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

MPC4 – AQA GCE Mark Scheme, 2006 January seriesMPC4 (cont)Q Solution Marks Total Comments(b)(i) l 2 has equationOr⎡4⎤ ⎡⎡4⎤ ⎡ 2⎤⎤⎡4⎤ ⎡2⎤M1A1 2 ⎡ 2 ⎤ ⎡2⎤⎢ ⎥r =⎢1⎥+ λ⎢1⎥−⎢− 3⎥=⎢1⎥+ λ⎢4⎥⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥⎥r =⎢–3⎥t⎢4⎥⎢ ⎥ ⎢ ⎥⎢ ⎥+⎢ ⎥M1 calculate and use⎢⎣1⎥⎦ ⎢⎣⎢⎣1⎥⎦ ⎢⎣−1⎥⎦ ⎥⎦⎢⎣1⎥⎦ ⎢⎣2⎥⎦⎢⎣ –1⎥⎦ ⎢⎣2⎥⎦direction vector A1 all correct(ii) ⎡1⎤ ⎡ 4⎤⎢2⎥ ⎢0⎥⎢ ⎥•⎢ ⎥= 4− 4=0⎢⎣1⎥⎦ ⎢⎣−4⎥⎦8(a)M1A1⇒ 90° ( or perpendicular )A1F 3∫dxdx= ∫−2dtx − 6Total 10M1Clear attempt to use directions of AC andl 2 in scalar productAccept a correct ft value of cosθAttempt to separate and integrate2 x − 6 = − 2t+ cA1A1 c on either sidet = 0 x=70 ⇒ c = 16m1A1F Follow on c from sensible attempt atintegrals ( not ln )t = 8− x− 6A1 6 CAO ( or AEF )(b)(i) The liquid level stops falling/flowing/ B1 1at minimum depthx= 22 t = 8 – 22 – 6M1 Use x = 22 in their equation providedthere is a cOr start again using limitsM1 2 64 – 2 16 =± 2 t, A1 t = 4t = 4 A1 2 CAOTotal 9Total 756