MS3
£3.00
GENERAL CERTIFICATE OF EDUCATION
TYSTYSGRIF ADDYSG GYFFREDINOL
MARKING SCHEME
MATHEMATICS - C1-C4 & FP1-FP3
AS/Advanced
SUMMER 2008
INTRODUCTION
The marking schemes which follow were those used by WJEC for the Summer 2008
examination in GCE MATHEMATICS. They were finalised after detailed discussion at
examiners' conferences by all the examiners involved in the assessment. The conferences
were held shortly after the papers were taken so that reference could be made to the full
range of candidates' responses, with photocopied scripts forming the basis of discussion.
The aim of the conferences was to ensure that the marking schemes were interpreted and
applied in the same way by all examiners.
It is hoped that this information will be of assistance to centres but it is recognised at the
same time that, without the benefit of participation in the examiners' conferences, teachers
may have different views on certain matters of detail or interpretation.
WJEC regrets that it cannot enter into any discussion or correspondence about these
marking schemes.
Mathematics C1 May 2008
Solutions and Mark Scheme
1.
(a)
Gradient of AB = increase in y
increase in x
Gradient of AB = –1/2
M1
(or equivalent)
A1
(b)
A correct method for finding the equation of AB using the candidate’s
value for the gradient of AB.
M1
1
Equation of AB :
y – 4 = – /2 [x – (–7)]
(or equivalent)
A1
(f.t. the candidate’s value for the gradient of AB)
Equation of AB :
x + 2y – 1 = 0
(f.t. one error if both M1’s are awarded) A1
(c)
A correct method for finding the length of AB
AB = √125
M1
A1
(d)
A correct method for finding E
E(–2, 1·5)
M1
A1
(e)
Either:
An attempt to find the gradient of a line perpendicular to AB using the
fact that the product of the gradients of perpendicular lines = –1. M1
An attempt to find the gradient of the line passing through C and D
M1
2 = 1 – (–15)
(Equating expressions for gradient)
M1
6–k
k = –2
(f.t. candidate’s gradient of AB)
A1
Or:
An attempt to find the gradient of a line perpendicular to AB using the
fact that the product of the gradients of perpendicular lines = –1. M1
An attempt to find the equation of line perpendicular to AB passing
through C(or D)
M1
–15 – 1 = 2(k – 6)
(substituting coordinates of unused point in the equation) M1
k = –2
(f.t. candidate’s gradient of AB)
A1
1
2.
(a)
(b)
3.
Either:
√75 = 5√3
9 = 3√3
√3
√6 × √2= 2√3
√75 – 9 + (√6 × √2) = 4√3
√3
Or:
√75 = 15
√3
√6 × √2 = 6
√3
√75 – 9 + (√6 × √2) = 12
√3
√3
√75 – 9 + (√6 × √2) = 4√3
√3
B1
B1
B1
(c.a.o.) B1
B1
B1
B1
(c.a.o.) B1
5√5 – 2 = (5√5 – 2 )(4 – √5)
M1
4 + √5
(4 + √5)( 4 – √5)
Numerator:
20√5 – 25 – 8 + 2√5
A1
Denominator:
16 – 5
A1
5√5 – 2 = 2√5 – 3
(c.a.o.) A1
4 + √5
Special case
If M1 not gained, allow B1 for correctly simplified numerator or
denominator following multiplication of top and bottom by 4 + √5
dy = 6x – 8
dx
(an attempt to differentiate, at least one non-zero term correct) M1
An attempt to substitute x = 2 in candidate’s expression for dy
m1
dx
(c.a.o.) A1
Value of dy at P = 4
dx
Gradient of normal =
–1
m1
candidate’s value for dy
dx
y-coordinate of P = 3
B1
Equation of normal to C at P: y – 3 = –1/4(x – 2)
(or equivalent) A1
(f.t. one slip in either candidate’s value for dy or candidate’s value for the
dx
y-coordinate at P provided M1and both m1’s awarded)
2
4.
5.
(a)
y + δy = 5(x + δx)2 + 3(x + δx) – 4
Subtracting y from above to find δy
δy = 10xδx + 5(δx)2 + 3δx
Dividing by δx and letting δx → 0
dy = limit δy = 10x + 3
dx δx→ 0 δx
(b)
dy = –8 × x–2 + 3 × 1 × x–1/2
dx
2
–2
Either 4 = 1 or 4–1/2 = 1
16
2
dy = 1
dx 4
B1
M1
A1
M1
(c.a.o.) A1
B1, B1
B1
(c.a.o) B1
(a)
a=3
b = –13
B1
B1
(b)
2b on its own or least (minimum) value = 2b, with correct explanation
or no explanation
B1
x = –a
B1
Note: Candidates who use calculus are awarded B0, B0
6.
(5 + 2x)3 = 125 + 150x + 60x2 + 8x3
7.
(a)
Use of f (2) = 0
M1
32 + 4p – 22 + q = 0
A1
Use of f (–1) = 9
M1
– 4 + p + 11 + q = 9
A1
Solving simultaneous equations for p and q
M1
p = – 4, q = 6
(convincing) A1
Note:
Candidates who assume p = – 4, q = 6 and then verify that x – 2 is a
factor and that dividing the polynomial by x + 1 gives a remainder of 9
may be awarded M1 A1 M1 A1 M0 A0
(b)
f (x) = (x – 2)(4x2 + ax + b) with one of a, b correct
M1
f (x) = (x – 2)(4x2 + 4x – 3)
A1
f (x) = (x – 2)(2x – 1)(2x + 3)
A1
(f.t. only for f (x) = (x – 2)(2x + 1)(2x – 3) from 4x2 – 4x – 3)
3
Two terms correct
Three terms correct
All 4 terms correct
B1
B1
B1
8.
(a)
y
(1, 4)
O
(–3, 0)
(5, 0)
x
Concave down curve with maximum (1, 4) or (5, 4)
Two of the three points correct
All three points correct
B1
B1
B1
(b)
y
(3, 7)
(0, 5)
x
O
Concave down curve with maximum (3, 7) or (3, 1)
Maximum point (3, 7)
Point of intersection with y-axis (0, 5)
4
B1
B1
B1
9.
10.
dy = –6x2 + 6x + 12
dx
Putting derived dy = 0
dx
x = 2, –1
(both correct)
B1
M1
(f.t. candidate’s dy)
dx
Stationary points are (2, 15) and (–1, –12) (both correct)
(c.a.o)
A correct method for finding nature of stationary points
(2, 15) is a maximum point
(f.t. candidate’s derived values)
(–1, –12) is a minimum point
(f.t. candidate’s derived values)
A1
M1
A1
A1
(a)
B1
(b)
Finding critical points x = –1·5, x = 3
A statement (mathematical or otherwise) to the effect that
x ≤ –1·5 or 3 ≤ x
(or equivalent)
(f.t. candidate’s critical points)
Deduct 1 mark for each of the following errors
the use of strict inequalities
the use of the word ‘and’ instead of the word ‘or’
(i)
(ii)
A1
B2
An expression for b2 – 4ac, with b = (–)6 and at least one of
a or c correct
M1
2
2
b – 4ac = [(–)6] – 4 × 3 × m
A1
b2 – 4ac < 0
m1
m>3
(c.a.o.) A1
3x2 – 4x + 7 = 2x + k
M1
2
No points of intersection ⇔ 3x – 6x + (7 – k) = 0 has no real
roots
(allow one slip in quadratic)
m1
4>k
A1
(if candidate uses (b)(i), f.t. candidate’s inequality for m)
5
Mathematics C2 May 2008
Solutions and Mark Scheme
1.
