06 Core Mathematics C3 Mock Paper Mark Scheme

7/23/2019 06 Core Mathematics C3 Mock Paper Mark Scheme

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EDEXCEL CORE MATHEMATICS C3 (6665)MOCK PAPER MARK SCHEME

Question

number Scheme Marks

1. 2 x2 +7 x +6 = ( x + 2)(2 x + 3) M1 A1

5

2

3

)23(7

)23)(2(

3

x

x

x x

x +×

++

= 3)2(

7

x x+ some correct algebraic cancelling M1 A1 ()

(4 marks)

2. (a) ! "1( x) = 21 x# x ∈ℝ $1 $1 (2)

(b) g! "1( x) = g( 21 x) =

3 x2 + 2 M1 A1 (2)

(c) %ange g! −1( x) ≥ 2 $1 (1)

(5 marks)

3. (i) e2 x + 3 = 6

2 x + 3 = ln 6 M1

x =2

1(ln 6 − 3) M1 A1 (3)

(ii) ln (3 x + 2) =

3 x + 2 = e M1

x =3

1(e − 2) M1 A1 (3)

(6 marks)

1




Question

number Scheme Marks

4. (i) u = x3=

x

u

&

&3 x2

v = e3 x =

x

v

&

&3e3 x

=

x

y

&

& 3 x2 e3 x + x33e3 x or e'ui M1 A1 A1 (3)

(ii) u = 2 x =

x

u

&

&2

v = cos x =

x

v

&

& "sin x

=

x

y

&

&

x

x x x

2cos

sin2cos2 +

or e'ui M1 A1 A1 (3)

(iii) u = tan x =

x

u

&

& sec2 x

y = u2 =

u

y

&

&2u

=

x

y

&

&2u sec2 x M1

=

x

y

&

&2 tan x sec2 x A1 (2)

(i) u = y2 =

y

u

&

&2 y

x = cos u =

u

x

&

&− sin u M1

=

y x&

& − 2 y sin y2 A1

=

x

y

&

&2sin2

1

y y

−

M1 A1 ()

(12 marks)

2




Question

number Scheme Marks

5. (a) (i) sin ( A + B) " sin ( A " B)

= sin A cos B + sin B cos A " sin A cos B + sin B cos A M1

= 2 sin B cos A () A1 cso (2)

(ii) cos ( A " B) " cos ( A + B)

= cos A cos B + sin A sin B " cos A cos B + sin A sin B M1

= 2 sin A sin B () A1 cso (2)

(b)( ) ( )

( ) ( ) B A B A

B A B A

+−−

−−+

sincos

sinsin =

B A

A B

sinsin2

cossin2M1

= A

A

sin

cosA1

= cot A () A1 cso (3)

(c) *et A = 75° an& B = 15° $1

o

oo

oo

75cot+,cos6,cos

6,sin+,sin=

−

−M1

cot 75° =

,2

12

31

−

−

= 2 − 3 M1 A1 ()

(11 marks)

3




Question

number Scheme Marks

6. (a)

O 1 $1 sha-e

$1 sha-e

$1 sha-e

$1 -oints ()

(b)

3

M1an. translation

M1 correct

&irection#

translation

$1 -oints

$1 as.m-totes

()

(c)

"1 , 1

$1 sha-e / ,

$1 sha-e 0 ,

$1 -oints

$1 as.m-totes

()

(12 marks)

(",5# 2)

(,# )

x 0 ,

, 0 x 0 1

x / 1

y

x

(25# "2)

(3# ")

y

x

(",# ") (,# ")

y

x




Question

number Scheme Marks

7. (a)

$1 sha-e

$1 xinterce-t

labelle&

(2)

(b) x x

y 1

&

&= so tangent line to (e# 1) is y =

e

1 x + C M1

the line -asses through (e# 1) so 1 = ee

1 + C an& C = , M1

he line -asses through the origin A1 (3)

(c) All lines y = mx -assing through the origin an& haing a gra&ient / , lie

aboe the xa4is

hose haing a gra&ient 0e

1 ill lie belo the line $1

y =e

x so it cuts y = ln x beteen x =1 an& x = e $1 (2)

(&) x, = 16

x1 = 3en x

= 15 M1

x2 = 15 A1

x3 = 15

x = 15

x5 = 157 A1 (3)

(e) hen x = 1575# ln x " x3

1

= ,,,, ,6 8 / , M1hen x = 1565# ln x = ",,,, 1,8 0 , A1

9hange o! sign im-lies there is a root beteen A1 (3)

(13 marks)

5

y = ln x

1

y

xO

1

y

x

y = ln x

y =

O




Question

number Scheme Marks

8. (a) sin θ " 3 cos θ = R sin θ cos α − R cos θ sin α

sin θ terms gie = R cos α

cos θ terms gie 3 = R sin α

tan α = ,75 M1

α = 36o A1

R2 = 2 + 32 = 25 ⇒ R = 5 M1 A1 ()

(b) 5 sin (θ " 36o) = 3

sin (θ " 36o) = ,6 M1

θ " 36o = 36o # 131 A1 M1

θ = 737o # 1,° art 7° A1 A1 (5)

(c) Ma4 alue 5 $1 (1)

(&) sin (θ "36o) = 1 M1

θ −

36°

= ,°

θ = ,° + 36° = 126° A1 (2)

(12 marks)

6




Question Specification

Section

AO1 AO2 AO3 AO4 AO5 Totas

Q1 11 2 2

Q2 12 3 2 5

Q3 31# 32 2 6

Q 1# 2# 3 5 6 1 12

Q5 23 3 5 3 11

Q6 13# 1 7 5 12

Q7 32# 52 2 3 2 3 3 13

Q 21# 23 2 2 3 2 3 12

TOTA! 26 2" 5 8 7 75

7

06 Core Mathematics C3 Mock Paper Mark Scheme

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