Question Answer 1 C 2 B 3 D 4 A 5 C 6 B 7 C 8 D 9 B 10 D 11 C 12 C 13 B 14 D 15 A 16 B 17 D 18 C 19 B 20 A

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1 Paper Section A Question Answer C B D 4 A 5 C 6 B 7 C 8 D 9 B 0 D C C B 4 D 5 A 6 B 7 D 8 C 9 B 0 A Summary A B 6 C 6 D 5 Page 5

2 Paper - Section B Question Generic Scheme Illustrative Scheme Ma Mark (a). know to use interpret result and state conclusion Method 6() 9() 4 0 ( ) is a factor. Method remainder 0 ( ) is a factor. Method ( ) is a factor. state quadratic factor 4 factorise completely 5 4 stated or implied by Page 6

3 . Communication at must be consistent with working at that stage i.e. a candidate s working must arrive legitimately at 0 before is awarded.. Accept any of the following for : f() 0 so ( )is a factor since remainder is 0, it is a factor the 0 from the table linked to the word factor by e.g. so, hence,,,. Do not accept any of the following for : double underlining the zero or boing the zero without comment is a factor, ( ) is a factor, is a root, ( ) is a root, ( ) is a root the word factor only, with no link 4 in any order. 4. At 4 the epression may be written as 5. An incorrect quadratic correctly factorised may gain Where the quadratic factor obtained is irreducible, candidates must clearly demonstrate that b 4ac 0 to gain must appear at or for to be awarded. 8. For candidates who do not arrive at 0 at the stage 4 are not available. 9. Do not penalise candidates who attempt to solve a cubic equation. However, within this working there may be evidence of the correct factorisation of the cubic. 0. Evidence for & 4 may appear in part (b). Page 7

4 Question Generic Scheme Illustrative Scheme Ma Mark (b)(i). 5 know to and differentiate 5 6 find gradient 6 7 state equation of tangent 7 y. 7 is only available if an attempt has been made to find the gradient from differentiation.. At mark 7 accept y ( ), y or any other rearrangement of the equation. Candidate A 5 6 using y m c y m c c 7 y (b)(ii). 8 set ycurve yline arrange equation in standard cubic form identify coordinate of B and calculate y coordinate 0 (4,9). 9 is only available if 0 appears in at least one arrangement of the equation. 4. Solutions at 0 must be consistent with working at 4 and Candidates who obtain three distinct factors at 4 can gain 8 and 9 but 0 is unavailable. 6. For 0 accept 4, y Do not penalise the appearance of (,). Page 8

5 Question Generic Scheme Illustrative Scheme Ma Mark. arrange in differentiable form f ( ) 4 start differentiation complete differentiation and set f( ) 0 4 evaluate f at stationary point 8 or 8 0 4, f ( ) 5 consider end-points 5 7 f() 5, f(4) (or 4 5) 4 6 state ma and min values 6 ma5, min 6. Candidates must attempt to differentiate a term with a ve or fractional power by for to be awarded.. is not available for simply stating f '( ) 0. A clear link between the candidates derivative and f '( ) 0 is required.. For candidates who integrate, but clearly believe they are finding the derivative are available (see CORs K, L, and M). In other instances where candidates have integrated then only and 5 are available. A numerical approach can only gain and 6 are not available to candidates who consider stationary points only. 5. Treat maimum (,5) and minimum (,) as bad form. 6. Vertical marking is not applicable to 5 and The appearance of (,) following any nd derivative or nature table gains If at the 4 stage a value of is obtained outwith the given interval then 4 is unavailable, but 6 may still be gained (see CORs F and G). 9. Candidates who consider the end values but do not evaluate the stationary value cannot gain 6. Candidate A Candidate B Candidate C Candidate D f ( ) 4 f '( ) 8 for stationary points f '( ) 0 With no further working f ( ) 4 ^ for stationary ^ points f '( ) 0 With no further working f ( ) 4 f '( ) 8 for stationary ^ dy points 0 d With no further working f ( ) 4 8 for stationary points f '( ) 0 ^ With no further working Page 9

