06a Further Maths - FP3 Mark Schemes v2 (1)

8/19/2019 06a Further Maths - FP3 Mark Schemes v2 (1)

1/68

FP3 mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers (back to June 2002)

Please note that the following pages contain mark schemes for questions from past papers.

Where a question reference is marked with an asterisk (*), it is a partial version of the original.

The standard of the mark schemes is variable, depending on what we still have man! are

scanned, some are handwritten and some are t!ped.

The questions are available on a separate document, originall! sent with this one.

"P# question mark schemes from old P$, P%, P& and "P', "P, "P# papers ersion arch ++


2/68

1 (a)

y #

$ x -losed shape #, $

' (1)

(b) b / a(' e) ⇒ / '&(' e) '

$

0=e oe awrt +.&&' 1' (2)

(c) "oci are at (±ae , +) use of ae '(√0, +) and ( √0, +) awrt .&%, + is required, ft their e 1' ft (2)

(5 marks) 2P% 3une ++ 4n '5

2

'

'

d

d

x x

y

−

= (/ √ at x / 6√) '

x x x

y.)' 6(

d

d 7#

−−= 6at

)'( 7#

==

−

= x x

x'

8se of

7#

d

d

d

d'

x

y

x

y

+

= ρ '

to obtain

7#

7#

)'(

'

''

x x

x

−

−+

= ρ oe

1'

−=

x

x 7#)(

ma! be unsimplified or implied correct numerical

answer if first ' clearl! gained

1t x / 6√ , #

#= ρ accept#

, or e9act equivalents ', 1'

(6 marks) 2P% 3une ++ 4n 5

3 ''

ee

ee'+ =

−+

+ −− x x x x '

+$''e& =+− x x x quadratic in e x ', 1'

+)$e#)('e( =−− x x ' e x / 6 and $7# 1'

x / ln 6 and ln $7# ', 1'



3/68

(! marks)

"lt 3 )cosh(sinhcosh'+ α +≡+ x R x x

R / √& and%

'tan =α ', 1'

&

'')cosh( =+α x

−

±=+ '

&

''

&

''lnα x either 1'

&

$ln= and

$

&ln both as single ln

1'

#ln6

$

&ln ,

#ln6

&

$ln −−= x

−

−=

#

$

&ln ,

#

&

$ln combine either into single ln.

:ependent on first two s

'

x / ln 6 and ln $7# 1', 1'

2;ne answer b! alt. method gains ma9. '1''+'1'1+5

(11 marks) 2P% 3une ++ 4n #5



4/68

4 (a) ∫ ∫ −−= x x xn x x x x x nnn dsinsindcos ' ', 1' 5dcos)'(cos2 ......

' x x xnn xnx nn ∫ −− −+−= '

8sing limits )'(

−−−

= n

n

n I nn I π () cso ', 1' (5)

(b) ∫ ==

+

+ dcos

π

x x I [ ] +

sin xπ

/ ' at an!

stage

'

$

&

& #+

I I −

= π

'

−

−

= &&

'

#+

I π π

+

$&

0+

#&+

#+

I −

+

−

= π π π '


5/68

#d

dsec =

x

y y 1'

'

#

tan'

#

d

d

x y x

y

+=

+= () ', 1' (4)

(b) dx x

x x x x x x ∫ ∫ +−=

'

#arctan#d#arctan& ', 1'

x x

x d'

'' .......

∫ + −+−= '

x x #arctan#

' ...... +−= 1'

,#

#

#

#

#

+

π π +−=

'

)##$(

' −= π () cso 1' (6)

(10 marks) 2P% 3une ++ 4n &5



6/68

6 (a) 6 x y = , 6 d

d x

x

y= ', 1'

x x

x x x

y d'

'$dd

d'

6...

...

6

...

...

+=

+ ∫ ∫ π π '

x x d'$

'

+∫ += π () 1' (4)

(b) ( ))'(

)+(

7#...

...

'#

$d'$

+=+= ∫ x x xS π π ', 1'

( )'#

= 7# −= π

or an! e9act

equivalent

1' (3)

(c) ⌡

⌠

+ x

x

yd

d

d'

'

/ ⌡

⌠

+ x

xd

''

'

'

x x

x

d

'

⌡⌠ +

1'

8sing s!mmetr!, x x

x s d

'

'

+

⌡⌠ += () 1' (3)

(d ) θ θ θ

θ coshsinhd

d ,sinh ==

x x oe '

θ θ θ θ

θ dcoshsinh.

sinh

sinh'

∫ +

= I '

∫ = θ θ dcosh$

∫ += θ θ d)cosh'( '

θ θ sinh += 1' >imits are + and )'(ln( 'arsinh += arsinh'

+

sinh ' sinh s θ θ θ = + +

( ) ( )arsinh' ' ' ln ' √ √ √ = + + = + + ' 1' (6)

6 (d) "lt The last four marks can be gained?

θ θ θ θ θ θ

d)ee(d

ee$

∫ ∫ ++=

+= I '

e

e θ θ

θ += 1'



7/68

( )( )

+−++=

'

''

''arsinh s

−−

+−+++=

#

#.#

''

' ....

)5'ln(2)##6()'ln( ++=+−+++= () ', 1'

6 (d) "lt The last two marks ma! be gained b! substituting back tothe variable x

[ ] [ ] .........