0
1
0·2
1·060596059
0·4
1·249358235
(2 values correct)
B1
0·6
1·586018915
(4 values correct)
B1
Correct formula with h = 0·2
M1
I ≈ 0·2 × {1 + 1·586018915 + 2(1·060596059 + 1·249358235)}
2
I ≈ 0·72059275
I ≈ 0·721
(f.t. one slip)
A1
Special case for candidates who put h = 0·15
0
1
0·15
1·033939138
0·3
1·137993409
0·45
1·318644196
0·6
1·586018915
(all values correct)
B1
Correct formula with h = 0·15
M1
I ≈ 0·15 ×{1 + 1·586018915 + 2(1·033939138 + 1·137993409 +
2
1·318644196)}
I ≈ 0·71753793
I ≈ 0·718
(f.t. one slip)
A1
6
2.
(a)
tan θ = 1·5
(c.a.o.) B1
θ = 56·31°
B1
θ = 236·31°
B1
Note: Subtract (from final two marks) 1 mark for each additional root
in range, ignore roots outside range.
(b)
3x = 25·84°, 334·16°, 385·84°, 694·16°
(one value) B1
x = 8·61°
(f.t candidate’s value for 3x)
B1
x = 111·39°, 128·61°
(f.t candidate’s value for 3x) B1, B1
Note: Subtract (from final two marks) 1 mark for each additional root
in range, ignore roots outside range.
(c)
sin2θ – 4(1 – sin 2θ ) = 8 sin θ
(correct use of cos2θ = 1 – sin 2θ ) M1
An attempt to collect terms, form and solve quadratic equation
in sin θ, either by using the quadratic formula or by getting the
expression into the form (a sin θ + b)(c sin θ + d),
with a × c = coefficient of sin2θ and b × d = constant
m1
5 sin2θ – 8 sin θ – 4 = 0 (5 sin θ + 2)(sin θ – 2) = 0
sin θ = –2,
(sin θ = 2)
(c.a.o.)
A1
5
θ = 203·58°, 336·42°
B1 B1
Note: Subtract (from final two marks) 1 mark for each additional root
in range from 5 sin θ + 2 = 0, ignore roots outside range.
sin θ = – , f.t. for 2 marks, sin θ = +, f.t. for 1 mark
3.
(a)
15 = 1 × x × (x + 4) × sin 150°
(correct use of area formula) M1
2
Either x(x + 4) = 60 or expressing the equation correctly in the form
A1
ax2 + bx + c = 0
x = 6 (negative value must be rejected)
(c.a.o.) A1
(b)
BC2 = 62 + 102 – 2 × 6 × 10 × cos 150°
(correct substitution of candidate’s derived values in cos rule)
M1
BC = 15·5 cm
(f.t. candidate’s derived value for x) A1
7
4.
5.
(a)
Sn = a + [a + d] + . . . + [a +(n – 2)d] + [a + (n – 1)d]
(at least 3 terms, one at each end) B1
Sn = [a + (n – 1)d] + [a + (n – 2)d] + . . . + [a + d] + a
2Sn = [a + a + (n – 1)d] + [a + a + (n – 1)d] + . . . + [a + a + (n – 1)d]
(reverse and add)
M1
2Sn = n[2a + (n – 1)d]
Sn = n[2a + (n – 1)d]
(convincing) A1
2
(b)
10 × [2a + 9d] = 320
B1
2
2a + 9d = 64
[a + 11(12)d] + [a + 15(16)d] = 166
M1
[a + 11d] + [a + 15d] = 166
A1
2a + 26d = 166
An attempt to solve the candidate’s two equations simultaneously by
eliminating one unknown
M1
d = 6, a = 5 (both values)
(c.a.o.) A1
a + ar = 7·2
a = 20
1–r
A valid attempt to eliminate a
20(1 – r) + 20(1 – r)r = 7·2
r = 0·8
(r = –0·8)
r = 0·8 and a = 4
B1
B1
M1
(a correct quadratic in r)
A1
(c.a.o.) A1
(f.t. candidate’s positive value for r) A1
8
6.
(a)
5 × x3/2 – 4 × x1/3
3/2
1/3
(b)
(i)
(ii)
( + c)
B1,B1
4 – x2 = 3x
An attempt to rewrite and solve quadratic equation
in x, either by using the quadratic formula or by getting the
expression into the form (x + a)(x + b), with a × b = – 4
(x – 1)(x + 4) = 0 x = 1 (– 4)
(c.a.o.)
A(1, 3) (f.t. candidate’s x-value, dependent on M1 only)
B(2, 0)
M1
m1
A1
A1
B1
Area of triangle = 3/2
(f.t. candidate’s coordinates for A)
B1
2
Area under curve = (4 – x2 ) dx
(use of integration)
M1
1
= [4x – (1/3)x3]
2
1
(correct integration) B2
= [(8 – 8/3) – (4 – 1/3)]
(substitution of candidate’s limits) m1
= 5/3
Use of candidate’s, xA, xB as limits and trying to find total area
by adding area of triangle and area under curve
m1
Total area = 3/2 + 5/3 = 19/6
(c.a.o.) A1
9
7.
(a)
(b)
(c)
8.
(a)
Let p = logax
Then x = ap
xn = apn
∴loga xn = pn
∴loga xn = pn = n logax
loga (3x + 4) – loga x = 3 loga2.
loga 3x + 4 = loga23
x
3x + 4 = 23
x
x = 0·8
(c)
(use of subtraction law)
B1
(use of power law)
B1
(removing logs)
(c.a.o) B1
(f.t. for 23= 6, 9 only) B1
Either:
(3y + 2) log 10 4 = log 10 7
(taking logs on both sides and using the power law)
y = log 10 7 ± 2 log 10 4
3 log 10 4
y = – 0·199
(c.a.o.)
Or:
3y + 2 = log 4 7
(rewriting as a log equation)
y = log 4 7 ± 2
3
y = – 0·199
(c.a.o.)
(i)
(ii)
(iii)
(b)
(relationship between log and power) B1
(the laws of indicies) B1
(relationship between log and power)
(convincing) B1
A(5, 3)
A correct method for finding radius
Radius = √65
Equation of C:
(x – 5)2 + (y – 3)2 = (√65)2
(f.t. candidate’s coordinates of A.)
Either:
An attempt to substitute the coordinates of R in the equation of C
Verification that L.H.S. of equation of C = 65 R lies on C
Or:
An attempt to find AR2
AR2 = 65 R lies on C
Either:
RQ = √26
(RP = √234)
cos QPR = √234
sin QPR = √26
2√65
2√65
QPR = 18·43°
Or:
RQ = √26 or RP = √234
(√26)2 = (√234)2 + (2√65)2 – 2 × (√234) × (2√65) × cos QPR
(correct use of cos rule)
QPR = 18·43°
10
M1
m1
A1
M1
m1
A1
B1
M1
A1
B1
M1
A1
M1
A1
B1
M1
A1
B1
M1
A1
9.
Let AÔB = θ radians
(a)
6 × θ = 5·4
θ = 0·9
Area of sector AOB = 1 × 62 × θ
2
Area of sector AOB = 16·2 cm2
(b)
M1
A1
M1
(convincing ) A1
Area of triangle AOB = 1 × 62 × sin θ
M1
2
(f.t. candidate’s value for θ ) A1
Area of triangle AOB = 14·1 cm2
Shaded area = 2·1 cm2
(f.t. candidate’s value for θ ) A1
11
A LEVEL MATHEMATICS
PAPER C3
Summer 2008
Marking Scheme
1.
h = 0·25
Integral
M1 (h = 0·25, correct formula)
0.25
3
[0·4142136 + 1·9282847 +
B1 (3 correct values)
4 (1·5112992 + 1·7655028)
B1 (2 correct values)
+ 2(1·6274893)]
1·642
A1 (F.T. one slip)
4
2.
(a)
θ=
π
6
B1 (choice of θ and one correct
, for example
evalution)
tan 2θ =
2 tan
1 + tan 2
tan
=
π
=
3
2
3
1 + 13
3
1·73.