6 Candidate E Solely a numerical attempt f() 5, f(), 7 f(), f(4) 9 4 Award only 5 For any similar attempt which includes the evaluation of f for a value outwith the range award 0. Candidate F f ( ) 4 f '( ) 8... f '( ) f( ) f() 5, f(4) min 4, ma Candidate G f ( ) 4 f '( ) 8... f '( )... 0 f( ) f() 5, f(4) 4 4 min 4, ma Candidate H Candidate I Candidate J f ( ) 4 f '( ) 8... f '( ) f ( ) 8 6 f() 5, f(4) 4 4 min 4, ma f ( ) 4 '( ) 0 0 f 0 f ( ) undefined 4 5 ^ 6 ^ 4 f '( ) 0 f() 5, f(4) ^ Candidate K Candidate L Candidate M f ( ) 4 f '( ) 4 f() f() 5, f(4) f ( ) f() f() 5, f(4) f() f() 5, f(4) min, ma 5 min, ma 5 min, ma 5 Page 0

7 Question Generic Scheme Illustrative Scheme Ma Mark. collect log terms use laws of logs use laws of logs Method log 7 log stated or implied by 7 log 4 solve for 7 4 Method log 7 log log stated or implied by log 7 log log ( ) log 7 log 8( ). For accept 7 log log 8. Candidate A 4 Candidate B 4 log 7 log log log 7 log log (7) log ( ) log (7) log ( ) log (7)( ) ( 7)( ) 45 0 ( 5)( ) 0 5 or discard as is not available for candidates who do not discard Page

8 Question Generic Scheme Illustrative Scheme Ma Mark 4. interpret the values of a, b and c and substitute know to use discriminant 0 Method simplify or factorise quadratic inequation 4 state range of values of k Method simplify or factorise quadratic epression 4k 9k... 0 Method 9 k k k or k Method 9 6k 0 k, 4 evidence and range of values of k 4 graph or other evidence leading to k. The 0 must appear at least once at the or stage for to be awarded.. If an appears in the candidate s discriminant only may be awarded.. The use of any epression masquerading as the discriminant can gain only at most. 4. Award to candidates who write BOTH 9 6k 0 AND 9 6k For candidates who at simplify or factorise an equation 4 can only be awarded if evidence of solving an inequation (for eample a graph) appears. 6. At stage, quoting b 4ac 0 is not sufficient. 7. At stage, in Method, solutions for k need not be simplified. 4 Page

9 Question Generic Scheme Illustrative Scheme Ma Mark 5(a) know to use Theorem of Pythagoras or distance formula process to obtain result Be wary of fudged solutions! Candidate A D t 5 t 0 Beware D 4t 0t 5 t 0t 00 5t 40t5 D y y ( 5) 0 0 ( 0) D t t D t t See note in the general instructions 5(b) write in differentiable form 4 start differentiation 5 complete differentiation A(t 5, 0) B(0, t 0) 0 t5 t5 AB t0 0 t0 D (t 5) ( t 0) D t t t t D 5t 40t 5 NB: This is an eception to note in the general instructions 5t 40t t t t 40 6 substitute t = 5 and interpret result 6 0 D(5) 0 increasing is only available for differentiating an epression of the form (trinomial) proper fraction.. Do not penalise the use of instead of t.. 6 is only available to candidates who substitute into a derivative. 4 Page

10 Candidate A Candidate B D t t t ( ) ( ) D t t t... (0t 40) 0 D(5) 0decreasing 50 Candidate C 4 see note. 5 6 D( t) 5t 40t 5 D( t) 5t 40t (0t 40) 5 D(5) 0decreasing 50 Candidate D D( t) 5t 40t 5 D( t) 5t 40t (0t 40) D (5) increasing Candidate E D( t) 5t 40t 5 D( t) 5t 40t 5 0t 40 D(5) 0 increasing Calculating Distance t 5 D 50 t 4 D 45 so distance is increasing Award 0 marks as answer is not from differentiation D( t) 5t 40t 5 D( t) 5t 40t 5 0t 40 D( t) 5t 5t 40t D(5) 40 0decreasing 50 Candidate F Alternative Method D t t dd 0t 40 dt dd t dt dd dt D is increasing Alternative Response 0 0 D is increasing D t t D 5( 8 5) 5[( t 4) 9] Graph together with a statement indicating that when t 5, D is increasing and therefore D is increasing. Min TP at Page 4