Mark Scheme (Pre-standardisation)

Transcription

1 Mark (Pre-standardisation) June 013 GCE Core Mathematics C3 (6665/01)

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3 General Marking Guidance All candidates must receive the same treatment. Examiners must mark the first candidate in exactly the same way as they mark the last. Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than penalised for omissions. Examiners should mark according to the mark scheme not according to their perception of where the grade boundaries may lie. There is no ceiling on achievement. All marks on the mark scheme should be used appropriately. All the marks on the mark scheme are designed to be awarded. Examiners should always award full marks if deserved, i.e. if the answer matches the mark scheme. Examiners should also be prepared to award zero marks if the candidate s response is not worthy of credit according to the mark scheme. Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and exemplification may be limited. When examiners are in doubt regarding the application of the mark scheme to a candidate s response, the team leader must be consulted. Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response.

4 1. (a) x + 3x + = ( x + )( x + 1) Attempt as a single fraction 6x + 1 ( x + 3x+ ) 6 ( x + 1) or x + 3x + x x = x + 1 (b)(i) Same shape intercept at (0,8) x intercept at (1,0) (b)(ii) Correct shape in quadrants 1& Both (0,) and (4,0) () (8 marks)

5 . (a) (b) cot = = 0 tan 40 p Attempts to use tan 40 = sec 40 0 sec40 = (1 + p ) (1) () tan 45 + tan 40 (c ) Attempts to use tan85 = tan( ) = tan45 tan p tan85 = 1 p () (5 marks)

6 3. (a) (.5,0) (0,-5) (b) 8 x 5= 3 x x= oe. 3 x 5= 3 x x = 8, () (5 marks)

7 4 (a) Shape for y = 10 x Shape for y = e x co- ordinates correct (0,10),(10,0)) and (0,1) (b) One solution as theree is one point of intersection (1) (c ) x Sub x= and x= =3 into f(x) = e 10 + x f() )=-0.61, f=(+)13.1 Both correct to 1sf, reason (change of sign) and conclusionn (hence root) (d) Substitutes x 1 = into x n + 1 = ln(10 x ) x =.0794, x 3 =.0695 x 4 =.0707 n, () (9 marks)

8 5. (a) (i) Applies vu ' + uv' to 1 x ln x = ln x x + x x ln x 1 = + x x * (ii) Sets ln x + 1 = 0 x x Multiplies (or factorises) by x, with correct ln work leading to x= P e e 1 = (, ) oe., (b) vu ' uv' x k Applies to y = with u = x k and v = x+ k v x + k d y ( x+ k) 1 ( x k) 1 = d x ( x+ k) dy k = d x ( x+ k) dy As k > 0 > 0 C has no turning points dx (4) (4) (11 marks)

9 6. (a) f( x) 3 (1) (b) f(0) = 5or attempts to put their f(0) into 8 e x 4 Correct answer ff(0)= e 4 () (c ) Either 5 x = 1 x = 8 Or 8 e x 4= 1 ln Correct order and ln work x = oe. ln5 + 4 (5) (d) f does not have an inverse as it is a many to one function Accept f does not have an inverse as it is not a one to one function (9 marks) (1)

10 7. (a) (b) cosx 1 sinx cosxcos x+ (1 sin x)(1 sin x) + = 1 sinx cos x (1 sin x)cosx cos x + sin x+ 1 sin x = (1 sin x) cos x sinx = (1 sin x) cos x (1 sin x) = (1 sin x) cos x = = secx cos x * cos x 1 sin x + = 8sin x 1 sinx cosx secx = 8sinx 1= 4sinx cosx asinxcosx=b (4) 1 sin x = sin x = sin xcos x 1 1 π x = arcsin( ) = 1, (4) (8 marks)

11 8. (a) 9cosθ sinθ = Rcos( θ + α) R = (9 + = 85 α = arctan( ) = = awrt (b) (i) 85 (ii) θ + α = π θ = awrt 6.06 dp (c ) Seeing (or implied by their working) πt H = 10 Rcos( + α ) for their R and α 5 Hmax = 10 + their R= ( = 19.m) πt πt Maximum occurs when cos( + α ) = 1 or ( + α ) = π 5 5 t= awrt 4.65 πt πt (d) Setting and solving = π (for 1 cycle) or = 4π (for cycles) 5 5 Two revolutions = 0 minutes (4) () (1 marks)

12 Question 9. (a) x=3 or (3,0) (b) 1 dx 1 = ( y y ) (16 4 y) oe dy (c ) Substitute y=0 into their d x dy or d y dx dx 8 = or d y 3 dy 3 dx = 8 (1) Uses their numerical d y and their 3 from (3,0) to find equation of tangent dx y = or y 0 = ( x 3) x (4) (8 marks)