FP2 MS Jun 2009

June 2009 6668 Further Pure Mathematics FP2 (new) Mark Scheme Question Number Q1

(a)

Scheme

1 1 1 = − r (r + 2) 2r 2( r + 2)

Marks

1 1 − 2r 2(r + 2)

B1 aef (1)

n

(b)

∑

4 = r (r + 2) r =1

n

⎛2

2 ⎞

∑ ⎜⎝ r − r + 2 ⎟⎠ r =1

⎛2 2⎞ ⎛2 2⎞ = ⎜ − ⎟ + ⎜ − ⎟ + ...... ⎝1 3⎠ ⎝2 4⎠ ⎛ 2 2 ⎞ ⎛2 2 ⎞ .............. + ⎜ − ⎟ + ⎜ − ⎟ ⎜ n −1 n +1 ⎟ ⎜n n + 2 ⎟ ⎝ ⎠ ⎝ ⎠

=

Includes the first two underlined terms and includes the final two M1 underlined terms.

2 2 2 2 + − − A1 1 2 n +1 n + 2

2 2 2 2 + ;− − 1 2 n +1 n + 2

= 3−

List the first two terms M1 and the last two terms

2 2 − n +1 n + 2

=

3( n + 1)(n + 2) − 2(n + 2) − 2(n + 1) (n + 1)(n + 2)

=

3n 2 + 9n + 6 − 2n − 4 − 2n − 2 (n + 1)(n + 2)

=

3n 2 + 5n (n + 1)(n + 2)

=

n (3n + 5) (n + 1)(n + 2)

Attempt to combine to an at least 3 term fraction to a single fraction and an attempt to take M1 out the brackets from their numerator.

Correct Result A1 cso AG (5) [6]

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Question Number Q2

(a)

Scheme

Marks

z3 = 4 2 − 4 2 i , − π < θ „ π y

O α

4 2

x

arg z 4 2

( 4 2 , − 4 2)

r=

(4 2 ) + (− 4 2 ) 2

θ = − tan −1

( )=− 4 2

π

4 2

4

2

=

32 + 32 =

64 = 8

A valid attempt to find the modulus and argument of M1 4 2 − 4 2 i.

z 3 = 8 ( cos ( − π4 ) + isin ( − π4 ) ) 1 ⎛ ⎛−π So, z = ( 8 ) 3 ⎜ cos ⎜ 4 ⎝ 3 ⎝

Taking the cube root of the modulus and dividing the M1 argument by 3.

⎞ ⎛ − π4 ⎞ ⎞ isin + ⎟ ⎜ 3 ⎟⎟ ⎠ ⎝ ⎠⎠

⇒ z = 2 ( cos ( − 12π ) + i sin ( − 12π ) )

2 ( cos ( − 12π ) + isin ( − 12π ) )

Also, z 3 = 8 ( cos ( 74π ) + isin ( 74π ) )

Adding or subtracting 2π to the argument for z 3 in order to find M1 other roots.

or

z 3 = 8 ( cos ( − 94π ) + isin ( − 94π ) )

⇒ z = 2 ( cos 712π + isin 712π )

Any one of the final two roots A1

( ( ) + isin ( ))

and z = 2 cos

− 3π 4

− 3π 4

Both of the final two roots. A1

Special Case 1: Award SC: M1M1A1M1A0A0 for ALL three of 2 ( cos 12 + isin 12 ) , π

π

( ( ) + isin ( )) .

2 ( cos 34π + isin 34π ) and 2 cos

− 7π 12

− 7π 12

Special Case 2: If r is incorrect (and not equal to 8) and candidate states the brackets ( ) correctly then give the first accuracy mark ONLY where this is applicable.

8371_9374 GCE Mathematics January 2009

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56

[6]

Question Number Q3

Scheme

sin x

Marks

dy − y cos x = sin 2 x sin x dx An attempt to divide every term in the differential equation by M1 sin x .

