4756
Mark Scheme
January 2006
1(a)(i
)
B1
For one loop in correct
quadrant(s)
B1
For two more loops
B1
3 Continuous and broken lines
Dependent on previous B1B1
(ii)
1π
Area is ∫
1
2
⌠6 1 2
r dθ = ⎮
a cos 2 3θ dθ
⌡− 1 π 2
M1
For ∫ cos 2 3θ dθ
A1
For a correct integral expression
including limits (may be implied
by later work)
2
6
1π
⌠6 1 2
=⎮
a (1 + cos 6θ ) dθ
⌡− 1 π 4
M1
6
⎡
=⎢
⎣
=
1π
1
4
⎤6
a 2 (θ + 16 sin 6θ ) ⎥
⎦− 1π
6
1
π
12
a
For ∫ cos 2 3θ dθ = 12 θ + 121 sin 6θ
A1
2
Accept 0.262a 2
B1
5
(b)
3
3
⎛ 2x ⎞ ⎤ 4
1
⎜⎜
⎟⎟ ⎥
arcsin
2
⎝ 3 ⎠ ⎥⎦ 0
⎡
⌠4
1
dx = ⎢
⎮
2
⌡0 3 − 4 x
⎣⎢
⎛ 3 ⎞
⎟⎟
= 12 arcsin⎜⎜
⎝2 3⎠
=
1
π
6
M1
For arcsin
A1A1
For
M1
Dependent on previous M1
1
2
and
2x
3
A1
5
(c)
Putting
3 x = tan θ
1π
⌠3
Integral is ⎮
⎮
⌡0
1π
⌠3
=⎮
⌡0
⎛ sec 2 θ
⎜
sec3 θ ⎜⎝ 3
1
⎞
⎟ dθ
⎟
⎠
M1
For any tan substitution
A1A1
For
1
(sec 2 θ )
3
2
and
sec 2 θ
3
1π
3
⎡ sin θ ⎤
dθ = ⎢
⎥
3
⎣ 3 ⎦0
cos θ
Including limits of θ
M1
1
=
2
A1
5
OR
Putting 2 x = 3 sin θ
1π
3
Integral is ⌠
⎮
⌡0
1
2
= 16 π
dθ
M1
A1
For any sine substitution
A1
For
M1
A1
For changing to limits of θ
Dependent on previous M1
∫ 2 dθ
1
4756
2 (i)
Mark Scheme
January 2006
w =
w*
jw
1
2
, arg w = 3θ
B1
arg w* = −3θ
= 12 ,
= 12 ,
B1 ft
arg jw = 3θ + 12 π
B1B1 ft
B2
(ii)
w * and jw in correct positions
6 relative to their w in first
quadrant
Give B1 for at least two points
in correct quadrants
(1 + w)(1 + w*) = 1 + 12 e 3 jθ + 12 e −3 jθ + ( 12 e 3 jθ )( 12 e −3 jθ ) M1
for w* = 12 e −3 jθ
for 1 + 14 correctly obtained
A1
= 1 + 12 (cos 3θ + j sin 3θ ) + 12 (cos 3θ − j sin 3θ ) +
=
5
4
+ cos 3θ
1
4
for w = 12 (cos 3θ + j sin 3θ )
M1
A1 (ag)
for cos 3θ correctly obtained
4
(iii) C + jS = e
=
=
=
=
2 jθ
− 12 e
5 jθ
+ 14 e
8 jθ
− ...
e 2 jθ
1+ e
1
2
3 jθ
M1
Obtaining a geometric series
M1
A1
Summing an infinite geometric
series
e 2 jθ (1 + 12 e − 3 jθ )
(1 + 12 e3 jθ )(1 + 12 e − 3 jθ )
e
e
2 jθ
(1 + e
1
2
− 3 jθ
5
4
+ cos 3θ
2 jθ
+ 12 e − jθ
5
4
+ cos 3θ
4 cos 2θ + 2 cos θ
5 + 4 cos 3θ
4 sin 2θ − 2 sin θ
S=
5 + 4 cos 3θ
M1
Using complex conjugate of
denom
)
2 jθ
⎛
+ 2e − jθ
⎜ = 4e
⎜
5 + 4 cos 3θ
⎝
⎞
⎟
⎟
⎠
A1
C=
M1
A1 (ag)
Equating real or imaginary parts
Correctly obtained
A1
8
4756
3 (i)
Mark Scheme
January 2006
(1 − λ ) [ (−3 − λ )(−4 − λ ) − 12 ]
− 2 [ − 2(−4 − λ ) − 12 ] + 3 [ − 4 − 2(−3 − λ ) ] = 0
Evaluating det(M − λ I) Allow
one omission and two sign
errors
det(M − λ I ) correct
M1
A1
(1 − λ )(λ + 7λ ) − 2( 2λ − 4) + 3(2λ + 2) = 0
2
λ3 + 6λ2 − 9λ − 14 = 0
A1 (ag)
3 Correctly obtained (=0 is
required)
(ii)
When λ = −1 , −1 + 6 + 9 − 14 = 0
B1
(λ + 1)(λ2 + 5λ − 14) = 0
(λ + 1)(λ − 2)(λ + 7) = 0
M1
Other eigenvalues are 2 , − 7
A1
or showing that (λ + 1) is a
factor, and deducing that −1 is a
root
for (λ + 1) × quadratic factor
3
(iii)
x + 2 y + 3z = − x
− 2x − 3 y + 6z = − y
2x + 2 y − 4z = −z
⎛ 1⎞
⎜ ⎟
z = 0 , x + y = 0 An eigenvector is ⎜ − 1⎟
⎜ 0⎟
⎝ ⎠
⎛ 2⎞
⎜ ⎟
⎛ − 2⎞
⎜
⎟
⎛ 18 ⎞
⎜
⎟
⎜ 3⎟
⎝ ⎠
⎜ 6⎟
⎝
⎠
⎜
⎝
OR ⎜ 2 ⎟ × ⎜ − 2 ⎟ = ⎜ − 18 ⎟
(iv)
(v)
⎛ 3⎞ ⎛ 6⎞
⎛ 3⎞
⎜ ⎟ ⎜ ⎟
⎜ ⎟
M ⎜0⎟ = ⎜ 0⎟ = 2 ⎜0⎟
⎜1⎟ ⎜ 2⎟
⎜1⎟
⎝ ⎠ ⎝ ⎠
⎝ ⎠
0 ⎟⎠
0⎞
⎛−1 0
⎜
⎟
D=⎜ 0 2
0⎟
⎜ 0 0 − 7⎟
⎝
⎠
At least two equations
M1
Solving to obtain an eigenvector
A1
3
Appropriate vector product
M1
M1
Evaluation of vector product
A1
⎛ 0⎞ ⎛ 0⎞
⎛ 0⎞
⎜ ⎟ ⎜
⎟
⎜ ⎟
M ⎜ 3 ⎟ = ⎜ − 21⎟ = −7 ⎜ 3 ⎟
⎜ − 2 ⎟ ⎜ 14 ⎟
⎜ − 2⎟
⎝ ⎠ ⎝
⎠
⎝ ⎠
0⎞
⎛ 1 3
⎜
⎟
3⎟
P = ⎜−1 0
⎜ 0 1 − 2⎟
⎝
⎠
M1
M1
Any method for verifying or
finding an eigenvector
A1A1
3
B1 ft
3
0⎞
⎛−1 0
⎜
⎟
=⎜ 0 8
0⎟
⎜ 0 0 − 343 ⎟
⎝
⎠
seen or implied (ft)
(condone eigenvalues in wrong
order)
M1
A1 ft
3 Order must be consistent with P
(when B1 has been awarded)
(vi) By CHT,
M 3 + 6M 2 − 9M − 14I = 0
M 2 + 6M − 9I − 14M −1 = 0
1
M −1 = 14
M 2 + 73 M − 149 I
B1
Condone omission of I
M1
Condone dividing by M
A1
3
4756
4 (a)
1
2
Mark Scheme
January 2006
( e x − e − x ) + 2( e x + e − x ) = 8
5e 2 x − 16e x + 3 = 0
(5e x − 1)(e x − 3) = 0
e x = 15 , 3
M1
Exponential form
M1
Quadratic in e x
M1
A1A1
A1 ft
x = − ln 5 , ln 3
Solving to obtain a value of e x
Exact logarithmic form from 2
positive values of e x
6 Dependent on M3
c 2 − 1 = 8 − 4c
OR
15c 2 − 64c + 65 = 0
M1
Obtaining quadratic in c (or s)
( 15s 2 + 16s − 48 = 0 )
Solving to obtain a value of c (or
s)
or s = 43 , − 125
M1
c = 53 ,
13
5
A1A1
x = ± ln 3 , ± ln 5
M1
A1
x = ln 3 , − ln 5
(b)
Logarithmic form (including ±
if c)
cao
2
⌠ 1 x x
−x
⎮ 2 e (e − e ) dx
⌡0
]
0
= ( 14 e 4 − 1 ) − (
1
4
=
[
1
4
e2x −
1
2
x
Exponential form
Integrating to obtain a multiple
of e 2 x
M1
M1
A1
2
)
= (e − 5)
4
1
4
A1
4
(c)(i)
⎛
2
⎜ =
⎜
9 + 4x2
⎝
2
3
1+ (
2
3
x)
2
⎞
⎟
⎟
⎠
B2
Give B1 for any non-zero
2 multiple of this
M1
(ii)
[ x arsinh(
2
3
x)
]
2
2
0
⌠
−⎮
⌡0
2x
9 + 4x2
= ⎡ x arsinh( 23 x ) −
⎢⎣
(
= 2 arsinh(
4
3
)−
5
2
dx
Integration by parts applied to
arsinh( 23 x ) × 1
A1 ft
2
1
2
9 + 4x2 ⎤
⎥⎦ 0
)− ( −
= 2 ln⎛⎜ 43 + 1 + 169 ⎞⎟ − 1
⎝
⎠
= 2 ln 3 − 1
3
2
⌠
for ⎮
x
⌡ 9 + 4x2
B1
)
dx =
1
4
9 + 4x2
Using both limits (provided both
give non-zero values)
Logarithmic form for arsinh
(intermediate step required)
M1
M1
A1 (ag)
6
4756
5 (i)
Mark Scheme
January 2006
x = 2 , x = −2
y=x+
4x − k
B1
3
M1
x2 − 4
Asymptote is y = x
Dividing out
or B2 for y = x stated
A1
3
k<2
(ii)
B1
for LH and RH sections
for central section, with positive
intercepts on both axes
B1
k>2
for LH and central sections
for RH section, crossing x-axis
B1
k<2
k>2
B1
4
(iii) dy
dx
=
=
( x 2 − 4)( 3x 2 ) − ( x 3 − k 3 )(2 x)
( x − 4)
2
M1
A1
2
Using quotient rule (or
equivalent)
Any correct form
x( 2k + x − 12 x)
3
3
( x 2 − 4) 2
dy
= 0 when x = 0
dx
When x ≈ 0 , 2k 3 + x 3 − 12 x > 0
A1 (ag)
Correctly shown
dy
dy
< 0 when x < 0 ,
> 0 when x > 0
dx
dx
Hence there is a minimum when x = 0
M1
or evaluating
A1 (ag)
5
(iv) Curve crosses y = x when x 3 − k 3 = x( x 2 − 4)
M1
x = 14 k 3
So curve crosses this asymptote
A1 (ag)
2
or
d2 y
dx 2
d2 y
dx 2
when x = 0
= 18 k 3 > 0 when x = 0
4756
(v)
Mark Scheme
January 2006
k<2
B2
Asymptotes shown
Intercepts 14 k 3 and k indicated
Minimum on positive y-axis
Maximum shown
Give B1 for minimum and
maximum on central section
k>2
Asymptotes shown
Intercepts 14 k 3 and k indicated
B2
Minimum on positive y-axis
4 RH section crosses y = x and
approaches it from above
Give B1 for RH section
approaching both asymptotes
correctly
4756
Mark Scheme
June 2006
1(a)(i)
B1
Correct shape for − 12 π < θ < 12 π
including maximum in 1st
quadrant
B1
(ii)
Area is
∫
1
2
M1
3π
⌠4 1 2
r dθ = ⎮
a ( 2 + 2 cos θ ) 2 dθ
⌡− 3 π 2
2
2 Correct form at O and no extra
sections
A1
For integral of ( 2 + 2 cos θ ) 2
For a correct integral expression
including limits (may be implied
by later work)
B1
Using 2 cos 2 θ = 1 + cos 2θ
B1B1 ft
Integration of cos θ and cos 2θ
4
3π
⌠4
a 2 (1 + 2 2 cos θ + 1 + cos 2θ ) dθ
=⎮
⌡− 3 π
4
3π
⎡
⎤4
= ⎢ a 2 (2θ + 2 2 sin θ + 12 sin 2θ ) ⎥
⎣
⎦− 3π
4
= 3(π + 1) a 2
Evaluation using sin 34 π = (± )
M1
A1
7
(b)(i) f ′ ( x) = sec 2 ( 1 π + x )
4
f ′′ ( x) =
2 sec 2 ( 14 π
B1
+ x ) tan( 14 π + x )
Any correct form
B1
f(0) = 1, f ′ (0) = 2 , f ′′ (0) = 4
Evaluating f ′ (0) or f ′′ (0)
M1
f( x) = 1 + 2 x + 2 x 2 + ...
B1A1A1
6
OR g′ (u ) = sec 2 u ( where g(u ) = tan u )
g′′ (u ) = 2 sec 2 u tan u
g( 14 π )
g′ ( 14 π )
= 1,
= 2,
Condone sec 2 x etc
B1
B1
g′′ ( 14 π )
2
=4
M1
f( x) = g( 14 π + x ) = 1 + 2 x + 2 x + ...
B1A1A1
Evaluating g′ ( 14 π ) or g′′ ( 14 π )
(ii) ⌠ h
2
2
⎮ x (1 + 2 x + 2 x + ...) dx
⌡− h
⎡
=⎢
⎣
h
1
3
x3 +
1
2
x4 +
2
5
⎤
x 5 + ... ⎥
⎦ −h
M1
A1 ft
≈ ( 13 h 3 + 12 h 4 + 52 h 5 ) − ( − 13 h 3 + 12 h 4 − 52 h 5 )
= 23 h 3 + 54 h 5
Using series and integrating
(ft requires three non-zero
terms)
A1 (ag)
3 Correctly shown
Allow ft from 1 + kx + 2 x 2 with
k≠0
1
2
4756
2
(a)(i)
Mark Scheme
zn +
1
zn
4
(ii)
zn −
= 2 cos nθ ,
1
zn
= 2 j sin nθ
B1B1
2
2
1 ⎞ ⎛
1 ⎞
⎛
⎜ z − ⎟ ⎜ z + ⎟ = 64 sin 4 θ cos 2 θ
z ⎠ ⎝
z ⎠
⎝
1
2
1
= z6 − 2z 4 − z 2 + 4 − 2 − 4 + 6
z
z
z
= 2 cos 6θ − 4 cos 4θ − 2 cos 2θ + 4
sin 4 θ cos 2 θ =
( A=
June 2006
1
32
1
32
cos 6θ − 161 cos 4θ −
1
32
cos 2θ + 161
1
1
1
, B = − 16
, C = − 32
, D = 16
)
(b)(i)
Expansion z 6 + ... + z −6
M1
A1
1
= 2 cos nθ with
zn
n = 2 , 4 or 6 . Allow M1 if used
M1
Using z n +
A1 ft
in partial expansion, or if 2
omitted, etc
A1
6
Accept 5.7 ; 0.79 , 45°
B1B1
arg(4 + 4 j) = 14 π
4 + 4 j = 32 ,
B1
2
1
(ii)
r= 2
θ = − 34 π , −
7
π
20
,
1
π
20
,
9
π
20
,
17
π
20
Accept 32 10 , 1.4 ,
B1
5
4 2 etc
Accept −2.4 , − 1.1, 0.16 , 1.4 , 2.7
Give B2 for three correct
Give B1 for one correct
Deduct 1 mark (maximum) if
degrees used
B3
(−135°, − 63°, 9°, 81°, 153°)
1
π
20
+ 52 kπ earns B2; with
k = −2 , − 1, 0 , 1 , 2 earns B3
B2
6
(iii)
2e
− 3π j
4
⎛
1
1 ⎞
= 2 ⎜⎜ −
−
j ⎟⎟
2
2 ⎠
⎝
= −1 − j
p = −1 , q = −1
Give B1 for four points correct,
or B1 ft for five points
M1
Exact evaluation of a fifth root
A1
Give B2 for correct answer
2 stated or obtained by any other
method
4756
Mark Scheme
M1
A1
3 (i)
M
−1
June 2006
⎛ 1 5k − 13 5 − 2k ⎞
⎟
1 ⎜
=
⎜ 1 52 − 8k 3k − 20 ⎟
5−k ⎜
5 ⎟⎠
− 12
⎝−1
M1
A1
M1
A1
Evaluating determinant
For (5 − k ) must be simplified
Finding at least four cofactors
At least 6 signed cofactors
correct
Transposing matrix of cofactors
and dividing by determinant
6 Fully correct
or elementary column
operations
OR Elementary row operations applied to M
(LHS)
and I (RHS), and obtaining at least two
zeros in LHS
M1
Obtaining one row in LHS consisting of two
zeros and a multiple of (5 − k )
A1
Obtaining one row in RHS which is a multiple
of a row of the inverse matrix
A1
Obtaining two zeros in every row in LHS M1
Completing process to find inverse
(ii)
M1A1
22 − 9 ⎞ ⎛12 ⎞
⎛ x⎞
⎛ 1
⎜ ⎟
⎟⎜ ⎟
1⎜
1⎟ ⎜ m ⎟
⎜ y⎟ = − ⎜ 1 − 4
2⎜
⎜z⎟
5 ⎟⎠ ⎜⎝ 0 ⎟⎠
⎝ ⎠
⎝ − 1 − 12
M1
Substituting k = 7 into inverse
M1
Correct use of inverse
M1
Evaluating matrix product
x = −11m − 6 , y = 2m − 6 , z = 6m + 6
A2 ft
Give A1 ft for one correct
5 Accept unsimplified forms or
solution left in matrix form
OR e.g. eliminating x ,
3 y − z = −24
M2
Eliminating one variable in two
different ways
y = 2m − 6
M1
x = −11m − 6 , y = 2m − 6 , z = 6m + 6
A2
Obtaining one of x , y , z
Give M3 for any other valid
method leading to one of x , y ,
z in terms of m
5 y − z = 4m − 36
Give A1 for one correct
M2
(iii)
Eliminating x ,
3 y + 3z = −24
5 y + 5 z = 4 p − 36
For solutions, 4 p − 36 = −24 ×
5
3
A1
Eliminating one variable in two
different ways
Two correct equations
M1
Dependent on previous M2
OR Replacing one column of matrix with
column
from RHS, and evaluating determinant
M2
determinant 12 + 12 p or − 12 − 12 p
A1
For solutions, det = 0
M1
Dependent on previous M2
4756
Mark Scheme
June 2006
OR Any other method leading to an equation
from
which p could be found
M3
Correct equation
A1
p = −1
Let z = λ ,
A1
M1 (or
M3)
x = 5 − λ , y = −8 − λ , z = λ
A1
Obtaining a line of solutions
Give M3 when M0 for finding p
or x = 13 + λ , y = λ , z = −8 − λ
or x = λ , y = −13 + λ , z = 5 − λ
7 Accept x = 5 − z , y = −8 − z
or x = y + 13 = 5 − z etc
4756
4 (i)
Mark Scheme
June 2006
1 + 2 sinh 2 x = 1 + 2 [ 12 ( e x − e − x ) ]2
= 1 + 12 ( e 2 x − 2 + e − 2 x )
= 12 ( e 2 x + e − 2 x )
= cosh 2 x
B1
For ( e x − e − x ) 2 = e 2 x − 2 + e −2 x
B1
For cosh 2 x = 12 ( e 2 x + e −2 x )
For completion
B1 (ag)
3
(ii)
2(1 + 2 sinh 2 x) + sinh x = 5
4 sinh 2 x + sinh x − 3 = 0
(4 sinh x − 3)(sinh x + 1) = 0
sinh x =
x = arsinh(
3
4
) = ln(
3
4
3
4
, −1
+
9
16
M1
Using (i)
M1
Solving to obtain a value of
sinh x
A1A1
+ 1 ) = ln 2
A1 ft
x = arsinh(−1) = ln( − 1 + 1 + 1 ) = ln( 2 − 1)
A1 ft
6 or − ln( 2 + 1)
SR Give A1 for
± ln 2 , ± ln( 2 − 1)
OR 2 e 4 x + e3 x − 10 e 2 x − e x + 2 = 0
x
x
(e − 2)(2 e + 1)(e
2x
x
+ 2 e − 1) = 0
x = ln 2 , ln( 2 − 1)
(iii)
1
2
ln 3
1
2
⎤
x ⎥
⎦0
1⎞ 1
1⎛
= ⎜ 9 − ⎟ − ln 3
9⎠ 2
8⎝
10 1
=
− ln 3
9 2
(iv)
Put x = 3 cosh u
when x = 3 , u = 0
when x = 5 , u = ar cosh 53 = ln 3
5
Expressing in integrable form
or ∫ 14 ( e 2 x − 2 + e −2 x ) dx
M1
(cosh 2 x − 1) dx
⎡
= ⎢ 14 sinh 2 x −
⎣
Obtaining a linear or quadratic
factor
For (e x − 2) and (e 2 x + 2 e x − 1)
A1A1 ft
ln 3
⌠
⎮
⌡0
M2
A1A1
A1A1
or ( 18 e 2 x − 18 e −2 x ) − 12 x
M1
For e 2 ln 3 = 9 and e −2 ln 3 =
M0 for just stating sinh(2 ln 3) =
A1 (ag)
etc
5 Correctly obtained
M1
Any cosh substitution
B1
For ln 3
arcosh
ln 3
⌠
⌠
2
⎮ x − 9 dx = ⎮ (3 sinh u )(3 sinh u du )
⌡0
⌡3
A1
Not awarded for
5
3
Limits not required
ln 3
⌠
= 9⎮ sinh 2 u du
⌡0
= 10 − 92 ln 3
1
9
A1
4
40
9
4756
Mark Scheme
June 2006
5 (i)
At least two cusps clearly shown
Give B1 for at least two arches
B2
Has cusps
Periodic / Symmetrical in y-axis / Has
maxima /
Is never below the x-axis
B1
B1
Any other feature
4
(ii)
At least two minima (zero
gradient) clearly shown
Give B1 for general shape
correct (at least two cycles)
B2
The curve has no cusps
For description of any difference
B1
3
(iii) (A)
B2
(B)
At least two loops
2 Give B1 for general shape
correct (at least one cycle)
M1
dy
sin θ
=
dx 1 − 2 cos θ
A1
2
Correct method of differentiation
Allow M1 if inverted
sin θ
Allow
1 − k cos θ
(C)
dy
is infinite when 1 − 2 cos θ = 0
dx
θ = 13 π
x=
1
π
3
− 2 sin
= −( 3 −
1
π
3
2π
=2 3−
3
k = 4.6
A1
Any correct value of θ
M1
1
π)
3
Hence width of loop is 2( 3 − 13 π )
(iv)
M1
M1
Finding width of loop
A1 (ag)
Correctly obtained
5 Condone negative answer
B2
Give B1 for a value between 4
2 and 5 (inclusive)
4756
Mark Scheme
Jan 2007
1(a)(i)
Correct shape for 0 ≤ θ ≤ 12 π
B1
B1
Correct shape for
1
2 2 π ≤ θ ≤ π Requires decreasing r
on at least one axis
Ignore other values of θ
Area is
(ii)
π
⌠
⌡0
∫ 12 r 2 dθ = ⎮
1
2
a 2 (e − kθ ) 2 dθ
⎤
⎥
⎥⎦ 0
For a correct integral expression
including limits (may be implied
by later work) (Condone
reversed limits)
Obtaining a multiple of e−2 kθ as
the integral
A1
4
1
1
⎡ 1
⎛ 2x ⎞ ⎤ 2
⌠2 1
⎜⎜
⎟⎟ ⎥
d
arctan
x
=
⎢
⎮
⌡0 3 + 4 x 2
⎝ 3 ⎠ ⎦⎥ 0
⎣⎢ 2 3
=
A1
M1
a2
=
(1 − e − 2 kπ )
4k
=
For ∫ (e − kθ ) 2 dθ
π
⎡ a 2 − 2 kθ
e
=⎢−
⎢⎣ 4k
(b)
M1
⎛ 1 ⎞
⎟⎟
arctan⎜⎜
2 3
⎝ 3⎠
M1
For arctan
A1A1
For