≈ 0 ⋅ 866
B1 (2 correct evaluations)
(Statement is false)
(b)
2 (1 + tan2 θ) = 8 – tan θ
M1 (use of correct formula
sec2θ = 1 + tan2θ)
2 tan2 θ + tan θ – 6 = 0
M1 (attempt to solve quadratic)
(2 tan θ – 3)(tan θ + 2) = 0
tan θ =
3
,–2
2
A1
θ = 56·3º, 236·3º, 116·6º, 296·6º
(allow to nearest degree)
B1 (56·3º, 236·3º)
B1 (116·6º)
B1 (296·6º)
8
12
3.
2x + sin y + x cos y
dy
dx
+ 3y2
dy
dx
=0
B1 (sin y + x cos y
dy
B1 (3y2
2+0−
dy
dx
+3
2
dy
dx
=0
dx
dy
)
dx
)
B1 (correct differentiation
of x2 ,
dy
−2
= 2
≈ 0 ⋅ 0699
dx 3π − 1
3
B1 (F.T.
(allow −0·07(0))
, 1)
d
dy
(sin y) = – cos y)
4
4.
(a)
dy
dt
= 2e 2 t
M1 (ke2t , k =
1
, 1, 2)
2
A1 (k = 2)
dy
dt
=
dy
B1
t
= 2te2t
dx
(b)
1
(o.e.)
A1 (C.A.O)
d dy
= 2te 2t + 4te 2t
dt dx
M1 (f (t) e2t + 2t g (t))
A1 (f(t) = 2,
g(t) = 2e2t)
d2y
dx
2
= 2te 2t (1 + 2t )
(o.e.)
M1 (use of
d2y
dx
2
=
d
dy
dt dx
÷
dx
dt
A1 (F. T candidates
)
d
dy
dt dx
)
8
13
5.
(a)
dy
1
=
dx
1 − x2
dy
=0
dx
1
− 3x 2
1= 3 x
B1, B1
1 − x2
M1 (move one term to other side,
and attempt to square both sides)
1 = 9 x (1 − x 2 )
9 x 3 − 9 x +1= 0
9 x3 − 9 x + 1
x
(b)
A1
0
0⋅2
M1 (attempt to check values
or signs)
1
− 0 ⋅ 728
Change of sign indicates
is between 0 and 0·2
x0 = 0·1, x1 = 1·1121111, x2 = 0·11252022, x3 = 0·1125357
x3
0·11254
A1 (correct values or signs and
conclusion)
B1 (x1 )
B1 (x3 )
Check 0·112535, 0·112545
x
0·112535
0·112545
9 x3 – 9x + 1
0·000011
– 0·00008
M1 (attempt to check values
or signs)
A1 (correct values or signs, F.T. x3)
Change of sign indicates α is 0·11254 correct to 5 dec places
A1
11
14
6.
(a)
B1 (2 correct x values)
B1 (y = – 12 for st pt)
B1 (all correct, correct shape)
(b)
4|x|=1
x=±
(c)
B1 (a | x | = b, a = 4, b = 1)
1
4
B1 (both values, F.T. a, b)
2x – 9 > 3
x>6
B1
2x – 9 < – 3
x>3
x > 6 or x > 3
M1
A1
A1 (answer must contain 'or') (o.e.)
or
(o.e.) e.g. (–
, 3) ∪ (6, )
alternatively
(2x – 9)2 > 9
4x2 – 36x + 72 > 0
x2 – 9 x + 18 > 0
B1 (for x > 6 )
(x – 3) ( x – 6) > 0
M1 A1 for x < 3
x>6
A1 for union of intervals
or
x<3
9
15
7.
(a)
−
(i)
cos 3 x
3
1
3
(+ C )
M1 (k cos 3x, k = ± ,–1, –3)
1
A1 (k = − )
3
2
(ii)
3
ln | 3x + 5 | (+ C)
M1 (k ln | 3x + 5 | )
e3 x + 4
(+ C )
3
(iii)
A1 (k =
2
3
)
M1 (ke3x+4) A1 (k =
−
(b)
1
1
6(2 x + 1)
M1 (a (2x + 1) –3,
3
0
a±
1
1 2 1
,− ,− – )
6 3 3 8
A1 (a = −
1
−
=
13
≈ 0 ⋅160
81
3
6×3
+
1
=
1
)
3
1
6
)
A1
6
A1 (C.A.O, either answer)
10
16
8.
(a)
– 2 cosec2 2x
M1 (k cosec2 2x) A1 (k = – 2)
(b)
x2 x + 2x ln x = x + 2x ln x
1
M1 (x2 f(x) + lnxg (x))
A1 (f(x) =
1
, g (x) = 2x
x
and simplification)
(c)
( x 2 − 2)(2 x) − ( x 2 + 1)(2 x )
M1
( x 2 − 2) 2
( x 2 − 2) f ( x ) − ( x 2 + 1) g ( x )
( x 2 − 2) 2
A1 (f(x) = 2x, g(x) = 2x)
=
−6 x
A1 (simplified form, C.A.O.)
( x − 2) 2
2
7
9.
(a)
Range is [– 2, ),
B1
(b)
Let y – (x + 1)2 – 2
y + 2 = (x + 1) 2
M1 (y ± 2 = (x + 1)2)
and attempt to isolate x
x = –1±
y+2
A2 (A1 for – 1 + y + 2
Take – ve square root since domain is x
–1
A1 (correct choice of sign)
x= –1– x+2
f – 1 (x) = – 1 –
x+2
B1 (F.T expression for x)
Domain is [– 2, ), Range is (–
(or x
– 2)
(or f – 1 (x)
, – 1]
B1 (both F.T. from (a))
– 1)
7
17
10.
(a)
f(–1) = 2 e–1
0·736
B1
gf–1 cannot be formed since domain
of g is x
(b)
1
B1
fg(x) = 2e 3lnx
B1
= 2e lnx3
B1
= 2x3
B1
Domain is [1, ), range is [2, ) (o.e.)
B1, B1
7
18
MATHEMATICS C4
1.
(a)
Let
A B
C
1
≡ + 2 +
2x − 1
x (2 x − 1) x x
M1 (Correct form)
2
1 ≡ Ax(2 x − 1) + B(2 x − 1) + Cx 2
x=0
1 = B(− 1)
x = 12
1 = C 14
x
M1 (correct clearing and
attempt to substitute)
∴ B = −1
∴C = 4
A1 (2 constants C.A.O.)
0 = 2 A + C ∴ A = −2
2
A1 (third constant, F.T.
one slip)
(no need for display)
(b)
−2 1
4
dx
− 2 +
x
2x − 1
x
1
+ 2 ln 2 x − 1
x
(+C )
= −2 ln x +
2x + x
2.
B1,B1.B1
7
dy
dy
+ y + 4y
=0
dx
dx
dy
+ y)
dx
dy
B1 (4 y )
dx
B1 ( x
dy
=5
dx
B1 (C.A.O.)
Gradient of normal = −
1
5
M1 (
−1
dy
candidate' s
dx
,
numerical value)
Equation of normal is y − 1 = − 15 ( x + 3)
A1 (F.T. candidate’s value)
5
3.
(a)
R sin α = 2 , R cos α = 3
R = 13 , α =
B1 ( R = 13 )
( ) = 33.7°or 34°
−1
tan 23
M1 (correct method for α )
A1 ( α = 34° )
(b)
cos( x − 33.7°) =
1
13
x − 33.7° = 73.9°,286.1°
x = 107.6°, 319.8°
B1 (one value)
B1, B1
19
6
4
Volume = π
4.
x+
3
x
1
2
dx
4
=π
x2 + 6 x +
1
B1
9
dx
x
M1 (attempt to square, at
least 2 correct terms)
A1 (all correct)
3
x3
=π
+ 4 x 2 + 9 ln x
3
4
A3 (integration of 3 terms,
1
F.T. similar work
Ax 2 + B x +
= π [49 + 9 ln x ] ~
− 193.1
or 61.48π
c
)
x
A1 (C.A.O.)
7
5. (a)
•
dy − 2 sin 2t
=
dx
4 cos t
dy y
M1 (
= )
dx •
x
B1 ( 4 cos t )
M1 ( k sin 2t , k = −1,±2,−
A1 ( k = −2 )
− 4 sin t cos t
=
= − sin t
4 cos t
1
)
2
M1 (correct use of formula)
A1 (C.A.O.)