Can be implied.

dy y cos x sin 2 x sin x − = dx sin x sin x dy y cos x − = sin 2 x dx sin x

cos x

Integrating factor = e

=

cos x

∫ − sin x dx

= e

e

− ln sin x

1 sin x

∫ ± sin x ( dx )

± their P( x ) ( dx ) or e ∫ e − ln sin x or eln cosec x

1 or (sin x) −1 or cosec x sin x

dM1 A1 aef A1 aef

sin 2 x ⎛ 1 ⎞ dy y cos x − = ⎜ ⎟ 2 sin x ⎝ sin x ⎠ dx sin x

d dx

d ( y × their I.F.) = sin 2 x × their I.F M1 dx

⎛ y ⎞ 1 ⎜ ⎟ = sin 2 x × sin x ⎝ sin x ⎠

d dx

d dx

⎛ y ⎞ ⎜ ⎟ = 2cos x ⎝ sin x ⎠

y = sin x

⎛ y ⎞ ⎜ ⎟ = 2cos x or ⎝ sin x ⎠ A1 y = 2cos x ( dx ) sin x

∫

∫ 2cos x dx

y = 2sin x + K sin x

A credible attempt to integrate dddM1 the RHS with/without + K

y = 2sin 2 x + K sin x

y = 2sin 2 x + K sin x

A1 cao [8]

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Question Number

Scheme

Q4

1 A= 2

Marks

1 Applies 2

2π

∫ ( a + 3cosθ ) dθ 2

2π

∫ r ( d θ ) with 2

0

B1

correct limits.

Ignore dθ .

0

(a + 3cosθ ) 2 = a 2 + 6a cosθ + 9cos 2 θ ± 1 ± cos 2θ M1 2 Correct underlined expression. A1

cos 2 θ =

⎛ 1 + cos 2θ ⎞ = a + 6a cosθ + 9 ⎜ ⎟ 2 ⎝ ⎠ 2

1 A= 2

2π

∫ 0

9 9 ⎛ 2 ⎞ ⎜ a + 6a cos θ + + cos 2θ ⎟ dθ 2 2 ⎝ ⎠ Integrated expression with at least 3 out of 4 terms of the form ± Aθ ± B sin θ ± Cθ ± D sin 2θ . M1* Ignore the 12 . Ignore limits.

2π

9 9 ⎛1⎞ ⎡ ⎤ = ⎜ ⎟ ⎢ a 2θ + 6a sin θ + θ + sin 2θ ⎥ 2 2 4 ⎝ ⎠⎣ ⎦0

=

a 2θ + 6a sin θ + correct ft integration. A1 ft 1 Ignore the 2 . Ignore limits.

1 ⎡( 2π a 2 + 0 + 9π + 0 ) − ( 0 ) ⎤ ⎦ 2⎣

= π a2 +

Hence,

9π 2

π a2 +

π a2 + 9π 107 = π 2 2

a2 +

Integrated expression equal to

9π 2

107 2

A1

π . dM1*

9 107 = 2 2 a 2 = 49 a=7

As a > 0, a = 7

A1 cso [8]

Some candidates may achieve a = 7 from incorrect working. Such candidates will not get full marks

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Question Number

Scheme

Marks

y = sec 2 x = (sec x) 2

Q5 (a)

Either 2(sec x)1 (sec x tan x)

dy = 2(sec x)1 (sec x tan x) = 2sec 2 x tan x dx Apply product rule: ⎧ u = 2sec 2 x ⎪ ⎨ du 2 ⎪ = 4sec x tan x ⎩ dx

or 2sec 2 x tan x

B1 aef

v = tan x ⎫ ⎪ dv 2 ⎬ = sec x ⎪ dx ⎭

Two terms added with one of either A sec 2 x tan 2 x or B sec 4 x M1 in the correct form.

