A-LEVEL MATHEMATICS 7357/2 Paper 2 Mark scheme June 2019.

A-LEVEL MATHEMATICS 7357/2 Paper 2 Mark scheme June 2019. AS/A-level Maths/Further Maths assessment objectives AO Description AO1 AO1.1a Select routine procedures AO1.1b Correctly carry out routine procedures AO1.2 Accurately recall facts, terminology and definitions AO2 AO2.1 Construct rigorous mathematical arguments (including proofs) AO2.2a Make deductions AO2.2b Make inferences AO2.3 Assess the validity of mathematical arguments AO2.4 Explain their reasoning AO2.5 Use mathematical language and notation correctly AO3 AO3.1a Translate problems in mathematical contexts into mathematical processes AO3.1b Translate problems in non-mathematical contexts into mathematical processes AO3.2a Interpret solutions to problems in their original context AO3.2b Where appropriate, evaluate the accuracy and limitations of solutions to problems AO3.3 Translate situations in context into mathematical models AO3.4 Use mathematical models AO3.5a Evaluate the outcomes of modelling in context AO3.5b Recognise the limitations of models AO3.5c Where appropriate, explain how to refine models 4 MARK SCHEME – A-LEVEL MATHEMATICS – 7357/2 – JUNE 2019 Examiners should consistently apply the following general marking principles No Method Shown Where the question specifically requires a particular method to be used, we must usually see evidence of use of this method for any marks to be awarded. Where the answer can be reasonably obtained without showing working and it is very unlikely that the correct answer can be obtained by using an incorrect method, we must award full marks. However, the obvious penalty to students showing no working is that incorrect answers, however close, earn no marks. Where a question asks the student to state or write down a result, no method need be shown for full marks. Where the permitted calculator has functions which reasonably allow the solution of the question directly, the correct answer without working earns full marks, unless it is given to less than the degree of accuracy accepted in the mark scheme, when it gains no marks. Otherwise we require evidence of a correct method for any marks to be awarded. Diagrams Diagrams that have working on them should be treated like normal responses. If a diagram has been written on but the correct response is within the answer space, the work within the answer space should be marked. Working on diagrams that contradicts work within the answer space is not to be considered as choice but as working, and is not, therefore, penalised. Work erased or crossed out Erased or crossed out work that is still legible and has not been replaced should be marked. Erased or crossed out work that has been replaced can be ignored. Choice When a choice of answers and/or methods is given and the student has not clearly indicated which answer they want to be marked, mark positively, awarding marks for all of the student's best attempts. Withhold marks for final accuracy and conclusions if there are conflicting complete answers or when an incorrect solution (or part thereof) is referred to in the final answer. 5 MARK SCHEME – A-LEVEL MATHEMATICS – 7357/2 – JUNE 2019 Q Marking instructions AO Mark Typical solution 1 Ticks the correct response 2.2a R1 Total 1 Q Marking instructions AO Mark Typical solution 2 Circles the correct response 1.1b B1 Total 1 Q Marking instructions AO Mark Typical solution 3 Circles the correct response 1.2 B1 Total 1 8 15 a ( ) 2 fx x = 6 MARK SCHEME – A-LEVEL MATHEMATICS – 7357/2 – JUNE 2019 Q Marking instructions AO Mark Typical solution 4 Explains how the factor theorem applies with reference to f(-2) = 0 for either function or Explains that either quadratic expression can be factorised in the form (x + 2) (x + p) as (x + 2) is a factor or Explains that on division by (x + 2) the remainder would be zero 2.4 E1 As is a factor, then when , f(x) = 0 ( ) 42 0 42 0 42 42 2 2 2 b c d e bc de d b ec db ec − += − += − +=− + − =− − =− Uses the factor theorem with substituted into one of the expressions to obtain a correct expression NB It is not necessary to equate to zero for this mark or Expands one of their factorised forms and equates coefficients correctly 2 ( 2)( ) ( 2) 2 x xp x p x p + +=++ + 2 2 p b p c + = = or Divides one of the expressions by (x + 2) to obtain a correct remainder. Either one of 4 2 4 2 b c d e − + − + 1.1a M1 Deduces both correct equations using factor theorem or division PI by 42 42 − +=− + bc de or Expands both of their factorised forms and equates coefficients to deduce the correct equations – must not use p in both 2.2a A1 Forms a single equation for b, c, d and e and completes rigorous argument to show the required result NB R1 can be awarded even if E1 was not awarded 2.1 R1 Total 4 ( x + 2) x = −2 x = −2 42 0 42 0 b c d e − += − += 7