FP2 MS Jan 2010

Mark Scheme (Results) January 2010

GCE

Further Pure Mathematics FP2 (6675)

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January 2010 Publications Code UA022799 All the material in this publication is copyright Š Edexcel Ltd 2010

January 2010 6675 Further Pure Mathematics FP2 Mark Scheme

Question Number Q1

Scheme

(a) Using “ b 2 = a 2 ( e 2 − 1) ”

x2 y2 − =1 4a 2 a 2 a 2 = 4a 2 ( e 2 − 1)

e=

Leading to

M1 A1

√5

A1

2

2a a x= "x = "⇒ e e ex 5 5 a= = 2 2

(b)

ρ=

Q2

At s = 0 , ψ =

Marks

π 4

ft their e

(2) [5]

ds = tanψ dψ

B1

s = ∫ tanψ dψ = ln secψ

( + A)

0 = ln √ 2 + A ⇒ A = − ln √ 2

⎛ ⎛ secψ ⎞ ⎞ ⎜ s = ln ⎜ 2 ⎟ ⎟ ⎝ √ ⎠⎠ ⎝

GCE Further Pure Mathematics FP2 (6675) January 2010

M1 A1ft

(3)

M1 A1 M1 A1

(5) [5]

1

Question Number

Q3

Scheme

(a)

(b)

Marks

⎛ e x − e − x ⎞⎛ e x + e − x ⎞ 2sinh x cosh x = 2 ⎜ ⎟⎜ ⎟ ⎝ 2 ⎠⎝ 2 ⎠ e 2 x − e −2 x = 2 = sinh 2x

M1

cso

x=0 2sinh x cosh x = 6 sinh 2 x + 7 sinh x 2cosh x = 6sinh x + 7 e x + e − x = 3e x − 3e − x + 7 2 e2 x + 7 e x − 4 = 0 ( 2 e x − 1)( e x + 4 ) = 0 ex =

(2)

B1 M1 M1 A1 M1

1 2

A1

x = − ln 2

GCE Further Pure Mathematics FP2 (6675) January 2010

A1

accept ln

2

1 2

A1

(7) [9]

Question Number Q4

Scheme

Marks

I n = ∫ 1. ( ln x ) dx n

(a)

= x ( ln x ) − ⌠ ⎮ x.n ( ln x ) ⌡ n

= x ( ln x ) − n ∫ ( ln x ) n

n −1

n −1

1 × dx x

M1 A1 A1

dx

= n ( ln x ) − nI n − 1 n

cso

I 3 = x ( ln x ) − 3I 2 3

(b)

(4)

M1

(

= x ( ln x ) − 3 x ( ln x ) − 2 I1 3

A1

2

)

M1

= x ( ln x ) − 3 x ( ln x ) + 6 I1 3

2

= x ( ln x ) − 3 x ( ln x ) + 6 ( x ln x − I 0 ) 3

2

= x ( ln x ) − 3 x ( ln x ) + 6 x ln x − 6 x 3

2

M1 A1

With the limits 1 and e I 3 = e − 3e + 6e − 6e + 6 = 6 − 2e

M1 A1

(6) [10]

If I1 is integrated directly I 3 = x ( ln x ) − 3I 2 3

(b)

M1

(

= x ( ln x ) − 3 x ( ln x ) − 2 I1 3

2

)

= x ( ln x ) − 3 x ( ln x ) + 6 I1 3

2

M1

1 I1 = ∫ 1.ln x dx = x ln x − ⌠ ⎮ x × dx = x ln x − x ⌡ x

(I

= x ( ln x ) − 3x ( ln x ) + 6 x ln x − 6 x 3

3

2

With the limits 1 and e I 3 = e − 3e + 6e − 6e + 6 = 6 − 2e

GCE Further Pure Mathematics FP2 (6675) January 2010

3

Allow if quoted

M1 A1

)

M1 A1

(6)

Question Number

Q5

Scheme

(a)

Marks

dy = 3sinh 3 x − 4 dx 3sinh 3x − 4 = 0 4 sinh 3 x = 3 4 3 x = arsinh 3 1 4 x = arsinh 3 3 1 ⎡ 2 = ln ⎢ 43 + √ ( 43 ) + 1 3 ⎣

(

(b)

