page 22 Notes: 4. For a numerical approach award 0/6. 5. •6 can be awarded for 2 32 4000 x x − = . 6. For candidates who integrate any term at the •5 stage, only •6 is available on follow through for setting their ‘derivative’ to 0. 7. •7, •8 and •9 are only available for working with a derivative which contains an index 2 ≤−. 8. 3 4000 32 must be simplified at •7 or •8 for •7 to be awarded. 9. •8 is not available to candidates who consider a value of 0 x ≤ in the neighbourhood of 5. 10. •9 is still available in cases where a candidate’s table of signs does not lead legitimately to a minimum at •8. 11. •8 and •9 are not available to candidates who state that the minimum exists at a negative value of x . See Candidates C and D. For the table of signs for a derivative, accept: x 5 − 5 5 + ( ) A x ′ − 0 + Shape or slope x → 5 → ( ) A x ′ − 0 + Shape or slope Arrows are taken to mean ‘in the neighbourhood of’ x a 5 b ( ) A x ′ − 0 + Shape or slope Where 0 5 a < < and 5 b > • For this question do not penalise the omission of ‘x’ or the word ‘shape’/‘slope’. • Stating values of ( ) A x ′ in the table is an acceptable alternative to writing ‘+’ or ‘−’ signs. Values must be checked for accuracy. • The only acceptable variations of ( ) A x ′ are: A′, dA dx and 2 32 4000 x x − − . Commonly Observed Responses: Candidate A – differentiating over multiple lines •4 ^ ( ) 1 32 4000 A x x x− ′ = + ( ) 2 32 4000 A x x x− ′ = − •5 8 2 32 4000 0 x x − − = •6 9 1 Candidate B – differentiating over multiple lines 2 1 16 4000 A x x − = + •4 9 ( ) 1 32 4000 A x x x− ′ = + ( ) 2 32 4000 A x x x− ′ = − •5 8 2 32 4000 0 x x − − = •6 9 1 Candidate C – only considers 5 2 1 16 4000 A x x − = + •4 9 2 32 4000 0 A x x − ′ = − = •5 9 •6 9 5 x = ± •7 8 x → 5 → A′ − 0 + ∴ minimum •8 9 1 1200 or min value 1200 A = = •9 9 1 Candidate D – considers 5 and negative 5 in separate tables 2 1 16 4000 A x x − = + •4 9 2 32 4000 0 A x x − ′ = − = •5 9 •6 9 5 x = ± •7 8 x → 5 → x → −5 → A′ − 0 + A′ − 0 + ∴ minimum when 5 x = •8 9 1 1200 or min value 1200 A = = •9 9 1 Ignore incorrect working in second table page 23 Question Generic scheme Illustrative scheme Max mark 12. Method 1 •1 state linear equation •2 introduce logs •3 use laws of logs •4 use laws of logs •5 state a and b Method 1 •1 log4 3 1 y x = − •2 log log log 4 4 4 3 4 4 y x = − •3 log log log 3 4 4 4 4 4 x y = − •4 log log 3 4 4 4 4 x y ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ or log log 1 3 4 4 4 4 x y − = •5 , 1 64 4 a b = = 5 Method 2 •1 state linear equation •2 convert to exponential form •3 use laws of indices •4 state a •5 state b Method 2 •1 log4 3 1 y x = − •2 3 1 4 x y − = •3 1 3 4 4 x y − = •4 1 4 a = •5 64 b = 5 Method 3 •1 introduce logs to x y ab = •2 use laws of logs •3 interpret intercept •4 interpret gradient •5 state a and b Method 3 The equations at •1, •2, •3 and •4 must be stated explicitly. •1 log log 4 4 x y ab = •2 log log log 4 4 4 y a x b = + •3 log4 1 a −= •4 log4 3 b = •5 , 1 64 4 a b = = 5 page 24 Question Generic scheme Illustrative scheme Max mark Method 4 •1 interpret point on log graph •2 convert from log to exponential form •3 interpret point and convert •4 substitute into x y ab = and evaluate a •5 substitute other point into x y ab = and evaluate b Method 4 •1 log4 3 8 x y = = and •2 8 3 4 x y = = and •3 log4 0 1 x y = =− and 1 0 4 x y − = = and •4 1 0 1 4 4 ab a −= ⇒ = •5 8 3 1 4 64 4 b b = ⇒ = 5 Notes: 1. In any method, marks may only be awarded within a valid strategy using x y ab = . 2. Accept 1 64 4 x y ⋅ = for •5. 3. Markers must identify the method which best matches the candidates approach; they must not mix and match between methods. 4. Penalise the omission of base 4 at most once in any method. 