M1
Dependent on first M1
1
π
2 3
2x
3
5
OR
Putting 2 x = 3 tan θ
M1
A1
For any tan substitution
A1
For
π
⌠6 1
Integral is ⎮
dθ
⌡0 2 3
=
and
A1
12 3
1
1
M1
π
1
3
dθ
For changing to limits of θ
Dependent on first M1
A1
12 3
∫2
(c)(i) f ( x) = tan x , f (0) = 0
f ′ ( x) = sec 2 x , f ′ (0) = 1
B1
f ′′ ( x) = 2 sec 2 x tan x , f ′′ (0) = 0
tan x = x +
A1
x3
(2) + ... ( = x + 13 x3 + ... )
3!
B1 ft
4h
(ii) ⌠ 4 h tan x
⌠
1 2
d
x
≈
⎮
⎮ (1 + 3 x ) dx
⌡h
x
⎡
=⎢ x+
⎣
= ( 4h +
Obtaining f ′′′ ( x)
For f ′′ (0) and f ′′′ (0) correct
M1
f ′′′ ( x) = 2 sec 4 x + 4 sec 2 x tan 2 x , f ′′′ (0) = 2
ft requires x3 term and at least
4 one other to be non-zero
Obtaining a polynomial to
integrate
M1
⌡h
4h
1
9
x
3
⎤
⎥
⎦h
64 3
h )
9
3
A1 ft
1
9
For x + 19 x 3
3
− (h + h )
= 3h + 7h
ft requires at least two non-zero
terms
A1 ag
3
32
4756
2(a)(i)
Mark Scheme
w = 3 , arg w = − 121 π
B1
z = 2 , arg z = − 13 π
B1B1
w
=
z
B1B1 ft
(ii) w
z
(b)(i)
− 12 jθ
3
2 2
1
+ e2
+
3
Accept degrees and decimal
approxs
M1
j
2 2
Deduct 1 mark if answers given
in form r (cos θ + jsin θ ) but
modulus and argument not
stated.
5
= 32 (cos 14 π + j sin 14 π )
=
e
3
2
w
1
, arg = (− 12
π ) − (− 13 π ) = 14 π
z
Jan 2007
Accept 1.125 + 1.125 j
A1
2
jθ
= (cos 12 θ − jsin 12 θ ) + (cos 12 θ + jsin 12 θ )
= 2 cos
1
jθ
1 + e jθ = e 2 ( e
=e
1
2
jθ
M1
1
θ
2
For either bracketed expression
A1
− 1 jθ
2
(2 cos
1
jθ
+ e2 )
M1
1
θ)
2
A1 ag
4
OR 1 + e jθ = 1 + cos θ + j sin θ
= 2 cos 2 12 θ + 2 j sin 12 θ cos 12 θ
M1
= 2 cos 12 θ (cos 12 θ + j sin 12 θ )
1
= 2e 2
(ii)
jθ
cos 12 θ
A1
⎛ n⎞
⎛n⎞
⎛ n⎞
C + jS = 1 + ⎜⎜ ⎟⎟ e jθ + ⎜⎜ ⎟⎟ e 2 jθ + ... + ⎜⎜ ⎟⎟ e njθ
⎝ n⎠
⎝ 2⎠
⎝1⎠
M1
= (1 + e jθ ) n
n
1
nθ
2
M1A1
j
=2 e
cos n 12 θ
C = 2 n cos( 12 nθ ) cos n 12 θ
S = 2 n sin( 12 nθ ) cos n 12 θ
M1
A1
n
n 1
1
sin( 12 nθ )
S 2 sin( 2 nθ ) cos 2 θ
= n
=
= tan( 12 nθ )
C 2 cos( 1 nθ ) cos n 1 θ cos( 12 nθ )
2
2
33
Using (i) to obtain a form from
which the real and imaginary
parts can be written down
A1
B1 ag
1
Allow ft from C + jS = e 2
7 real function of n and θ
nθ j
× any
4756
3 (i)
Mark Scheme
det P = 1(6 − k ) − 1(4 − 2)
=4−k
2
⎛−1
1 ⎜
−1
P =
⎜ 4 −4−k
4−k ⎜
2
⎝−1
M1
A1
6−k ⎞
⎟
k − 12 ⎟
2 ⎟⎠
4 ⎞
⎛−1 2
⎟
1⎜
−1
When k = 2 , P = ⎜ 4 − 6 − 10 ⎟
2⎜
2 ⎟⎠
⎝−1 2
(ii)
Jan 2007
M1
M1
A1 ft
Evaluating at least three
cofactors
Fully correct method for inverse
Ft from wrong determinant
B1 ag
6 Correctly obtained
⎛ 4⎞ ⎛ 0⎞
⎛ 4⎞
⎛ 2⎞ ⎛ 2⎞
⎛ 2⎞
⎜ ⎟ ⎜ ⎟
⎜ ⎟
⎜ ⎟ ⎜ ⎟
⎜ ⎟
M ⎜ 1 ⎟ = ⎜ 0 ⎟ = 0 ⎜ 1 ⎟ M ⎜ 1⎟ = ⎜ 1⎟ = 1 ⎜ 1⎟
⎜ 1⎟ ⎜ 0⎟
⎜1⎟
⎜ 0⎟ ⎜ 0⎟
⎜ 0⎟
⎝ ⎠ ⎝ ⎠
⎝ ⎠
⎝ ⎠ ⎝ ⎠
⎝ ⎠
2
4
2
⎛ ⎞ ⎛ ⎞
⎛ ⎞
⎜ ⎟ ⎜ ⎟
⎜ ⎟
M⎜ 3 ⎟=⎜ 6 ⎟=2⎜ 3 ⎟
⎜ − 1⎟ ⎜ − 2 ⎟
⎜ − 1⎟
⎝ ⎠ ⎝ ⎠
⎝ ⎠
For one evaluation
M1
Eigenvalues are 0, 1, 2
A1A1A1
4
OR
Obtaining an eigenvalue (e.g. by
solving −λ 3 + 3λ 2 − 2λ = 0 )
Give A1 for one correct
Verifying given eigenvectors,
linking with eigenvalues
correctly
M1
Eigenvalues are 0, 1, 2
A2
A1
⎛0 0 0 ⎞
⎛4 2 2 ⎞
⎟
⎜
⎟
⎜
For ⎜ 1 1 3 ⎟ and ⎜ 0 1 0 ⎟
⎜ 0 0 2n ⎟
⎜ 1 0 − 1⎟
⎠
⎝
⎠
⎝
(iii)
B1B1
4 ⎞
⎛4 2 2 ⎞ ⎛0 0 0 ⎞ ⎛− 1 2
⎟⎜
⎜
⎟
⎟ 1⎜
M n = ⎜ 1 1 3 ⎟ ⎜ 0 1 0 ⎟ ⎜ 4 − 6 − 10 ⎟
2
⎜ 1 0 − 1⎟ ⎜ 0 0 2 n ⎟ ⎜ − 1 2
2 ⎟⎠
⎠⎝
⎝
⎠ ⎝
⎛ 0 2 2 n +1 ⎞ ⎛ − 1 2
4 ⎞
⎟
⎟
1⎜
n ⎜
= ⎜ 0 1 3 × 2 ⎟ ⎜ 4 − 6 − 10 ⎟
2⎜
⎜ 0 0 − 2 n ⎟⎟ ⎜⎝ − 1 2
2 ⎟⎠
⎝
⎠
⎛ 4−2
−6+2
− 10 + 2 ⎞⎟
⎜
n −1
n
= ⎜2 − 3× 2
− 3 + 3 × 2 − 5 + 3 × 2n ⎟
⎜⎜
⎟⎟
n −1
2
− 2n
− 2n
⎝
⎠
4
6
10
2
4
4
−
−
−
⎞
⎞
⎛
⎛
⎟
⎟
⎜
⎜
6 ⎟
= ⎜ 2 − 3 − 5 ⎟ + 2 n −1 ⎜ − 3 6
⎜ 1 − 2 − 2⎟
⎜0 0
0 ⎟⎠
⎠
⎝
⎝
n
n +1
seen
(for B2, these must be
consistent)
For S Dn S −1 (M1A0 if order
wrong)
M1A1
B1 ft
⎛4 2 2 ⎞ ⎛ 0
⎟⎜
1⎜
or ⎜ 1 1 3 ⎟ ⎜ 4
2⎜
⎟⎜ n
⎝ 1 0 − 1⎠ ⎝ − 2
n +1
34
M1
A1
0
−6
2 n +1
0 ⎞
⎟
− 10 ⎟
2 n +1 ⎟⎠
Evaluating product of 3 matrices
Any correct form
A1 ag
8
4756
Mark Scheme
OR Prove M n = A + 2n−1 B by induction
When n = 1, A + B = M
k
Assuming M = A + 2
M
k +1
= AM + 2
k −1
k −1
Jan 2007
B1
B,
BM
M1A2
= A + 2k −1 (2 B)
A1A1
k
= A+2 B
True for n = k ⇒ True for n = k + 1 ;
hence
true for all positive integers n
or M k +1 = M A + 2k −1 M B
A1
A1
35
Dependent on previous 7 marks
4756
4 (i)
Mark Scheme
M1
If y = arcosh x , x = cosh y = 12 (e y + e − y )
e
2y
y
− 2x e + 1 = 0
ey =
Jan 2007
1
2
and + must be correct
M1
2
2x ± 4x − 4
2
M1
= x ± x2 − 1
A1
Since y ≥ 0 , e y ≥ 1, so e y = x + x 2 − 1
arcosh x = y = ln( x + x 2 − 1 )
A1 ag
5
(ii)
3.9
1
For arcosh (or any cosh
substitution)
M1
3.9
⎡ 1
⎛ 2x ⎞ ⎤
dx = ⎢ arcosh⎜ ⎟ ⎥
2
2
⎝ 3 ⎠ ⎦ 2.5
⎣
4x − 9
⌠
⎮
⌡2.5
A1A1
1
2x
and
2
3
(or 2 x = 3cosh u and
For
5
3
1
2
= ( arcosh 2.6 − arcosh )
= 12 ⎛⎜ ln(2.6 + 2.6 2 − 1) − ln( 53 +
⎝
25
9
− 1) ⎞⎟
⎠
M1
1
2
1
2
(or limits of u in logarithmic
form)
= (ln 5 − ln 3)
= ln 53
∫ 12 du )
A1
5
OR
M2
For ln(kx + k 2 x 2 − ... )
Give M1 for ln(k1 x + k2 2 x 2 − ...)
⎡
⎢⎣
1
2
3.9
ln(2 x + 4 x 2 − 9 ) ⎤
⎥⎦ 2.5
For
A1A1
= 12 ln 53
dx
=
=
A1
(2 + sinh x) sinh x − (cosh x)(cosh x )
(2 + sinh x)
2 sinh x − 1
and ln(2 x + 4 x 2 − 9 )
(or ln( x + x 2 − 94 )
= 12 ln 15 − 12 ln 9
(iii) dy
1
2
2
M1
A1
Using quotient rule
Any correct form
M1
Quadratic in sinh x (or product of
two quadratics in e x )
M1
Solving quadratic to obtain at
least one value of sinh x (or e x )
M1
Obtaining x in logarithmic form
(must use a correct formula for
arsinh)
(2 + sinh x) 2
dy 1
= when 18 sinh x − 9 = (2 + sinh x) 2
dx 9
2
sinh x − 14 sinh x + 13 = 0
sinh x = 1, 13
When sinh x = 1, cosh x = 2 , x = ln(1 + 2 )
⎛
Point is ⎜ ln(1 + 2 ) ,
⎜
⎝
2
3
⎞
⎟
⎟
⎠
A1 ag
When
sinh x = 13 , cosh x = 170 , x = ln(13 + 170 )
⎛
Point is ⎜⎜ ln(13 + 170 ) ,
⎝
170
15
SR B1B1 for verifying y = 13 2
⎞
⎟
⎟
⎠
and
A1A1
8
36
dy 1
= when x = ln(1 + 2 )
dx 9
4756
Mark Scheme
Jan 2007
Alternatives for Q4 (i)
cosh ln( x + x 2 − 1) = 12 (eln( x +
x 2 −1)
+ e− ln( x +
= 12 ( x + x 2 − 1 +
x 2 −1)
1
)
M1
)
M1
= 12 ( x + x 2 − 1 + x − x 2 − 1 )
M1
x + x2 − 1
=x
Since ln( x + x 2 − 1) > 0, arcosh x = ln( x + x 2 − 1)
A1
A1
5
If y = arcosh x then
ln( x + x 2 − 1) = ln(cosh y + cosh 2 y − 1)
M1
= ln(cosh y + sinh y )
M1
A1
since
sinh y > 0
M1
A1
= ln(e y )
=y
5
37
4756
Mark Scheme
Jan 2007
5 (i)
k =1
B1
General shape correct
B1
Cusp at O clearly shown
B1
General shape correct
B1
‘Dimple’ correctly shown
k = 1.5
k=4
B1
5
(ii) Cusp
B1
1
(iii) When k = 1 , there are 3 points
When k = 1.5 , there are 4 points
When k = 4 , there are 2 points
Give B1 for two cases correct
B2
2
(iv)
B1
x = k cos θ + cos 2 θ
dx
= − k sin θ − 2 cos θ sin θ
dθ
= − sin θ (k + 2 cos θ )
B1
= 0 when θ = 0 , π , or cos θ = − 12 k
M1
For just two points, k ≥ 2
Allow k > 2
A1
4
(v)
M1
d 2 = r 2 + 12 − 2r cos θ
= (k + cos θ ) 2 + 1 − 2( k + cos θ ) cos θ
2
2
= k + 1 − cos θ
2
A1
2
( = k + sin θ )
2
Since 0 ≤ cos θ ≤ 1 ,
2
2
or 0 ≤ sin 2 θ ≤ 1
M1
A1 ag
2
k ≤ d ≤ k +1
4
M1
(vi) When k is large, k 2 + 1 ≈ k , so d ≈ k
Curve is very nearly a circle,
with centre (1, 0) and radius k
A1
2
38
4756
Mark Scheme
June 2007
1(a)(i)
B2
Must include a sharp point at O and
2 have infinite gradient at θ = π
Give B1 for
r increasing from zero for
0 < θ < π , or decreasing to zero for
−π < θ < 0
(ii)
M1
A1
For integral of (1 − cosθ ) 2
For a correct integral expression
including limits (may be implied by
later work)
B1
⎡
⎤
= 12 a 2 ⎢ 32 θ − 2sin θ + 14 sin 2θ ) ⎥
⎦0
⎣
Using cos 2 θ = 12 (1 + cos 2θ )
B1B1 ft
Integrating a + b cos θ and k cos 2θ
= 12 a 2 ( 43 π − 2)
B1
Accept 0.178a 2
1
∫
Area is
1
2
⌠2
r dθ = ⎮
⌡0
2
1π
⌠2
= 12 a 2 ⎮⎮
⌡0
π
1
2
2
2
a (1 − cos θ ) dθ
( 1 − 2 cosθ + 12 (1 + cos 2θ ) ) dθ
1π
2
6
(b)
Put x = 2 sin θ
1
π
6
⌠
Integral is ⎮
1
3
⌡0 (4 − 4 sin 2 θ ) 2
1
6
=⌠
⎮
⌡0
⎡
=⎢
⎣
=
(c)(i)
(ii)
f ′ ( x) =
π
⌠6
dθ = ⎮
3
8 cos θ
⌡0
1
1
4
1
2 cos θ
⎤6
tan θ ⎥
⎦0
(2 cos θ ) dθ
π
1
4
M1
or x = 2 cos θ
A1
Limits not required
M1
For ∫ sec 2 θ dθ = tan θ
sec 2 θ dθ
π
1 1
1
×
=
4
3 4 3
SR If x = 2 tanh u is used
M1 for 14 sinh( 12 ln 3)
A1 ag
4 A1 for 1 ( 3 −
8
1
)
3
=
1
4 3
(max 2 / 4)
−2
B2
1 − 4x2
2
f ′ ( x) = −2(1 − 4 x 2 )
−1
2
= −2(1 + 2 x 2 + 6 x 4 + ...)
f ( x) = C − 2 x −
Give B1 for any non-zero real
4
3
x 3 − 125 x 5 + ...
f ( x) = 12 π − 2 x −
4
3
x 3 − 125 x 5 + ...
Binomial expansion (3 terms, n = − 12 )
A1
Expansion of (1 − 4 x 2 ) 2 correct
(accept unsimplified form)
Integrating series for f ′ ( x)
A1
27
−2
etc)
sin y
M1
M1
f (0) = 12 π ⇒ C = 12 π
multiple of this (or for
−1
Must obtain a non-zero x5 term
4 C not required
4756
Mark Scheme
OR by repeated differentiation
Finding f (5) ( x)
Evaluating f (5) (0) ( = −288 )
2
4
f ′ ( x) = −2 − 4 x − 12 x + ...
f ( x ) = 12 π − 2 x −
4
3
June 2007
M1
M1
A1 ft
x 3 − 125 x 5 + ...
A1
28
Must obtain a non-zero value
ft from (c)(i) when B1 given
4756
2 (a)
Mark Scheme
June 2007
(cos θ + j sin θ ) 5
= c 5 + 5 jc 4 s − 10c 3 s 2 − 10 jc 2 s 3 + 5cs 4 + j s 5
Equating imaginary parts
sin 5θ = 5c 4 s − 10c 2 s 3 + s 5
= 5(1 − s 2 ) 2 s − 10(1 − s 2 ) s 3 + s 5
M1
M1
A1
M1
= 5s − 10 s 3 + 5s 5 − 10 s 3 + 10 s 5 + s 5
= 5 sin θ − 20 sin 3 θ + 16 sin 5 θ
A1 ag
5
(b)(i)
− 2 + 2j = 8 ,
arg(−2 + 2 j) = 34 π
r= 2
θ = 14 π
B1B1
Accept 2.8 ; 2.4 , 135°
B1 ft
(Implies B1 for 8 )
One correct (Implies B1 for
B1 ft
Adding or subtracting
M1
A1
11
θ = 12
π , − 125 π
6
3
π
4
)
2
π
3
Accept θ = 14 π + 32 kπ , k = 0, 1, − 1
(ii)
B2
Give B1 for two of B, C, M in the
2 correct quadrants
Give B1 ft for all four points in the
correct quadrants
(iii)
w =
1
2
Accept 0.71
B1 ft
2
arg w = 12 ( 14 π +
11
π
12
) = 127 π
Accept 1.8
B1
2
(iv)
w6 = (
1
2
2 )6 =
M1
1
8
arg( w 6 ) = 6 × 127 π = 72 π
6
1
8
w = ( cos
7
π
2
+ j sin
7
π
2
Obtaining either modulus or
argument
Both correct (ft)
A1 ft
)
= − 18 j
A1
3
Allow from arg w = 14 π etc
SR If B, C interchanged on
diagram
(ii) B1
(iii) B1 B1 for − 121 π
(iv) M1A1A1
29
4756
3 (i)
Mark Scheme
det(M − λ I ) = (3 − λ )[(3 − λ )(−4 − λ ) − 4]
− 5 [5(−4 − λ ) + 4] + 2 [−10 − 2(3 − λ )]
June 2007
M1
A1
Obtaining det(M − λ I)
Any correct form
M1
A1 ag
Simplification
= (3 − λ )(−16 + λ + λ2 ) − 5(−16 − 5λ ) + 2(−16 + 2λ )
= −48 + 19λ + 2λ2 − λ3 + 80 + 25λ − 32 + 4λ
= 48λ + 2λ2 − λ3
Characteristic equation is λ3 − 2λ2 − 48λ = 0
4
(ii)
λ (λ − 8)(λ + 6) = 0
M1
A1
Solving to obtain a non-zero value
M1
Two independent equations
M1
A1
Obtaining a non-zero eigenvector
5 x + 3 y − 2 z = −6 y
M1
Two independent equations
⎛ 1 ⎞
⎜
⎟
y = − x , z = −2 x ; eigenvector is ⎜ − 1 ⎟
⎜ − 2⎟
⎝
⎠
M1
A1
Obtaining a non-zero eigenvector
Other eigenvalues are 8 , − 6
When λ = 8 , 3x + 5 y + 2 z = 8 x
( 5x + 3 y − 2 z = 8 y )
2 x − 2 y − 4 z = 8z
⎛1⎞
⎜ ⎟
y = x and z = 0 ; eigenvector is ⎜ 1 ⎟
⎜0⎟
⎝ ⎠
( −5 x + 5 y + 2 z = 8 x etc can earn
M0M1 )
When λ = −6 , 3x + 5 y + 2 z = −6 x
(iii)
8
⎛ 1 1 1 ⎞
⎟
⎜
P = ⎜ − 1 1 − 1⎟
⎜ 1 0 − 2⎟
⎠
⎝
B0 if P is clearly singular
B1 ft
2
⎛0 0 0 ⎞
⎟
⎜
D = ⎜0 8 0 ⎟
⎜ 0 0 − 6⎟
⎠
⎝
0
0
0
⎞
⎛
⎟
⎜
= ⎜ 0 64 0 ⎟
⎜ 0 0 36 ⎟
⎠
⎝
(iv)
M1
A1
Order must be consistent with P
3 when B1 has been earned
M1
M 3 − 2M 2 − 48M = 0
M 3 = 2M 2 + 48M
M 4 = 2M 3 + 48M 2
= 2(2M 2 + 48M ) + 48M 2
M1
2
= 52M + 96M
A1
3
30
4756
4 (a)
Mark Scheme
1
⌠
⎮
⎮
⎮
⌡0
M1
1
3x ⎤
⎡1
dx = ⎢ arsinh
2
3
4 ⎥⎦ 0
⎣
9 x + 16
1
= 13 arsinh 34
= 13 ln( 43 +
9
16
June 2007
+1 )
A1
For arsinh or for any sinh
substitution
For 34 x
or for 3 x = 4 sinh u
A1
For
1
3
or for
∫ 13 du
M1
= 13 ln 2
A1
5
OR
For ln(kx + k 2 x 2 + ... )
M2
[ Give M1 for ln(ax + bx 2 + ... ) ]
1
⎡ 1 ln(3 x + 9 x 2 + 16 ) ⎤
⎢⎣ 3
⎥⎦ 0
or
A1A1
1
ln( x
3
+ x 2 + 169 )
= 13 ln 8 − 13 ln 4
= 13 ln 2
A1
(b)(i) 2 sinh x cosh x = 2 × 1 ( e x − e − x ) 1 ( e x + e − x )
2
2
= 12 ( e 2 x − e − 2 x )
M1
= sinh 2 x
A1
( e x − e − x ) ( e x + e − x ) = ( e 2 x − e −2 x )
For completion
2
(ii) dy
dx
= 20 sinh x − 6 sinh 2 x
B1B1
When exponential form used, give
B1 for any 2 terms correctly
differentiated
M1
Solving
A1
sinh x , cosh x or e x ( or x = 0
For stationary points,
20 sinh x − 12 sinh x cosh x = 0
4 sinh x(5 − 3 cosh x) = 0
sinh x = 0 or cosh x =
x = 0 , y = 17
x = ( ± ) ln(
5
3
⎛
y = 10⎜ 3 +
⎝
1⎞ 3⎛
1 ⎞ 59
⎟ − ⎜9 + ⎟ =
3⎠ 2 ⎝
9⎠ 3
+
25
9
5
3
− 1 ) = ln 3
59
x = − ln 3 , y =
3
dy
= 0 to obtain a value of
dx
A1 ag
stated)
Correctly obtained
A1 ag
Correctly obtained
The last A1A1 ag can be replaced
by B1B1 ag for a full verification
B1
7
ln 3
(iii)
3
⎡
⎤
⎢ 20 sinh x − 2 sinh 2 x ⎥
⎣
⎦ − ln 3
When exponential form used, give
B1 for any 2 terms correctly
integrated
Exact evaluation of sinh(ln 3) and
B1B1
⎧ ⎛
1⎞ 3⎛
1⎞ ⎫
= ⎨ 10⎜ 3 − ⎟ − ⎜ 9 − ⎟ ⎬ × 2
3⎠ 4 ⎝
9⎠ ⎭
⎩ ⎝
⎛ 80 20 ⎞
=⎜
−
⎟ × 2 = 40
3 ⎠
⎝ 3
M1
sinh( 2 ln 3)
A1 ag
4
31
4756
Mark Scheme
June 2007
5 (i)
B1
k = −2
Maximum on LH branch and
minimum on RH branch
Crossing axes correctly
B1
B1
Two branches with positive
gradient
Crossing axes correctly
B1
k =1
B1
k = −0.5
B1
(ii)
( x + k )( x − 2k ) + 2k 2 + 2k
x+k
2k (k + 1)
= x − 2k +
x+k
Straight line when 2k (k + 1) = 0
y=
k = 0 , k = −1
Maximum on LH branch and
minimum on RH branch
Crossing positive y-axis and
6 minimum in first quadrant
M1
Working in either direction
A1 (ag)
For completion
B1B1
4
(iii)(A) Hyperbola
B1
1
(B) x = − k
B1
B1
y = x − 2k
2
32
4756
Mark Scheme
June 2007
(iv)
33
B1
Asymptotes correctly drawn
B1
Curve approaching asymptotes
correctly (both branches)
B1
Intercept 2 on y-axis, and not
crossing the x-axis
B1
Points A and B marked, with
minimum point between them
B1
Points A and B at the same height
5 ( y = 1)
4756
Mark Scheme
4756 (FP2)
January 2008
Further Methods for Advanced Mathematics
M1
1(a)
For ∫ (1 − cos 2θ ) 2 dθ
π
⌠ 1 2
a (1 − cos 2θ ) 2 dθ
2
⌡0
Area is ⎮
π
(
)
⌠
= ⎮ 12 a 2 1 − 2 cos 2θ + 12 (1 + cos 4θ ) dθ
⌡0
= 12 a 2
[ 32 θ − sin 2θ + 18 sin 4θ ] 0
= πa
2
3
4
Correct integral expression
including limits (may be implied
by later work)
A1
B1
π
B1B1B1
ft
A1
For cos 2 2θ = 12 (1 + cos 4θ )
Integrating a + b cos 2θ + c cos 4θ
[ Max B2 if answer incorrect and
no mark has previously been
7 lost ]
d
1
arctan u =
du
1+ u2
dy
1
=
or
dx sec2 y
(b)(i)
Applying
M1
1
f ′ ( x) =
A1
1 + ( 3 + x) 2
f ′′ ( x) =
− 2( 3 + x )
(1 + (
3 + x) 2
)
2
4
(ii) f (0) = 13 π
f ′ (0) =
1
4
Applying chain (or quotient) rule
