe − 3xe − 2x ∈ R, x ≠ ln 2.Apply quotient rule:xx⎧ u = e − 3 v = e − 2 ⎫⎪⎪⎨dux dvx⎬⎪ = e = e ⎪⎩ dxdx⎭g ′( x)==x x x xe (e − 2) − e (e − 3)x 2(e − 2)xex(e − 2)2M1 A1A1 cso (3)xe(c) g ′( x) = 1 ⇒ = 1x 2(e − 2)xe (e 2)x 2= − M1x 2x x xe = e − 2e − 2e + 42xxe 5e 4 0− + = A1xx(e − 4)(e − 1) = 0M1xxe = 4 or e = 1x = ln 4 or x = 0A1 (4)(12 marks)6

EDEXCEL CORE MATHEMATICS C3 (6665) – JUNE 2009FINAL MARK SCHEMEQuestionSchemeMarksNumber8. (a) sin 2x = 2sin x cos xB1 (1)(b) cosec x − 8cos x = 0 , 0 < x < π18cos x 0sin x − = M118cos xsin x =1 = 8sin xcosx( x x)1 = 4 2sin cos1 = 4sin 2xM11sin 2x =4A1{ }{ }Radians 2x= 0.25268..., 2.88891...Degrees 2x= 14.4775..., 165.5225...{ }{ }Radians x = 0.12634..., 1.44445...Degrees x = 7.23875..., 82.76124...A1A1 cao(5)(6 marks)7