2

d y = 4sec 2 x tan 2 x + 2sec4 x dx 2

Correct differentiation A1 = 4sec 2 x(sec 2 x − 1) + 2sec 4 x

Applies tan 2 x = sec 2 x − 1 leading to the correct result. A1 AG

d2 y = 6sec 4 x − 4sec 2 x dx 2

Hence,

(4) (b)

yπ = 4

( 2)

2

⎛ dy ⎞ = 2, ⎜ ⎟ = 2 ⎝ dx ⎠ π4

⎛d y⎞ ⎜ 2⎟ = 6 ⎝ dx ⎠ π 2

( 2)

4

−4

( 2)

2

( 2 ) (1) = 4

⎛ dy ⎞ Both y π = 2 and ⎜ ⎟ = 4 4 ⎝ dx ⎠ π4

2

B1

Attempts to substitute x = π4 into both terms in the expression for M1 d2 y . dx 2

= 24 − 8 = 16

4

Two terms differentiated with either 24sec 4 x tan x or M1 − 8sec2 x tan x being correct

d3 y = 24sec3 x(sec x tan x) − 8sec x(sec x tan x) 3 dx

= 24sec 4 x tan x − 8sec 2 x tan x ⎛ d2 y ⎞ ⎜ 2 ⎟ = 24 ⎝ dx ⎠ π4

( 2 ) (1) − 8 ( 2 ) (1) = 96 − 16 = 80

sec x ≈ 2 + 4 ( x −

4

2

⎛ d3 y ⎞ ⎜ 3 ⎟ = 80 ⎝ dx ⎠ π

B1

4

π

{sec x ≈ 2 + 4 ( x −

4

) + 162 ( x − π4 ) π

4

2

) + 8( x − 4 )

π 2

+

80 6

(x − )

+

40 3

(x − 4 )

π 3 4

π 3

Applies a Taylor expansion with at least 3 out of 4 terms ft M1 correctly.

Correct Taylor series expansion. A1

+ ...

}

(6)

+ ...

[10]

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Question Number Q6

Scheme

w= (a)

z , z = −i z+i Complete method of rearranging M1 to make z the subject. iw z= A1 aef (1 − w)

w( z + i) = z ⇒ wz + i w = z ⇒ i w = z − wz iw ⇒ i w = z (1 − w) ⇒ z = (1 − w) iw =3 1− w

z =3 ⇒

Putting z in terms of their w = 3

⎧⎪ i w = 3 1 − w ⇒ w = 3 w − 1 ⇒ w 2 = 9 w − 1 2 ⎨ 2 2 ⎪⎩ ⇒ u + i v = 9 u + i v − 1

2 2 2 2 ⎪⎧ ⇒ u + v = 9u − 18u + 9 + 9v ⎨ 2 2 ⎪⎩ ⇒ 0 = 8u − 18u + 8v + 9

⇒ 0 = u 2 − 94 u + v 2 + ⇒ (u −

9 2 8

)

−

⇒ (u −

9 2 8

)

+ v2 =

81 64

{Circle} centre

Simplifies down to dddM1 u 2 + v 2 ± α u ± β v ± δ = 0.

9 8

=0

9 64

( 98 , 0 ) , radius

One of centre or radius correct. A1 Both centre and radius correct. A1

3 8

(8)

Circle indicated on the Argand diagram in the correct position B1ft in follow through quadrants.

Ignore plotted coordinates.

v

O

⎫⎪ ⎬ ⎪⎭

⎫⎪ ⎬ ⎪⎭

9 8

+ v2 +

dM1

Applies w = u + i v , and uses Pythagoras correctly to get an ddM1 equation in terms of u and v without any i 's.

Correct equation. A1

⇒ u 2 + v 2 = 9 ⎡⎣ (u − 1) 2 + v 2 ⎤⎦

(b)

Marks

Region outside a circle indicated B1 only.

u

(2) [10]

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Question Number

Scheme

Marks

y = x2 − a2 , a > 1

Q7 (a)

y a2

−a

O

Correct Shape. Ignore cusps. B1 Correct coordinates. B1 a

x

(2) (b)

x − a = a − x , a >1 2

2

2

{ x > a},

x2 − a2 = a2 − x

x2 − a2 = a2 − x

M1 aef

⇒ x 2 + x − 2a 2 = 0 −1 ±

⇒ x=

−1 ±

⇒ x=

{ x < a},

Applies the quadratic formula or completes the square in order to M1 find the roots.