M1
A1
, f ′′ (0) = − 18 3
arctan( 3 + x) = 13 π + 14 x − 161 3 x 2 + ...
B1
Stated; or appearing in series
Accept 1.05
M1
Evaluating f ′ (0) or f ′′ (0)
A1A1 ft
For
4
1
4
1
x and − 16
3 x2
ft provided coefficients are
non-zero
(iii) ⌠ h
3
1
1 2
1
⎮ ( 3 π x + 4 x − 16 3 x + ...) dx
⌡− h
⎡
⎤
1 3
1
x − 64
3 x 4 + ... ⎥
= ⎢ 16 π x 2 + 12
⎣
⎦
≈
( 16
2
πh +
1
12
3
h −
1
64
M1
h
A1 ft
−h
4
Integrating (award if x is
missed)
for
3h )
1
12
x3
2
1 3
1
h − 64
− ( 16 π h − 12
3 h4 )
= 16 h 3
A1 ag
2
1
3 Allow ft from a + 4 x + cx
provided that a ≠ 0
Condone a proof which neglects
h4
20
4756
2(a)
Mark Scheme
1π j
are r e jθ where
4th roots of 16 j = 16 e 2
1
r=2
θ = 18 π
2kπ
+
8
4
7
θ = − 8 π , − 83 π ,
θ=
January 2008
π
5
π
8
B1
B1
Accept 16 4
M1
Implied by at least two correct
(ft) further values
or stating k = −2 , − 1 , (0) , 1
A1
Points at vertices of a square
centre O
or 3 correct points (ft)
or 1 point in each quadrant
M1
A1
6
(b)(i)
jθ
(1 − 2 e )(1 − 2 e
− jθ
) = 1− 2 e
jθ
−2e
= 5 − 2(e
jθ
− jθ
+e
+4
− jθ
For e jθ e − jθ = 1
M1
A1
)
= 5 − 4 cos θ
A1 ag
3
OR
(1 − 2 cos θ − 2 j sin θ )(1 − 2 cos θ + 2 j sin θ )
2
M1
2
= (1 − 2 cos θ ) + 4 sin θ
2
A1
2
= 1 − 4 cos θ + 4(cos θ + sin θ )
= 5 − 4 cos θ
A1
(ii) C + jS = 2 e jθ + 4 e 2 jθ + 8 e 3 jθ + ... + 2 n e njθ
=
=
=
(
2 e jθ 1 − ( 2 e jθ ) n
)
1 − 2 e jθ
Obtaining a geometric series
M1
A1
Summing (M0 for sum to
infinity)
2 e jθ (1 − 2 n e njθ )(1 − 2 e − jθ )
(1 − 2 e jθ )(1 − 2 e − jθ )
2e
jθ
n +1 ( n +1) jθ
−4−2 e
+2
5 − 4 cos θ
M1
n+2
e
njθ
A2
2 cos θ − 4 − 2 n +1 cos(n + 1)θ + 2 n + 2 cos nθ
C=
5 − 4 cos θ
S=
M1
Give A1 for two correct terms in
numerator
M1
A1 ag
2 sin θ − 2 n +1 sin( n + 1)θ + 2 n + 2 sin nθ
5 − 4 cos θ
Equating real (or imaginary)
parts
A1
9
21
4756
3 (i)
Mark Scheme
Characteristic equation is
(7 − λ )(−1 − λ ) + 12 = 0
January 2008
M1
2
λ − 6λ + 5 = 0
λ = 1, 5
3 ⎞⎛ x⎞ ⎛ x⎞
⎛ 7
⎟⎟ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟
When λ = 1 , ⎜⎜
⎝ − 4 − 1⎠ ⎝ y ⎠ ⎝ y ⎠
A1A1
7x + 3y = x
−4 x − y = y
⎛ 1 ⎞
y = −2 x , eigenvector is ⎜⎜ ⎟⎟
⎝ − 2⎠
3 ⎞⎛ x⎞
⎛ 7
⎛ x⎞
⎟⎟ ⎜⎜ ⎟⎟ = 5 ⎜⎜ ⎟⎟
When λ = 5 , ⎜⎜
⎝ − 4 − 1⎠ ⎝ y ⎠
⎝ y⎠
3 ⎞ ⎛ x ⎞ ⎛0⎞
⎛ 6
⎟⎜ ⎟ = ⎜ ⎟
−
4
−
2 ⎟⎠ ⎜⎝ y ⎟⎠ ⎜⎝ 0 ⎟⎠
⎝
M1
or ⎜⎜
M1
can be awarded for either
eigenvalue
Equation relating x and y
A1
7 x + 3 y = 5x
−4 x − y = 5 y
or any (non-zero) multiple
M1
⎛ 3 ⎞
y = − 23 x , eigenvector is ⎜⎜ ⎟⎟
⎝ − 2⎠
A1
8
(ii)
3 ⎞
⎛ 1
⎟⎟
P = ⎜⎜
⎝ − 2 − 2⎠
⎛1 0⎞
⎟⎟
D = ⎜⎜
⎝ 0 5⎠
B1 ft
B1 ft
22
SR (M − λ I )x = λ x can earn
M1A1A1M0M1A0M1A0
B0 if P is singular
For B2, the order must be
2 consistent
4756
(iii)
Mark Scheme
n
M = PD P
−1
M1
⎛ 1 0 ⎞ −1
⎟
= P ⎜⎜
n ⎟P
⎝0 5 ⎠
3 ⎞ ⎛ 1 0 ⎞ 1 ⎛ − 2 − 3⎞
⎛ 1
⎟⎟ ⎜⎜
⎟ ⎜
⎟
= ⎜⎜
n⎟ ⎜
1 ⎟⎠
⎝ − 2 − 2⎠ ⎝0 5 ⎠ 4 ⎝ 2
⎛ 1
3 × 5 n ⎞⎟ 1 ⎛ − 2 − 3 ⎞
⎜
⎟
= ⎜⎜
n⎟ ⎜
1 ⎟⎠
⎝− 2 − 2 × 5 ⎠ 4 ⎝ 2
=
1 ⎛⎜ − 2 + 6 × 5 n
4 ⎜⎝ 4 − 4 × 5 n
B1 ft
For P −1
− 3⎞
⎟
5 n ⎟⎠
Obtaining at least one element
in a product of three matrices
M1
A1 ag
b = − 34 + 34 × 5 n
3
2
Dependent on M1M1
3 ⎞1⎛ −2
⎛ 1
⎟⎟ ⎜⎜
n
⎝ − 2 − 2⎠ 4 ⎝ 2 × 5
− 3 + 3 × 5 n ⎞⎟
6 − 2 × 5 n ⎟⎠
d=
A1 ft
or ⎜⎜
a = − 12 + 32 × 5 n
c = 1 − 5n
May be implied
M1
M = P D P −1
n
January 2008
− 12 × 5 n
A2
8
Give A1 for one of b, c, d
correct
SR If M n = P −1 D n P is used,
max
marks are
M0M1A0B1M1A0A1
(d should be correct)
SR If their P is singular, max
marks
are M1M1A1B0M0
23
4756
Mark Scheme
1
2
4 (i)
(e x + e − x ) = k
e 2 x − 2k e x + 1 = 0
ex =
January 2008
M1
or cosh x + sinh x = e x
M1
or k ± k 2 − 1 = e x
A1
One value sufficient
M1
or cosh x is an even function
(or equivalent)
2k ± 4k 2 − 4
= k ± k 2 −1
2
x = ln(k + k 2 − 1) or ln(k − k 2 − 1)
2
2
2
2
(k + k − 1)(k − k − 1) = k − (k − 1) = 1
1
2
ln(k − k − 1) = ln(
2
k + k 2 −1
) = − ln(k + k − 1)
x = ± ln(k + k 2 − 1)
A1 ag
5
For arcosh or
M1
(ii)
ln(λx + λ2 x 2 − ... )
or any cosh substitution
A1
2
⌠
⎮
⌡1
For arcosh 2 x or 2 x = cosh u or
2
⎡ 1
⎤
dx = ⎢ arcosh 2 x ⎥
2
⎣
⎦1
4x −1
1
= 12 ( arcosh 4 − arcosh 2 )
=
1
2
( ln(4 +
15 ) − ln(2 + 3 )
ln(2 x + 4 x 2 − 1) or ln( x + x 2 − 14 )
A1
2
For
M1
)
1
2
or
∫ 12 du
Exact numerical logarithmic
form
A1
5
(iii) 6 sinh x − 2 sinh x cosh x = 0
M1
M1
cosh x = 3 (or sinh x = 0 )
x=0
B1
x = ± ln(3 + 8 )
A1
Obtaining a value for cosh x
or x = ln(3 ± 8 )
4
OR e 4 x − 6 e 3 x + 6 e x − 1 = 0
(e 2 x − 1)(e 2 x − 6 e x + 1) = 0
or (e x − e − x )(e x + e − x − 6) = 0
M2
B1
A1
x=0
x = ln(3 ± 8 )
(iv) dy
= 6 cosh x − 2 cosh 2 x
dx
dy
= 5 then 6 cosh x − 2(2 cosh 2 x − 1) = 5
If
dx
B1
Using cosh 2 x = 2 cosh 2 x − 1
M1
4 cosh 2 x − 6 cosh x + 3 = 0
Discriminant D = 6 2 − 4 × 4 × 3 = −12
Considering D, or completing
square, or considering turning
point
M1
Since D < 0 there are no solutions
A1
4
24
4756
Mark Scheme
OR Gradient g = 6 cosh x − 2 cosh 2 x
January 2008
B1
g ′ = 6 sinh x − 4 sinh 2 x = 2 sinh x(3 − 4 cosh x)
= 0 when x = 0 (only)
M1
g ′′ = 6 cosh x − 8 cosh 2 x = −2 when x = 0 M1
Max value g = 4 when x = 0
So g is never equal to 5
A1
25
Final A1 requires a complete
proof showing this is the only
turning point
4756
5 (i)
Mark Scheme
λ = −1
λ =0
January 2008
λ =1
B1B1B1
cusp
(ii)
B1B1
loop
x =1
Two different features (cusp,
5 loop, asymptote) correctly
identified
B1
1
(iii) Intersects itself when y = 0
(iv)
M1
t = (±) λ
A1
⎞
⎛ λ
, 0⎟
⎜
1
+
λ
⎝
⎠
A1
3
dy
= 3t 2 − λ = 0
dt
λ
t=±
x=
M1
3
λ
3
1+ λ
=
3
λ
3+ λ
A1 ag
λ
⎛ λ
y = ± ⎜ ( ) −λ ( )
3
⎝ 3
3
2
⎛ 1
1
= ± λ ⎜⎜
−
3
⎝ 3 3
3
2
=±
1
2
⎞
⎟
⎠
3
⎞
⎟ = ± λ2
⎟
⎠
⎛
2
⎜ −
⎜
⎝ 3 3
One value sufficient
M1
⎞
⎟
⎟
⎠
4λ 3
27
A1 ag
4
(v)
From asymptote, a = 8
From intersection point,
λ=
B1
aλ
=2
1+ λ
M1
1
3
From maximum point, b
A1
4λ 3
=2
27
M1
b = 27
A1
5
26
4756
Mark Scheme
June 2008
4756 (FP2) Further Methods for Advanced Mathematics
1(a)(i) x = r cos θ , y = r sin θ
2
2
2
(M0 for x = cos θ , y = sin θ )
M1
2
2
( r cos θ + r sin θ ) = 3(r cos θ )(r sin θ )
4
3
2
A1
2
r = 3r cos θ sin θ
r = 3 cos θ sin 2 θ
A1 ag
3
(ii)
B1
Loop in 1st quadrant
B1
Loop in 4th quadrant
B1
(b)
1
⎡ 1
3x ⎤
⌠
1
dx = ⎢
arcsin
⎥
⎮
2
2 ⎥⎦
⌡0 4 − 3 x
⎣⎢ 3
=
=
1
3
arcsin
1
0
3
2
π
M1
For arcsin
A1A1
For
M1
Exact numerical value
Dependent on first M1
(M1A0 for 60 / 3 )
A1
3 3
3 x = 2 sin θ
=
and
3x
2
Any sine substitution
A1
For ⎮
π
1
3
3
1
4
⌠ 1
dθ
⌡ 3
M1A1
3 3
4
1
5
M1 dependent on first M1
B1
(c)(i) ln(1 + x) = x − 1 x 2 + 1 x 3 − 1 x 4 + 1 x 5 − ...
2
3
4
5
2
3
M1
A1
π
1
⌠
1
⌠3 1
dx = ⎮
dθ
⎮
2
⌡0 3
⌡0 4 − 3 x
1
2
1
5
OR
Put
Fully correct curve
3 Curve may be drawn using
continuous or broken lines in
any combination
Accept unsimplified forms
5
ln(1 − x) = − x − x − x − x − x − ...
B1
2
(ii)
⎛1+ x ⎞
ln⎜
⎟ = ln(1 + x) − ln(1 − x)
⎝ 1− x ⎠
M1
= 2 x + 23 x 3 + 52 x 5 + ...
A1
25
Obtained from two correct
2 series
Terms need not be added
If M0, then B1 for 2x + 23 x3 + 52 x5
4756
(iii)
Mark Scheme
∞
1
r =0
(2r + 1) 4 r
∑
= 1+
1
1
+
+ ...
3× 4 5× 42
June 2008
B1
Terms need not be added
= 2 × 12 + 23 × ( 12 ) 3 + 52 × ( 12 ) 5 + ...
B1
For x =
⎛1+ 2 ⎞
⎟ = ln 3
= ln⎜⎜
1 ⎟
⎝ 1− 2 ⎠
B1 ag
1
2
seen or implied
1
2 (i)
Satisfactory completion
3
z = 8 , arg z = 14 π
B1B1
Must be given separately
Remainder may be given in
exponential or r cjs θ form
z * = 8 , arg z* = − 14 π
B1 ft
(B0 for
z w = 8 × 8 = 64
B1 ft
arg( z w) = 14 π + 127 π = 56 π
)
B1 ft
z
8
= =1
8
w
(B0 if left as 8 / 8 )
B1 ft
z
7
) = 14 π − 12
π = − 13 π
w
arg(
7
π
4
B1 ft
7
(ii)
z
= cos(− 13 π ) + j sin(− 13 π )
w
=
a=
(iii)
M1
If M0, then B1B1 for
1
3
−
j
2 2
1
2
A1
, b=−
1
2
3
B1 ft
r=38 =2
θ=
1
π
12
2kπ
θ= +
12
3
θ = − 127 π ,
w* = 8 e
M1
3
4
−7π j
12
−1π j
4
3π j
So 2 e 4
π
A1
, so 2 e
−7π j
12
=
1
4
w*
8
= −8 e
Matching w* to a cube root with
argument − 127 π and k1 = 14 or ft
3π j
= −8 e 4
= − 14 z *
( 1π + 7 π ) j
2
Implied by one further correct
(ft) value
Ignore values outside the
4 required range
ft is
k 2 = − 14
jw = 8e
3
Accept
B1 ft
1
4
k1 =
z* = 8 e
3
B1
π
(iv)
1
2 2 and − 2
12
1π j
12
= 8e
M1
Matching z* to a cube root with
argument 34 π May be implied
A1 ft
ft is −
M1
Matching jw to a cube root with
argument 121 π May be implied
, so 2 e
r
z*
OR M1 for arg( jw) = 12 π + arg w
13
πj
12
1π j
12
r
8
(implied by
= − 14 j w
ft is −
A1 ft
k 3 = − 14
5
26
r
8
13
π
12
11
or − 12
π)
4756
Mark Scheme
M1
3 (i)
A1
M1
−1 ⎞
⎛ −1 k + 2
1 ⎜
⎟
−
− 2⎟
1
4
3
k
k
k − 3 ⎜⎜
−5
1 ⎟⎠
⎝1
⎛ −1 6 −1⎞
When k = 4, Q −1 = ⎜⎜ 1 −8 2 ⎟⎟
⎜ 1 −5 1 ⎟
⎝
⎠
Q −1 =
(ii)
A1
M1
A1
⎛ 2 −1 4 ⎞
⎛1 0 0⎞
⎜
⎟
⎜
⎟
P = ⎜ 1 0 1 ⎟ , D = ⎜ 0 −1 0 ⎟
⎜ 3 1 2⎟
⎜ 0 0 3⎟
⎝
⎠
⎝
⎠
June 2008
Evaluation of determinant
(must involve k)
For (k − 3)
Finding at least four cofactors
(including one involving k)
Six signed cofactors correct
(including one involving k)
Transposing and dividing by det
Dependent on previous M1M1
Q −1 correct (in terms of k) and
6 result for k = 4 stated
After 0, SC1 for Q −1 when k = 4
obtained correctly with some
working
B1B1
M = P D P −1
For B2, order must be
consistent
B2
⎛ 2 −1 4 ⎞ ⎛ 1 0 0 ⎞ ⎛ −1 6 −1⎞
⎜
⎟⎜
⎟⎜
⎟
= ⎜ 1 0 1 ⎟ ⎜ 0 −1 0 ⎟ ⎜ 1 −8 2 ⎟
⎜ 3 1 2 ⎟ ⎜ 0 0 3 ⎟ ⎜ 1 −5 1 ⎟
⎝
⎠⎝
⎠⎝
⎠
⎛ 2 1 12 ⎞ ⎛ −1 6 −1⎞
⎜
⎟⎜
⎟
= ⎜ 1 0 3 ⎟ ⎜ 1 −8 2 ⎟
⎜ 3 −1 6 ⎟ ⎜ 1 −5 1 ⎟
⎝
⎠⎝
⎠
Give B1 for M = P −1 D P
⎛ 2 −1 4 ⎞ ⎛ −1 6 −1 ⎞
⎟⎜
⎟
0 1 ⎟ ⎜ −1 8 −2 ⎟
⎜ 3 1 2 ⎟ ⎜ 3 −15 3 ⎟
⎝
⎠⎝
⎠
or ⎜⎜ 1
M1
⎛ 11 −56 12 ⎞
⎜
⎟
= ⎜ 2 −9 2 ⎟
⎜ 2 −4 1 ⎟
⎝
⎠
Good attempt at multiplying two
matrices (no more than 3
errors), leaving third matrix in
correct position
A2
7
(iii) Characteristic equation is
(λ − 1)(λ + 1)(λ − 3) = 0
λ 3 − 3λ 2 − λ + 3 = 0
B1
In any correct form
(Condone omission of =0 )
M1
A1
M satisfies the characteristic
equation
Correct expanded form
(Condone omission of I )
M3 = 3 M 2 + M − 3 I
M 4 = 3 M3 + M 2 − 3 M
2
M1
2
= 3(3 M + M − 3 I ) + M − 3 M
A1
= 10 M 2 − 9 I
5
a = 10, b = 0, c = −9
27
Give A1 for five elements
correct
Correct M implies B2M1A2
5-8 elements correct implies
B2M1A1
4756
Mark Scheme
4 (i) cosh 2 x = [ 1 (e x + e− x ) ]2 = 1 (e 2 x + 2 + e−2 x )
2
4
B1
sinh 2 x = [ 12 (e x − e− x ) ]2 = 14 (e2 x − 2 + e−2 x )
B1
cosh 2 x − sinh 2 x = 14 (2 + 2) = 1
B1 ag
June 2008
3 For completion
OR
cosh x + sinh x = 12 (e x + e − x ) + 12 (e x − e − x ) = e x
B1
cosh x − sinh x = 12 (e x + e− x ) − 12 (e x − e − x ) = e− x B1
cosh 2 x − sinh 2 x = e x × e− x = 1
B1
(ii) 4(1 + sinh 2 x) + 9sinh x = 13
4sinh 2 x + 9sinh x − 9 = 0
sinh x =
3
4
, −3
x = ln 2, ln(−3 + 10 )
Completion
M1
(M0 for 1 − sinh 2 x )
M1
A1A1
Obtaining a value for sinh x
A1A1 ft
Exact logarithmic form Dep on
6 M1M1
Max A1 if any extra values given
OR 2 e4 x + 9 e3 x − 22 e2 x − 9 e x + 2 = 0
(2 e 2 x − 3e x − 2)(e 2 x + 6 e x − 1) = 0
e x = 2, − 3 + 10
Quadratic and / or linear factors
Obtaining a value for e x
Ignore extra values
A1A1 ft
x = ln 2, ln(−3 + 10 )
(iii)
M1
M1
A1A1
Dependent on M1M1
Max A1 if any extra values given
Just x = ln 2 earns
M0M1A1A0A0A0
B1
dy
= 8cosh x sinh x + 9 cosh x
dx
= cosh x(8sinh x + 9)
= 0 only when sinh x = − 98
B1
cosh 2 x = 1 + ( − 98 ) 2 = 145
64
M1
145
9
17
y = 4×
+ 9× ( − ) = −
64
8
16
A1
Any correct form
or y = (2sinh x + 94 )2 + ... ( − 17
)
16
Correctly showing there is only
one solution
Exact evaluation of y or cosh 2 x
or cosh 2x
Give B2 (replacing M1A1) for
4 −1.06 or better
ln 2
(iv)
⌠
⎮ (2 + 2 cosh 2 x + 9sinh x) dx
⌡0
= [ 2 x + sinh 2 x + 9 cosh x
⎧
1⎛
1
= ⎨ 2 ln 2 + ⎜ 4 −
2
4
⎝
⎩
33
= 2 ln 2 +
8
]0
ln 2
1 ⎞⎫
⎞ 9⎛
⎟+ 2 ⎜ 2+ 2 ⎟ ⎬−9
⎠
⎝
⎠⎭
M1
Expressing in integrable form
A2
Give A1 for two terms correct
M1
sinh(2 ln 2) = 12 (4 − 14 )
A1 ag
28
Must see both terms for M1
Must also see cosh(ln 2) = 12 (2 + 12 )
5 for A1
4756
Mark Scheme
June 2008
ln 2
⌠
(e2 x + 2 + e −2 x + 92 (e x − e− x )) dx
⌡0
OR ⎮
=⎡
⎣
1 2x
e
2
ln 2
+ 2 x − 12 e−2 x + 92 e x + 92 e− x ⎤
⎦0
1
9
⎛
= ⎜ 2 + 2 ln 2 − + 9 +
8
4
⎝
33
= 2 ln 2 +
8
M1
Expanded exponential form
(M0 if the 2 is omitted)
A2
Give A1 for three terms correct
⎞ ⎛ 1 1 9 9 ⎞
⎟ − ⎜ 2 − 2 + 2 + 2 ⎟ M1
⎠ ⎝
⎠
e2ln 2 = 4 and e−2ln 2 =
1
4
Must also see
A1 ag
eln 2 = 2 and e− ln 2 =
1
2
for A1
λ = 0.5
5 (i)
λ =3
λ =5
B1B1B1
3
(ii) Ellipse
B1
1
M1
(iii) y = 2 cos(θ − 1 π )
4
Maximum y = 2 when θ =
1
π
4
or
2 sin(θ + 14 π )
A1 ag
2
OR
dy
= − sin θ + cos θ = 0 when θ = 14 π
dθ
y=
(iv)
1
2
1
2
+
M1
A1
= 2
x 2 + y 2 = λ 2 cos 2 θ − 2 cos θ sin θ +
1
sin 2 θ
λ2
+ cos 2 θ + 2 cos θ sin θ + sin 2 θ
2
1
2
= (λ + 1)(1 − sin θ ) + (
= 1+ λ2 + (
1
λ2
λ2
M1
2
+ 1 ) sin θ
− λ 2 ) sin 2 θ
A1 ag
When sin 2 θ = 0, x 2 + y 2 = 1 + λ 2
M1
1
When sin 2 θ = 1, x 2 + y 2 = 1 +
λ2
Since 0 ≤ sin θ ≤ 1 , distance from O,
M1
2
x 2 + y 2 , is between
1+
1
λ2
Using cos 2 θ = 1 − sin 2 θ
M1
and
1+ λ2
A1 ag
6
M1
(v) When λ = 1, x 2 + y 2 = 2
Curve is a circle (centre O) with radius
2
A1
2
29
both seen
4756
Mark Scheme
June 2008
(vi)
B4
A, E at maximum distance
4 from O
C, G at minimum distance
from O
B, F are stationary points
D, H are on the x-axis
Give ½ mark for each point,
then round down
Special properties must be clear
from diagram, or stated
Max 3 if curve is not the correct
shape
30
4756
Mark Scheme
January 2009
4756 (FP2) Further Methods for Advanced
Mathematics
1 f (x) = cos x
(a)(i)
f ′(x) = −sin x
f ″(x) = −cos x
f ″′(x) = sin x
f ″″(x) = cos x
f (0) = 1
f ′(0) = 0
f ″(0) = −1
f ″′(0) = 0
f ″″(0) = 1
1
1
cos x = 1 − x 2 + x 4 ...
2
24
⇒
cos x × sec x = 1
(ii)
⎛ 1 2 1 4⎞
x ⎟ (1 + ax 2 + bx 4 ) = 1
⎜1 − x +
24 ⎠
⎝ 2
1⎞
1
1 ⎞
⎛
⎛
1 + ⎜ a − ⎟ x2 + ⎜ b − a + ⎟ x4 = 1
2
2
24
⎝
⎠
⎝
⎠
1
1
1
a− = 0, b− a+
=0
2
2
24
1
a=
2
5
b=
24
⇒
⇒
⇒
⇒
M1
Derivatives cos, sin, cos, sin, cos
A1
Correct signs
A1
A1 (ag)
4
E1
Correct values. Dep on previous A1
www
M1
Multiply to obtain terms in x2 and x4
A1
Terms correct in any form
(may not be collected)
B1
Correctly obtained by any method:
must not just be stated
B1
Correctly obtained by any method
o.e.
5
y = arctan
(b)(i)
x
a
⇒ x = a tan y
M1
(a) tan y = and attempt to
differentiate both sides
⇒
dx
= a sec2y
dy
A1
Or sec2y
⇒
dx
= a(1 + tan2y)
dy
A1
Use sec2y = 1 + tan2y o.e.
⇒
dy
a
=
dx a 2 + x 2
A1 (ag)
www
dy 1
=
dx a