(b)
Equation of tangent is y − cos 2 p = − sin p ( x − 4 sin p )
M1 ( y − y1 = m( x − x1 ) )
x sin p + y = cos 2 p + 4 sin 2 p
= 1 − 2 sin 2 p + 4 sin 2 p
M1 (attempt to use correct
formula)
A1
= 1 + 2 sin 2 p
9
20
6.
(a)
(3x + 1) e 2 x dx = (3x + 1) e
2x
2
−
3 2x
e dx
2
M1 ( f ( x )(3 x + 1) − 3 f ( x )dx )
1
2
A1 ( f ( x ) = ke 2 x , k = 1, ,2 )
A1 ( k =
= (3 x + 1)
e 2 x 3e 2 x
−
(+ C )
2
4
1
)
2
A1 (F.T. one slip)
π
(b)
2
π
9 − 9 sin 2 .3 cos d
B1 (for first line, unsimplified)
6
B1 (simplified without limits)
π
=
2
π
9 cos 2 dθ
B1 (limits)
6
π
=
2
π
1 + cos 2
2
d
M1 ( cos 2 θ = a + b cos 2θ )
6
A1 ( a = b =
9θ 9 sin 2θ
=
+
2
4
1
)
2
π
2
M1 ( k sin 2θ , k = ±
π
b
,2b, b )
2
6
A1 ( k =
=
3π 9 3
−
2
8
or
2.764
b
)
2
A1 (C.A.O.)
12
21
7.
(a)
dW
= kW
dt
(b)
dW
=
W
(k > 0)
B1
kdt
M1 (attempt to separate
variables)
ln W = kt + C
A1 (allow absence of C)
t = 0, W = 0.1, C = ln 0.1
B1 (value of C)
ln
W
= kt
0 .1
M1 (use of logs or exponentials)
W
= e kt
0 .1
k = 3.0007
B1 (value of k)
W = 0.1e 3t
A1
7
8.
(a)(i)
(ii)
(b)
AB = i + 2j – 2k
B1 (AB)
Equation of AB is r = 4i – j +k + λ (i + 2j – 2k)
M1 (reasonable attempt to write
equations)
A1 (must contain r, F.T.
candidate’s AB)
(Point of intersection is on both lines)
Equate coeffs of i and j (any two of i,j,k)
1+ µ = 4 + λ
λ=
1
3
µ=
10
3
M1 (attempt to solve)
kkkk
jjjj
iiii
kkkk
jjjj
iiii
= 3,
− +
− 2 )(
. − +
)=
A1 (F.T. value of λ or µ )
and − +
is required
= 3
kkkk
jjjj
iiii
kkkk
jjjj
iiii
( +2
kkkk
jjjj
iiii
+2 −2
kkkk
jjjj
angle between + 2 − 2
A1 (C.A.O.)
kkkk
13 1 1
− +
3
3 3
iiii
(c)
jjjj
iiii
Position vector is
M1 (attempt to write equations
using candidate’s equation)
A1 (2 correct equations, F.T.
candidate’s equations)
×
B1 (coeffs of λ and µ )
B1 (for one modulus)
cos θ
1 − 2 − 2 = 3 3 cos θ
M1 (use of correct formula)
B1 (l.h.s. unsimplified)
θ = 125.3°
A1 (C.A.O.)
13
22
9.
(1 + 3x )(1 − 2 x )
−
1
2
(
(
2
)
= (1 + 3x ) 1 + x + x + L
= 1 + 4x +
3
2
2
9x 2
+L
2
B1, B1 (unsimplified)
B1 ( 1 + 4 x )
B1 (
Exparsim is valid for x <
10.
(x
)
= (1 + 3 x ) 1 + (− 12 )(− 2 x ) + (− 12 )(− 32 ) 21! 1(− 2 x ) + L
1
2
9x 2
)
2
B1
5
+ 49 < 14 x )
2
x − 14 x + 49 < 0
( x − 7 )2 < 0
x − 7 is not real
2
B1
B1
B1
B1 (accept impossibility)
contradiction
∴x+
49
≥ 14
x
for all real and positive x
4
23
A/AS Level Mathematics - FP1 – June 2008 – Markscheme – Draft 4 (Post
Stand)
Sn =
1
n
r3 +
r =1
n
r2
M1
r =1
n 2 (n + 1) 2 n(n + 1)(2n + 1)
+
4
6
2
n(n + 1)(3n + 3n + 4n + 2)
=
12
2
n(n + 1)(3n + 7n + 2)
=
12
n(n + 1)(n + 2)(3n + 1)
=
12
=
2
(a)
A1A1
M1
A1
A1
detA = 2(2 – 2) + 4( 2 – 1) +2(1 – 2) = 2
0
1 −1
M1A1
2
Cofactor matrix = − 2 0
4 −2 0
[Accept matrix of minors]
0 −2 4
M1A1
1
0 −2
−1 2
0
[Award A0A0 if rule of signs not used]
0 −2 4
1
Inverse matrix =
1
0 −2
2
−1 2
0
Adjugate matrix =
0 −2 4 8
1
y =
1
0 −2 8
2
z
0 5
−1 2
A1
A1
x
(b)
M1
2
= −1
4
[FT from their inverse]
3
A1
(2 − i) 2 = 4 − 4i + i 2 = 3 − 4i
M1A1
7 − 4i (7 − 4i)(2 − i)
=
= 2 – 3i
M1A1
2+i
(2 + i)(2 − i)
A1
z = 3 – 4i + 2 – 3i – 8 = − 3 − 7i (cao)
B1
(b)
Mod(z) = 58 (7.62)
−1
tan (7 / 3) = 1.17 or 67°
B1
Arg(z) = 4.31 or 247°
B1
nd
nd
rd
[FT from (a) but only for the 2 B1 in arg if their z is in the 2 or 3 quadrant]
(a)
24
2 x + y + 3z = 5
4
(a)
(b)
5
− 5y + z = 7
5 y − z = k − 10
[Or equivalent row operations]
M1A1A1
2 x + y + 3z = 5
− 5y + z = 7
A1
0 = k −3
[Award this A1 for -7 = k – 10]
k = 3 (cao)
A1
Let z = α .
α−7
y=
5
16 − 8α
x=
5
[FT from (a)]
Other possible solutions are
Let x = α
− 8 −α
16 − 5α
y=
;z =
8
8
Let y = α
z = 5α + 7; x = −8α − 8
M1
A1
A1
The statement is true for n = 1 since 7 + 5 is divisible by 6.
B1
k
Let the statement be true for n = k, ie 7 + 5 is divisible by 6 (so that
7 k + 5 = 6N).
M1
k +1
k
M1A1
Consider
7 + 5 = 7×7 +5
= 7(6 N − 5) + 5
A1
A1
= 42 N − 30
This is divisible by 6 so true for n = k true for n = k + 1, hence proved by
induction.
A1
25
6
(a) Let the roots be α,β,γ. Then
b
α+β+γ= −
a
c
βγ + γα + αβ =
a
d
αβγ = −
a
[Award this B1 if candidates go straight to what follows]
Now use the fact that the roots are α, 2α and 4α.
Then
b
(1 + 2 + 4)α = −
a
c
(2 + 4 + 8) α 2 =
a
d
(1 × 2 × 4)α 3 = −
a
[Award -1 for each error]
It follows that
7α × 14α 2 − b c − a
=
× ×
a a d
8α 3
98 bc
=
8 ad
4bc = 49ad
42
3
(b)
so α =
7α =
8
4
3 3
The roots are , ,3
4 2
26
B1
M1
A2
M1A1
A1
M1A1
A1
0 −1 0
7
(a)
Rotation matrix = 1
0
0
0
0
1
B1
1 0 1
Translation matrix = 0 1 2
0 0 1
B1
1 0 1 0 −1 0
T= 0 1 2 1 0 0
0 0 1 0 0 1
[Award this B1 for writing their matrices in the right order]
0 −1 1
= 1
0
(b)
(c)
0
0
Fixed points satisfy
0 −1 1 x
B1
2
1
x
1 0 2 y = y
0 0 1 1
1
giving
− y +1 = x
x+2= y
1 3
The solution is ( x, y ) = − , .