... coshsinhsinh θ θ θ θ θ +=+= s

[ ]'+'arsinh x x x ++= )'ln('arsinh =+=+=

'ln( ++= () ', 1'

2P% 3une ++ 4n =5



8/68

!

ee)sinh(

)i(i θ π θ π

θ π −−− −

=−i '

)'(ee)'( −÷−−=− θ θ

'

ee θ θ +−=−

1'

θ sinh= () cso 1'"lt θ π θ π θ π sinhicoshcoshisinh)isinh( −=− '

θ sinh)'(+ −−= ', 1' θ sinh= () cso 1'

(4 marks)

2P& 3une ++ 4n '5



9/68



10/68

$ (a)

−=

−=

−

'

#

#

&

&

'

#$$

$%+

$+'

, ∴eigenvalue is # '1', 1' (3)

(b) @ither +&$&$)'&($)'&$(=

&$$

$$+

$+=

=+−=+−−=−

−− ' 1' (2)

1lt(b)or +

#$$

$%+

$+'

=−

−−

λ

λ

λ

⇒

+)%('&)'('&)#)(%)('( =−−−−−−− λ λ λ λ λ ⇒ +)#)()(#( =+−− λ λ λ ⇒ λ is an eigenvalue ' 1'

z

y

x

eigenvector ⇒ x A $ z / x, % y A $ z / y, $ x A $ y A # z A z

1t least two of these equations

'

1ttempt to solve z / x, z / y, x A y / # z 1'

∴

'

1' (3)

(c) ake e.vectors unit to obtain P /

'

# # #

'

# # #

'

# # #

÷− ÷ ÷−

columns in an! order ', 1'ft

% /

#++

++

++#

λ

, where λ # / #, P and % consistent ', 1', '

(5)

Alt P /

−−

'

'

'

, % /

− 0+++='+

++0

, P and % consistent

'1'ft,

'1',

' 2P& 3une ++ 4n %5

# (a) = AB %& A # ' AC = #& A ' k or BC = & ' k

'#

+#%

−=×

k '&

AC AB / #& A % ' A k ', 1'



11/68

∴r / & A ' A k A λ (#& A % ' A k ) ' ft (3)

(b) olume / ).(&

' AC AB AD × AD /& A # ' A k '

/ '7&(& A # ' A k ).(#& A % ' A k ) '

/ ''7& 1' (3) (c) r(#& A % ' A k ) / (& A ')(#& A % ' A k ) ', 1' ft

/ ' 1' (3) (d ) 2&(' #λ ) A '( A %λ ) A k (' A λ )5(#& A % ' A k ) / ' ', 1'ft

# A λ A '+ A %λ A # A λ / '

#%λ A '+ / ' ⇒ λ / #%'' '

∴ @ is

#%

$,

#%

'%,

#%

&=1' (4)

(e) :istance /#%

#%''%#

#%

''=++−− k ji () ' 1' (2)

( f )#%

#%

'' −=

−×=λ '

r:′ / & A ' A k A #%

(#& A % ' A k )'

:′ is

−

#%

=#,

#%

$+,

#%

'+'1' (3)

(1$ marks) 2P& 3une ++ 4n 05

10

ee

ee$

x x x x −− −+

+ / = '

%e x '&e x A # / + ' 1'

(%e x ')(e x #) / + 1'

e x / %' , #

x / ln( %' ), ln # accept ln % ' 1' (&)

(6 marks)

2P% 3une ++# 4n '5

11 (a) y / artanh x

tanh y / x

sech y x

y

d

d / ' ' 1'

x

y

d

d / '

'

tanh'

'

x y −=

− (*)1' (#)



12/68

cso

(b) ∫ '.artanh x d x / x artanh x ∫ − ' x x

d x ' 1'

/ x artanh x A '

ln(' x) ( A c) ' 1' ($)

(! marks)

1lt.

+−−≡− x x x

x

'

'

'

'

'

'

∫ − ' x x

d x / ' ln(' x)

' ln(' A x)

This is acceptable (with the rest correct) for final ' 1' 2P% 3une ++# 4n 5

12 ∫ + $

'+

xd x / ∫ +

$

'+

'

xd x '

/

#

arsinh

'+ x

++= '

$

#

ln% x x

' 1'

[ ] %

+ / % arsinh #

'+

≈

+= %$.

'+

#

'+ln%

ft on %

' 1' ft

1rea / .%$ B '++ / &+ (m) ' 1'

(! marks)

8sing a substitution

(i) x / # sinh θ C d x / # cosh θ dθ

∫ + $'+ x

d x /#

cosh#'+ ×∫ θ cosh θ dθ complete

subs.

'

/ % ∫ dθ / % arsinh#

x' 1'

then as before,

or changing limits to + and arsinh#

'+ (or

+

'+

#

'+ln ) can

gain this 1'

(ii) x / # tan θ C d x / # sec

θ dθ

∫ + $

'+

xd x /

#

tan

'+

×

+∫

θ secθ dθ '

/ % ∫ sec θ dθ / % ln (sec θ A tan θ ) '

>imits are + and arctan#

'+1'



13/68

[ ]

++=#

'+'

'++ln%#

'+arctan

+ etc ' 1'ft

2P% 3une ++# 4n #5

13 (a) x

y

d

d / sinh

a

x'

∫

+

d

d'

x

y d x / ∫

+

a

xsinh' d x / ∫ a x

cosh d x / sinha

x', 1'

>ength / ka

a

xa

+

sinh

/ a sinh k (*) ' 1' (%)

