Practice Paper A5 Ms

EDEXCEL CORE MATHEMATICS C3

PRACTICE PAPER A5 MARK SCHEME

Question Number

Scheme y = tan x =

1.

Marks

sin x cos x

M1

cos x cos x − sin x(− sin x) dy = dx cos 2 x =

1 2

cos x

(use of quotient rule)

= sec2 x

*

M1 A1

A1

(4 marks) 2.

(a)

2+

3 x+2

 2 ( x + 2) + 3   =  x + 2  

3 (b) y = 2 + x + 2 y–2=

x=

y =

3 y−2

x=




,x ≠2

M1

7 − 2y y−2 f-−1 (x ) =

x

(1)

yx –2x = 7 – 2y M1

x (y – 2) = 7 – 2y

3 −2 x−2

Domain of f−1(x) is

B1

2x +7 x+2

y (x + 2) = 2x + 7 =

3 −2 y−2

∴ f−1 (x) = (c)

or

3 x+2

x+2=

2x+7 2( x + 2) + 3 or x+2 x+2

=

7 − 2x x−2

o.e

[NB x ≠ + 2 ]

A1

(3)

B1

(1)

(5 marks) 3.

(a)

2 13 + x − 3 ( x − 3) ( x + 7) =

(b)

2( x + 7) + 13 2 x + 27 = ( x − 3)( x + 7) ( x − 3)( x + 7)

2x + 27 = x2 + 4x − 21

M1

M1 A1

(3)

M1

x2 + 2x − 48 = (x + 8)(x − 6) = 0 x = −8, 6

M1 A1

(3)

(6 marks)

EDEXCEL CORE MATHEMATICS C3 Question Number

4.

(a)

PRACTICE PAPER A5 MARK SCHEME Scheme

x 2 + 4 x + 3 ( x + 3)( x + 1) = x( x + 1) x2 + x =

Marks

Attempt to factorise numerator or denominator

x+3 3 or 1 + x x

A1

 x 2 + 4x + 3  (b) LHS = log2   2  x +x 

Use of log a – log b

RHS = 24 or 16

3 15

or

1 5

(2)

M1 B1

x + 3 = 16x x=

M1

Linear or quadratic equation in x

or 0.2

M1 A1

(4)

(6 marks) 5.

(i) Choosing values of A and B and attempting to evaluate LHS and RHS of statement Showing that LHS ≠ RHS + conclusion (ii) Using tan θ ≡

to obtain

sin θ cos θ sin 2 θ + cos 2 θ sin θ cos θ

M1 A1

(2)

M1

A1

Using cos2θ + sin2 θ ≡ 1

M1

Using 2 sin θ cos θ ≡ sin 2θ

M1

Leading without any error or fudge to 2 cosec 2θ

A1 cso

(5)

(7 marks)

EDEXCEL CORE MATHEMATICS C3 Question Number

6.

PRACTICE PAPER A5 MARK SCHEME Scheme

(a) LHS =

Marks

2 sin 2 θ 2 sin θ cos θ =

M1 A1

sin θ = tan θ = RHS cos θ

(b) From (a)

A1 c.s.o

1 − cos 2θ 1 = tan θ 2 sin 2θ =

π

θ=

M1

1 2

A1

5π 6 6

2θ =

π

(3)

,

B1 M1

5π 12 12

M1 A1 cao

,

(6)

(9 marks) 7. (a) (i)

x = ay

i.e ln x = ln a y

(ii) In both sides of (i)

dy 1 1 × ∗ ,= dx ln a x

1 × ln x , ⇒ ln a

(b)

y=

(c)

  dy 1 dy 1 = ⋅ ln a , ⇒ = ∗ or x dx dx x ln a   log10 10 = 1 ⇒ A is (10, 1)

equ of target i.e

(d)

m=

y = 0 in (c ) ⇒ 0 =

y A =1

B1 B1

M1

1 (x − 10) or y = 1 x + 1 − 1 10 ln 10 10 ln 10 ln 10

x = 10 − 10 ln10 or 10 (1 − ln10) or 10 ln 10(

(1)

(2)

1 1 or or 0.043 (or better) 10 ln a 10 ln 10

1 x +1− 10 ln 10 ln10

B1 cso

M1, A1 cso

y − 1 = m(x − 10) y −1=

(1)

ln x ln a

⇒ y ln a = ln x

= y ln a *

from(b )

or ( y =) loga x =

B1

(o.e)

A1

 1  ⇒ x , = 10 ln10  − 1  ln10 

M1

1 − 1) ln 10

A1

(4)

(2)

(10 marks)

EDEXCEL CORE MATHEMATICS C3 Question Number

8.

PRACTICE PAPER A5 MARK SCHEME Scheme

Marks

(a)

Shape B1

y

p=

O

1 3

or { 13 , 0} seen B1

(2)

1 3

(b) Gradient of tangent at Q =

1 q

B1

Gradient of normal = −q

M1

Attempt at equation of OQ [y = −qx] and substituting x = q, y = ln 3q

or attempt at equation of tangent [y – 3 ln q = −q(x – q)] with x = 0, y = 0 or equating gradient of normal to (ln 3q)/q

M1

q2 + ln 3q = 0 (*)

A1

(4)

M1; A1

(2)

2

(c) ln 3x = −x2 ⇒ 3x = e − x ; ⇒ x = 13 e − x

2

(d) x1 = 0.298280; x2 = 0.304957, x3 = 0.303731, x4 = 0.303958 Root = 0.304 (3 decimal places)

M1; A1 A1

(3)

(11 marks)

EDEXCEL CORE MATHEMATICS C3 Question Number

9.

(a)

PRACTICE PAPER A5 MARK SCHEME Scheme

Marks

y

shape B1 intersections with axes (c, 0), (0, d) B1

(0, d)

O (b)

(c, 0)

x shape B1

y

x intersection ( 12 d, 0) B1

(0, 3c)

y intersection (0, 3c) B1 O

(2)

( 12 d, 0)

x

(c)(i) c = 2

B1

(ii) −1 < f(x) ≤ (candidate’s) c value (d) 3(2−x) = 1 ⇒ 2−x =

(3)

1 3

B1 B1 ft (3)

and take logs; −x =

ln 13

M1; A1

ln 2

d (or x) = 1.585 (3 decimal places) − log x (e) fg(x) = f[log2 x] = [ 3(2 2 ) − 1 ]; = [ 3(2

=

log 2 1x

3 −1 x

) − 1 ] or

3 2 log 2

x

−1

A1

(3)

M1; A1 A1

(3)

(14 marks)