SC1: Use derivative of arctan x and
Chain Rule (properly shown)
4
2
2
(ii)(A)
1
x⎤
⎡1
∫−2 4 + x 2 dx = ⎢⎣ 2 arctan 2 ⎥⎦ −2
=
π
4
M1
arctan alone, or any tan substitution
A1
1
x
and , or
2
2
A1
Evaluated in terms of π
1
∫ 2 dθ
without limits
3
1
2
(ii)(B)
4
∫ 1 + 4x
dx =
1
2
∫
1
dx
+ x2
M1
arctan alone, or any tan substitution
= ⎡⎣ 2 arctan ( 2x ) ⎤⎦ 2 1
A1
2 and 2x, or ∫ 2dθ without limits
=π
A1
−
1
2
2
1
1 4
−
2
1
−
2
Evaluated in terms of π
3
24
19
4756
Mark Scheme
2 (i) Modulus = 1
(ii)
B
B1
B1
π
Argument =
January 2009
3
Must be separate
Accept 60°, 1.05c
2
G2: A in first quadrant, argument ≈
2
A
2
B
G2,1,0
jπ
4
a = 2e
arg b =
b = 2e
−
B1
π
±
4
jπ
12
π
M1
3
7 jπ
A1ft
, 2 e 12
4
B in second quadrant, same mod
B′ in fourth quadrant, same mod
Symmetry
G1: 3 points and at least 2 of above, or
B, B′ on axes, or BOB′ straight
line, or BOB′ reflex
Must be in required form
(accept r = 2, θ = π/4)
Rotate by adding (or subtracting) π/3 to
(or from) argument. Must be π/3
Both. Ft value of r for a. Must be in
required form, but don’t penalise twice
1
-1
π
5
jπ
3
⎛
⎞
6
(iii) z16 = ⎜ 2e ⎟ =
⎝
⎠
( 2)
6
e 2 jπ
( 2)
M1
=8
6
= 8 or
π
3
× 6 = 2π seen
A1 (ag) www
and θ = −
π
2
M1
“Add”
2π
π
2π
, − , 0,
,π
3
3
3
A1
Correct constant r and five values of θ.
Accept θ in [0, 2π] or in degrees
Others are re jθ where r =
3
to argument more than once
(iv)
1
G1
G1
-1
6 points on vertices of regular hexagon
Correctly positioned (2 roots on real
axis). Ignore scales
SC1 if G0 and 5 points correctly plotted
6
(iv) w = z1e
=
jπ
−
12
jπ
3
=
2e e
jπ
−
12
=
2e
jπ
4
π
−
π
M1
arg w =
A1
G1
Or B2
Same modulus as z1
3
12
π
π⎞
⎛
2 ⎜ cos + j sin ⎟
4
4⎠
⎝
=1+j
3
⎛
jπ
4
⎞
6
(v) w6 = ⎜ 2e ⎟ = 8e
⎝
3 jπ
2
M1
⎠
= −8j
Or z16 e
A1
jπ
2
= 8e
−
jπ
2
cao. Evaluated
2
25
−
18
4756
Mark Scheme
January 2009
3(a)(i)
Region for (ii)
G1
G1
G1
r increasing with θ
Correct for 0 ≤ θ ≤ π/3 (ignore extra)
Gradient less than 1 at O
3
π
(ii) Area =
π
4
1 2
1 24
θ
=
r
d
a tan 2 θ dθ
∫0 2
2 ∫0
M1
Integral expression involving tan2θ
M1
Attempt to express tan2θ in terms of
sec2θ
A1
tan θ − θ and limits 0,
A1
A0 if e.g. triangle − this answer
G1
Mark region on graph
π
1 4
= a 2 ∫ sec2 θ − 1 dθ
2 0
π
1 2
a [ tan θ − θ ]04
2
1 2⎛ π ⎞
= a ⎜1 − ⎟
2 ⎝ 4⎠
=
π
4
5
(b)(i) Characteristic equation is
(0.2 − λ)(0.7 − λ) − 0.24 = 0
⇒ λ2 − 0.9λ − 0.1 = 0
⇒ λ = 1, −0.1
M1
A1
⎛ −0.8 0.8 ⎞⎛ x ⎞ ⎛ 0 ⎞
When λ = 1, ⎜
⎟⎜ ⎟ = ⎜ ⎟
⎝ 0.3 −0.3 ⎠⎝ y ⎠ ⎝ 0 ⎠
⇒ −0.8x + 0.8y = 0, 0.3x − 0.3y = 0
⎛ 1⎞
⇒ x − y = 0, eigenvector is ⎜ ⎟ o.e.
⎝ 1⎠
⎛ 0.3 0.8 ⎞ ⎛ x ⎞
A1
⎛ 0⎞
When λ = −0.1, ⎜
⎟⎜ ⎟ = ⎜ ⎟
⎝ 0.3 0.8 ⎠ ⎝ y ⎠ ⎝ 0 ⎠
⇒ 0.3x + 0.8y = 0
M1
A1
⎛8⎞
⇒ eigenvector is ⎜ ⎟ o.e.
⎝ −3 ⎠
⎛1 8 ⎞
(ii) Q = ⎜
⎟
⎝1 −3 ⎠
⎛1
(M −λI)x = x M0 below
At least one equation relating x and y
M1
At least one equation relating x and y
6
B0 if Q is singular. Must label
correctly
B1ft
0 ⎞
B1ft
B1
D= ⎜
⎟
⎝ 0 −0.1⎠
If order consistent. Dep on B1B1
earned
3
26
17
4756
Mark Scheme
4 cosh2x = ⎡ 1 e x + e − x ⎤ 2 =
)⎦
⎣2 (
(a)(i)
sinh2x = ⎡⎣ 12 ( e x − e − x ) ⎤⎦ =
2
1
4
1
4
(e
(e
+ 2 + e −2 x )
2x
2x
Both expressions (M0 if no “middle”
term) and subtraction
A1 (ag) www
2
− 2 + e −2 x )
M1
cosh2x − sinh2x = 14 ( 2 + 2 ) = 1
OR cosh x + sinh x = e x
cosh x − sinh x = e − x
cosh2x − sinh2x = e x × e − x = 1
cosh x = 1 + sinh x = 1 + tan y
M1
= sec y
A1
A1 (ag) www
3
⇒ tanh x =
sinh x tan y
=
= sin y
cosh x sec y
M1
arsinh x = ln(x + 1 + x 2 )
(ii)(B)
Both, and multiplication
Completion
Use of cosh2x = 1 + sinh2x and sinh x
= tan y
A1
2
2
(ii)(A)
January 2009
⇒ arsinh(tan y) = ln(tan y + 1 + tan y )
⇒ x = ln(tan y + sec y)
e x − e− x
= tan y
2
e 2 x − 2e x tan y − 1 = 0
Attempt to use ln form of arsinh
A1
A1 (ag) www
3
2
OR sinh x = tan y ⇒
⇒
M1
Arrange as quadratic and solve for e x
A1
A1
o.e.
www
tanh y = and attempt to differentiate
⇒ e x = tan y ± tan 2 y + 1
⇒ x = ln(tan y + sec y)
(b)(i)
y = artanh x ⇒ x = tanh y
M1
⇒
dx
= sech2y
dy
⇒
dy
1
1
1
=
=
=
2
2
dx sech y 1 − tanh y 1 − x 2
Or sech2y
dy
=1
dx
A1
Or B2 for
1
www
1 − x2
M1
artanh or any tanh substitution
1
Integral = [ artanh x ]2 1
−
= 2 artanh
2
A1 (ag) www
4
1
2
1
1
A
B
=
=
+
2
1− x
(1 − x )(1 + x ) 1 − x 1 + x
(ii)
⇒ 1 = A(1 + x) + B(1 − x)
M1
⇒ A = ½, B = ½
A1
⇒
1
∫ 1− x
2
dx = ∫
1
2
1
2
+
dx
1− x 1+ x
= − 12 ln 1 − x + 12 ln 1 + x + c or 12 ln
1+ x
1− x
+ c o.e.
Correct form of partial fractions and
attempt to evaluate constants
M1
Log integrals
A1
www. Condone omitted modulus
signs and constant
After 0 scored, SC1 for correct
answer
4
1
2
(iii)
1
1
1
1
2
dx
ln
1
x
ln
1
x
=
⎡−
−
+
+
⎤
2
⎣ 2
⎦ − 1 = ln 3
∫1 1 − x 2
2
−
Substitution of ½ and –½ seen
anywhere (or correct use of 0, ½)
M1
2
1
2
1
2
1
2
⇒ 2 artanh = ln 3 ⇒ artanh = ln 3
A1 (ag) www
2
27
18
4756
5 (i)
Mark Scheme
2
January 2009
y
1
2
-1
G1
G1
G1
-2
Symmetry in horizontal axis
(3, 0) to (0, 0)
(0, 0) to (0, 1)
3
(ii)(A) a > 0.5
a < −0.5
(ii)(B) Circle: r is constant
(ii)(C) The two loops get closer together
The shape becomes more nearly circular
(ii)(D) Cusp
a = −0.5
B1
B1
B1
B1
B1
B1
B1
Shape and reason
7
1
(iii) 1 + 2a cos θ = 0 ⇒ cos θ = −
2a
1
If a > 0.5, −1 < − < 0 and there are two values
2a
B1
Equation
of θ in [0, 2π],
⎛ 1 ⎞
⎛ 1 ⎞
⎟ and π + arccos ⎜ ⎟
⎝ 2a ⎠
⎝ 2a ⎠
1
⎛ ⎞
These differ by 2 arccos ⎜ ⎟
⎝ 2a ⎠
π − arccos ⎜
M1
A1 (ag)
M1
⎛ 1 ⎞
2
⎟ = arctan 4a − 1
2a
⎝ ⎠
arccos ⎜
Relating arccos to arctan by triangle
or tan2θ = sec2θ − 1
A1
A1
Tangents are y = x 4a − 1
2
A1ft
and y = −x 4a 2 − 1
Negative of above
E1
4a − 1 is real for a > 0.5 if a > 0
2
8
28
18
4756
Mark Scheme
June 2009
4756 (FP2) Further Methods for Advanced
Mathematics
x 2 x3 x 4 x5
1
ln(1 + x) = x − + − + ...
(a)(i)
2 3 4 5
ln(1 – x) = − x −
x 2 x3 x 4 x5
− − − ...
2 3 4 5
⎛1+ x ⎞
ln ⎜
⎟ = ln(1 + x) – ln(1 – x)
⎝ 1− x ⎠
2 x3 2 x5
= 2x +
+
...
3
5
Series for ln(1 – x) as far as x5
s.o.i.
M1
Seeing series subtracted
A1
Valid for –1 < x < 1
(ii)
B1
B1
4
Inequalities must be strict
M1
Correct method of solution
A1
B2 for x = ½ stated
1+ x
=3
1− x
⇒ 1 + x = 3(1 – x)
⇒ 1 + x = 3 – 3x
⇒ 4x = 2
1
2
3
5
1 2 ⎛1⎞ 2 ⎛1⎞
ln 3 ≈ 2 × + × ⎜ ⎟ + × ⎜ ⎟
2 3 ⎝2⎠ 5 ⎝2⎠
⇒ x=
= 1+
Substituting their x into their
series in (a) (i), even if outside
range of validity.
Series must have at least two
terms
SR: if >3 correct terms seen in (i),
allow a better answer to 3 d.p.
Must be 3 decimal places
M1
1 1
+
12 80
= 1.096 (3 d.p.)
A1
4
(b)(i)
y
1
G1
G1
G1
0.5
–1
–0.5
0.5
1
(ii) r + y = r + r sin θ
M1
a
× (1 + sin θ)
= r(1 + sin θ) =
1 + sin θ
r(0) = a, r(π/2) = a/2 indicated
Symmetry in θ = π/2
Correct basic shape: flat at θ =
π/2, not vertical or horizontal at
ends, no dimple
3 Ignore beyond 0 ≤ θ ≤ π
Using y = r sin θ
=a
⇒ r=a–y
A1 (ag)
⇒ x2 + y2 = (a – y)2
M1
A1
Using r2 = x2 + y2 in r + y = a
Unsimplified
A1
A correct final answer, not
spoiled
⇒ x2 + y2 = a2 – 2ay + y2
⇒ 2ay = a2 – x2
⇒ y=
a2 − x2
2a
5
24
16
4756
Mark Scheme
⎛3− λ
⎜
M
−
I
=
λ
2 (i)
⎜ 0
⎜ 2
⎝
1
−1 − λ
0
June 2009
−2 ⎞
⎟
0 ⎟
1 − λ ⎟⎠
det(M – λI) = (3 – λ)[(–1 – λ)(1 – λ)] + 2[2(–1
– λ)]
= (3 – λ)(λ2 – 1) + 4(–1 – λ)
Attempt at det(M – λI) with all
elements present. Allow sign
errors
Unsimplified. Allow signs
reversed. Condone omission of =
0
M1
A1
⇒ λ3 – 3λ2 + 3λ + 7 = 0
det M = –7
B1
3
(ii) f (λ) = λ3 – 3λ2 + 3λ + 7
f (–1) = –1 – 3 – 3 + 7 = 0 ⇒ –1 eigenvalue
Showing –1 satisfies a correct
characteristic equation
Obtaining quadratic factor
www
(M – λI)s = (λ)s M0 below
B1
f (λ) = (λ + 1)(λ2 – 4λ + 7)
M1
λ2 – 4λ + 7 = (λ – 2)2 + 3 ≥ 3 so no real roots A1
(M – λI)s = 0, λ = –1
⇒
⎛ 4 1 −2 ⎞ ⎛ x ⎞ ⎛ 0 ⎞
⎜
⎟⎜ ⎟ ⎜ ⎟
⎜ 0 0 0 ⎟⎜ y ⎟ = ⎜0⎟
⎜ 2 0 2 ⎟⎜ z ⎟ ⎜0⎟
⎝
⎠⎝ ⎠ ⎝ ⎠
⇒ 4x + y – 2z = 0
2x + 2z = 0
M1
⇒ x = –z
y = 2z – 4x = 2z + 4z = 6z
M1
Obtaining equations relating x, y
and z
Obtaining equations relating two
variables to a third. Dep. on first
M1
⎛ −1⎞
⇒ s = ⎜⎜ 6 ⎟⎟
⎜1⎟
⎝ ⎠
⎛ 3 1 −2 ⎞ ⎛ x ⎞ ⎛ −0.1⎞
⎜
⎟⎜ ⎟ ⎜
⎟
⎜ 0 −1 0 ⎟ ⎜ y ⎟ = ⎜ 0.6 ⎟
⎜ 2 0 1 ⎟ ⎜ z ⎟ ⎜ 0.1 ⎟
⎝
⎠⎝ ⎠ ⎝
⎠
⇒ x = 0.1, y = –0.6, z = –0.1
A1
Or any non-zero multiple
M1
Solution by any method, e.g. use
of multiple of s, but M0 if s itself
quoted without further work
Give A1 if any two correct
A2
9
C-H: a matrix satisfies its own characteristic
(iii)
equation
⇒ M3 – 3M2 + 3M + 7I = 0
B1
⇒ M3 = 3M2 – 3M – 7I
B1 (ag)
⇒ M2 = 3M – 3I – 7M–1
⇒ M–1 = − 17 M 2 + 73 M − 73 I
M1
A1
Idea of λ ↔ M
Must be derived www. Condone
omitted I
Multiplying by M–1
o.e.
4
1 −2 ⎞ ⎛ 3 1 −2 ⎞ ⎛ 5
⎟⎜
⎟ ⎜
−1 0 ⎟ ⎜ 0 −1 0 ⎟ = ⎜ 0
0 1 ⎟⎠ ⎜⎝ 2 0 1 ⎟⎠ ⎜⎝ 8
−8 ⎞
⎛ 3 1 −2 ⎞
⎛1
⎟ 3⎜
⎟ 3⎜
0 ⎟ + ⎜ 0 −1 0 ⎟ − ⎜ 0
7⎜
⎟ 7 ⎜0
−3 ⎟⎠
⎝2 0 1 ⎠
⎝
1
2
1
1
2
⎞
⎛
⎞
7
7
1⎜
⎟
⎟
−1 0 ⎟ or ⎜ 0 −7 0 ⎟
7
3
⎜ −2 −2 3 ⎟
− 72 7 ⎟⎠
⎝
⎠
⎛3
⎜
2
M
=
(iv)
⎜0
⎜2
⎝
⎛5 2
1⎜
− ⎜0 1
7⎜
⎝8 2
⎛ 17
⎜
= ⎜ 0
⎜− 2
⎝ 7
2 −8 ⎞
⎟
1 0⎟
2 −3 ⎟⎠
0 0⎞
⎟
1 0⎟
0 1 ⎟⎠
25
M1
Correct attempt to find M2
M1
Using their (iii)
A1
SC1 for answer without working
4756
Mark Scheme
June 2009
⎛ −1 0
2⎞
⎟
2⎟
M1
⎜ −2 0 −3 ⎟
⎝
⎠
⎛ −1 −1 −2 ⎞
Adjugate matrix ⎜⎜ 0 7 0 ⎟⎟ : det M = –7 M1
⎜ 2 2 −3 ⎟
⎝
⎠
OR Matrix of cofactors: ⎜⎜ −1 7
Finding at least four cofactors
Transposing and dividing by
determinant. Dep. on M1 above
3
26
19
4756
Mark Scheme
June 2009
3(a)(i)
1
–1
G1
Correct basic shape (positive
gradient, through (0, 0))
1
y = arcsin x ⇒ sin y = x
M1
sin y = and attempt to diff. both
sides
Or cos y
⇒
dx
= cos y
dy
A1
⇒
dy
1
1
=
=
dx cos y
1 − x2
A1
Positive square root because gradient
positive
dy
=1
dx
www. SC1 if quoted without
working
Dep. on graph of an increasing
function
B1
4
1
(ii)
∫
0
=
arcsin function alone, or any sine
substitution
M1
1
x ⎤
⎡
dx = ⎢arcsin
⎥
2 ⎦0
⎣
2 − x2
1
x
A1
π
2
A1
4
, or ∫ 1 dθ www without limits
Evaluated in terms of π
3
(b)
C + jS = e jθ + 13 e3 jθ + 19 e5 jθ + ...
M1
This is a geometric series
M1
with first term a = e jθ , common ratio r = 13 e2 jθ A1
Sum to infinity =
a
e jθ
3e jθ
(=
)
=
1 2 jθ
1− r 1− 3 e
3 − e 2 jθ
A1
=
3e jθ
3 − e −2 jθ
×
2 jθ
3−e
3 − e −2 jθ
M1*
=
9e jθ − 3e − jθ
9 − 3e −2 jθ − 3e 2 jθ + 1
M1
=
9 ( cos θ + j sin θ ) − 3 ( cos θ − j sin θ )
M1
10 − 3 ( cos 2θ − j sin 2θ ) − 3 ( cos 2θ + j sin 2θ )
6 cos θ + 12 j sin θ
10 − 6 cos 2θ
6 cos θ
⇒ C=
10 − 6 cos 2θ
=
Forming C + jS as a series of
powers
Identifying geometric series and
attempting sum to infinity or to n
terms
Correct a and r
Sum to infinity
Multiplying numerator and
denominator by 1 − 13 e−2 jθ o.e.
Or writing in terms of trig
functions and realising the
denominator
Multiplying out numerator and
denominator. Dep. on M1*
Valid attempt to express in terms
of trig functions. If trig functions
used from start, M1 for using the
compound angle formulae and
Pythagoras
Dep. on M1*
A1
M1
27
Equating real and imaginary
parts.
Dep. on M1*
4756
Mark Scheme
3cos θ
5 − 3cos 2θ
6sin θ
S=
5 − 3cos 2θ
=
June 2009
A1 (ag)
A1
o.e.
11
28
19
4756
Mark Scheme
June 2009
eu + e − u
2
4 (i) cosh u =
e 2u + 2 + e −2u
2
e 2u + e −2u
2
2 cosh u – 1 =
2
⇒ 2 cosh2u =
B1
(e
⇒
B1
cosh 2u =
= cosh 2u
B1 (ag)
u
+ e − u ) = e 2 u + 2 + e −2 u
2
e 2u + e −2u
2
Completion www
3
(ii) x = arsinh y
⇒ sinh x = y
⇒ y=
e x − e− x
2
Expressing y in exponential form
( 12 , – must be correct)
M1
e2x – 2yex – 1 = 0
(ex – y)2 – y2 – 1 = 0
(ex – y)2 = y2 + 1
ex – y = ± y 2 + 1
⇒
⇒
⇒
⇒
⇒ ex = y ± y 2 + 1
Reaching ex by quadratic formula
or completing the square.
Condone no ±
Or argument of ln must be
positive
Completion www but
independent of B1
M1
Take + because ex > 0
B1
⇒ x = ln(y + y 2 + 1 )
A1 (ag)
4
(iii) x = 2 sinh u ⇒
∫
dx
= 2 cosh u
du
x + 4 dx = ∫ 4sinh u + 4 × 2 cosh u du
2
dx
and substituting for all
du
M1
2
elements
Substituting for all elements
correctly
A1
= ∫ 4 cosh u du
2
= ∫ 2 cosh 2u + 2 du
= sinh 2u + 2u + c
M1
Simplifying to an integrable form
A1
Any form, e.g. 12 e2u − 12 e−2u + 2u
Condone omission of + c
throughout
M1
Using double “angle” formula and
attempt to express cosh u in
terms of x
A1 (ag)
Completion www
= 2 sinh u cosh u + 2u + c
= x 1+
=
x2
x
+ 2 arsinh + c
4
2
1
x
x 4 + x 2 + 2 arsinh + c
2
2
6
(iv) t2 + 2t + 5 = (t + 1)2 + 4
1
1
∫
2
t + 2t + 5 dt =
−1
B1
∫ ( t + 1)
2
+ 4 dt
−1
M1
2
=
Completing the square
∫
x 2 + 4 dx
0
⎡1
A1
x⎤
2
= ⎢ x 4 + x 2 + 2arsinh ⎥
2 ⎦0
⎣2
29
Simplifying to an integrable form,
by substituting x = t + 1 s.o.i. or
complete alternative method
Correct limits consistent with
their method seen anywhere
4756
Mark Scheme
=
8 + 2 arsinh 1
= 2 2 + 2 ln(1 +
2)
2) +
2)
= 2(ln(1 +
June 2009
M1
Using (iii) or otherwise reaching
the result of integration, and
using limits
A1 (ag)
Completion www.
Condone 8 etc.
5
30
18
4756
Mark Scheme
5 (i) If a = 1, angle OCP = 45°
so P is (1 – cos 45°, sin 45°)
1
⇒ P (1 –
2
,
1
2
June 2009
M1
)
A1 (ag)
2
Completion www
2
OR Circle (x – 1) + y = 1, line y = –x + 1
(x – 1)2 + (–x + 1)2 = 1
⇒ x=1±
Q (1 +
1
and hence P
2
1
1
,–
2
2
Complete algebraic method to
find x
M1
A1
)
B1
3
(ii) cos OCP =
sin OCP =
a
a2 + 1
1
a2 + 1
M1
Attempt to find cos OCP and sin
OCP in terms of a
A1
Both correct
A1 (ag)
Completion www
P is (a – a cos OCP, a sin OCP)
a2
⇒ P (a –
a
,
a2 + 1
a2 + 1
)
1
a
OR Circle (x – a)2 + y2 = a2, line y = − x + 1
⎛ 1
⎝ a
⎞
⎠
2
(x – a)2 + ⎜ − x + 1⎟ = a2
2
2a +
⇒ x=
⇒ x=a±
Q (a +
2
2⎞
⎛
± ⎜ 2a + ⎟
a
a⎠
⎝
1
⎛
2 ⎜1 + 2
⎝ a
a2
a2 + 1
a2
a2 + 1
1 ⎞
⎛
− 4 ⎜1 + 2 ⎟
⎝ a ⎠
⎞
⎟
⎠
and hence P
a
,–
a2 + 1
M1
Complete algebraic method to
find x
A1
Unsimplified
A1
)
B1
4
(iii)
1
0.5
–5
Locus of P (1st & 3rd quadrants)
through (0, 0)
Locus of P terminates at (0, 1)
Locus of P: fully correct shape
Locus of Q (2nd & 4th quadrants:
dotted) reflection of locus of P in
y-axis
Stated separately
Stated
Attempt to consider y as a → –∞
Completion www
G1
G1
G1
G1ft
5
–0.5
–1
As a → ∞, P → (0, 1)
As a → –∞, y co-ordinate of P → –1
a
2
a +1
→
B1
B1
M1
A1
a
= −1 as a → –∞
−a
8
(iv) POQ = 90°
Angle in semicircle
Loci cross at 90°
B1
B1
B1
o.e.
3
31
18
4756
Mark Scheme
January 2010
4756 (FP2) Further Methods for Advanced
Mathematics
1 (a)
y = arctan x
u = x , y = arctan u