(b) a sinh k / =a

sinh k / $ '

x / ka / a arsinh $ / a ln ($ A √'0) '

y / a cosha

ka / a√(' A sinh k ) / a√'0 ' 1' ($)

(# marks) 2P% 3une ++# 4n %5



14/68

14. (a) sec y / e x '

sec y tan y x

y

d

d / e x ( / sec y) ' 1'

x

y

d

d /

y y

x

tansec

e /

'sec

'

− y / 'e

'

−

x (*)

cso' 1' (%)

(b)

y

π

O

Dhape, curve → (+, +)

1s!mptote, ( y / )

π

'

' ()

(c) ( x / ln ) y / arcsec /#

π

'

x

y

d

d / '$

'

− / #'

'

tangent is y #

π

/#

'( x ln ) '

x / +, y /#

π

#

' ln e9act answer

onl!

1' ($)

(11 marks)

1lt to (a) cos y / e x '

sin y x

y

d

d /

y

x

cos'

e

−

−

' 1'

/ x

x

e'

e

−

−

− /

'e

'

− x (*) cso

' 1' (%)

2P% 3une ++# 4n &5



15/68

15 (a) I n /''

'

+ +e en x n x x n x − − ∫ d x / e nI n ' (*) cso ' 1' ()

(b) J n /''

'

+ +e en x n x x n x− − − − + ∫ d x ' 1'

/ e ' A nJ n ' 1' (#)

(c) J / e ' A J '

J ' / e ' A J + J and J ' '

/ e ' A ∫ −'

+

e x d x

/ e ' A (' e ') ( / ' e ') 1'

J / e ' A (' e ') /

e

% (*) 1' (#)

(d )'

+cosh dn x x x∫ /

'

+

e ed

x x

n x x− +

÷

⌠ ⌡

/ ' ( I n A J n) (*) ' (')

(e) I / e I ' / e (e I +) / I + e

/ '

+e d x x∫ e / 2e '5 e ( / e ) ' 1'

'

( I A J ) / '

(e A e

%) /

'(e e

%) ' 1' ($)

(13 marks)2P% 3une ++# 4n 05



16/68

16 (a)t

x

d

d / a sec t tan t ,

t

y

d

d / b sect ' 1'

x

y

d

d

/

= t ab

t t a

t b

sin tansec

sec

' 1'

gradient of normal is b

t a sin

y b tan t / b

t a sin( x a sec t ) '

ax sin t A by / (a A b) tan t (*) cso 1' (&)

(b) y / + ⇒ x /

+=

+t a

ba

t a

t ba

cos

sin

tan)(

'

b / a(e ') ⇒ b /$

% a'

OS / ae and OA / # AS '

a A$

% a / #a B

#B cos t

cos t / ' ' 1'

t /#

%,

#

π π 1'

! s!mmetr! or (as OA /t a

ba

cos

+)

t a

ba

cos

+ / #ae

t /#

$,

#

π π ' 1' (=)

(14 marks)1lt. to

(a)

b

y

a

x−

x

y

d

d / + ' 1'

x

y

d

d/

t ba

t ab

ya

xb

tan

sec

= E ' 1'

then as before 2P% 3une ++# 4n =5


1!

1$


17/68

2P& 3une ++# 4n '5

2P& 3une ++# 4n #5



18/68

2P& 3une ++# 4n &5


1#


19/68

2P& 3une ++# 4n 05

21 (a)

(b)

' '

$ $cosh sinh ( ) ( ) x x x x x x e e e e− −− = + − −

/ '

$( ) x x x xe e e e− −+ + − + −

/' ∗

' cosh sinh sinh

x

x x− = )

' ( x x x x

e e e e− −∴ − + = − ' xe∴ =

ake x the subFect of the formula, 'ln( ) ln x = = −

1

"1

"1

(3)

1

"1

1, "1

($) 2P% 3une ++$ 4n '5


3020


20/68

22 (a)

(b)

, ', '&a b c= = =

'.%

+.%

'd

( ') '& x

x− + +⌠ ⌡

'.%

+.%

' 'arc tan

= $

x

−

+ = ÷

#

π =

1, 1, 1

(3)

1

1 "1

1 (4)

2P% 3une ++$ 4n 5

23 (a)

(b)

1s %

$ (' ) ,e e= − ∴ =

8ses ae to obtain that the foci are at ( %,+)±

( ) PS PS e PM PM ′ ′+ = + ' for single statement e.g, PD / eP

/a

ee

× ' needs complete method

/ a = &

1, "1

1 "1

(4)

1

1

"1

(3)

2P% 3une ++$ 4n #5



21/68

24 (a)

(b)

(c)

8sing product rule ( ')sinh cosh sinhn n

dyn x x x

dx

−= − +

8sing cosh ' sinh x x= + in derived e9pression

to obtain ( ')sinh (' sinh ) sinhn n

dyn x x x

dx

−= − + +

and( ')sinh sinhn ndy n x n x

dx

−= − + ∗

sinh' sinh'

' sinh'

+

+ +

2sinh cosh 5 ( ')sinh d sinh dar ar

n ar n n x x n x x n x x− −= − +∫ ∫ Do cosh( sinh') ( ') n nar n I nI −= − +

Gf sinh α /' then cosh α / ' sinh α + /

( ')

n nnI n I −∴ = − − ∗

*+ sinh' sinh'

sinh'' '