∫ ( cosh 3x − 4 x ) dx = 1

ln 3

⎡ sinh 3 x ⎤3 A=⎢ − 2x2 ⎥ ⎣ 3 ⎦0

M1

M1

)⎤⎥⎦ = 13 ln 3

A1

sinh 3 x − 2x2 3

sinh ( ln 3) 2 ( ln 3 ) = − 3 9

=

M1 A1

M1 A1 2

M1 A1

3 − 13 eln 3 − e − ln 3 − …= − …

6 6

4 2 ( ln 3) = − 9 9 2 2 = ⎡ 2 − ( ln 3) ⎤ ⎣ ⎦ 9

(5)

M1

2

GCE Further Pure Mathematics FP2 (6675) January 2010

4

cso

A1

(6) [11]

Question Number

Q6

Scheme

x=

(a)

Marks

a dx a ⇒ =− 2 u du u

⌠ 1 ⌠ = x d ⎮ ⎮ 2 2 ⌡ x √ (a − x ) ⎮a ⎮u ⌡

⎛ a⎞ ⎜ − 2 ⎟ du ⎛ 2 a2 ⎞ ⎝ u ⎠ ⎜a − 2 ⎟ u ⎠ ⎝ 1

√

1⌠ 1 =− ⎮ du a ⌡ √ ( u 2 − 1) 1 = − arcosh u ( + A ) a 1 ⎛a⎞ = − arcosh ⎜ ⎟ ( + A ) a ⎝x⎠

(b)

B1

⌠ 1 1 ⎛5⎞ dx = − arcosh ⎜ ⎟ ⎮ 2 5 ⎝ x⎠ ⌡ x √ ( 25 − x )

M1 A1

M1 M1 A1

(6)

M1

4

⎡ 1 1⎛ ⎛ 5 ⎞⎤ ⎛5⎞ ⎛ 5 ⎞⎞ ⎢ − 5 arcosh ⎜ x ⎟ ⎥ = 5 ⎜ arcosh ⎜ 3 ⎟ − arcosh ⎜ 4 ⎟ ⎟ ⎝ ⎠⎦ 3 ⎝ ⎠ ⎝ ⎠⎠ ⎣ ⎝ 1 = [ ln 3 − ln 2] 5 1 3 = ln 5 2

M1 A1

M1 A1

(5) [11]

Alternative for part (a) x=

a dx sinh u ⇒ =− cosh u du cosh 2 u

⌠ ⌠ ⎮ 1 ⎮ dx = ⎮ ⎮ 2 2 ⎮ x √ (a − x ) ⎮ a ⎮ ⎮ cosh u ⌡ ⌡

⎛ a sinh u ⎞ ⎜− ⎟ du 2 ⎛ 2 ⎛ a ⎞2 ⎞ ⎝ cosh u ⎠ ⎜⎜ a − ⎜ ⎟ ⎟ ⎝ cosh u ⎠ ⎟⎠ ⎝ 1

√

1 1 1du = − u ( + A) ∫ a a 1 ⎛a⎞ = − arcosh ⎜ ⎟ ( + A) a ⎝ x⎠

=−

GCE Further Pure Mathematics FP2 (6675) January 2010

B1

5

M1 A1

M1 M1 A1

(6)

Question Number

Q7

Scheme

(a)

u = k arcsin 2 x ,

Marks

du 2k = dx √ (1 − 4 x 2 )

dy = cos u du dy 2k = cos u × 2 dx √ (1 − 4 x )

M1 A1

cos u = √ (1 − sin 2 u ) = √ (1 − y 2 )

dy 2k = √ (1 − y 2 ) × 2 dx √ (1 − 4 x )

√ (1 − 4 x ) dx = 2k √ (1 − y 2

dy

2

A1

)

2

⎛ ⎞ (1 − 4 x ) ⎜ ddxy ⎟ = 4k 2 (1 − y 2 ) ⎝ ⎠ 2

cso M1 A1

(5)

Using u = arcsin 2 x is marked similarly.