dy
1
du
1
=
,
=
dx 2 x du 1 + u 2
dy
1
1
=
×
2
dx 1 + u 2 x
1
1
1
=
×
=
1 + x 2 x 2 x ( x + 1)
M1
A1
Using Chain Rule
Correct derivative in any form
A1
Correct derivative in terms of x
OR tan y = x
 sec2y
dy
1
=
dx 2 x
Rearranging for x or x and
differentiating implicitly
M1A1
sec2y = 1 + tan2y = 1 + x


dy
1
=
dx 2 x ( x + 1)
1

0
A1
1
1
dx =  2 arctan x 
x ( x + 1)
0
= 2 arctan 1 − 2 arctan 0
π π
=2× =
4
(b)(i)
 r2 = r2 cos θ sin θ + 1
 r2 = ½r2 sin 2θ + 1
 2r2 = r2 sin 2θ + 2
 r2(2 − sin 2θ) = 2
(ii) Max r is 2
Occurs when sin 2θ = 1
π 5π
 θ= ,
4
k=2
M1
Using at least one of these
A1
A1
LHS
RHS
6
3
B1
Occurs when sin 2θ = −1
3π 7π
 θ= ,
4
A1
Clearly obtained
A1 (ag) SR: x = r sin θ, y = r cos θ used
M1A1A0A0 max.
4
B1
M1
Attempting to solve
Both. Accept degrees.
A1
A0 if extras in range
2
2 − sin 2θ
4
2
Min r =
3
Integral in form k arctan x
A1 (ag)
2
x = r cos θ, y = r sin θ, x2 + y2 = r2
x2 + y2 = xy + 1
 r2 =
M1
Attempting to solve (must be −1)
Both. Accept degrees.
A0 if extras in range
M1
A1
4
6
24
4756
(iii)
Mark Scheme
January 2010
y
1
0.5
x
–1
–0.5
0.5
1
–0.5
G1
–1
G1
M1
2 (a) cos 5θ + j sin 5θ = (cos θ + j sin θ)5
= cos5θ + 5 cos4θ j sin θ + 10 cos3θ j2 sin2θ
+ 10cos2θ j3 sin3θ + 5 cos θ j4 sin4θ + j5 sin5θ M1
= cos5θ − 10cos3θ sin2θ + 5 cos θ sin4θ + j(…)
A1
cos 5θ = cos5θ − 10cos3θ sin2θ + 5 cos θ sin4θ
= cos5θ − 10cos3θ(1−cos2θ) + 5 cos θ(1−cos2θ)2
= 16cos5θ − 20cos3θ + 5 cos θ
M1
M1
A1
Closed curve, roughly elliptical, with
no points or dents
Major axis along y = x
2
18
Using de Moivre
Using binomial theorem appropriately
Correct real part. Must evaluate
powers of j
Equating real parts
Replacing sin2θ by 1 − cos2θ
a = 16, b = −20, c = 5
6
(b) C + jS
= e jθ + e
 2π 
j θ +

n 

+ ... + e
 ( 2 n − 2 )π 
j θ +

n


This is a G.P.
a = e jθ , r = e
M1
Forming series C + jS as exponentials
A1
Need not see whole series
Attempting to sum finite or infinite
G.P.
Correct a, r used or stated, and n terms
Must see j
M1
j
2π
n
A1
  j 2π  n 
e 1 −  e n  
 
 

Sum =
2π
j
1− e n
Numerator = e jθ (1 − e2π j ) and e 2π j = 1
jθ
A1
so sum = 0
 C = 0 and S = 0
E1
E1
Convincing explanation that sum = 0
C = S = 0. Dep. on previous E1
Both E marks dep. on 5 marks above
7
t
2
(c) e ≈ 1 + t + 12 t
t
t
≈
et − 1 t + 12 t 2
t
1 2
2
t+ t
OR
=
1 − 1 t 1 − 12 t
1
1
=
× 12 =
1
1
1 + 2 t 1 + 2 t 1 − 2 t 1 − 14 t 2
Hence
OR

1
−1
= (1 + 12 t ) = 1 − 12 t + ...
1
1+ 2 t
(e
M1
A1
Substituting Maclaurin series
Suitable manipulation and use of
binomial theorem
M1
A1 (ag)
− 1) (1 − 12 t ) = ( t + 12 t 2 + ...) (1 − 12 t )
≈ t + terms in t
t
≈ 1 − 12 t
t
e −1
Ignore terms in higher powers
M1
t
≈ 1 − 12 t
et − 1
t
B1
3
M1
A1
M1
Substituting Maclaurin series
Correct expression
Multiplying out
A1
Convincing explanation
5
25
18
4756
3 (i)
 2 −2 − 2a 2 + a 
1 

M =
2 2 − 3a 2a − 2 

4−a
5
−3 
 −1
−1
Mark Scheme
M1
A1
M1
A1
M1
2 0 1
1
When a = −1, M =  2 5 −4 
5

 −1 5 −3 
−1
January 2010
Evaluating determinant
4−a
Finding at least four cofactors
Six signed cofactors correct
Transposing and dividing by det
M−1 correct (in terms of a) and result
for a = −1 stated
A1
SR: After 0 scored, SC1 for M−1 when
a = −1, obtained correctly with some
working
6
(ii)  x 
 2 0 1   −2 
  1
 
 y  = 5  2 5 −4   b 
z
 −1 5 −3   1 
 

 

3
8
1
x = − , y =b− , z =b−
5
5
5
OR 4x + y = b − 4
x − y = 1 − b o.e.
M2
Attempting to multiply (−2 b 1)T by
given matrix (M0 if wrong order)
M1
Multiplying out
A2
A1 for one correct
M1
M1
Eliminating one unknown in 2 ways
Or e.g. 3x + z = b − 2, 5x = −3
Or e.g. 3y − 4z = −b − 4, 5y − 5z = −7
Solve to obtain one value.
Dep. on M1 above
One unknown correct
After M0, SC1 for value of x
Finding the other two unknowns
A1
Both correct
M1
 x= −

3
5
A1
8
1
y =b− , z =b−
5
5
5
(iii) e.g. 3x − 3y = 2b + 2
5x − 5y = 4
Consistent if
 b=
2b + 2 4
=
3
5
1
5
M1
A1A1
Eliminating one unknown in 2 ways
Two correct equations
Or e.g. 3x + 6z = b − 2, 5x + 10z = −3
Or e.g. 3y + 6z = −b − 4, 5y + 10z = −7
M1
Attempting to find b
A1
Solution is a line
B2
7
26
18
4756
Mark Scheme
x
4 (i) sinh x = e − e
−x
 sinh2x =
(e
x
−e
2
e − 2 + e −2 x
=
4
)
4
2x
 2 sinh2x + 1 =
=
January 2010
−x 2
B1
e 2 x − 2 + e −2 x
B1
Correct completion
B1
B1
Both correct derivatives
Correct completion
e 2 x − 2 + e −2 x
+1
2
e 2 x + e −2 x
= cosh 2x
2
 2 sinh 2x = 4 sinh x cosh x
 sinh 2x = 2 sinh x cosh x
4
(ii)




2 cosh 2x + 3 sinh x = 3
2(1 + 2 sinh2x) + 3 sinh x = 3
4 sinh2x + 3 sinh x − 1 = 0
(4 sinh x − 1)(sinh x + 1) = 0
sinh x = ¼, −1
1 + 17
)
4
x = arsinh(−1) = ln( −1 + 2 )
 x = arsinh(¼) = ln(

Using identity
Correct quadratic
Solving quadratic
Both
M1
Use of arsinh x = ln( x + x 2 + 1 ) o.e.
Must obtain at least one value of x
A1
Must evaluate x 2 + 1
A1
OR 2e 4 x + 3e3 x − 6e 2 x − 3e x + 2 = 0
 ( 2e2 x − e x − 2 )( e2 x + 2e x − 1) = 0