++ +

sinh sinh d sinh cosh ( ') cosh sinh d

ar ar

ar n n n

x x x x x n x x x− − − = − − ∫ ∫

and use cosh ' sinh x x= +

coect ( ')n n I n I + − to obtain ( ')n nnI n I −= − − ∗

+ arsinh' I = +

I I = −

$ $ # I I = − and use with previous results to obtainE '

=(#ar sinh' )= − / +.'%$ (either answer acceptable)

'

'

1' (#)

'

1' ()

'

1' ()

'

'

'

1' ($)

2P% 3une ++$ 4n %5



22/68



23/68

25 (a) # cos sin # sin cosdx dy

a ad d

θ θ θ θ θ θ

= − =

$ $ ( ( )) s a c s s c d θ = +∫ /

#a c s d θ ∫ /# cos sina d θ θ θ ∫

Total length /

+

#$ 2sin 5

a π θ ×

/&a

'

'

'

1'

'

' 1' (0)

(b) # sin # cos sin A a a d π θ θ θ θ = ×∫

$

& sin cosa d π θ θ θ = ∫ /

%

+

&sin

%

a π π θ ×

/'

%

aπ

' 1'

'

' 1' (%)

2P% 3une ++$ 4n 05



24/68

26 (a) AB /

−

−

$

#

$

AC /

−#

'

AB × AC /

=− −− $

'

#$#$

k '&

1'? ;ne value correct, 1'? 1ll correct ' 1' 1' ($)

(b) =$#

$

'

$

'

#

$

'

+−=

−=

r 0

$

'

=

r ' 1'ft ()

(c) AD AB × AC (1ttempt suitable triple scalar product) '

−'#

or

−−

'

#

(if using AD) '

olume / ( ) '&

'

$

'

'

#

&

'=−+=

− ' 1'(cso)($)

10

2P& 3une ++$ 4n #5

2!

=

−

−=

k

k

k

c !

b

a

cba

!

++

++

++

'

+$

#'

+#

'$',



25/68

(a) ( ) #++#

'$' =⇒=

− !

!

' 1' ()

(b) ( ) '='

$

'

'$' =⇒=

−− k k (ft on their !, if used) ' 1'ft ()

(c) equations? a A $b c / + #a A #c / + '

a and b in terms of c (or equiv.)? a / c b / c (ft on their !) ' 1'ft

"sin# )'=('= =++=

cba

c

b

a

c

b

a

- a / √, b / √, c / √ ' 1(',+) (&)

(d) det / (#√) $('√) '(#√) / %$√ ' 1'(cso) ()

12

1lternatives?

(c) Hequire

c

b

a

parallel to

×

− #+

#

'

$

'

,

−−='

&

'

', ' 1'

(Then as in main scheme, scaling to give a, b and c.) ' 1(',+) (&)

(d) det#

'=) =,( , det / det , , det / '=√'= (/%$√) ' 1' ()

2P& 3une ++$ 4n %5

2$ (a) det " / + +')#( =−− λ '

$,+)$)((+=&

===−−=+− λ λ λ λ λ λ 1'

−=+=

=

'

'r@igenvecto,+,+

''

''? y x

y

xλ (or equiv.) ' 1'

=+−=

−

−=

'

'r@igenvecto,+,+

''

''?$ y x

y

xλ (or equiv.) 1' (%)



26/68

(b) P /

−''

''k ? eigenvectors as columns,

'=k ', 1'

−

==−''

''

'' ,PP

=

−

−== −

+

+$

''

''

'

#'

'#

,''

''

''

"PP% ', ' 1' (%)

(c) '. Hotation of$

π

clockwise (about (+, +)).

. Dtretch, × $ parallel to xIa9is, × parallel to yIa9is.

# #. Hotation of$

π

anticlockwise (about (+, +)).

'. and #. both rotation, or both reflection. '

Correct angles, opposite sense or correct lines (reflection). A1

Dtretch. '

1ll correct, including order. 1' ($) 14

2P& 3une ++$ 4n &5



27/68

2"P7P% 3une ++% 4n '5

2"P7P% 3une ++% 4n 5


2#

31

30

30


28/68

2"P7P% 3une ++% 4n #5

2"P7P% 3une ++% 4n $5


3131

32


29/68

2"P7P% 3une ++% 4n &5


33


30/68

2"P7P% 3une ++% 4n =5


34


31/68

2"P#7P& 3une ++% 4n 5


35


32/68

2"P#7P& 3une ++% 4n #5


36


33/68

2"P#7P& 3une ++% 4n 05


3!


34/68

2"P7P% 3anuar! ++& 4n '5

2"P7P% 3anuar! ++& 4n 5


3$

3#


35/68

2"P7P% 3anuar! ++& 4n #5

2"P7P% 3anuar! ++& 4n $5


40

41


36/68

2"P7P% 3anuar! ++& 4n 05


42


37/68

2"P7P% 3anuar! ++& 4n =5


43


38/68

2*"P#7P& 3anuar! ++& 4n #5


44(a)

(b)


39/68

2"P#7P& 3anuar! ++& 4n $5


45

3


40/68

2"P#7P& 3anuar! ++& 4n 05


46


41/68

4! ''

ee

ee% =

−−

+ −− x x x x'

+0ee# =+− x x ? Dimplif! to form quadratic in xe

' 1'

0e,#

'e+)0e)('e#( ===−− x x x x ? Dolve # term quadratic. ' 1'

( ) 0ln#lnor#

'ln =−= x x 1' (&)

