GCE Mathematics Written Mark Scheme January 2006 - Gosford Hill ...

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MPC4 – AQA GCE Mark Scheme, 2006 January seriesMPC4 (cont)Q Solution Marks Total Comments3(a) R = 13 Or 3.6B1 1(b)sinα2tan α α 33.7cosα 3–2 3Allow M1 for tan α = or ±3 2AG convincingly obtained(c) maximum value = 13B1Fcos ( θ + 33.7) = 1 ( θ = –33.7)M1θ = 326.3A1 3 AWRT 326Total 64(a) A = 80B1 1(b)5000 80 k56= × M1 SC1 Verification. Need 62.51 or betterk = 5000 56 ≈ 1.0766480M1A1 3⎧⎨⎩⎛5000⎞Or using logs: M1 ln ⎜ ⎟ = 56 ln k⎝ 80 ⎠⎛62.5⎞ln⎜⎟⎝ 56 ⎠A1 k = eOr 3/3 for k = 1.076636Or 1.076637 seen106(c)(i) V = 80× k = 200707M1A1 2 200648 using full register k(ii) ln10000 = ln ktM1ln10000t = 124.7 2024ln k= ⇒ M1A1 3 M1 t ln k = ln 10000A1 CAOOr trial and improvement M1expressionM1 125, 124, A1 2024Total 9−1 ( −1)( −2)21− x = 1+ −1 ( − x) + ( − x)M1 First two terms + kx222= 1+ x + xA1 25(a)(i) ( ) ( )(ii)( x)−11 1⎛213 2 3 3 x ⎞= ⎜ − ⎟B1− ⎝ ⎠Or directly substitute into formula;2⎛ 2 2 ⎞M1 power of 3⎛ ⎞≈ * 1+ x+⎜ x⎜⎟3 ⎝ 3 ⎠ ⎟M1M1 other coefficients (allow one error)⎝⎠A1 CAO1 2 4 2≈ + x + xA1 3 AG convincingly obtained3 9 27(b) ( 1 x) 1 ( 2)( x)( )( )( x) 2−2 −2 −3−− = + − − +22= 1+ 2x+ 3xM1A1 2First two terms +2kx4
AQA GCE Mark Scheme, 2006 January series – MPC4MPC4 (Cont)Q Solution Marks Total Comments5(c)22x − 3=2A 1− x + B 3−2x 1− x + C 3− 2xM1 Or by equating coefficients(d)6(a)( ) ( )( ) ( )3 3 1x= 1 –1 = C× 1 x= = A×2 2 4C =− 1 A= 6A1 Follow on A and Cx= 0 − 3 = 6 + 3B−3( )M1or other value ⇒ equation in A, B, C m1B =− 2A1 56 2 1− −3−2x 1−x 1−x( ) 2( )M1 sameA1 collect termsM1 equate coefficientsA1 2 correctA1 3 correct6⎛2 4 2⎞2≈ 1 x x 2 1 x x3⎜ + + − + +3 9⎟⎝⎠M1A1F Follow on A B C and expansions2 8 37 2− ( 1+ 2x + 3 x ) ≈−1− x−x 3 9A1 3 CAOTotal 152cos 2x= 2cos x − 1B1B1 2(b)7(a)(i)cos x= (cos 2x+1)2π22 11 ⎡1x ⎤cos 2 1 d sin 22∫x+ x= ⎢ x+4 2⎥0⎣ ⎦π=4⎡6⎤ ⎡2⎤ ⎡4⎤!!!"AB =⎢5⎥ ⎢1⎥ ⎢4⎥⎢ ⎥−⎢ ⎥=⎢ ⎥⎢⎣3⎥⎦ ⎢⎣3⎥⎦ ⎢⎣0⎥⎦π20M1A1A1Attempt to expresscos 2x2cos x in terms ofM1A1F 5 Use limits. Ft on integer a.Total 7Penalise use of co-ordinates at firstM1occurrence onlyA1 2(ii) ⎡4⎤⎡1⎤⎢4⎥4⎢1⎥⎢ ⎥=⎢ ⎥⇒ parallel⎢⎣0⎥⎦⎢⎣0⎥⎦E1 1Needs comment “same direction”Or “same gradient”Or by scalar product( )(iii) ⎡ 2⎤ ⎡ 6⎤ ⎡1⎤⎢3⎥ ⎢1⎥λ⎢1⎥⎢−⎥=⎢ ⎥+⎢ ⎥M1⎢⎣−1⎥⎦ ⎢⎣−1⎥⎦ ⎢⎣0⎥⎦is satisfied by λ =− 4A1 2 λ = – 4satisfies 2equations5
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