M1
A1
M1
A1
1 ± 17
or −1 ± 2
4
1 + 17
x = ln(
) or ln( −1 + 2 )
4
ex =
M1A1
Factorising quartic
M1A1
Solving either quadratic
Using ln (dependent on first M1)
M1A1A1
7
(iii)



5
et + e − t 5
cosh t = 
=
4
2
4
2e 2t − 5et + 2 = 0
( 2e
t
et =
M1
− 1)( et − 2 ) = 0
M1
1
,2
2
A1
 t = ±ln 2
5
A1 (ag) Convincing working
5
x

dx =  arcosh 
4 4

x − 16
4
5
= arcosh − arcosh 1
4

1
B1
2
= ln 2
5
OR

4
1
(
)
5
dx = ln x + x 2 − 16 

 4
2
−
x 16
= ln 8 − ln 4
= ln 2
Forming quadratic in et
Solving quadratic
M1
Substituting limits
A1
A0 for ±ln 2
B1
M1
A1
Substituting limits
7
27
18
4756
Mark Scheme
5 (i) Horz. projection of QP = k cos θ
Vert. projection of QP = k sin θ
Subtract OQ = tan θ
January 2010
B1
B1
B1
Clearly obtained
3
(ii) k = 2
k=1
y
y
2
2
1
1
x
0.5
1
1.5
x
2
0.5
–1
–1
–2
–2
k=½
1
1.5
2
G1
G1
Loop
Cusp
k = −1
y
y
2
4
1
2
x
–0.1
0.1
0.2
0.3
0.4
0.5
x
0.6
–1
–0.5
–1
–2
–2
–4
G1
G1
4
(iii)(A) for all k, y axis is an asymptote
(B) k = 1
(C) k > 1
B1
B1
B1
Both
3
(iv) Crosses itself at (1, 0)
k = 2  cos θ = ½  θ = 60°
 curve crosses itself at 120°
M1
A1
Obtaining a value of θ
Accept 240°
2
y = 8 sin θ − tan θ
(v)



dy
= 8 cos θ − sec2θ
dθ
1
8 cos θ −
= 0 at highest point
cos 2 θ
1
1
cos3θ =  cos θ = ±  θ = 60° at top
8
2
 x=4
y= 3 3
M1
A1
Complete method giving θ
A1
Both
3
(vi) RHS =
( k cos θ − 1)
=
( k cos θ − 1)
2
k 2 cos 2 θ
(k
2
− k 2 cos 2 θ )
M1
Expressing one side in terms of θ
M1
Using trig identities
2
k 2 cos 2 θ
× k 2 sin 2 θ
= ( k cos θ − 1) tan 2 θ
2
= ( ( k cos θ − 1) tan θ )
2
= ( k sin θ − tan θ ) = LHS
2
E1
3
28
18
GCE
Mathematics (MEI)
Advanced GCE 4756
Further Methods for Advanced Mathematics (FP2)
Mark Scheme for June 2010
Oxford Cambridge and RSA Examinations
4756
Mark Scheme

f (t) = arcsin t
1
f ′(t) =
= 1− t2
2
1− t

f ″(t) = − 12 1 − t 2
1 (a)(i)
(
(
=
)
− 32
)
− 12
× −2t
June 2010
B1
Any form
M1
Using Chain Rule
t
(1 − t )
2
A1 (ag)
3
2
3
f (x) = arcsin (x + ½)
(ii)

f (0) = arcsin (½) =
f ′(0) = 1 −

and f ″(0) =
( 12 )



2
π
− 12
1
2
(1− ( ) )
2
1
2
3
2
=
=
f (x) = f (0) + x f ′(0) +
2
3
6
obtained clearly from f (0) www
M1
A1 (ag)
Clear substitution of x = 0 or t = ½
M1
Evaluating f ″(0) and dividing by 2
A1
Accept 0.385x2 or better
4 3
9
x2
f ″(0) + …
2
2 3 2
x
9
term in x2 is

π
B1 (ag)
6
5
(b)
π
Area =

1
2
G1
G1
Complete spiral with r(2π) < r(0)
r(0) = a, r(2π) = a/3 indicated
or r(0) > r(π/2) > r(π) > r(3π/2) > r(2π)
Dep. on G1 above
Max. G1 if not fully correct
M1
Integral expression involving r2
r 2 dθ
0
π
=
π 2a2
 2 (π + θ )
2
dθ =
0
π 2a2
2
π
1
 (π + θ )
2
dθ
0
π
2 2
=
π a  −1 
2  π + θ  0
A1
Correct result of integration with correct
limits
=
π 2 a 2  −1 1 
+
2  2π π 
M1
Substituting limits into an expression of
k
the form
. Dep. on M1 above
π +θ
=
1
π a2
4
A1
6
3
2
(c)
1
 9 + 4x
2
dx =
0
=
=
1
4
3
2

9
0 4
1
+ x2
dx =
3
2
1 2
2x 
×  arctan 
4 3
3 0
1
arctan 1
6
π
24
M1
arctan
A1A1
1 2
2x
× and
4 3
3
M1
Substituting limits. Dep. on M1 above
A1
Evaluated in terms of π
5
1
19
4756
Mark Scheme
1
1
= 2 j sin nθ
B1
Both
10 5
1
1

5
3
 z − z  = z − 5 z + 10 z − z + 3 − 5
z
z


M1
1

Expanding  z − 
z

2 (a) z n +
z
= 2 cos nθ , z n −
June 2010
n
zn
5
= z5 −
1 
1


− 5  z 3 − 3  + 10  z − 
z
z
z 


1
5
M1
32j sin5θ = 2j sin 5θ − 10j sin 3θ + 20j sin θ

5

5
sin θ =
1
16
sin 5θ
− 165
5
8
sin 3θ + sin θ
A1
Introducing sines (and possibly cosines)
of multiple angles
RHS
A1ft
Division by 32(j)
A = 85 , B = − 165 , C = 161
5
th
(b)(i) 4 roots of −9j = 9e
r= 3
3π
2
j
jθ
are re where
3π
8
3π 2kπ
θ=
+
8
4
7π 11π 15π
θ=
,
,
8
8
8
θ=


1
B1
Accept 9 4
B1
M1
Implied by at least two correct (ft)
further values
A1
Or stating k = (0), 1, 2, 3
Allow arguments in range −π ≤ θ ≤ π
w
2
–2
2
Points at vertices of a square centre O
or 3 correct points (ft)
or 1 point in each quadrant
M1
–2
A1
6
(ii) Mid-point of SP has argument
and modulus of
π
B1
8
3
2
Argument of w = 4 ×
B1
π
8
=
π
2
4
M1
 3 9
and modulus = 
=
 2  4


A1
G1
2
Multiplying argument by 4 and modulus
raised to power of 4
Both correct
w plotted on imag. axis above level of P
5
16
4756
Mark Scheme
3 (a)(i) 2λ3 + λ2 − 13λ + 6 = 0  (λ − 2)(2λ2 + 5λ − 3) = 0
 λ = 2 or 2λ2 + 5λ − 3 = 0
 (2λ − 1)(λ + 3) = 0
 λ = ½, λ = −3
June 2010
B1
M1
Substituting λ = 2 or factorising
Obtaining and solving a quadratic
A1A1
4
3
3  6
 
   
(ii) M  −3  = 2  −3  =  −6 
1
1  2
 
   
B1
1  4
   
M v = 2 v = 4  −1 =  −4 
 1   4 
 3   3 
B2
Give B1 for one component with the
wrong sign
 32 
 32   3 
 3
   
M  − 2  = 2  − 32  =  −3 
 1 
 1  1
 2 
 2   
M1
Recognising that the solution is a
multiple of the given RHS

A1
2
2
x = 32 , y = − 32 , z = 12
Correct multiple
5


2λ3 + λ2 − 13λ + 6 = 0
2M3 + M2 − 13M + 6I = 0
M3 = − 12 M2 + 132 M − 3I

M4 = − 12 M3 + 132 M2 − 3M
(iii)
4
2

M = − (− M

M4 = 27
M2 − 25
M + 32 I
4
4
A=
1
2
27
4
1
2
,B=−
25
4
+ 132
,C=
M − 3I)
+ 132
2
M − 3M

Multiplying by M
M1
Substituting for M3
4
B1
1  1 1
P−1 = 

3  −2 1
1  1 −1  −1 0   1 1
N= 



3  2 1   0 2   −2 1
1  −1 −2   1 1
= 


3  −2 2   −2 1
 a c  −1
 −1

  = 2  
b
d
1

 
1
a + 2c = −1, −a + c = −2
b + 2d = −2, −b + d = 2
a = 1, c = −1; b = −2, d = 0
Order must be correct
B1
B1
For B1B1, order must be consistent
B1ft
Ft their P
M1
Attempting matrix product
A1
 a c  1 
1

  = −1 
 b d  2 
 2

M1
A1
1  3 −3   1 −1
= 
=

3  −6 0   −2 0 
a c 
OR Let N = 

b d 

Using Cayley-Hamilton Theorem
3
2
(b) N = PDP−1
 −1 0 
where D = 

 0 2
 1 −1
and P = 

2 1 

M1
c 1 0
 a +1
Or 
  =  
d + 1  2   0 
 b
c   −1  0 
a −2
Or 
  =  
b
d
− 2 1  0

B1
B1
B1
B1
M1A1
Solving both pairs of equations
6
3
19
4756
Mark Scheme
4 (i) 2 sinh x cosh x
e x + e− x e x − e− x
=2×
×
2
2
2x
−2 x
e −e
=
2
= sinh 2x
Differentiating,
2 cosh 2x = 2 cosh2x + 2 sinh2x
 cosh 2x = cosh2x + sinh2x
June 2010
Using exponential definitions and
multiplying or factorising
M1
A1 (ag)
B1
B1
One side correct
Correct completion
4
(ii)
y
2
Volume = π
 ( cosh x − 1)
2
dx
G1
Correct shape and through origin
M1
 ( cosh x − 1)
A1
A correct expanded integral expression
including limits 0, 2 (may be implied by
later work)
M1
Attempting to obtain an integrable form
Dep. on M1 above
A2
Give A1 for two terms correct
0
2

= π cosh 2 x − 2 cosh x + 1 dx
0
2
=π

1
2
cosh 2 x − 2 cosh x + 32 dx
0
= π  14 sinh 2 x − 2sinh x + 23 x 
2
0
2
dx
= π  14 sinh 4 − 2sinh 2 + 3
= 8.070
A1
3 d.p. required. Condone 8.07
7
(iii)
y = cosh 2x + sinh x
dy

= 2 sinh 2x + cosh x
dx
At S.P. 2 sinh 2x + cosh x = 0
B1

4 sinh x cosh x + cosh x = 0
M1

cosh x(4 sinh x + 1) = 0
M1


cosh x = 0 (rejected)
sinh x = − 14

 1
17 
x = ln  − +

 4
4 

A1
A1
M1
A1
4
Any correct form
Setting derivative equal to zero and
using identity
dy
=0 to obtain value of sinh x
Solving
dx
Repudiating cosh x = 0
Using log form of arsinh, or setting up
and solving quadratic in ex
A0 if extra “roots” quoted
7
18
4756
Mark Scheme
5(i)(A) Circle
(B)
June 2010
B1
1
G1
G1
B1
B1
B1
–1
(C) Square
(D) −1 ≤ x ≤ 1
−1 ≤ y ≤ 1
Sketch of circle, centre (0, 0)
Sketch of “squarer” circle on same axes
Give B1B0 for not all non-strict or
unclear
6
(ii)(A) Odd roots exist for all real numbers
(B) Line
(C)
2
B1
B1
Any equivalent explanation
Sketch insufficient
y
1
–2
2
–1
G1
B1
Asymptote: x + y = 0
(D)
2
y
1
–2
2
–1
Line x + y = 0 outside unit square
Lines y = 1 and x = 1 on unit square
G1
G1
6
(iii)
G1
B1
1
0 ≤ x, y ≤ 1
G0 if curve beyond (1, 0) or (0, 1)
Accept strict, or indication on graph
2
(iv)(A)
1
–1
1
G2ft
Give G1 for a partial attempt. Ft from
(iii) on shape
–1
(B) Limit is a “plus sign”
where x → 0 for −1 ≤ y ≤ 1 and vice versa
B1
B1
4
5
18
GCE
Mathematics (MEI)
Advanced GCE
Unit 4756: Further Methods for Advanced Mathematics
Mark Scheme for January 2011
Oxford Cambridge and RSA Examinations
4756
Mark Scheme
1 (a)(i)




x = r cos θ, y = r sin θ, x2 + y2 = r2
r = 2(cos θ + sin θ)
r2 = 2r(cos θ + sin θ)
x2 + y2 = 2x + 2y
x2  2x + y2  2y = 0
(x  1)2 + (y  1)2 = 2
which is a circle centre (1, 1) radius 2
January 2011
M1
Using at least one of these
A1 (ag)
Working must be convincing
M1
Recognise as circle or appropriate
algebra leading to (x − a)2 + (y − b)2 = r2
2
1
G1
Attempt at complete circle with centre in
first quadrant
A circle with centre and radius indicated,
or centre (1, 1) indicated and passing
through (0, 0), or (2, 0) and (0, 2)
indicated and passing through (0, 0)
G1
2
5
(ii) Area =


1
2
2
0
r d
2

 cos   sin  
0
= 2
= 2
2

2
0
 cos
2
2
d
  2sin  cos   sin 2   d

= 2 1  2sin  cos   d
2
0
M1
Integral expression involving r2 in terms
of θ
M1
Multiplying out
A1
cos2θ + sin2θ = 1 used
2
= 2   12 cos 2  02 or 2   sin 2   etc.
0
A2
=2
M1
Correct result of integration with correct
limits. Give A1 for one error
Substituting limits. Dep. on both M1s
=  2
A1
Mark final answer


2  12   0  12 
7
(b)(i)
f ( x) 
1
2
(ii)
f ( x) 
1
2
1 
1 
1
1
4
x2

1
4
2

x
2
=
4  x2
1

1
2
1 
M1
A1
1
4
x 
2
1
16
x  ...
4
Using Chain Rule
Correct derivative in any form
2

 12  18 x 2  321 x 4  ...
1
 f  x   12 x  241 x3  160
x 5  ...  c
But c = 0 because arctan(0) = 0
M1
Correctly using binomial expansion
A1
M1
A1
A1
Correct expansion
Integrating at least two terms
Independent
5
1
19
4756
Mark Scheme
2 (a)(i) zn + z–n = 2 cos nθ
zn – z–n = 2j sin nθ
January 2011
B1
B1
2
(ii) (z + z–1)6 = z6 + 6z4 + 15z2 + 20 + 15z2 + 6z4 + z6
= z6 + z–6 + 6(z4 + z–4) + 15(z2 + z–2) + 20
 64 cos6θ = 2 cos 6θ + 12 cos 4θ + 30 cos 2θ + 20
M1
Expanding (z + z–1)6
M1
Using zn + z–n = 2 cos nθ with n = 2, 4 or
6. Allow M1 if 2 omitted, etc.
 cos6θ = 321 cos 6  163 cos 4  15
cos 2  165
32
 cos6θ = 321  cos 6  6 cos 4  15cos 2  10 
A1 (ag)
3
(iii) (z – z–1)6 = z6 + z–6 – 6(z4 + z–4) + 15(z2 + z–2) – 20
 64 sin6θ = 2 cos 6θ  12 cos 4θ + 30 cos 2θ  20
B1
M1
A1
Using (i) as in part (ii)
Correct expression in any form
 sin6θ = 321 cos 6  163 cos 4  15
cos 2  165
32
M1
A1
15
 cos6θ  sin6θ = 161 cos 6  16
cos 2
OR cos2θ = 12 (cos 2θ + 1)
Attempting to add or subtract
B1

16 cos4θ = 2 cos 4θ + 8 cos 2θ + 6
cos4θ = 18 cos 4θ + 12 cos 2θ + 83

cos6θ  sin6θ = 2 cos6θ − 3 cos4θ + 3 cos2θ − 1
15
= 161 cos 6  16
M1A1
cos 2
This used
Obtaining an expression for cos4θ
Correct expression in any form
M1
A1
Attempting to add or subtract
5
2
(b)(i) z 1 = 8e
j
3
 z 1 = 2 2e
 2 2e
j
Correctly manipulating modulus and
argument
7
5
8,
or 
. Condone r(c + js)
6
6
Correctly manipulating modulus and
argument
13
5
2,
or 
. Condone r(c + js)
9
9
M1
j 7
6
A1
z 2 3 = 8e 3  z 2 = 2e
 2e


j   
6

  4 
j 

9 3 
M1
j13
9
A1
w
z1
Moduli approximately correct
Arguments approximately correct
Give G1G0 for two points approximately
correct
G1
G1
z2
6
(ii) z 1 z 2 = 2 2e
= 4 2e
j 7
6
× 2e
j13
9
 7 13 
j


9 
 6
j11
= 4 2e 18
Lies in second quadrant
M1
Correctly manipulating modulus and
argument
A1
Accept any equivalent form
A1
3
2
19
4756
Mark Scheme
3 (i) det(M – λI) = (1 – λ)[(3 – λ)(1 – λ) + 8]
+ 4[2(1 – λ) – 2] + 5[8 + (3 – λ)]
= (1  λ)(λ2  4λ + 11) + 4(2λ) + 5(11  λ)
= λ3 + 5λ2  15λ + 11  8λ + 55  5λ = 0
 λ3  5λ2 + 28λ  66 = 0
January 2011
M1
A1
Obtaining det(M – λI)
Any correct form
M1
A1 (ag)
Simplification
www, but condone omission of = 0
4
(ii) λ3  5λ2 + 28λ  66 = 0
Factorising and obtaining a quadratic.
If M0, give B1 for substituting λ = 3
Correct quadratic
Considering discriminant o.e.
Conclusion from correct evidence www
M1
 (λ  3)(λ2  2λ + 22) = 0
λ2  2λ + 22 = 0  b2  4ac = 84
so no other real eigenvalues
A1
M1
A1
4
 2 4 5   x   0 

   
λ
=
3

(iii)
 2 0 2   y    0 

   
 1 4 2   z   0 
 2x  4y + 5z = 0
2x  2z = 0
x + 4y  2z = 0
 x = z = k, y = 34 k

 4
 
eigenvector is  3 
 4
 

eigenvector with unit length is v 
Magnitude of Mnv is 3n
M1
M1
Two independent equations
Obtaining a non-zero eigenvector
A1
 4
1  
3
41  
 4
B1
B1
Must be a magnitude
5
(iv) λ3  5λ2 + 28λ  66 = 0
 M3 – 5M2 + 28M – 66I = 0
 M2 – 5M + 28I – 66M−1 = 0
1
 M−1 =
(M2 – 5M + 28I)
66
M1
Use of Cayley-Hamilton Theorem
M1
A1
Multiplying by M–1 and rearranging
Must contain I
3
3
16
4756
Mark Scheme
4 (i)

sinh t + 7 cosh t = 8
1
(et – e–t) + 7 × 12 (et + e–t) = 8
2
t
January 2011
M1
Substituting correct exponential forms
Obtaining quadratic in et
Solving to obtain at least one value of et
Condone extra values
–t




4e + 3e = 8
4e2t − 8et + 3 = 0
(2et − 1)(2et − 3) = 0
et = 12 or 32
M1
M1
A1A1

t = ln( 12 ) or ln( 32 )
A1
These two values o.e. only. Exact form
6
(ii)
dy
= 2 sinh 2x + 14 cosh 2x or 8e2x + 6e−2x
dx
2 sinh 2x + 14 cosh 2x = 16  sinh 2x + 7 cosh 2x = 8

2x = ln( 12 ) or ln( 32 )  x =
1
2
ln( 12 ) or
x = 12 ln( 12 )  y = −4
( 12 ln( 12 ), −4)
x = 12 ln( 32 )  y = 4
( 12 ln( 32 ), 4)
1
2
ln( 32 )
dy
= 0  2 sinh 2x + 14 cosh 2x = 0
dx
 tanh 2x = −7 or e4x =  34 etc.
x
No solutions because −1 < tanh 2x < 1 or e > 0 etc.
B1
M1
A1
Complete method to obtain an x value
Both x co-ordinates in any exact form
B1
Both y co-ordinates
M1
A1 (ag)
Any complete method
www
G1
Curve (not st. line) with correct general
shape (positive gradient throughout)
Curve through (0, 1). Dependent on last
G1
40
20
(0,1)
–20
–40
G1
8
  cosh 2 x  7 sinh 2 x  dx 
a
(iii)
0

a
 12 sinh 2 x  72 cosh 2 x  0 
1
2



sinh 2a + 7 cosh 2a = 8

2a = ln( 12 ) or ln( 32 )  a =

a=
1
2
1
2
M1
Attempting integration
A1
Correct result of integration
sinh 2a  72 cosh 2a  − 72 = 12
1
2
ln( 32 )
1
2
ln( 12 ) or
1
2
ln( 32 )
( 12 ln( 12 ) < 0)
Using both limits and a complete method
to obtain a value of a
Must reject 12 ln( 12 ), but reason need not
M1
A1
be given
4
4
18
4756
Mark Scheme
January 2011
5 (i) a = 1
2
y
1.5
1
0.5
x
–6
–4
–2
2
4
2
4
6
G1
a=2
3
y
2
1
x
–6
–4
–2
6
G1
–1
a = 0.5
2
y
1.5
1
0.5
x
–6
–4
–2
2
4
6
(A) Loops when a > 1
(B) Cusps when a = 1
G1
M2
A1
A1
(ii) If x → −x, t → −t
but y(−t) = y(t)
Curve is symmetrical in the y-axis
M1
A1 (ag)
B1
Evidence s.o.i. of further investigation
7
Considering effect on t
Effect on y
3
dy
a sin t
=
(iii)
dx 1  a cos t
dy
= 0  a sin t = 0  t = 0 and ±π
dx
t = 0  T.P. is (0, 1 − a)
t = ±π  T.P. are (±π, 1 + a)
M1
A1
Using Chain Rule
A1
Values of t
A1
A1
Both, in any form
5
(iv) a = 2 : both t = 2 and 32 give the point (π, 1)