6 arks2"P 3une ++& 4n '5

4$ (a)

8sing )'( eab −= or equiv. to find e or ae? (a / and b / ')

#=e

' 1'

8sing xae y )($

= x y #$

= ( requires values for a ande)

' 1' ($)

(b) #−= x 'ft (')5 arks

2"P 3une ++& 4n 5



42/68

4# (a) '$hsec$d

d −= x x

y'

Put ( )($cosh$$cosh+

d

d (===

x x x

y

')#ln($ ±= x or )#$0ln(= ±= x or #e$ ±= x or

#$0e$ ±= x (± or A) 1'

)#ln($

'+= x or )#$0ln(

=

'+= x (or equiv.) 1' ($)

(b) tanh(....))#ln(

$

'++−= y (Dubstitute for

x)

'

#$tanh,$tanh'

'$hsec =−== x x x '

( ){ }#ln#$')#ln($'# +−=+−= y (*) 1' (#)! arks

(a) JDecond solutionK, if seen, must be reFected to score the final

mark.

(b) nd requires an e9pression in terms of √# without h!perbolics,e9ponentials and logarithms.

2"P 3une ++& 4n %5



43/68

50 (a)

'

d

d''

d

d −=−= t

t

y

t t

x ' '

t t t t

t

''

'',

''

'

+=++=

+

−

−or

t

t '+ ', 1'

[ ] ( ) $ln#'$ln$lnd''>ength $'$

'+=−+=+= += ∫

t t t t

(*) ' ' 1'(0)

(b) Durface area /

+=

+

− ∫ ∫

−−t t t t t

t t d=d

''$

$

'

'

'$

'

'

π π

'

#

'&+

#

$

#

'&)=(

#

)=(

$

'

'

#

π π π =

+−

+=

+= t t

π #

'%#

' ' 1'

($)

(11 marks)2"P 3une ++& 4n &5

51 x x

x x

x x x x d

'#

arsinh#

d arsinh(

##

( ∫ ∫ +−= ' 1' 1'

,arsinh#/arsinh#

#

+

#

x

x'+#ln,or + '

>et x x

u xu

d

d' =+=

=+= x

x

uu xu

d

d' '

uuuu

u

uu

xu

xd

&

'd'

&

'd

.#

'

'

'

'

'

#

∫ ∫ ∫

−=

−

=

−

( )

−∫ uu d'#'

'

−=

'

#

#

&

'u

u

−= u

u

##

' #

'

When x / +, u / ' and when x / #, u / '+ '+...... =u

−−

−=

−

#

'+

#

'++

&

'

#

&

'

'+

'

'

#

uu ' 1'

1rea / ( ) ( )'+0,

''+#ln,

#

$

#

'+'$

&

'#arsinh, +−+=

+−

(*)

1'cso ('+)

:ependent marks?

? -hoose an appropriate substitution L find x

u

d

d or JDet upK integration b!



44/68

parts.? Met all in terms of JuK or 8se integration b! parts.

? Dound integration.

? Dubstitute both limits (for the correct variable) and subtract.

51 1lternative solution?

>et θ θ

θ coshd

dsinh ==

x x '

θ θ θ θ d coshsinhd sinhar (( ∫ ∫ = x x x '

( ) θ θ θ θ θ d'coshsinh#

'

#

sinh #

∫ −−

= ' 1' 1'

arsinh#/#

sinh#arsinh

+

#

θ θ '

( )

−=−

∫ θ

θ θ θ θ cosh

#

cosh

#

'd'coshsinh

#

' #'

−−

−=

− '

#

''+

#

'+'+

#

'cosh

#

cosh

#

'#arsinh

+

#

θ θ

' 1'

1rea / ( ) ( )'+0,

''+#ln,

#

#

'+0

#

'#arsinh, +−+=

+−

(*)

1'cso('+)

10 arks

51 1 few alternatives for? x x

xd'

#

∫ +

.

(i) >et x x

u xu

d

d ==



45/68

uu

uu

uu

ud

'

'd

'

'

'

#

∫ ∫ +=⋅+ No marks !etE needs another substitution, or parts, or perhapsE

uu

u

u

+−+=

+ ''

''

'

uu

uu d'

'

'd'

' ∫ ∫ +−+ '

( ) ( ) '

#

''#

'uu +−+ '

>imits (+ to ) '

(ii) >et θ θ

θ coshd

dsinh ==

x x '

( ) θ θ θ θ θ θ

θ d'coshsinhdcosh

cosh

sinh #

∫ ∫ −=⋅ '

Then, as in the alternative solution,

( )

−=−∫ θ θ θ θ θ cosh#

cosh

#

'd'coshsinh

#

' #'

>imits (+ to arsinh#) '



46/68

51 (iii) >et θ θ

θ secd

dtan ==

uu '

( )∫ ∫ −=⋅ θ θ θ θ θ θ θ

θ d'secsectandsec

sec

tan #

'

( ) ( ) θ θ

θ θ θ θ θ θ θ sec#

secdtansecdtansecsec

#

−=−= ∫ ∫ '>imits ( '+secto'sec == θ θ '

(iv) (! partsE must be the Jright wa! roundK, not integrating x )

',

'd

d

d

d, x$

x

x

x

$ x

x

u xu +=

+=== '

x x x x x d'' ∫ +−+ '

( ) #

'

#

' +−+ x x x '

>imits '

(v) (! parts) x$

x x

$ x

x

u xu arsinh,

'