2

(b)

⎛ dy ⎞ dy d 2 y dy −8 x ⎜ ⎟ + (1 − 4 x 2 ) 2 = − 8k 2 y 2 dx dx dx ⎝ dx ⎠ dy d2y −4 x + (1 − 4 x 2 ) 2 = −4k 2 y dx dx

(1 − 4 x 2 )

d2y dy − 4 x + 4k 2 y = 0 2 dx dx

M1 A1 A1 M1

cso

A1

(5) [10]

Alternative for part (a) arcsin y = k arcsin 2 x 1 dy 2k = 2 dx 2 √ (1 − y ) √ (1 − 4 x )

√ (1 − 4 x ) dx = 2k √ (1 − y 2

dy

2

M1 A1 A1

)

2

⎛ ⎞ (1 − 4 x ) ⎜ ddxy ⎟ = 4k 2 (1 − y 2 ) ⎝ ⎠ 2

GCE Further Pure Mathematics FP2 (6675) January 2010

6

cso M1 A1

(5)

Question Number Q8

Scheme

Marks

(a) Eliminating y between y = mx + c and y 2 = 4ax

( mx + c ) = 4ax m2 x 2 + 2 ( mc − 2a ) x + c 2 = 0 2

For a tangent

“ b 2 − 4ac = 0 ” 2 4 ( mc − 2a ) − 4m 2 c 2 = 0 c=

Leading to

(b)

M1 A1 M1

a m

cso A1

a passes through ( 4a, 5a ) m a 5a = 4ma + m 2 4 m − 5m + 1 = 0 ( 4m − 1)( m − 1) = 0

(4)

y = mx +

M1 A1 M1

1 m= , 1 4 y=

1 x + 4a, 4

A1

y = x+a

both

(c) Obtaining the points of contact between the tangents and the curve 1 e.g. Substituting m = , c = 4a into the quadratic in part (a) 4 2 1 2 ⎛x ⎞ x − 2ax + 16a 2 = ⎜ − 4a ⎟ = 0 ⇒ x = 16a 16 ⎝4 ⎠ Point has coordinates (16a, 8a )

A1

(5)

M1 A1

Substituting m = 1, c = a into the quadratic in part (a) x 2 − 2ax + a 2 = ( x − a ) = 0 ⇒ x = a 2

Point has coordinates ( a, 2a )

A1

With all methods, M1 is for finding the x- or y-coordinate of one point of contact. A1 is for any two coordinates correct. The second A1 is for all four coordinates correct.

RQ 2 = (16a − a ) + ( 8a − 2a ) = 261a 2 2

RQ = a √ 261

2

( = 3a √ 29)

M1 A1

(5) [14]

GCE Further Pure Mathematics FP2 (6675) January 2010

7

Question Number Q8

Scheme

Marks

Alternatives for part (a) 1 Let the point of contact of l and C be ( 2at , at 2 ) d y y& 2a 1 = = = dx x& 2at t 1 l has the form y = x+c t 1 2at = at 2 + c ⇒ c = at ( 2at , at 2 ) lies on l t a c = at = m a 2 Eliminate y between y = mx + and y 2 = 4ax m 2 a⎞ ⎛ ⎜ mx + ⎟ = 4ax m⎠ ⎝ m=

M1 A1 M1

cso A1

a2 = 4ax m2 a2 m 2 x 2 − 2ax + 2 = 0 m 2 a⎞ ⎛ ⎜ mx − ⎟ = 0 m⎠ ⎝

(4)

m 2 x 2 + 2ax +

M1 A1 M1

a cso m Note: Alternative 2 allows the coordinates of R and Q in part (c) to be a written down using x = 2 . m

As this is a complete square, l is a tangent to C when c =

A1

(4)

Alternative for part (b) y = mx +

a x = + at passes through ( 4a, 5a ) m t

4a + at t t 2 − 5t + 4 = 0 ( t − 1)( t − 4 ) = 0 t = 1, 4 1 y = x + 4a, y = x + a 4 5a =

GCE Further Pure Mathematics FP2 (6675) January 2010

8

M1 A1 M1 A1

both

A1

(5)

Further copies of this publication are available from Edexcel Publications, Adamsway, Mansfield, Notts, NG18 4FN Telephone 01623 467467 Fax 01623 450481 Email publications@linneydirect.com January 2010 For more information on Edexcel qualifications, please visit www.edexcel.com/quals Edexcel Limited. Registered in England and Wales no.4496750 Registered Office: One90 High Holborn, London, WC1V 7BH