2 2
Under T,
x′ = − y + 1
y′ = x + 2
x = y′ − 2
or
y = − x′ + 1
So y = 2x – 1 becomes
− x ′ + 1 = 2( y ′ − 2) − 1
x′ + 2 y ′ = 6
[FT their x,y]
M1
A1
M1A1
M1A1
A1
M1
A1
Alternatively, they might show that the image of (x, 2x – 1) is (2 – 2x, x + 2)
for which award M1A1A1, then M1A1 for eliminating x.
27
8
(a)
ln f ( x) = cos x ln x
f ′( x)
cos x
= − sin x ln x +
f ( x)
x
f ′( x) = x cos x − sin x ln x +
M1
A1A1
cos x
x
(b)
A1
At a stationary point,
cos x
M1
sin x ln x =
x
1
A1
tanxlnx =
x
leading to the printed result.
(c)
When α = 1.27, αlnαtanα = 0.978.. or αlnαtanα - 1 = -0.021..
M1
When α = 1.28, αlnαtanα = 1.055.. or αlnαtanα - 1 = 0.055..
Since these values are either side of 1 (or 0), the root lies between 1.27 and 1.28.
A1
[The actual values and an appropriate comment are required for A1]
28
9
(a)
(b)
1
u − iv
= 2
u + iv u + v 2
Equating real and imaginary parts,
u
−v
;
=
x +1 = 2
y
u + v2
u 2 + v2
Substituting
x + iy + 1 =
Simplifying,
2
−v
+ 2
u + v2
1
u2 + v2 =
4
u
2
u + v2
29
M1A1
M1
A1
2
=4
M1A1
M1A1
A/AS Mathematics -FP2 – June 2008 – Markscheme – Final Draft
1
(a) Odd because f (− x) =
−x
= − f ( x)
(− x) 2 + 1
B1B1
(b) Neither because f (− x) = e − x + 1 which is neither f(x) or f(-x).
B1B1
2
For x < 2, f ′( x) = 3ax 2
For x ≥ 2, f ′( x) = 2bx
1 + 8a = 4b – 3, 12a = 4b
The solution is a = 1, b = 3 cao
B1
B1
B1B1
M1A1
3
(a)
u = x2
du = 2 xdx
and [0,√3] → [0, 3]
3
1
du
I=
2 0 9 + u2
1
u
=
arctan
6
3
=
(b)
1
0
B1
B1
M1
3
A1
0
π
24
dx
25 − 9 x 2
A1
=
1
3
1
0
dx
M1
25 / 9 − x 2
1
3x
sin −1
=
3
5
1
A1
0
1 −1
sin 0.6
3
= 0.2145
[FT on minor arithmetic error, do not FT the omission of the factor 1/3]
=
4
(a)
(b)
Substituting,
2t
1− t2
+ 3×
=1
2×
1+ t2
1+ t2
4t + 3 − 3t 2 = 1 + t 2
2t 2 − 2t − 1 = 0
2 ± 12
t=
(-0.366, 1.366)
4
Either
θ
θ
= −0.350879.. or
= 0.93888..
2
2
General solution is
θ
θ
= −0.350879.. + nπ or
= 0.93888.. + nπ
2
2
θ = −0.702 / 5.58 + 2nπ , θ = 1.88 + 2nπ
[FT from their particular solution]
30
A1
A1
M1
A1
A1
AG
B1
B1
M1
A1A1
5
dy dy / dp
2a
1
=
=
=
dx dx / dp 2ap p
Gradient of normal = − p
Equation of normal is
y − 2ap = − p( x − ap 2 )
whence the printed result.
(b)(i) Putting y = 0,
x = a(2 + p 2 )
1
1
Coords of R are (x,y) =
(ap 2 + a{2 + p 2 }), (0 + 2ap )
2
2
= a(1 + p 2 ), ap
(ii) Eliminating p,
y2
x − a = a × 2 or y 2 = a( x − a )
a
a
5a
Coordinates of focus = (a + ,0) , ie ( ,0)
4
4
(a)
(
6
(a)
(b)
M1A1
A1
M1
M1
A1
M1
)
A1
M1A1
M1A1
z − n = cos(− nθ) + isin(− nθ)
= cos nθ − i sin nθ
n
−n
z − z = cos nθ + isinnθ − (cos nθ − i sin nθ)
= 2isinnθ
z−
1
z
3
= z 3 − 3z 2 .
1
1
+ 3 z.
z
z
1
1
−3 z −
3
z
z
3
(2isinθ) = 2i sin 3θ − 6isinθ
3
1
sin 3 θ = sin θ − sin 3θ
4
4
= z3 −
31
2
−
1
z
M1
A1
M1
3
M1A1
m1
A1
A1
7
5 − 3x
A
B
A( x − 3) + B( x − 1)
≡
+
=
M1
( x − 1)( x − 3) x − 1 x − 3
( x − 1)( x − 3)
x = 1 gives A = - 1 ; x = 3 gives B = -2
A1A1
1
2
B1
+
(b)
f ′( x) =
2
( x − 1)
( x − 3) 2
This is the sum of 2 positive quantities which can never be zero so no
stationary points.
M1A1
[Accept a solution which obtains the derivative from the original form of f(x)
and then shows that the numerator has no real zeroes]
(a) Let
(c)
y
G2
O
1
3
x
[-1 each region omitted]
(d)
8
(a)
(i)
(5/3, 0); (0,5/3)
(ii)
x = 1, x = 3, y = 0
Consider
5 − 3x
=1
( x − 1)( x − 3)
5 − 3x = x 2 − 4 x + 3
x2 − x − 2 = 0
x = -1, 2
−1
f ( A) = (−∞,−1) ∪ (5 / 3, 2)
B1B1
B1B1B1
M1
A1
A1
A1A1
Modulus = 8 , Argument = 90 o
First cube root = 3 8 , ∠90 / 3
B1B1
M1A1
= 3 +i
Second cube root = 3 8 , ∠(90 / 3 + 120)
A1
M1A1
=− 3 +i
Third cube root = 3 8 , ∠(90 / 3 + 240)
= − 2i
32
A1
M1
A1
A/AS Level Mathematics – FP3 – June 2008 – Markscheme – Final Draft
1
(a)
y
G2
x
(b)
2
[Award G1 for a graph of the difference]
The two graphs intersect at the two points shown, one being x = 0 and the
other x > 0.
B1
The Newton-Raphson iteration is
(cosh x − 1 − sin x)
M1A1
x→ x−
(sinh x − cos x)
Starting with x = 1.5, we obtain
1.5
1.327589340
M1A1
1.295884299
1.294832874
1.294831731
A1
So root = 1.2948 (correct to 4 dps)
A1
dx = coshθdθ ; [1,2] → [ 0, sinh −1 1] (0.881)
B1B1
sinh −1 1
1 + 2 sinh θ + sinh 2 θ − 2 − 2 sinh θ + 2 coshdθ
I=
M1
0
sinh −1 1
cosh 2 θ dθ
=
A1
0
1
=
2
sinh −1 1
(1 + cosh 2θ) dθ
M1
0
1
1
=
θ + sinh 2θ
2
2
sinh −1 1
(FT on minor integration error]
A1
0
1
1
sinh −1 1 + sinh(2 sinh −1 1)
2
2
= 1.15 (cao)
=
33
A1
A1
3
(a)
(b)
f(1) = 1
1
1
f ′( x) = − x −3 / 2 , f ′(1) = −
2
2
3
3
f ′′( x) = x −5 / 2 , f ′′(1) =
4
4
1
1
3
= 1 − ( x − 1) + ( x − 1) 2 + ...