2
B1 (ag)
Verification
M1
A1
Complete method for angle
Accept 115° (or 65°)

2
Gradients are a and −a (or and − )
Hence angle is 2 arctan( 2 ) ≈ 2.01 radians
3
5
18
GCE
Mathematics (MEI)
Advanced GCE
Unit 4756: Further Methods for Advanced Mathematics
Mark Scheme for June 2011
Oxford Cambridge and RSA Examinations
4756
Mark Scheme
June 2011
4756 (FP2) Further Methods for Advanced Mathematics
1
(a)(i)
G1
Correct general shape including
symmetry in vertical axis
Correct form at O and no extra sections.
Dependent on first G1
For an otherwise correct curve with a
sharp point at the bottom, award G1G0
G1
2
2
1 2
2
(ii) Area = a  1  sin   d
2 0
2
=
1 2
1  2sin   sin 2  d
a
2 0
=
1 2 3
1

a    2sin   cos 2  d
2 0 2
2



M1
Integral expression involving (1 − sin θ)2
M1
A1
Expanding
Correct integral expression, incl. limits
(which may be implied by later work)
1 1
Using sin 2    cos 2
2 2
2
M1
2
1 3
1

= a 2    2 cos   sin 2 
2 2
4
0
3
=  a2
2
A2
Correct result of integration.
Give A1 for one error
A1
Dependent on previous A2
7
1
2
1
1
1
1 2 1
1
dx =  1
dx =  2 arctan 2 x 2 1
(b)(i) 
2
2
2
4 1 4  x
1  4x
4
1
2
2
M1
arctan alone, or any tan substitution
A1
1
 2 and 2x
4
A1
Evaluated in terms of π
1     
=    
2  4  4 

=
4
3
(ii) x = 12 tan θ
M1
Any tan substitution
2
 dx = sec θ dθ
1
2

1
4

 4
 sec  
2

4
=
3
2

sec 2 
d
2
1
A1A1
 sec  
2
3
2
,
sec 2 
2
1
 2 cos  d
 4
Integrating a cos bθ and using consistent
limits. Dependent on M1 above
a
sin b
b
M1

1
4
=  sin  
2
  4
A1ft
1  1  1 

= 

2 2 
2  
1
=
2
A1
6
1
18
4756
Mark Scheme
June 2011
5
2 (a) cos 5θ + j sin 5θ = (cos θ + j sin θ)
= c5 + 5c4js − 10c3s2 − 10c2js3 + 5cs4 + js5
M1
A1
A1
Expanding
Separating real and imaginary parts.
Dependent on first M1
Alternative: 16c5 − 20c3 + 5c
Alternative: 16s5 − 20s3 + 5s
M1
Using tan θ =
M1
5
3 2
4
 cos 5θ = c − 10c s + 5cs
sin 5θ = 5c4s − 10c2s3 + s5
5c 4 s  10c 2 s 3  s 5
 tan 5θ = 5
c  10c 3 s 2  5cs 4
5t  10t 3  t 5
=
1  10t 2  5t 4
=

t t 4  10t 2  5

A1 (ag)
5t  10t  1
4
sin 
and simplifying
cos 
2
6
(b)(i) arg( 4 2 ) = π
 fifth roots have r = 2
and θ =
B1

B1
5
 z = 2e
1
j
5
, 2e
3
j
5
j
, 2e , 2e
7
j
5
, 2e
No credit for arguments in degrees
2
5
All correct. Allow −π ≤ θ < π
Adding (or subtracting)
M1
9
j
5
A1
4
(ii)
G1
Points at vertices of “regular” pentagon,
with one on negative real axis
Points correctly labelled
G1
2
1   3
(iii) arg(w) =  
2 5 5
w = 2 cos
 2
 5

B1

M1
A1ft
5
Attempting to find length
F.t. (positive) r from (i)
3

n


 wn =  2 cos  e
5
5

2 n
which is real if sin
=0n=5
5
(iv) w = 2 cos
e
2
j
5


w5 =  2 cos 
5

5
 a = 2 2 cos5
2
 nj
5
B1
M1
Attempting the nth power of his modulus
in (iii), or attempting the modulus of the
nth power here
A1
Accept 1.96 or better
5

5
3
2
18
4756
Mark Scheme
3 (i) det(M) = 1(16 − 12) + 1(20 − 18) + k(10 − 12)
= 6 − 2k
 no inverse if k = 3
June 2011
M1
A1
A1
Obtaining det(M) in terms of k
Accept k ≠ 3 after correct determinant
Evaluating at least four cofactors
(including one involving k)
Six signed cofactors correct
(including one involving k)
Transposing and dividing by det(M).
Dependent on previous M1M1
M1
 4 4  2k
1 
−1
M =
2 4  3k
6  2k 
5
 2
6  4k 

5k  6 
9 
A1
M1
A1
7
 1 1 3  3   3 

   
(ii)  5 4 6  3    3 
 3 2 4  1   1 

   
M1
Setting k = 3 and multiplying
A1
2
 3 
 
(iii)  3  is an eigenvector
1
 
corresponding to an eigenvalue of 1
For credit here, 2/2 scored in (ii)
Accept “invariant point”
B1
B1
2
(iv) 3x + 6y = 1 − 2t, x + 2y = 2, 5x + 10y = −4t
(or 9x + 18z = 4t + 1, 5x + 10z = 2t, x + 2z = −1)
(or 9y − 9z = 1 − 5t, 5y − 5z = −3t, 2y − 2z = 3)
For solutions, 1 − 2t = 3 × 2
5
 t =
2
M1
Eliminating one variable in two different
ways
A1
M1
Two correct equations
Validly obtaining a value of t
A1
M1
x = λ, y = 1 − 12 λ, z = − 12 − 12 λ
A1
Straight line
B1
3
Obtaining general solution by setting one
unknown = λ and finding other two in
terms of λ (accept unknown instead of λ)
Accept “sheaf”. Independent of all
previous marks
7
18
4756
Mark Scheme
4 (i) cosh y = x  x =
1 y
e  e y
2


June 2011
B1
Using correct exponential definition
 2 xe y  1  0
M1
Obtaining quadratic in e y
2 x  4 x2  4
= x  x2  1
2
M1
Solving quadratic
A1
x  x2  1
M1
Validly attempting to justify ± in printed
answer
 2x = e y  e  y
 
2
 ey
 ey 
 y = ln( x  x 2  1 )
x 


x2  1 x  x2  1  1
 y = ±ln( x  x 2  1 )
A1 (ag)
arcosh(x) = ln( x  x 2  1 ) because this is the
principal
value
B1
Reference to arcosh as a function, or
correctly to domains/ranges
7
1
(ii)

4
5
1
1
25 x  16
2
dx =
1
1
dx
5 4 x 2  16
25
5
1
1
 5x  
= arcosh   
5
 4   54
M1
arcosh alone, or any cosh substitution
A1A1
1 5x
,
5 4
M1
Substituting limits and using (i) correctly
at any stage (or using limits of u in
logarithmic form). Dep. on first M1

1
5
=  arcosh    arcosh 1 
5
4

2

1 5
5
= ln      1  − 0

5 4
4


1
= ln 2
5
A1

ln kx  k 2 x 2  ...
1
1 
16  
OR = ln  x  x 2 


5  
25   4
M


Give M1 for ln k1 x  k2 2 x 2  ...
5

1
16 
, ln  x  x 2 
 o.e.

25 
5

A1A
1 8 1 4
= ln  ln
5 5 5 5
1
= ln 2
5

A
5
(iii) 5 cosh x − cosh 2x = 3
 5 cosh x − (2 cosh2x − 1) = 3
M1
 2 cosh2x − 5 cosh x + 2 = 0
 (2 cosh x − 1)(cosh x − 2) = 0
M1
 cosh x =
A1
A1
1
2
(rejected)
or cosh x = 2

 x = ln 2  3


A1ft


x = −ln 2  3 or ln 2  3

A1ft
4
Attempting to express cosh 2x in terms
of cosh x
Solving quadratic to obtain at least one
real value of cosh x
Or factor 2 cosh x − 1
F.t. cosh x = k, k > 1
F.t. other value. Max. A1A0 if additional
real values quoted
6
18
4756
Mark Scheme
June 2011
5 (i) (A) m = 1, n = 1
0.3
y
0.2
0.1
x
0.5
1
0.5
1
0.5
1
0.5
1
G1
Negative parabola from (0,0) to (1,0),
symmetrical about x = 0.5
G1
Bell-shape from (0,0) to (1,0),
symmetrical about x = 0.5; flat ends,
and obviously different to (A)
x
G1
Skewed curve from (0,0) to (1,0),
maximum to left of x = 0.5
x
G1
Skewed curve from (0,0) to (1,0),
maximum to right of x = 0.5
(B) m = 2, n = 2
0.1
y
x
(C) m = 2, n = 4
0.03
y
0.02
0.01
(D) m = 4, n = 2
0.03
y
0.02
0.01
4
(ii) When m = n, the curve is symmetrical
Exchanging m and n reflects the curve
B1
B1
(iii) If m > n, the maximum is to the right of x = 0.5
As m increases relative to n, the maximum point moves
further to the right
dy
n
n
n 1
= mx m 1 1  x   nx m 1  x 
y = x m 1  x  
dx
B1
2
= x m 1 1  x 
n 1
o.e.
Give B1B0 if the idea is correct but
vaguely expressed
Using product rule
Any correct form
B1
M1
A1
 m 1  x   nx 
Setting derivative = 0 and solving to
find a value of x other than 0 or 1
M1
dy
m
= 0  maximum at x =
dx
mn
A1
6
5
4756
Mark Scheme
June 2011
(iv) y′(0) = 0 provided m > 1
y′(1) = 0 provided n > 1
B1
B1
(v) For large m and n, the curve approaches the x-axis
B1
Comment on shape
B1
Independent
2
1
 x 1  x  dx  0 as m, n  

m
n
0
2
(vi) e.g. m = 0.01, n = 0.01
y
1
0.8
0.6
0.4
0.2
x
0.5
1
M1
A1
The curve tends to y = 1
Evidence of investigation s.o.i.
Accept “three sides of (unit) square”
2
6
18
4756
Mark Scheme
Question
1
(a)
(i)
Marks
Answer
sin y = x  cos y

dy
=1
dx
M1
dy
1
=
dx cos y
June 2012
Guidance
dy
= cos y
Differentiating w.r.t. x or y
dx
A1
dy
1

= 
dx
1  x2
A1(ag)
Taking + sign because gradient is positive
B1
Completion www, but
independent of B1
dy
1
=±
or ± not
dx
1  x2
considered scores max. 3
Validly rejecting − sign.
Dependent on A1 above
Or 

2
 y

2
 0 ≤ cos y ≤ 1
[4]
1
1
(a)
(ii)
(A)
x 

dx =  arcsin

2  1

2 x
1

2
1
=
1

M1
arcsin alone, or any appropriate
substitution
A1
arcsin
x
2
or  1 d www
Condone omitted or incorrect
limits
A1
2
[3]
1
2
(B)

 12
1
2
1
1  2x
2
dx =
1

2  12
1
1
2
 x2
dx
M1
1
1 
2
=
arcsin
2
x
  12
2
A1
M1
=

A1
2 2
[4]
5
arcsin alone, or any appropriate
substitution
1
1
and 2x or 
d www
2
2
Using consistent limits in order
and evaluating in terms of π.
Dependent on M1 above
e.g. 

4
with sub. x =
1
sin 
2
4756
Mark Scheme
Question
(b)
1
Marks
Answer
(a)
(i)
Guidance
r = tan θ
sin 
× cos θ = sin θ
cos 
sin 2 
sin 2 
x2
 r2 =
=
=
cos 2  1  sin 2  1  x 2
x2
r2 = x2 + y2  x2 + y2 =
1  x2
2
x
 y2 =
− x2
2
1 x
x 2  x 2 1  x 2 
x4
2
 y =
=
1  x2
1  x2
2
x
 y=
1  x2
Asymptote x = 1
 x = r cos θ =
2
June 2012
M1
A1(ag)
Using x = r cos θ o.e.
M1
A1(ag)
Obtaining r2 in terms of x
M1
Obtaining y2 in terms of x
Ignore discussion of 
A1(ag)
1
 2cos n
zn
1
z n  n  2 jsin n
z
zn 
B1
[7]
Condone x = ±1
B1
Mark final answer
B1
Mark final answer
x ≠ 1, x2 = 1 B0
[2]
4
2
(a)
(ii)
1
4 1
1

 2 1 
4
2
4
 z   = z  4z  6  2  4 = z  4  4 z  2   6
z
z
z
z 
z


M1
M1
 (2 cos θ )4 = 2 cos 4 θ + 8 cos 2 θ + 6
A1
3 1
1
 cos4 θ =  cos 2  cos 4
8 2
8
A1ft
[4]
6
Expanding by Binomial or
complete equivalent
Introducing cosines of multiple
angles
RHS correct
Dividing both sides by 16.
F.t. line above
Condone lost 2s
Both As depend on both Ms
3
1
1
A= ,B= ,C=
8
2
8
Give SC2 for fully correct
answer found “otherwise”
4756
Mark Scheme
Question
2
(a)
Marks
Answer
3 1
1
(iii) cos4 θ =   2cos 2   1  cos 4
8 2
8
M1
June 2012
Guidance
Using (ii), obtaining cos 4 θ and
expressing cos 2 θ in terms of
Condone cos 2 θ = 1  2 cos2 θ
2
cos θ
 cos4 θ = cos 2   18  18 cos 4
 cos 4 θ = 8 cos4 θ − 8 cos2 θ + 1
A1
[2]
c.a.o.
j
2
(b)
(i)
z = 4e 3 and w2 = z: let w = re j  w2 = r 2 e 2 j
 r2 = 4  r = 2
 7
and θ = ,
6 6
4
y
Condone r = 2
B1
B1B1
Or 
5
6
1

Award B2 for   k  
6

z
2
Ignore annotations and scales
w1
-4
-2
2
B1
4
B1
w2
Roots with approx. equal
moduli and approx. correct
argument
Dependent on first B1
z in correct position
≤ π/4
Modulus and argument bigger
[5]
j
2
(b)
(ii)
z = 4e 3  zn = 4n e
j n
3
n
=πn=3
3
n 
3
=  k  n =  3k
Imaginary if
3 2
2
so real if
B1
M1
which is not an integer for any k
j
6
3
A1(ag)
j
2
w 1 = 2e  w 1 = 8e = 8j
w 2 = 2e
7 j
6
 w 2 3 = 8e
7 j
2
= −8j
n
= 0 or
n 
= ...
3
3 2
An argument which covers the
positive and negative im. axis
cos
M1
Attempting their w3 in any form
A1
8j, −8j
[5]
7
Ignore other larger values
Must deal with mod and arg
4756
Question
(i)
3
Mark Scheme
Marks
M1A1
Answer
det (M) = 1(2a + 8) − 2(−2 − 12) + 3(2 − 3a)
= 42 − 7a
 no inverse if a = 6
(ii)
M1
A1
M1
 x
 8 10 8  1 
  1 
 
 y   42  14 7 7  2 
2
z

8
2 
 

 1 
 x=
Guidance
Obtaining det(M) in terms of a
Accept unsimplified
A1
 2a  8 10 8  3a 
1 

−1
14
M =
7
7 
42  7 a 
a  2 
 2  3a 8
3
June 2012
6
1
2
, y = , z =
7
2
7
At least 4 cofactors correct
(including one involving a)
Six signed cofactors correct
Transposing and ÷ by det(M).
Dependent on previous M1M1
Accept a ≠ 6 after correct det
M0 if more than 1 is multiplied
by the corresponding element
A1
[7]
M1
Substituting a = 0
M1
Correct use of inverse
One correct element.
Condone missing determinant
A2
Dependent on both M marks.
Give A1 for one correct
SC1 for x = 6, y = 3.5, z = −2
After M0, give SC2 for correct
solution and SC1 for one correct
Answers unsupported score 0
Eliminating one variable in two
different ways
Or 7x − 10y = 2b + 2
Mark final answer
[4]
3
(iii)
e.g. 7x − 10y = 10, 7x − 10y = 3b − 2
M1
(or e.g. 4x + 5z = 5, 4x + 5z = b + 1)
(or e.g. 8y + 7z = −1, 8y + 7z = 3 − b)
For solutions, 10 = 3b − 2
 b=4
A1
M1
A1
b=4
A1
A1
M1
x = λ , y = 0.7 λ − 1, z = 1 − 0.8 λ
A1
Straight line
B1
[7]
8
Or 8x + 10z = 3b − 2
Or 16y + 14z = b − 6
A method leading to an equation E.g. setting z = 0, augmented
from which b could be found
matrix, adjoint matrix, etc.
A correct equation
M2
OR
Two correct equations
Validly obtaining a value of b
Obtaining general soln. by e.g.
setting one unknown = λ and
finding other two in terms of λ
Any correct form
Accept “sheaf”, “pages of a
book”, etc.
Accept unknown instead of λ
x = 107  + 107 , y = λ , z =  87  − 17
x = 54 − 54  , y =  78  − 18 , z = λ
Independent of all previous
marks. Ignore other comments
4756
Mark Scheme
Question
4
4
(i)
(ii)
Marks
Answer
e e
2
u
2 u
e 2e
4
2u
2 u
e 2e
e 2u  e 2u
 2 sinh2u + 1 =
1=
2
2
= cosh 2u
u
sinh u =
 sinh2u =
2u
B1
B1
B1
[3]
e y  e y
2
 2u  e 2 y  2ue y  1  0
If cosh y = u, u =
 e y  e y

e
y
Guidance
e
u
 e  u   e 2u  2  e 2u
2
Accept other or mixed variables
e 2u  e 2u
2
Completion www
cosh 2u =
M1
Expressing u in exponential
form
1
, + must be correct
2
M1
Reaching e y
Condone omitted ±
 u   u2 1  0
2
 ey  u  u2 1

 y = ln u  u 2  1

A1(ag)
 
OR ln u  u 2  1 = ln cosh y  cosh 2 y  1
Completion www; indep. of B1


y = ln u  u 2  1 or ± not
considered scores max. 3
B1
y ≥ 0  ey  u  u2 1

June 2012

Validly rejecting − sign
Dependent on A1 above
M1
Substituting u = cosh y
= ln  cosh y  sinh y 
since sinh y > 0
= ln  e y 
B1
M1
Rejecting −ve square root
=y
A1
y
Reaching e
Completion www; indep. of B1
[4]
9
Dependent on A1
4756
Mark Scheme
Question
4
(iii)
Marks
Answer
x=
1
dx 1
cosh u 
 sinh u
2
du 2
M1
1
4 x 2  1 dx =  cosh 2 u  1  sinh u du
2
1
=  sinh 2 u du
2
1
1
=  cosh 2u  du
4
4
1
1
= sinh 2u  u  c
8
4
1
1
= sinh u cosh u  u  c
4
4
June 2012
Guidance
Reaching integrand equivalent
to k sinh2u

=
A1
M1
A1A1
1
1
4 x 2  1  2 x  arcosh 2 x  c
4
4
M1
1
1
= x 4 x 2  1  arcosh 2 x  c
2
4
1
a=
2
Simplifying to integrable form.
Dependent on M1 above
1
1
For sinh 2u o.e. and  u seen
8
4
1 2u 1 1 2u
e   e
8
4 8
1 2u 1
1
e  u  e 2u  c
Or
16
4
16
Condone omission of + c
throughout
Or
Clear use of
sinh 2u = 2 sinh u cosh u
Dependent on M1M1 above
a, b need not be written
separately
A1
[7]
1
4
(iv)
1

1
1

4 x 2  1 dx =  x 4 x 2  1  arcosh 2 x 
2
4

 12
M1
Using their (iii) and using limits
correctly
=
3 1
1
 arcosh 2  arcosh1
2 4
4
A1ft
May be implied
F.t. values of a and b in (iii)
1
2


M1
Using (ii) accurately
Dependent on M1 above


A1
c.a.o. A0 if ln 1 retained
Mark final answer
3 1
1
 ln 2  3  ln1
2 4
4
3 1
 ln 2  3
=
2 4
=
[4]
10
a 3  b arcosh 2. No decimals.
Must have obtained values for a
and b
Correct answer www scores 4/4
4756
Mark Scheme
Question
5
(i)
5
(ii)
Marks
Answer
Undefined for θ =

2
and
June 2012
3
2
Guidance
B1B1
[2]
4
2
-4
-2
2
4
-2
B1
-4
r = sec θ  r cos θ = 1
x=1
5
(iii)
M1
A1
[3]
Vertical line through (1, 0)
(indicated, e.g. by scale)
Use of x = r cos θ
a = 1:
4
2
-4
-2
2
4
B1
-2
B2
-4
a = −1 gives same curve
a = 1, 0 < θ < π corresponds to a = −1, π < θ < 2π
a = −1, 0 < θ < π corresponds to a = 1, π < θ < 2π
B1
B1
B1
[6]
11
Section through (2, 0)
(indicated)
Section through (0, 0)
(give B1 for one error)
If asymptote included max. 2/3
4756
Mark Scheme
Question
5 (iv)
Answer
June 2012
Marks
B1
Loop
e.g. a = 2
Guidance
4
2
-4
-2
2
4
-2
-4
5
(v)
r = sec θ + a
r
 r  a
x
 1
 r 1    a
 x
 x 1

x2  y 2 
a
 x 


x 2  y 2  x  1  ax
x
2
 y 2   x  1  a 2 x 2
2
B2
[3]
Give B1 for one error
M1
Use of x = r cos θ
M1
Use of r = x 2  y 2
M1
Correct manipulation
A1(ag)
[4]
12
GCE
Mathematics (MEI)
Advanced GCE
Unit 4756: Further Methods for Advanced Mathematics
Mark Scheme for January 2013
Oxford Cambridge and RSA Examinations
4756
Mark Scheme
Question
1
(a)
Answer
(i)
a tan y  x  a sec 2 y
Marks
dy
1
dx
dy
1

dx a sec 2 y
dy
1


dx

x2 
a 1  2 
 a 
dy
a


dx a 2  x 2

Guidance
M1
Differentiating with respect to x or y
A1
For
A1(ag)
January 2013
dy
dx
dx
 a sec 2 y
dy
dy
1
Or a

dx sec 2 y
Completion www with sufficient detail
[3]
1
(a)
(ii)
x  4x  8   x  2  4
2
2