'

d

d#

d

d,

# =+

=== No progress +

(vi)

#

''

)'(

'

)'(

' x

x

x

x x

x

x x x

x

x

+−

+

+=

+

−+=

+'

x x

x x x x d

'd'

∫ ∫ +

−+ '

( ) ( ) '

#

''#

' x x +−+= '

>imits '2"P 3une ++& 4n05



47/68

52 (a) x xnx x x x x xnnn dsinhsinhdcosh

'

∫ ∫ −−= ' 1' x x xnn xnx x x

nnn

dcosh)'(coshsinh ' ∫ −− −+−= '

' )'(coshsinh −− −+−= n

nn

n I nn xnx x x I (*) 1' ($)

(b) #$

$ 'cosh$sinh I x x x x I +−= '( )

+

#$

$ coshsinh'cosh$sinh I x x x x x x x x I +−+−= '(This ma! also be scored b! finding I b! integration.)

∫ +== k x x x I sinhd cosh+ '( ) ( ) ( )C x x x x x x I +−−+++= cosh$$,sinh$' #$$ 1', 1'(%)

(c) ( ) ( )[ ]'+#$ cosh$$sinh$' x x x x x x −−+++/ #0sinh ' =cosh ' ? x / ' substituted throughout (at some stage) '

+−

−=

−−

ee=

ee#0

''

'

? 8se of e9p. :efinitions (can be in terms of x)

( )'e&%e

' −−= 1' (#)

12 arks

(b) .nte/rat&on constant m&ss&n/ throu/hout loses the mark

2"P 3une ++& 4n =5



48/68

53(a) ( )

'

=+

+b

c%x

a

x )( bac%xa xb =++ '

( ( + =−+++ bca%cxa x%ab (*) 1' ()(b) ( ) ( ) ( ) $ bca%ab%ca −+= '

( $$ $$ b%ac%abcbac%a −+−= %abc +=(*)

1' ()

(c) "ind height and base of triangle (perhaps in terms of c). '

OB %abc +== and AO /

+=

%

%ab

%

c

1'

1rea of triangle OAB /%

%ab

%

c

+= ? "ind area and subs. for c. ' 1' ($)

(d)

%a

%b

%

%ab

'

+=+

=∆ −

+d

d

=+−=∆ − a%

b

%

a%

b=

a

b% =

' 1'

ababa

bab =

+

=∆

(*) 1' (#)

(e)Hoot of quadratic?

%ab

%ca x

+−

= (Dhould be correct if quoted

directl!)

'

8singa

b% = and %abc += ?

a x −= ' 1' (#)

(The nd is dependent on using the quadratic equation). 14 arks

(d) 1lternative? ba%%ab ≥+ (since +)( ≥− a%b ) 2'5

ab%

%ab ≥+

21'5

-onclusion 21'5

(e) 1lternative?

egin with full eqn. ( ) ( ) + =−+++ bca%cxa x%ab .

Gn the eqn., use conditionsa

b% = and )( b%abc =+= 2'5

Dimplif! and solve eqn., e.g.

+ a

xa xa x −==++

2"P 3une ++& 4n 5

54

( )''

' ' ' # ' '

+ ' ' + ' '

+ + ' + + '

+ ÷ ÷= = = ÷ ÷

÷ ÷

" " '

(


49/68

( )''

' # ' '

. + ' . + ' '

+ + ' + + '

k k

k k k

k +

+ ÷ ÷= = ÷ ÷ ÷ ÷ ÷

" " " '

( )'' ' #

+ ' '+ + '

k k k k

k

+ + + + ÷

= + ÷ ÷ ÷

( ) ( ) ( ) ' ' ' # % $ ' # #k k k k k k k k + + + = + + = + + + + ' :ep

( ) ( )( )

'

' # 'k k = + + + 1'

(


50/68

12

Aternati$e to (c), using characteristic pol!nomial of '−

( ) ( )' ' '& # # & +λ λ − − + × = '

>eading to ( ) ( ) ' '

#& % ' # ' ' + ,λ λ λ λ λ − + = − − = ⇒ = 1', 1' (3)

Aternati$e to (d)$

.

' '

x x

%x %x

′− = ÷ ÷ ÷

′ $ , x %x x x %x %x′ ′− = + = both '

'

$

%%

%

+ =−

1'

>eading to ( ) ( ) ' # ' ' ' + ,'% % % % %− + = − − = ⇒ = ' ' , y x y x= = both 1' (4)

2"P# 3une ++& 4n %5



51/68

56 (a) ( ) ( ) + #

$ % '

− × − = −− −

& ' k

b a c a '

'% '+ '+= − − −& ' k 1'A1'A1' (4) 1llow ' 1' for negative of above

(b) ( ) ( ) ( ). # # . #+ + = + + + −r & ' k & ' k & ' k or equivalent '

( ). # 0+ + =r & ' k or multiple 1' (2)

(c) >et , # x z λ λ = = − , then ( ) ' ' 0 # # y yλ λ λ = − − − ⇒ = − x, y and z in terms of a single parameter '

The direction of is an! multiple of ( ) − −& ' k '

( )( ) ( )' # − + × − − =r ' k & ' k 0 or equivalent ' 1' (4)

Possible equivalents are ( )( ) ( ) − + × − + + =r & k & ' k 0

and ( )( ) ( )'#− − × − + + =r & ' & ' k 0 The general form is

( ){ } ( )' c c− + + − − × − − = r & k & ' k & ' k 0

(d) ( ) ( )( ) ( )' ' # . +λ λ λ + − + − − − =& ' k & ' k '