5. Do not accept 1 4 a − = . Commonly Observed Responses: page 25 Question Generic scheme Illustrative scheme Max mark 13. •1 interpret information given •2 integrate any two terms •3 complete integration •4 interpret information given and substitute •5 process for c and state expression for ( ) f x •1 ( ) 2 3 16 11 f x x x ′ = − + or ( ) ( ) 2 3 16 11 f x x x dx = − + ∫ •2 eg ... 3 2 3 16 3 2 x x − •3 11x c …+ + •4 3 2 0 7 8 7 11 7 c = − × + × + •5 ( ) 3 2 8 11 28 f x x x x = − + − 5 Notes: 1. For candidates who make no attempt to integrate to find ( ) f x award 0/5. 2. Do not penalise the omission of ( ) f x or dx or the appearance of +c at •1. 3. If any two terms have been integrated correctly •1 may be implied by •2. 4. For candidates who omit c + , only •1 and •2 are available. 5. For candidates who differentiate any term, •3 •4 and •5 are not available. 6. Candidates must attempt to integrate both terms containing x for •4 and •5 to be available. See Candidate B. 7. Accept 3 2 8 11 28 y x x x = − + − at •5 since ( ) y f x = is defined in the question. 8. Candidates must simplify coefficients in their final line of working for the last mark available in that line of working to be awarded. Commonly Observed Responses: Candidate A – incomplete substitution ( ) 3 2 8 11 f x x x x c = − + + •1 9 •2 9 •3 9 ( ) 3 2 7 8 7 11 7 f x c = −× + × + •4 ^ 28 c = − ( ) 3 2 8 11 28 f x x x x = − + − •5 9 1 Candidate B - partial integration ( ) 3 2 8 11 f x x x c = − + + •1 9 •2 9 •3 8 3 2 0 7 8 7 11 c = − × + + •4 9 1 38 c = ( ) 3 2 8 49 f x x x = − + •5 9 1 page 26 Question Generic scheme Illustrative scheme Max mark 14. •1 expand •2 evaluate u.u •3 determine equation in cos θ •4 evaluate angle •1 . . + uu uv •2 16 •3 cos 20 5 θ = or cos 5 20 θ = •4 75 5 ⋅…° or 1 31 ⋅ … radians 4 Notes: 1. Do not accept 2 u for •1, however •2, •3 and •4 are still available. 2. Where there is no evidence for •1, then •2, •3 and •4 are not available, however see Candidates C and D. 3. Where candidates use 4 ≠ u , then •3 and •4 are not available. 4. Where there is no evidence of using 2 u , •3 is not available. See Candidate A. 5. Do not penalise omission of units in final answer. 6. Ignore the appearance of 284 5⋅°. 7. Accept answers which round to 76° or 1·3 radians. Commonly Observed Responses: Candidate A ( ) . . . + = + u u v uu u v •1 9 cos 4 20 21 θ + = •2 8 cos 17 20 θ = •3 9 2 31 7 θ = ⋅…° •4 9 1 Candidate B . 16 21 + = u v •1 9 •2 9 . 5 = uv cos 5 20 θ = •3 9 75 5 θ = ⋅° •4 9 Candidate C – missing working . 16 = u u •2 9 . 21 16 = − u v cos 5 20 θ = •1 9 •3 9 75 5 θ = ⋅° •4 9 Candidate D – missing working 21 16 5 − = •1 ^ cos 5 20 θ = •2 9 •3 9 75 5 θ = ⋅° •4 9 page 27 Question Generic scheme Illustrative scheme Max mark 15. (a) •1 find gradient of radius •2 state gradient of tangent •3 state equation of tangent •1 1 3 − •2 3 •3 3 2 y x = − 3 Notes: 1. Do not accept 3 2 1 y x = − for •3. 2. •3 is only available as a consequence of trying to find and use a perpendicular gradient. 3. At •3 accept, 3 2 0 y x − + = or any other rearrangement of the equation where the constant terms have been simplified. Commonly Observed Responses: (b) (i) •4 find coordinates of T •4 (0,−2) 1 (ii) •5 find midpoint CT •6 find radius of circle with diameter CT •7 state equation of circle •5 (4,5) •6 65 stated or implied by •7 •7 ( ) ( ) 2 2 4 5 65 x y − + − = 3 Notes: 4. Answers in part (b)(i) must be consistent with answers from part (a). 5. Accept 0, 2 x y = = − for •4. 6. ( ) ( ) ( ) 2 2 2 4 5 65 x y − + − = does not gain •7. 7. •7 is not available to candidates who use a line other than CT as the diameter of the circle. Commonly Observed Responses: [END OF MARKING INSTRUCTIONS]