4
0


B1
4
1
1
x  2
dx   arctan
2
x  4x  8
2
2  0
M1
A1
Integral of form a arctan bu or any
appropriate substitution
Correct integral with consistent limits
A1
Evaluated in terms of π
2
1
u
arctan 

2
2  2
1
 arctan 1  arctan  1 
2

4
[4]
1
(a)
(iii)
1 arctan x dx
 x arctan x  
M1
x
dx
1  x2
1
 x arctan x  ln 1  x 2   c
2
Using parts with u = arctan x and v = 1
A1
x
M1
1 x
A1
a
[4]
5
2
dx  a ln 1  x 2 
1
. Condone omitted c
2
Allow one other error
4756
Mark Scheme
Question
1
(b)
(i)
Answer
Marks
r  2cos   r  2r cos 
 x2  y 2  2 x
2
  x  1  y 2  1
2
OR
(i)
Explaining that the curve is a circle
x in terms of cos 2
2
A1
Explaining that the curve is a circle
A1(ag)
B1
B1
[5]
Independent
Independent
M1
A1
[2]
Using r 2  x 2  y 2 and y  r sin 
e.g. writing as  x      y     r 2
2
2
2
For answer alone www:
B1 for r  k sin  , B1 for k = 4
1  e j2  1  cos 2  jsin 2
 1   2cos 2   1  2 jsin  cos 
M1
Using e 2 j  cos 2  jsin 2
and double angle formulae
 2cos 2   2 jsin  cos 
 2cos   cos   jsin  
OR
A1(ag)
1  e j2  e j  e-j  e j 
OR
1  e j2  1   cos   jsin  
“Factorising” and complete replacement by
trigonometric functions
Completion www
M1
  cos   jsin    2cos 
Completion www
A1(ag)
2
 1  cos 2   sin 2   2 jsin  cos 
 2cos 2   2 jsin  cos 
M1
 2cos   cos   jsin  
A1(ag)
Using e j  cos   jsin 
and 1  sin 2   cos 2 
Completion www
[2]
6
2
e.g. writing as  x      y     r 2
x2   y  2  4  x2  y 2  4 y
 r 2  4r sin 
 r  4sin 
(a)
A correct cartesian equation in any form
2
cos 2  2cos   1  x  cos 2  1
Centre (1, 0)
Radius 1
2
A1
2
Using x  r cos  , y  r sin  and linking
  x  1  y  1
(ii)
Using r  x  y and x  r cos 
2
x  r cos   x  2cos 2 
y  r sin   y  2cos sin   sin 2 M1
2
(b)
Guidance
M1
A1(ag)
2
1
January 2013
Allow one error
2
4756
Mark Scheme
Question
2
(a)
(ii)
Answer
Marks
n
n
C  jS  1    e j2    e j4  ...  e j2n
1
 2
 1  e j2 
n
 2n cos n   cos   jsin  
(i)
e
j
2
3
 cos
M1
Forming C + jS
M1
A1
Recognising as binomial expansion
M1
A1
A1(ag)
 C  2n cos n  cos n
and S  2n cos n  sin n
(b)
Guidance
n
 2n cos n   cos n  jsin n 
2
Applying (i) and De Moivre o.e.
Dependent on M1M1 above
Completion www
Need to see e jn  cos n  jsin n o.e.
A1
[7]
2
2
1
3
 jsin
  j
3
3
2
2
B1
Must evaluate trigonometric functions
[1]
2
(b)
(ii)
Other two vertices are  2  4 j e
 1
3
  2  4 j    j

2 
 2

 
 1  2 3  j 2  3
and  2  4 j e
j
4
3

 
 1  2 3  j 2  3
2
j
3
M1
Award for idea of rotation by
2
3
e.g. use of arctan 2 
2
(3.202 rad)
3
(must be 2)

  2  4 j e
 1
3
  2  4 j    j

2 
 2
January 2013
A1A1
-j
2
3
M1
May be given as co-ordinates
Award for idea of rotation by 
2
3
e.g. use of arctan 2 
4
(5.296 rad)
3
(must be 2)

A1A1
May be given as co-ordinates
[6]
7
If A0A0A0A0 award SC1 for awrt
−4.46 − 0.27j and 2.46 − 3.73j
4756
Mark Scheme
Question
2
(b)
(iii)
Answer
Marks
So length of side = 2 20 cos

6
 2 20 
3
2
 2 15
3
(i)
Guidance
20
Length of (2 + 4j) =
1  

M  I   3
 0

det  M   I 
Alternative: finding distance between
M1
Complete method
A1(ag)
[2]
Completion www
0 

2   1 
1
1   
M1
Forming det  M   I 
A1
Any correct form
 1      2    3  9 1   
  3  13  12  0
(ii)


(2, 4) and 1  2 3, 2  3 o.e.
3
 1     2   1     1  3 3 1    
3
January 2013
   1   2    12   0
    1   3   4   0
 eigenvalues are 1, 3, −4
 0 3 0  x   0 
  
  1:  3 3 1
 y    0 
 0 1 0  z   0 

   
 y  0, 3 x  z  0
1
 
 eigenvector is  0 
 3
 
 2 3 0  x   0 
  
  3:  3 5 1 
 y    0 
 0 1 2  z   0 

   
 2 x  3 y  0,  y  2 z  0
A1(ag)
[3]
M1
Sarrus: 1   
 2     10 1   
or e.g.   1  1      2    11
2
Condone omission of 0
Factorising as far as quadratic
Allow one error
A1
A1
M2
For any one of λ = 1, 3, −4
Obtaining two independent equations
M1
Obtaining a non-zero eigenvector
A1
o.e.
A1
A1
o.e.
8
From which an eigenvector could be found
Allow e.g. 3 y  0, 3 x  3 y  z  0
4756
Mark Scheme
Question
Answer
Marks
January 2013
Guidance
 y  2 z , x  3 z
 3 
 
 eigenvector is  2 
1
 
 5 3 0  x   0 
  
  4 :  3 2 1
 y    0 
 0 1 5  z   0 

   
 5 x  3 y  0,  y  5 z  0
 y  5 z , x  3 z
 3 
 
 eigenvector is  5 
1
 
A1
A1
o.e.
A1
[12]
3
(iii)
E.g.
 1 3 3 


P   0 2 5 
3 1 1 


1 0
0 


n
0 
D  0 3

n

 0 0  4  
B1
Use of eigenvectors (ft) as columns
M1
Use of 1, 3, −4 (ft) in correct order
n not required for M1
A1
Power n
4 n A0
[3]
9
4756
Mark Scheme
Question
4
(i)
Answer
January 2013
Marks
Guidance
1 x 5 x
e  e
2
2
1 x 5 x
e  e
2
2
y  3sinh x  2cosh x
dy
 3cosh x  2sinh x
dx
dy
3
At TPs,
 0  tanh x 
dx
2

which has no (real) solutions
y  0  tanh x 
2
3
1 1  23
 x  ln
2 1  23
1
 x  ln 5
2
2
d y
 3sinh x  2cosh x  y
dx 2
d2 y
so y  0  2  0
dx
B1
M1
A1(ag)
Considering
dy
0
dx
e 2 x  5 ; e x  0 and e  x  0
Showing no real roots www
3
Solving y = 0 as far as e 2 x or tanh x etc.
e 2 x  5 ; cosh x 
M1
Solving as far as x
Attempt to verify
Award M1 for substituting x  12 ln 5
A1(ag)
Completion www
and M1 for clearly attempting to evaluate
exactly
5
; sinh x 
2
M1
5
3sinh  12 ln 5   2cosh  12 ln 5   0 must
B1(ag)
be explained, e.g. connected with y = 0
[7]
4
(ii)
B2
For a curve with the following features:
 increasing
 intersecting the positive x-axis
 (0, −2) indicated
 gradient increasing with large x

one point of inflection
[2]
10
Award B1 for a curve lacking one of these
features
4756
Mark Scheme
Question
4
(iii)
Answer
 3sinh x  2cosh x 
Marks
1
0
ln 5
B1
y 2 dx
1
5 2
13
   sinh 2 x  3cosh 2 x  x 
2 0
4
M1
Using double “angle” formulae or complete
alternative
Condone sign errors but need
A1
Accept unsimplified
1 2 x 25 2 x 5
e  e 
4
4
2
M1
Attempting to integrate their y2
(ignore limits)
A2
Correct results and limits c.a.o.
Ignore omitted π
ln 5
M1
13 5
13 12

     3   ln 5  3
4
5
5
4


M1
1
ln 5
2
OR
Guidance
2
 9sinh 2 x  12sinh x cosh x  4cosh 2 x
9
  cosh 2 x  1  6sinh 2 x  2  cosh 2 x  1
2
13
5
 cosh 2 x  6sinh 2 x 
2
2
V 2
January 2013
25
5 
1
   e 2 x  e 2 x  x 
8
2 0
8
5 5 5

     ln 5  3
8 8 4

 5

  3  ln 5
 4

Substituting both of their limits
Obtaining exact values of sinh  ln 5  and
cosh  ln 5 
A2
Correct results and limits
M1
Substituting both of their limits
M1
Obtaining exact values of e 2 x and e 2 x
A1(ag)
1
s
2
Give A1 for one error, or for all three
terms correct and incorrect limits
sinh  ln 5  
12
13
, cosh  ln 5  
5
5
Give A1 for one error, or for all three
terms correct and incorrect limits
e2 x  5 , e2 x 
1
5
Completion www
[9]
5
(i)
4
y
2
x
−6
−4
−2
2
4
6
−2
B2
B1
[3]
Three curves of correct shape
Correctly identified
11
Give B1 for two correct curves
a = 0, a = 1, a = 2 from left to right
4756
Mark Scheme
Question
5
(ii)
Answer
2
Marks
January 2013
Guidance
y
1
x
−2
2
−1
5
(iii)
Asymptote
5
(iv)
a = −1: cusp
a = −2: loop
5
(v)
r  sec  a cos   r cos  1  a cos 2 
 x2 
 x 1 a 2 
r 
 x2 
 x 1  a  2
2 
x y 
 x2 
 x2  2
2
 x2  y 2  a 

y

a


x
 x 1
 x 1
Hence asymptote at x = 1
5
(vi)
Curve exists for y2 ≥ 0
 1 
 a
 1  0
 x 1
If a > 0 then x – 1 > 0 and so a ≥ x – 1
i.e. 1 < x ≤ 1 + a
If a < 0 then x – 1 < 0 and so a ≤ x – 1
i.e. 1 + a ≤ x < 1
B1
B1
[2]
B1
[1]
B1
B1
[2]
Curve for a = −1
Curve for a = −2
M1
Using x  r cos 
M1
Using r 2  x 2  y 2
M1
A1(ag)
Making y2 subject
B1
[5]
M1
Considering y2 ≥ 0
M1
A1(ag)
M1
A1
[5]
12
Curve with cusp
Curve with loop
4756
Mark Scheme
Question
1
Answer
f  x   1  2 x 
(a)
Marks
June 2013
Guidance
2

f   x   2 1  2 x   2  4 1  2 x 

f   x   24 1  2 x 

f   x   192 1  2 x 

f  0   1, f   0   4,
3
4
5
3
M1
A1
A1
Derivative in the form k(1 − 2x)−3 o.e.
Any correct form www
Any correct form www
A1
Any correct form www
M1
Using Maclaurin series with derivatives
evaluated at x = 0
For first derivative
f   0   24, f   0   192


24 2 192 3
x 
x  ...
2!
3!
f  x   1  4 x  12 x 2  32 x 3  ...
f  x  1  4x 
Valid for 1  2 x  1
1
1
  x
2
2
SR: after M0M0 B2 for correct
binomial
A1
B1
Must have r! in denominator
Strict inequalities
[7]
1
1
(b)
(b)
(i)

(ii)
B2
[2]

6
ra

a 3
6
2
 a
and y  a sin 
6 2

x  a cos

For a complete loop correct at the origin
and at the extremity
B1
s.o.i.
B1
s.o.i.
Using x  r cos and y  r sin  with a
value of θ
M1
A1
Both. Condone 0.87a
[4]
5
Ignore beyond 0 ≤ θ ≤ π/3.
Incomplete loop B0.
Give B1 for wrong shape at one of
origin or extremity
4756
Mark Scheme
Question
Answer

1
(b)
(iii)
1
A   3 a 2 sin 2 3 d
0 2

1
 a 2  3 1  cos 6 d
0
4
Marks
M1
A1
M1
Guidance
An integral expression including sin 3
Correct integral expression with limits
1 1
Using sin 2 3   cos 6 and
2 2
attempting integration. Dep. on 1st M1
2

1 
1
3
 a 2   sin 6 
4 
6
0
1
  a2
12
June 2013
Limits may be inserted below
Allow sign and factor errors, but must
be cos 6θ
A1
Correct result of integration
1
1
i.e.  sin 2 3 d    sin 6
2
12
A1
Dependent on previous A1
Allow awrt 0.26a 2
[5]
6
4756
Mark Scheme
Question
2
(a)
(i)
Answer
June 2013
Marks
cos5  jsin 5   cos   jsin  
Guidance
5
 c 5  5c 4 js  10c 3 j2 s 2  10c 2 j3 s 3  5cj4 s 4  j5 s 5
M1
Expanding  c  js  (real terms only)
Allow one error. Must get beyond 5C2.
Must collect terms
M1
Separating real part and replacing s 2
with 1  c 2
Independent of M1
5
 c 5  10c 3 s 2  5cs 4  j  5c 4 s  10c 2 s 3  s 5 

cos5  c 5  10c 3 1  c 2   5c 1  c 2 
2
 c 5  10c 3  10c 5  5c 1  2c 2  c 4 
 16cos5   20cos3   5cos 
2
(a)
(ii)
A1(ag)
[3]
  18  cos5  0 *
 16cos5   20cos3   5cos   0
cos  0  16cos 4   20cos 2   5  0
20  20  4  16  5
2  16
2

cos 2  

 5  5 2
 5  5 2
cos    
 or  

 8 
 8 
1
Completion www in real part
B1
This equation s.o.i.
M1
A1
Solving a 3-term quadratic
Unsimplified values of cos2θ
Allow one error
1
SC Answers unsupported www B1
1
 5  5 2
cos18 is closest to 1  cos18  

 8 
cos 2 18  sin 2 18  1
5 5
 sin 2 18  1

8
3 5
 sin 2 18 
and sin18  0
8


A1(ag)
Justifying selection of this root
M1
Using cos 2   sin 2   1
A1
Must have this form
1

 3  5 2
sin18  

 8 

[6]
7
To include *
4756
Mark Scheme
Question
2
(b)
(i)
Answer
4 3  4 j  8e
j
Marks

6
June 2013
Guidance
Condone decimal equivalents for
arguments throughout (to 2 s.f.).

B1B1
8,
B1ft
3
B1ft
1

of their
6
3
Radians only
11
18
Radians only
6
Cube roots are re j
r3  8  r  2
3 

6


 
2
3


18
their 8
M1
 13 25
,
18 18
2
,
18
A1
Accept 
B1
Approx. order 3 rotational symmetry.
1st root in 0 < arg z < π/4
2nd root in 2nd quadrant
3rd root in 5π/4 < arg z < 3π/2
Im
1
Re
−2
−1
1
−1
2
−2
[7]
2
(b)
(ii)
1   13
arg w   
2  18 18
n  18
 7

 18
B1
B1
[2]
8
Ignore numbers etc. on diagram
4756
Mark Scheme
Question
3
(i)
Answer
Marks
det  A   k  4  9   7  4  3  4  6  2 
M1A1
Obtaining det  A  in terms of k
 13k  65
 no inverse if k = 5
B1(ag)
May be verified separately
At least 4 cofactors correct
(including one involving k)
Six signed cofactors correct
Transposing and ÷ by det  A  .
Dependent on previous M1M1
M1
26
13 
 13
1


A 
7 2k  4 3k  8 
13k  65 

 4 3k  7 2k  14 
A1
1
3
(ii)
M1
When k = 4,
 x
 13 26 13  p 
1 
 
 
 y   13  7 12 4  1 
z
 4 5

6 
 

 2 
OR e.g.
6 x  13 y  p  4
4 x  13 y  3 p  4
 x  p  4
A1
[7]
M1
Substituting k = 4
M2
Correct use of inverse

Guidance
Allow one error. isw unsimplified
65 − 13k M1A0 and allows B1 below
M0 if more than 1 is multiplied by the
corresponding element
Mark final answer
One correct element.
Condone missing determinant.
M0 if wrong order
Eliminating one unknown in two
different ways and reaching one
unknown in terms of p
Finding the other two unknowns
M2
M1
 x
 13 p  52 
1 
 

 y   13  7 p  20 
z
 4 p  17 
 


June 2013
A2
Dependent on all M marks.
Terms must be collected.
Give A1 for one correct
[5]
9
x  p  4 , y  
7
20
p
,
13
13
4
17
p
13
13
λ × correct vector (λ ≠ 0) A1
z
4756
Question
3 (iii)
Mark Scheme
Answer
e.g. 7 x  13 y  p  4, 7 x  13 y  3 p  4
(or 4 x  13 z  7  2 p, 4 x  13 z  1 )
(or 8 y  14 z  p  10, 4 y  7 z  3 )
Marks
For solutions, p  4  3 p  4
M2

A1
p4
M2
OR
p4
June 2013
Guidance
Or 7 x  13 y  8
Or 8 x  26 z  3 p  14
Or 4 y  7 z  5  2 p
Eliminating one unknown in two
different ways & obtaining a value of p
A method leading to an equation from
which p could be found
E.g. setting z = 0, augmented matrix,
adjoint matrix, etc.
Accept unknown instead of λ
x  137   87 , y   , z   74   73
A1
7
8
4
1
x  , y    , z    
13
13
13
13
M1
A1
Obtaining general soln. by e.g. setting
one unknown = λ and finding equations
involving the other two and λ
Any correct form
Straight line
B1
Accept “sheaf”, “pages of a book”, etc.
[6]
10
x   134   14 , y   74   34 , z  
Independent of all previous marks.
Ignore other comments
4756
Mark Scheme
Question
4
4
(i)
(ii)
Answer
u
Marks
2u
dy
1
1
 
 
2
dx
1  sinh y
1  x2
e

e y  x  1  x2

y  ln x    1  x 2
y
B1(ag)
Numerators of both expressions
Accept other variables
Completion www
Both expressions s.o.i. and multiplication
Completion www
M1
sinh y = … and differentiating w.r.t. y or
x
A1
o.e.
A1(ag)
y is an increasing function so take + sign
e y  e y
x  sinh y  x 
2
y
y
 e  e  2x
 e 2 y  2 xe y  1  0

B1
[2]
y  arsinh x  x  sinh y
dx

 cosh y
dy
dy
1


dx cosh y

Guidance
2 u
e e
e 2e
 cosh 2 u 
2
4
u
u
2u
e e
e  2  e 2u
sinh u 
 sinh 2 u 
2
4
 cosh 2 u  sinh 2 u  1
OR cosh u  sinh u  eu
cosh u  sinh u  e  u
 cosh 2 u  sinh 2 u  eu  e  u
B1
 cosh 2 u  sinh 2 u  1
B1(ag)
cosh u 
u
June 2013
Completion www with valid
intermediate step
B1
Validly rejecting negative value
B1
x in exponential form
M1
Obtaining quadratic in e y
M1
Solving to reach e y . Dep. on M1 above
 x   1  x2
Or cosh y
dy
 1 or differentiating (*)
dx
dy
1

as final answer or 
dx
1  x2
not considered scores max. 3/4
Or cosh y ≥ 1, or cosh y > 0
2


x  1  x 2  0 so take + sign
(*)
A1(ag)
B1
[9]
Allow one slip
Completion www
Validly rejecting negative root
11
e.g. e y > 0
4756
Mark Scheme
Question
4
(iii)

2
0
Answer
1
1 2
1
dx  
dx
2
0
2
4
3
4  9x
9  x
Marks
M1
2
1
3x 
  arsinh 
3
2 0
A1A1
Guidance
Integral involving arsinh
1 3x
,
o.e.
3 2
1
 arsinh3
3
2
1 
4 
OR  ln  x  x 2   
3  
9  
0
2
dx 2
 cosh u
OR x  sinh u 
3
du 3
2
ln  3 10  1
1
du
0 4  9 x 2 dx  0
3

1
 ln 3  10
3

June 2013

Integral in form ln kx  k 2 x 2  ...
M1

1
4
, x  x 2  or 3x  9 x 2  4
3
9
Using a sinh substitution
Correct substitution
A1A1
M1
A1
Or
3x
9 x2

1
2
4
1
 3 du
A1
A1(ag)
Completion with valid intermediate
step(s)
[4]
12
Condone omitted brackets
4756
Mark Scheme
Question
4
(iv)
Answer

1
0
1
1  x2
Marks
1
1
OR inspection

Guidance
arsinh x dx
1
2
  arsinh x    
arsinh x dx

0 0 1  x2
1
arsinh x dx   u du
OR 
M1
1  x2

June 2013
1
2
 arsinh x 
2
1
I  ln 1  2
2
 
M1
M1

2
Parts with u  arsinh x , v 
1
1  x2
Allow one error
Allow equivalent form
Substitution with u = arsinh x
or x = sinh u
Must reach u du
Recognising integrand as k(arsinh x)2
k≠0
A1
A correct indefinite integrand
A1
This answer only
[3]
13

Mark final answer