A-level MATHEMATICS 7357/2 Paper 2 - Mark Scheme

A-level

MATHEMATICS 7357/2

Paper 2: MARK SCHEME – A-LEVEL MATHEMATICS – 7357/2 – JUNE 2020

Uses the product of (2x)5 and  33

 ± x  terms condone sign error 

and/or omission of nCr term Or
Obtains any two correct terms (unsimplified) term

( )5  33 Multiplies their 2x and− x  by

 For this mark condone(2x)3 and

8C3 or 8C5 or 56 OE

 − 3 5

x

Obtains correct coefficient of x2 −48384

condone inclusion x2

Uses or states small angle approximationfor tan5x≈5x

Uses or states small angle approximation for

cos4x ≈1− (4x)2 2

Condone omission of bracket

1.1b

B1

Substitutes their expressions Oftheform tan5x≈mxand

cos4x≈1−nx2 2

into xtan5x

cos4x−1
Condone correct extra terms

Deduces A = −8 from a reasoned

argument CSO

du u=4x1+⇒ =4

dx x =− ⇒ u= 0

1

4
x = 6 ⇒ u = 25

x = u −1 4

∫6 x 4x+1dx=∫25u−1 u 1du −1 044

4

22
= 1 ∫25u3 u1du

16 1
1 2u5 2u3 25

=2 2− 165 3

  0

1255 253  =2 2−

853 

 = 875

12

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u2 −u2  or B(u4 −u2 )



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

−

8

MARK SCHEME – A-LEVEL MATHEMATICS – 7357/2 – JUNE 2020

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

5y +12x = 298
y=−12x 298+

55

y−9=5(x 7)− 12

12 y − 5x = 73 x = 19

y =14 (19,14)

Uses (7, 9) to find the equation of the radius
Or
Uses (7, 9) correctly in their equation of circle

Or −12
Uses 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  298−12x 2 (x−7) + 5 −9 =k

 2  253−12x 2

⇒(x−7) + 5  =k 

 298−5y 2 2
 12 −7 +(y−9) =k


 214 − 5 y 2 2

⇒ 12  +(y−9) =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