AQA_A_level_Mathematics_7357

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 MARK SCHEME – A-LEVEL MATHEMATICS – 7357/2 – JUNE 2019 Q Marking instructions AO Mark Typical solution 5 Separates the variables – one side correct Condone missing integral signs PI by correct integration 3.1a M1 ( ) 2 2 2 1 1 2 2 2 1 ln d d d 2 ln 1 1 1 ln d 1 1 ln d 1 1 ln 1 1 ln 2 2, 1 1 2 3 1 ln 6 2 xx tt x t tt c u x u x v x v x x xx x x x x x x x x x t x c x x tx c c x t x − − − = = + = ′ = ′ = = − − −− − + − − − −= + = = ⇒− = + = −   + = −     ∫ ∫ ∫ ∫ ∫ Integrates their t t d ∫ correctly 1.1b A1F Obtains 1 u x ′ = and 1 v x = − OE 1.1b B1 Integrates 2 1 ln dx x ∫ x Substitutes their u, u’, v and v’ into the correct formula for integration by parts Condone sign errors in formula 1.1a M1 Obtains 1 1 ln x x x − − 1.1b A1 Substitutes t = 2 and x = 1 into their integrated equation to find their +c 1.1a M1 Obtains correct solution m