1306 fp2 june 2013 - withdrawn paper mark scheme

Mark Scheme (Results) June 2013 GCE Further Pure Mathematics FP2 (6668/01) Original Paper

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June 2013 Further Pure Mathematics FP2 6668

Mark Scheme

Question Number Scheme Marks

1 (a) ''' '' ' 2siny xy y x+ + = − so ''' '' ' 2siny xy y x= − − − M1, A1, A1

(3)

(b) ''' 3 0y + = so ''' 3y = − B1 (1)

(c) Substitute to give

2

2

d 2d

yx

= B1

Use Taylor Expansion to give

321 3 ...

2xy x x= + + −

M1 A1

(3) (7)

2.

(a)

B1 (V shape) B1 (Parabola) B1 (positions correct)

(3)

(b) Put 2 24 2 3 or 4 2 3x x x x− = − − = − +

M1

Solve 2 2 7 0x x+ − = , to give 2 4 28 1 2 2

2x − + += = − +

M1 A1

Solve 2 2 1 0x x− − = , to give 2 4 4 1 2

2x − += = −

M1 A1

So 1 2 2 2 1x− < < −

B1

(6)(9)

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3.

(a) cosf ( )1 sink kxx

kx′ =

+

M1 A1

(2)

(b)

2

2

(1 sin ) sin cos ( cos )f ( )(1 sin )

kx k kx k kx k kxxkx

− + −′′ =+

M1

2 2 2 2

2

sin (sin cos )so f ( )(1 sin )

k kx k kx kxxkx

− − +′′ =+

and use 2 2sin coskx kx+ =1 M1

2

2

(1 sin )f ( )(1 sin )k kxx

kx− +′′ =

+

2

1 sink

kx−

=+

*A1cao

(3)

(c) 3

2

cosf ( )(1 sin )

k kxxkx

′′′ =+

B1

Finds f(0), f (0), f (0)′ ′′ and f (0)′′′ M1

Uses MacLaurin Expansion to obtain f(x) = 2 3

2 30 ...2 6k kkx x x+ − +

M1 A1 (4)

Alternative method for (c) Uses ln (1 + y) =

2 3

2 3y yy − + + with y = sinkx B1

Use sinkx = 3 3

6 ..k xkx − in ln expansion M1

Obtains 2 3

2 30 ...2 6k kkx x x+ − + M1 A1

(4)

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4. IF = 1 cot dx x x

x+

∫ B1

Obtains ln ln sinx x+ M1 A1

{So IF = xsinx}

A1ft

sin(IF ) IFd xydx x

= × M1

yxsinx = 2 1 cos 2sin d d2

xx x x−=∫ ∫ 1 sin 2 ( )

2 4x x c= − +

M1 A1

So y = 1 1cosec cos2 2 sin

cx xx x x

− + M1 A1

(9)

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5 (a) 1 2 1

( 1) ( 2)r r r− +

+ +

M1 A1 A1

(3)

(b) LHS = 1 2 11 2 3− +

+ 1 2 12 3 4− +

+ 1 2 13 4 5− +

…………

1 2 1( 2) ( 1)n n n

+ − +− −

+ 1 2 1( 1) ( 1)n n n

− +− +

+ 1 2 1( 1) ( 2)n n n

− ++ +

M1

A1

= 1 1 11

2 1 2n n− − +

+ +

2 3 2 2 4 2 22( 1)( 2)

n n n nn n

+ + − − + +=

+ +

M1 dep

2 32( 1)( 2)

n nn n

+=

+ + ( 3)

2( 1)( 2)n n

n n+

=+ +

∗ A1

(4)

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6

Modulus = 32

B1

Argument = 1 5arctan( )

63π

− = M1A1

1 15 5

5 532 (cos( ) i sin( )) 2(cos isin )6 6 6 6

z π π π π⎛ ⎞ ⎛ ⎞= + = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

M1 A1

OR 2 22(cos isin )6 5 6 5

n nπ π π π⎛ ⎞ ⎛ ⎞+ + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

M1

= 17 172(cos isin )30 30π π⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠, 29 292(cos isin )

30 30π π⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠, A1

7 72(cos isin )30 30π π− −⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠, 19 192(cos isin )

30 30π π− −⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ A1

(8)

7(a)

Differentiate twice and obtaining

2 2d 2 e ed

x xy xx

λ λ= + and 2

2 2 22

d 4 e 2 e 2 ed

x x xy xx

λ λ λ= + +

M1 A1

Substitute to give 3

2λ =

M1 A1

(4) (b)

Complementary function is 2 2e ex xy A B −= +

M1 A1

So general solution is 2 2e ex xy A B −= + + 23 e

2xx

A1

(3)

(7)

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8. (a)

B1

(1)

(b) w(z-2i) = z+7i so ( 1) 7i 2iz w w− = + and z = M1

7i 2i

1w

w+−

A1

So 7 2 1i iw w+ = − M1

Using w = u + iv, 2 2 2 2( 2 ) (2 7) ( 1)v u u v− + + = − +

M1

So 2 23 3 30 48 0u v u+ + + = , which is a circle equation A1 (5)

As 2 2 2( 5) 3u v+ + = So centre is –5 and radius is 3

M1 A1

(2) (8)

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Q9 (a)

2 2sin 1θ− = → sin 0.5θ = 5 or 6 6π πθ ⎛ ⎞∴ = ⎜ ⎟

⎝ ⎠,

M1 A1,

Points are 6(1, )π and 56(1, )π

A1

(3)

(b)

Uses Area = 212 (2 2sin ) dθ θ⎡ ⎤

−⎢ ⎥⎣ ⎦∫

M1

= 1

2 (4 8sin (2 2cos 2 ))dθ θ θ⎡ ⎤

− + −⎢ ⎥⎣ ⎦∫

M1

= [ ]12 (6 8cos sin 2 ))θ θ θ+ − = A1

Uses limits 2

π and 6π to give 7 3

4π − or 7 32

2π −

M1 A1

Finds area of sector of circle 23π B1

So required area is 8 7 33 2π−

M1 A1 (8)

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1306 fp2 june 2013 - withdrawn paper mark scheme

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