2
8
x
8
1 1 3 1
f
≈ 1+ × + ×
9
2 9 8 81
3
3 2 687
687
(or
)
≈
4
648
2 2 648
324
229
or (
) cao
2≈
229
162
4
≈
B1
B1
B1
B1
M1
M1A1
A1
(a) Consider
1 − tanh 2 x = 1 −
(e x − e − x ) 2
(e x + e − x ) 2
M1
(e x + e − x ) 2 − ( e x − e − x ) 2
(e x + e − x ) 2
4
= x
(e + e − x ) 2
=
(b)
= sec h 2 x
5 − 5 tanh 2 x = 11 − 13 tanh x
5 tanh 2 x − 13 tanh x + 6 = 0
3
tanh x = , 2
5
tanhx = 2 has no solution.
1 1+ 3/5
x = ln
2 1− 3/ 5
= ln2
A1
M1A1
AG
M1
A1
M1A1
B1
M1A1
A1
Alternative solution:
Substitute for sech and tanh in terms of exponential functions,
e 4 x − e 2 x − 12 = 0
M1A1
2x
e = 4 or − 3
M1A1
2x
e = −3 has no solution
B1
2x
e = 4 gives 2x = ln4
M1A1
x = ln2
A1
34
2
5
1 2
x
2
I n = (ln x) n d
(a)
1
M1 A1
2
2
1 2
1
1
x (ln x) n − x 2 .n(ln x) n −1 . dx
2
2
x
1
1
n
= 2(ln 2) n − I n −1
2
2
I 2 = 2(ln 2) − I 1
1
= 2(ln 2) 2 − 2 ln 2 − I 0
2
=
1
x2
= 2(ln 2) − 2 ln 2 + ×
2
2
A1A1
A1
M1
A1
2
2
= 2(ln 2) 2 − 2 ln 2 +
A1
1
3
4
A1
= 0.325
6
(a)
dx
dθ
2
A1
2
dy
+
dθ
= (−3 cos 2 θ sin θ) 2 + (3 sin 2 θ cos θ) 2
9 cos 4 θ sin 2 θ + 9 sin 4 θ cos 2 θ
=
A1
= 9 sin θ cos θ(cos θ + sin θ)
= 3sinθcosθ
3
= sin 2θ
2
2
π/2
dx
dθ
(b)(i) Length =
0
π/2
2
2
0
3
= − cos 2θ
4
3
=
2
π/2
dx
dθ
2
A1
A1
AG
2
dθ
3
sin 2θ dθ
2
=
(ii)
dy
+
dθ
2
M1A1
M1
A1
π/2
A1
0
A1
2
dy
+
dθ
2
dθ
M1
= 2π sin 3 θ.3 sin θ cos θ dθ
A1
y
CSA =
0
π/ 2
0
= answer given
π/2
6π
sin 5 θ 0
=
5
6π
(3.77)
=
5
[
]
M1A1
A1
35
(b)
7
(a)(i)
dy
d
=
[(1 − θ ) sin θ ]
dθ dθ
= (1 − θ ) cos θ − sin θ
= 0 when (1 - θ) = tanθ
leading to the printed answer.
M1
A1
A1
1
(ii)
Area =
1
(1 − θ) 2 dθ
20
1 − (1 − θ) 3
=
2
3
M1A1
1
A1
0
1
6
(b)(i) At the point of intersection, 1 − θ = 2θ 2
2θ 2 + θ − 1 = 0
(2θ − 1)(θ + 1) = 0
1
θ=
2
1
r=
2
=
(c)
1
Area =
2
M1
A1
A1
A1
1/ 2
( 2θ 2 ) 2 dθ
M1A1
0
[ ]
2 5 1/ 2
A1
θ 0
5
1
A1
=
80
[Award M1A1 for evaluating this integral but then mis-using it subsequently]
=
1.Mathematics C1 – C4 & FP1 – FP3 MS (Summer 2008)/LG
13 August 2008
36
WJEC
245 Western Avenue
Cardiff CF5 2YX
Tel No 029 2026 5000
Fax 029 2057 5994
E-mail: exams@wjec.co.uk
website: www.wjec.co.uk/exams.html
ERROR: invalidrestore
OFFENDING COMMAND: --restore-STACK:
-savelevel17764
(˜d)
(˜d)
17763
(˜c)
(˜c)
17762
(˜b)
(˜b)
17761
(˜a)
(˜a)
17760
(˜‘)
(˜‘)
17759
(˜_)
(˜_)
17758
(˜^)
(˜^)
17757
(˜])
(˜])
17756
(˜\)
(˜\)
17755
(˜[)
(˜[)
17754
(˜Z)
(˜Z)
17753
(˜Y)
(˜Y)
17752
(˜X)
(˜X)
17751
(˜W)
(˜W)
17750
(˜V)
(˜V)
17749
(˜U)
(˜U)
17748
(˜T)
(˜T)
17747
(˜S)
(˜S)
17746
(˜R)
(˜R)
17745
(˜Q)
(˜Q)
17744
(˜P)
(˜P)
17743
(˜O)
(˜O)
MS3
£3.00
GENERAL CERTIFICATE OF EDUCATION
TYSTYSGRIF ADDYSG GYFFREDINOL
MARKING SCHEME
MATHEMATICS - M1-M3 & S1-S3
AS/Advanced
SUMMER 2008
INTRODUCTION
The marking schemes which follow were those used by WJEC for the Summer 2008
examination in GCE MATHEMATICS. They were finalised after detailed discussion at
examiners' conferences by all the examiners involved in the assessment. The conferences
were held shortly after the papers were taken so that reference could be made to the full
range of candidates' responses, with photocopied scripts forming the basis of discussion.
The aim of the conferences was to ensure that the marking schemes were interpreted and
applied in the same way by all examiners.
It is hoped that this information will be of assistance to centres but it is recognised at the
same time that, without the benefit of participation in the examiners' conferences, teachers
may have different views on certain matters of detail or interpretation.
WJEC regrets that it cannot enter into any discussion or correspondence about these
marking schemes.
Mathematics M1(June 2008)
Markscheme
1.(a)
Using v = u + at with v = 22, u = 12, a = 0.5
22 = 12+ 0.5t
22 − 12
t =
0 .5
= 20 s
M1
cao
A1
1.(b)
attempt at v-t graph M1
Straight line joining (0, 18) to (9, 12) A1
Straight line joining c's (9, 12) to ((c's 29, 22)
A1
Straight line joining (c's 29, 22) to (c's 29+31, 22) + units and labels
A1
2.(a)
Using v = u + at with v = 0, a = (-)9.8, t = 2.5
0 = u + (-9.8) × 2.5
u = 24.5 ms-1
M1
2.(b)
Using s = ut + 0.5at2 with u = 24.5, a = (-)9.8, t = 4
s = 24.5 ×4 + 0.5 × (-9.8) × 42
= 19.6 m
cao
M1
A1
A1
2.(c)
Using v2 = u2 + 2as with u = 24.5, a = (-)9.8, s = (-)70
v2 = 24.52 + 2 × (-9.8) × (-70)
= 1972.25
v = 44.41 ms-1
cao
M1
A1
1
A1
A1
3.
(a)
(b)
4.(a)
4.(b)
5.(a)
5.(b)
N2Lapplied to lift and man
-R + 500g = 500a
500a = 500 × 9.8 – 4800
a = 0.2 ms-2
M1
A1
N2L applied to man
70g – R = 70a
R = 672 N
cao
A1
ft a
M1
A1
A1
Apply N2L to B
9g – T = 9a
Apply N2l to A
T – 5g = 5a
M1
B1
A1
Adding
m1
4g = 14a
4 × 9 .8
a =
14
= 2.8 ms-2
cao
A1
T = 5(g + a)
= 5(9.8 + 2.8)
= 63 N
cao
A1
Impulse = change in momentum
= 0.7(2- 5)
= -2.1 Ns
Direction of impulse is opposite to motion.