' ' & +λ λ λ − + − + = >eading to '#λ = ' 1' ( )'# '$ ? , , P − 1' (4)

14 Aternati$e to (d)

( ) ( ) ( )( ) ' ' '

$# & #0OP λ λ λ λ λ = + − + − = − + '

( ) '#d

+

d

OP λ

λ

= ⇒ = ' 1'

( )'# '$ ? , , P − 1' (4)

2"P# 3une ++& 4n 05

5! )(%$ −+=−+ x x x '



52/68

( )( )

' d arcosh

#

x x

x

+=+ −√∫ ' 1'ft

ft their completing the square, requires arcosh

( )#

'

%arcosh arcosh arcosh '

# #

x + = −

% % % $ln ' ln ln #

# # #

= + − = + = ÷ ÷ ÷ ' 1' (5)

5

Aternati$e

)(%$ −+=−+ x x x '

>et #sec x θ + = ,d

#sec tand

xθ θ

θ =

( )( ) ( )

' #sec tand dsec

x x

θ θ θ θ

=−√+ −√ ∫ ∫ '

sec dθ θ = ∫ 1'ft ( )

%arcsec

#

arcsec'

% $ln sec tan ln ln #

# #θ θ

+ = + = ÷ ' 1' (5)

2"P 3une ++0 4n '5

5$ (a) y #

& D



53/68

O % x

;ne ellipse centred at O '

1nother ellipse, centred at O, touching on yIa9is '

Gntersections? 1t least , %, and # shown correctl! ' (3)

(b) 8sing ( ) 'b a e= − , or equivalent, to find e or ae for D or & . '

"or S ? %a = and #b = ,$

%e = , $ae = ignore sign with ae 1'

"or ' ? #a′ = and b′ = ,%

#e

√′ = , a e′ ′ = 5√ ignore sign with a e′ ′ 1'

( )'& % 'S' = + =√ √ ' 1' (5) $

2"P 3une ++0 4n 5

5#

−=

x x

x

y '$

$

'

d

d'

' '

d '' d ' dd $ y x x x x x

÷+ = + − ÷ ÷ ÷ ÷ ÷ ∫ ∫

'

' '

' ' ' '' d d d

'& $ $ x x x x x x

x x x

= + + − = + = + ÷ ÷ ÷ ÷ ÷ ∫ ∫ ∫ ' 1'



54/68

ln

$

x x= + 1'

ln

'

=

'%

$

%.+ln

=

'

$

ln

$

ln

%.+

+=−−+=

+

x x ' 1'

(!)

'% '

,

=

a b = = ÷

(! marks)

2"P 3une ++0 4n #5

60 (a)

e e e e e e e ecosh cosh sinh sinh

A A B B A A B B

A B A B

− − − − + + − −− = − ÷ ÷ ÷ ÷

'

( ) B A B A B A B A B A B A B A B A −−−+−+−−−+−+ −++−+++= eeeeeeee$

'

( )( )

)cosh(eeee

$' B A

B A B A

B A B A −=+=+=−−−

−+−

cso

' 1' (3)

(b) x x x sinh'sinhsinh'coshcosh =− '

'sinh'

'coshtanh)'sinh'(sinh'coshcosh

+=⇒+= x x x '

'

'

' '

e ee e e 'tanh

e e e e e e ''

x

−

−

− −

++ +

= = =− + − + −+

cso ' 1' (4)

! Aternati$e for (b)

( )''e e e e

x x x x− −− −+ −= '

>eading to

e eee '

x +=−

'

'ee

'e

)'(eee

)'e(ee

'e

'e

ee

eetanh

−++

=−++−−+

=+−

=+−

= −−

x

x

x x

x x

x cso ' 1' (4)

2"P 3une ++0 4n $5

61 (a)=

$ $'# #

+

##

$ $(= ) (= ) dn n

n I x x nx x x− = − − + − ∫ ' 1'

$

' ##

$(= ) dnnx x x

−= −∫ ft numeric constants onl! 1'ft



55/68

' ' '

' ' '# # #(= )(= ) d =(= ) d (= ) dn n nnx x x x nx x x nx x x x− − −− − = − − −∫ ∫ ∫ ' 1' ' '

$#&

# $$n n n n n

n I nI nI I I

n− −= − ⇒ = +

cso 1' (6)

(b) ( )=

= $ $'# ##

++ +

# #= d (= ) = '

$ $ I x x x

= − = − − = × = ∫

' 1'

= '

# '

+

( %)(= ) d % I x x x x I I = + − = +∫ '

' +$

0 I I = , ' + +

$= $= $ %0&

'+ '+ 0 #% I I I I

= = × = ÷ ' 1'

' +'&=

The previous line can be implied b! %%

I I I I = + = ÷

( )%0& $ +'&

% ' $+#.#% 0 %

I = + × × = = ÷

1' (6)

(12 marks)2"P 3une ++0 4n &5



56/68

62 (a) ( )( ) ( )

' '

d ' ' 'arsinh

d ' ' x x

x x x x

− = × = ÷ ÷+ 2 +√ √ √ ' 1'

1t $ x = ,d '

d $ %

y

x=

√ accept equivalents 1' (3)