1 − 4(1)(−2a 2 ) 2

Both correct A1 “simplified down” solutions.

1 + 8a 2 2

− x 2 + a 2 = a 2 − x or

− x2 + a2 = a2 − x

{⇒ x

2

x2 − a2 = x − a2

M1 aef

− x = 0 ⇒ x( x − 1) = 0} x=0 x =1

⇒ x = 0 ,1

B1 A1 (6)

(c)

x2 − a2 > a2 − x , a > 1

x<

−1 −

{or}

1 + 8a 2 2

{or} x >

−1 +

x is less than their least value B1 ft x is greater than their maximum B1 ft value

1 + 8a 2 2

For { x < a} , Lowest < x < Highest

0 < x <1

0 < x <1

M1 A1 (4) [12]

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Question Number Q8

Scheme

Marks

d2 x dx dx + 5 + 6 x = 2e − t , x = 0, = 2 at t = 0.

2 dt dt dt (a)

AE , m 2 + 5m + 6 = 0 ⇒ (m + 3)(m + 2) = 0 ⇒ m = − 3, − 2.

So, xCF = Ae

−3 t

+ Be

Ae m1 t + Be m2 t , where m1 ≠ m2 .

−2 t

Ae

−3 t

+ Be

−2 t

M1 A1

⎧ ⎫ dx d2 x −t = − k e−t ⇒ 2 = k e− t ⎬ ⎨ x = ke ⇒ dt dt ⎩ ⎭

⇒ k e − t + 5(− k e − t ) + 6 k e − t = 2e − t ⇒ 2 k e − t = 2e− t ⇒ k =1

{ So,

Substitutes k e − t into the differential equation given in the M1 question.

Finds k = 1. A1

xPI = e − t }

their xCF + their xPI

So, x = Ae−3t + Be −2t + e − t Finds

dx = = − 3 Ae −3t − 2 Be −2t − e − t dt

t = 0, x = 0 ⇒ dx t = 0, =2 ⇒ dt

M1*

dx by differentiating dM1* dt their xCF and their xPI

Applies t = 0, x = 0 to x dx dx = 2 to to ddM1* and t = 0, dt dt form simultaneous equations.

0 = A+ B +1 2 = − 3 A − 2B − 1

⎧ 2 A + 2 B = − 2⎫ ⎨ ⎬ ⎩−3 A − 2 B = 3 ⎭ ⇒ A = − 1, B = 0

So, x = − e−3t + e− t

x = − e −3t + e− t

A1 cao (8)

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Question Number

Scheme

Marks

x = − e −3t + e− t (b)

Differentiates their x to give

dx = 3e − 3t − e − t = 0 dt

dx and puts equal to 0. dt

So, x = − e x = −3 =−

− 32

+e

− 12 ln 3

1 2

= −e

ln 3

−3 2

+e

−1 2

ln 3

Substitutes their t back into x and an attempt to eliminate out the ln’s. ddM1

− 12

+3

1 3 3

1 2 2 3 = = 9 3 3 3

+

uses exact values to give

d2 x = − 9e − 3t + e− t dt 2

At t = 12 ln 3, = − 9(3)

As

− 32

− 12

=−

2 3 9

d2 x dt 2 d2 x and substitutes their t into 2 dt

A1 AG

Finds

d2 x − 3 ln 3 − 1 ln 3 = − 9e 2 + e 2 2 dt

+3

M1

A credible attempt to solve. dM1* t = ln 3 or t = ln 3 or awrt 0.55 A1

3 − e 2t = 0 ⇒ t = 12 ln 3 − 32 ln 3

dx dt

9 3 3

+

dM1*

1 3 1 = − + 3 3 3

d2 x 9 1 ⎧ 2 ⎫ = − + = ⎨− ⎬ <0 2 dt 3 3 3 ⎩ 3⎭ then x is maximum.

−

9 3 3

+

1 < 0 and maximum 3 conclusion.

A1 (7) [15]

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