Speed after impact = 2 × 0.6
= 1.2 ms-1
oe
M1
A1 B1
B1
M1
A1
2
6.(a)
Resolve perpendicular to slope
R = 6.5 g cos α
= 6.5 × 9.8 ×
M1
A1
12
13
= 58.8
1
× 58.8
5
= 11.76 N
ft
F =
N2l down slope
6.5g sin α - F = 6.5 a
3 terms, dimensionally correct
B1
M1
A1 A1
5
− 11.76
13
6.5
6.5 × 9.8 ×
a =
= 1.96 ms-2
cao
A1
6.(b)
N2L up slope
4 terms, dimensionally correct
T – F – 6.5 g sin α = 6.5 a
Constant speed, therefore a = 0
s.i.
5
T = 11.76 + 6.5 × 9.8 ×
13
= 36.26 N
cao
3
M1
A1
B1
A1
7.
Momentum
M1
A1
7 × 5 + 5 × (-3) = 7vA + 5vB
Restitution
M1
A1
vB – vA = - 0.2 (-3 – 5)
5vB – 5vA = 8
Subtract
12 vA = 12
vA = 1 ms-1
vB = 2.6 ms-1
cao
cao
m1
A1
A1
Sphere A is moving in the same direction as before impact. Sphere B has
reversed its direction of motion.
A1
8.
Moments about P to obtain equation
5(x – 3) = 6 × 3 – 6 × RQ
5x – 15 = 18 – 6 RQ
6 RQ = 33 – 5x
For limiting equilibrium
RQ
33 – 5x
5x
x
= 0
=0
= 33
= 6.6
4
M1
A1 B1
si
M1
cao
A1
9.
Resolve perpendicular to 12 N
X = 9 – 8 cos30°
3
= 9–8×
2
= 2.0718
M1
A1
Resolve parallel to 12 N
Y = 12 – 8 sin30°
= 12 – 8 × 0.5
=8
M1
A1
Magnitude of resultant = 8 2 + 2.0718 2
= 8.264 N
M1
A1
θ = tan −1
ft X, Y
8
2,0718
M1
cao
= 75 48°
A1
10.(a)
triangle ABC
rectangle DEFG
lamina
Area
108
20
88
from AB
6
5
x
Moments about AB to obtain equation
20 × 5 + 88 x = 108 × 6
x = 6.23
Moments about AC to obtain equation
20 × 2.5 + 88 y = 108 × 4
y = 4.34
10.(b)
4.34
6.23
θ = 34.9°
from AC
4
2.5
y
B1
B1
B1
ft
cao
M1
A1
A1
ft
cao
M1
A1
A1
tan θ =
M1
ft x, y
5
A1
Mathematics M2 (June 2008)
Markscheme
1.(a)
Using T =
λx
l
with T = 12, x = 0.55 – 0.3, l = 0.3
12 =
λ (0.55 − 0.3)
A1
0 .3
12 × 0.3
λ =
0.25
= 14.4 N
1.(b)
M1
1 14.4 × 0.252
×
2
0.3
= 1.5 J
Using EE =
cao
A1
ft λ
M1 A1
ft λ
A1
2.
32 × 1000
16
= 2000 N
T =
si
N2L up plane
M1
A1
T – F – 900 g sin α = 0
Resistive force F = 2000 – 900 × 9.8 ×
= 560 N
6
M1
8
49
cao
A1
3.(a)
3.(b)
3.(c)
5a = 15t2 – 60t
a = 3t2 – 12t
When t = 2
a = 12 – 24
= - 12
Therefore magnitude of acceleration = -12 ms-1
Using F = ma
v
=
M1
A1
3t 2 − 12t dt
M1
= t3 – 6t2 (+ C)
When t = 0, v = 35
C = 35
v = t3 – 6t2 + 35
A1
m1
A1
Least value of v when a = 0
3t(t – 4) = 0
t = (0 or) 4
Therefore least value of v = 43 - 6× 42 + 35
= 3 ms-1
M1
ft v
A1
ft v
A1
8
3.(d)
t 3 − 6t 2 + 35 dt
Required distance =
attempt to integrate v
M1
correct integration
A1
2
=
t4
− 2t 3 + 35t
4
8
2
= (16 × 64 – 16 × 64 + 35 × 8) – (4 – 16 + 70)
= 280 – 58
cao
= 222 m
4.(a)
4.(b)
Difference in PE = 2232 × 90 g - 2128 × 90 g +
= (1968624 – 1876896)
= 91728 J
Difference in KE = 0.5 × 90 × 352 - 0.5 × 90 × 22
= 55125 – 180
= 54945 J
Work done against resistance = 36783 J
Work done = 1335 R = 36783
R = 27.55 N
A1
PE
M1 A1
KE
M1 A1
ft WD if M's gained in (a)
7
m1
M1 A1
M1
A1
5.(a)
5.(b)
Using s = ut + 0.5at2 with u = 14, a = (-)9.8, s = 8.4
8.4 = 14t – 4.9t2
attempt to solve
7t2 – 20t + 12 = 0
(7t – 6)(t – 2) = 0
t = 6/7, 2
As particle is on the way down, t = 2
Therefore horizontal distance of wall = 2 × 12
cao
= 24 m
M1
A1
m1
Using v = u + at with u = 14, a = (-)9.8, t = 2
v = 14 -9.8 × 2
= - 5.6 ms-1
ft t
ft t
M1
A1
A1
ft v
M1
A1
Therefore speed of motion = 5.6 2 + 12 2
= 13.24 ms-1
5 .6
12
= 25° to the horizontal.
θ = tan −1
6.(a)
6.(b)
7.(a)
7.(b)
A1
M1
A1
M1
ft v
A1
AB = (3i – j + 2k) – ( 2i + j + k)
= i – 2j + k
cao
M1
A1
Work done by F =
=
=
=
ft
ft
M1
m1
A1
used
M1
F.AB
(i – 4j + k).(i – 2j + k)
1+8+1
10 J
dv
dt
= 3 cos 3t i – 10 sin 5t j + 9t2 k
a =
A2
rA = (-8t -2)i + (3t + 3)j
rB = (-16t + 11)i + (9t – 8)j
rA – rB = (8t – 13)i + (-6t + 11)j
2
rA − rB = (-13 + 8t)2 + (11 – 6t)2
= 169 – 208t + 64t2 + 121 – 132t + 36t2
= 290 – 340t + 100t2
Minimum when 200t = 340
t = 1.7
Minimum distance = 290 − 340(1.7) + 100(1.7) 2
= 1
8
M1
M1 A1
cao
m1
A1
cao
M1
A1
8.
(a)
N2L towards centre
M1
2
mv
r
4 × 22
112
=
=
3
3
7
T sin θ =
A1
Resolve vertically
T cos θ = 4 × 9.8
= 39.2
112
Dividing
tan θ =
3 × 39.2
M1
A1
m1
∠AOP = θ = tan-1
cao
= 43.6°
(b)
(c)
112
3 × 39.2
A1
39.2
cos(43.6 o )
= 54.13 N
ft angle
A1
3
7 sin (43.6 o )
= 0.62 m
ft angle
B1
T =
l =
9
9.