(b)sinh , x θ =

dsinh cosh

d

xθ θ

θ =

arsinh d sinh cosh d x x θ θ θ θ = ×√∫ ∫ ' 1'

cosh cosh sinh d d

θ θ θ θ θ θ θ = = −∫ ∫ ' 1' A 1'

sinh

...$

θ = − '

arsinh

+

cosh sinh

$

θ θ θ − = E attempt at substitution '

( )

( )' sinh sinh cosh ' $ %

arsinh ' = $ $

θ θ θ θ + √ = − = × + −

' 1'

( )

ln % %

= + −√ √ 1' (10)

13

Aternati$e for (a)

dsinh , sinh cosh '

d

y x y y y

x= = '

( ) ( )

d ' ' '

d sinh cosh 'sinh sinh '

y

x y y x x y y

= = = ÷ ÷2 ++ √ √√

1'

1t $ x = ,d '

d $ %

y

x=

√ accept equivalents 1' (3)

An aternati$e for (b) is #i$en on t(e next !a#e



57/68

62 Aternati$e for (b)

( )

'' arsinh d arsinh d

' x x x x x x

x x× = − ×√ √ 2 +√ √∫ ∫ ' 1' A 1'

( )arsinh d

'

x x x x

x

√= −√ 2 +√∫

>etsinh , x θ =

dsinh cosh

d

xθ θ

θ =

( )

sinhd sinh cosh d

cosh'

x x

x

θ θ θ θ

θ

√ = ×+√∫ ∫ ' 1'

cosh ' sinh sinh d d ,

θ θ θ θ θ θ

−= = = −∫ ∫ ', '

arsinh arsinh

+ +

sinh sinh cosh %

arsinh

θ θ θ

θ θ

× × √− = − = − ' 1'

( )$

+

' % arsinh d $arsinh arsinh ln % %

x x

× ×√= − − = + −√ √ √ ÷

∫ 1' (10)

'(e ast ) %arks of t(e aternati$e soution can be #ained as foo*s

>ettan , x θ =

d tan sec

d

xθ θ

θ =

( )

tand tan sec dsec'

x x

x

θ θ θ θ

θ

√

= ×+√∫ ∫ dependent on first ' ' 1'

sec tan dθ θ θ = ∫

( ) #sec tan tan d sec tan sec dθ θ θ θ θ θ θ θ = −∫ ∫ ' ( )sec tan sec ' tan dθ θ θ θ θ = − +∫


58/68

63 (a)

# $ + +

' $ . ' '

' ' # ' '

!

+ λ

÷ ÷ ÷− − = ÷ ÷ ÷ ÷ ÷ ÷− −

Third row ' # λ λ − = − ⇒ = ' 1' (2)

(b)

# $ + $ +

' $ . ' $ '

' ' # ' '

! !

+ +

− ÷ ÷ ÷ ÷− − = + = ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷− − −

' 1'

"irst row $ + $ ! !− = ⇒ = ethod for either ' Decond row $ + ++ = ⇒ = − oth correct 1' ft (4)

(c)

# $ $ '+

' $ . $

' ' # #

%

n

÷ ÷ ÷− − − = − ÷ ÷ ÷

÷ ÷ ÷ # $ $ '+ % n+ + = $ $ % n− − − = − # # % n+ + = ;btaining # linear equations '

& %+ = # = %+ = Heducing to a pair of equations and ' solving for one variable

, ', + % n= = = Dolving for all three variables. ' 1' (4) 10

1lternative to (c)

'

= ='

' % =&

' '

−

÷= − − − ÷ ÷− −

" ' '

= = '+ '

' % = . $ '&

' ' # +

÷ ÷ ÷− − − − = ÷ ÷ ÷ ÷ ÷ ÷− −

' 1' (4)

2"P# 3une ++0 4n #5



59/68

64 (a) # # , $ AB AC = − + + = +& ' k ' k an! two '

' # # & $

+ $

AB AC × = − = − + −& ' k

& ' k uuur uuur

' 1' 1'

Mive 1' for an! two components correct or the negative of the correct answer. (4)

(b) -artesian equation has form # x y z !− + = ( ), ', + & ' !− ⇒ + = or use of another point ' # 0 x y z − + = or an! multiple 1' (2)

(c) Parametric form of line is

%

% '

#

λ

÷ ÷= + − ÷ ÷ ÷ ÷−

r or equivalent form ' 1'

Dubstituting into equation of plane

( ) ( ) ( )# % % # 0λ λ λ + − − + − = '>eading to #λ = − 1' ( )? ', =, ' − 1' (5)

(d) # , & & A' B' = − + + = − + +& ' k & ' k both ' These are parallel and hence A, B and ' are collinear (b! the a9iom of parallels) ' 1' (3)

141lternative to (d)

The equation of AB? ( ) # # µ = − + − + +r & ' & ' k or equivalent &? ' # µ µ − = − ⇒ = ' # = O' µ = ⇒ = − + +& ' k '


60/68

2"P 3une ++= 4n '5


65


61/68

2"P 3une ++= 4n 5


66


62/68

2"P 3une ++= 4n #5


6!


63/68

2"P 3une ++= 4n $5


6$


64/68

2"P 3une ++= 4n %5


6#


65/68

2"P 3une ++= 4n &5


!0


66/68

2"P 3une ++= 4n 05


!1


67/68

2"P# 3une ++= 4n 5


!2


68/68

2"P# 3 ++= 4 05

!3

06a Further Maths - FP3 Mark Schemes v2 (1)

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Transcript of 06a Further Maths - FP3 Mark Schemes v2 (1)