AQA A-level MATHEMATICS PAPER 2 Mark Scheme 2020.

AQA A-level MATHEMATICS PAPER 2 Mark Scheme 2020. Q Marking instructions AO Marks Typical solution 1 Circles correct answer 2.2a B1 ( ) 1 f x x e − = − Total 1 Q Marking instructions AO Marks Typical solution 2 Ticks correct box 2.2a B1 sec x = 0 Total 1 Q Marking instructions AO Marks Typical solution 3 Uses the product of ( ) 5 2x and 3 3 x    ±    terms condone sign error and/or omission of nCr term Or Obtains any two correct terms (unsimplified) term 3.1a M1 ( ) 3 5 8 2 3 3 2 56 32 27 coefficient is 48384 C x x x   × −  = × × −   ∴ − Multiplies their ( ) 5 2x and 3 3 x    −    by 8C3 or 8C5 or 56 OE For this mark condone( ) 3 2x and 5 3 x    −    1.1a M1 Obtains correct coefficient of x 2 −48384 condone inclusion x 2 1.1b A1 Total 3 MARK SCHEME – A-LEVEL MATHEMATICS – 7357/2 – JUNE 2020 7 Q Marking instructions AO Marks Typical solution 4 Uses or states small angle approximation for tan 5x ≈5x 1.1b B1 ( ) 2 2 2 tan 5 5 cos 4 1 4 1 1 2 5 8 5 8 x x x x x x x x × ≈ − − − ≈ − ≈ − Uses or states small angle approximation for ( ) 2 4 cos 4 1 2 x x ≈ − Condone omission of bracket 1.1b B1 Substitutes their expressions Of the form tan 5x≈mx and 2 cos 4 1 2 nx x ≈ − into tan 5 cos 4 1 x x x − Condone correct extra terms 1.1b M1 Deduces 5 8 A = − from a reasoned argument CSO 2.2a R1 Total 4 MARK SCHEME – A-LEVEL MATHEMATICS – 7357/2 – JUNE 2020 8 Q Marking instructions AO Marks Typical solution 5 Uses a suitable substitution u = 4x +1 oru = 4x +1 OE 3.1a M1 6 25 1 0 4 25 3 1 2 2 1 25 5 3 2 2 0 5 3 2 2 d 4 1 4 d 1 0 4 6 25 1 4 1 1 4 1 d d 4 4 1 d 16 1 2 2 16 5 3 1 25 25 8 5 3 875 12 u u x x x u x u u x u x x x u u u u u u u − = + ⇒ = = − ⇒= = ⇒= − = − + = = −     = −           = −       = ∫ ∫ ∫ Differentiates their substitution correctly 1.1b A1 Completes substitution to obtain correct integrand for their suitable substitution. Can be unsimplified. 1.1a M1 Correctly integrates their simplified integrand provided it is of the form 3 1 A u 2 2 u    −    or ( ) 4 2 B u −u 1.1a A1 Substitutes correct limits for their substitution or 6 and -1/4 for x 1.1a M1 Completes rigorous argument to show the required result. AG 2.1 A1 Total 6 MARK SCHEME – A-LEVEL MATHEMATICS – 7357/2 – JUNE 2020 9 Q Marking instructions AO Marks Typical solution 6(a) Begins a valid method to find the coordinates Uses gradient of L to find gradient of perpendicular radius Or Forms equation of circle with unknown radius and solves simultaneously with equation of L Or differentiates equation of circle implicitly 3.1a M1 ( ) 5 12 298 12 298 5 5 5 9 7 12 12 5 73 y x y x y x y x + = − = + − = − − = 19 14 x y = = (19,14) Uses (7, 9) to find the equation of the radius Or Uses (7, 9) correctly in their equation of circle Or Uses 12 5 − after their implicit differentiation 1.1a M1 Obtains 12y −5x = 73OE Or Correctly eliminates a variable to obtain a quadratic in x or y for example obtain a quadratic in x or y ( ) ( ) 2 2 2 2 298 12 7 9 5 253 12 7 5 x x k x x k  −  − +  −  =    −  ⇒ − +   =   ( ) ( ) 2 2 2 2 298 5 7 9 12 214 5 9 12 y y k y y k  −   −  + − =    −  ⇒   + − =   1.1b A1 Equates discriminant to zero PI By correct answer or Solves their simultaneous equations of tangent and radius PI by correct answer 3.1a M1 Obtains correct values for x and y (19,14) 1.1b A1 Subtotal 5