(a)
(b)
Energy considerations
-2 g × 0.5 cos θ + 0.5 mv2 = -2 g × 0.5 cos 60° + 0.5 × 2 × 42
v2 = g cos θ - 0.5 g + 16
v2 = 16 + g (cos θ - 0.5)
= g cos θ + 11.1
N2L towards centre
T - mg cos θ =
M1
A1A1
A1
M1
mv
r
2
A1
2
16 + g cosθ
0 .5
= 2g cos θ + 64 + 4g cos θ = 6g cos θ - 2g + 64
T = 58.8 cos θ + 44.4
T = 2g cos θ +
10
1
g
2
2g
-
m1
A1
Mathematics M3 (June 2008)
Markscheme
1,(a)
R = 35g
Friction F = µ R
1
× 35 g
=
7
= 5g = 49 N
B1
B1
N2L
dim. correct
-F - 0.7 v = 35 a
dv
35
+ 49 + 0.7v = 0
dt
dv
50
Divide by 0.7
+ 70 + v = 0
dt
(b)
M1
dv
dt
m1
convincing
A1
a=
50
dv =
dt
70 + v
- 50 ln (70 + v) = t (+ C)
−
M1
A1 A1
Either limits or initial conditions used
t = 0, v = 28
C = -50 ln 98
t = 50 ln 98 - 50 ln (70 + v)
m1
A1
Body come to rest when v = 0
m1
used
t = 50 ln (98/70)
= 50 ln(7/5)
= 16.82 s
11
cao
A1
2.(a)
(i)
N2L
Multiply by 2.5
(ii)
0.8 x – 1.2v = 2a
d2 x
dx
a= 2 , v=
dt
dt
2
dx
d x
+ 3 − 2x = 0
5
dt
dt 2
dim correct
M1
m1
A1
Auxiliary equation
5m2 + 3m -2 = 0
(5m – 2)(m + 1) = 0
m = -1, 0.4
General solution is x = Ae0.4t + Be-t
When t = 0, x = 0, v = 7
A+B = 0
dx
v=
= 0.4 Ae 0.4 t − Be −t
dt
0.4 A – B = 7
B1
ft m values
used
B1
M1
ft x
B1
ft x, v both
A1
Solving simultaneously
1.4 A = 7
A = 5
B = -5
x = 5e0.4t - 5e-t
both cao
A1
(iii)
e0.4t increases with t; e-t decreases with t, therefore e0.4t - e-t increases with t.
B1 B1
Hence x = 5e0.4t - 5e-t increases with t.
2.(b)
For PI try x = at + b
3a – 2(at + b) = 20t – 70
M1
A1
Compare coefficients
-2a = 20
a = -10
3a – 2b = -70
2b = 40
b = 20
Therefore general solution is x = Ae0.4t + Be-t – 10t + 20
m1
12
both cao
ft a, b
A1
A1
v2 = ω2(a2 – x2)
25 = ω2(a2 – 9)
14.0625 = ω2(a2 – 16)
3.(a)
At A
At B
Therefore
(b)
25(a2 – 16) = 14.0625(a2 – 9)
10.9375 a2 = 273.4375
a2 = 25
a = 5
25
25
ω2 =
=
25 − 9
16
ω = 1.25
2π
8π
=
Period =
ω
5
4.(a)
convincing
convincing
M1 A1
B1
Minimum speed = ωa
= 5 ×1.25
= 6.25 ms-1
cao
M1
A1
used
M1
si ft ω
A1
6.25 × 0.4 = 6.25 cos(1.25t)
cos (1.25t) = 0.4
t = 0.93 s
cao
dv v 2
+
+ 10 = 0
dx 90
dv
1 2
(v + 900)
v
= −
dx
90
v
90 2
dv = − dx
v + 900
45 ln(v2 + 900) = -x (+ C)
a=v
v
dv
dx
M1 B1
A1
M1
M1
A1 A1
When t = 0, x = 0, v = 15
C = 45 ln(275 + 900) = 45 ln(1125)
1125
x = 45 ln 2
v + 900
(b)
A1
A1
5
t
4
x = 5 sin (1.25 × 2)
= 2.99 m
We require t such that
M1
A1
A1
m1
x = 5 sin
Let
When t = 2,
(c)
used
Greatest height when v = 0
1125
x = 45 ln
900
= 45 ln(1.25)
= 10.04 m
m1
oe
A1
M1
A1
13
5.
(a)
(b)
Moments about A
S × 2.5 = 20 g ×1.5cos θ
2
4
=
cos θ =
2 .5
5
20 × 9.8 × 1.5 × 0.8
S =
2 .5
= 94.08 N
dim correct
B1
cao
Resolve horizontally
S sin θ = F
F = 94.08 × 0.6
= 56.448 N
6.
A1
M1
A1
ft S
Resolve vertically
R = 20 g + S cos θ
= 20 × 9.8 + 94.08 × 0.8
= 120 736 N
Limiting friction
M1
A1 A1
A1
M1
A1
ft S
F = µR
56.448
µ =
120.736
= 0.47
s.i.
A1
used
M1
cao
A1
Using v2 = u2 + 2as with u = 0, a = (-)9.8, s =0
v2 = 2 × 9.8 × 1.6
v = 5.6 ms-1
M1
A1
A1
Impulse = change in momentum
For A
J = 7 v'
For B
J = 3 × 5.6 - 3 v'
M1
B1
A1
Solving
applied to both particles
ft v
7v' = 16.8 – 3v'
10v' = 16.8
v' = 1.68
J = 11.76 Ns
m1
ft v
ft v
14
A1
A1
A/AS Mathematics - S1 – June 2008
Mark Scheme
1
(a)
2
2
2 1
P(2 red) = × or
9
9 8
2
(1/36)
B1
3
2
9
2
(1/12)
B1
P(2 grn) =
(b)
2
(a)
(b)
(c)
3 2
× or
9 8
4
2
4 3
(1/6)
P(2 yel) = × or
9
9 8
2
1
1 1 5
+ + =
P(same) =
36 12 6 18
13
P(diff) = 1 – P(same ) =
18
Using P ( A ∪ B ) = P ( A) + P ( B ) − P ( A ∩ B )
0.4 = 0.2 + P(B) – 0.2P(B)
0.8P(B) = 0.2
P(B) = 0.25
EITHER
P(A only) = 0.15
P(B only) = 0.2
Reqd prob = 0.35
OR
P(Exactly one) = P(A∪B) – P(A∩B)
= 0.4 – 0.2 × 0.25
= 0.35
0.2 × 0.75
Reqd prob =
0.35
= 3/7
[FT for P(B) in (b) and (c)]
15
B1
B1
M1A1
M1
A1
A1
A1
B1
B1
B1
M1
A1
A1
B1B1
B1
3
(a)(i)
(ii)
(b)
4
5
(a)
Prob = 0.9884 – 0.5697 or 0.4303 – 0.0116
= 0.4187 (cao)
3.255
(b)(i) Prob = e −3.25 ×
= 0.117
5!
3.25 2
−3.25
(ii)
P(X ≤ 2) = e
1 + 3.25 +
2
= 0.370
[Special case : Award B1 for 0.370 with no working]
(a)
(b)
6
Total number of poss = 36
[Award if any evidence is seen that there are 36 possibilities]
Number giving equality = 6
Reqd prob = 6/36 = 1/6
Number giving smaller score = 5 + 4 + 3 + 2 + 1 = 15
Reqd prob = 15/36 = 5/12
EITHER
Reduce the sample space to (5,1), (4,2), (3,3), (2,4), (1,5)
Reqd prob = 1/5
OR
P(3,3)
Prob =
P(Sum = 6)
1 / 36
=
5 / 36
= 1/5
(a)
(b)
(c)
P(B) = 0.25, P(C) = 0.25, P(D) = 0.5
[Award M1A0 for ¼, ¼, ¼ with no working]
P(Win) = 0.25 × 0.3 + 0.25 × 0.4 + 0.5 × 0.6
= 0.475
0 .5 × 0 .6
P(Dwin) =
0.475
= 0.632
[FT their initial probs in (b) but not in (a)]
Σp = 1
9k = 1
k = 1/9
2
3
4
E(X) = × 1 + × 2 + × 3
9
9
9
= 2.22 (20/9)
3 1 4 1
1
2
= ×1 + × + ×
E
X
9
9 2 9 3
= 0.537 (29/54)
16
B1
B1
B1
M1A1
A1
M1A1
A1
M1
A1
A1
B1B1
B1
M1A1
M1A1
A1
M1A1
M1A1
A1
B1B1
B1
M1
A1
AG
M1A1
A1
M1A1
A1
7
(a)
The number of sales, X, is B(50, 0.2) si
B1
50
(i)P(X = 12) =
× 0.212 × 0.838 or 0.8139 – 0.7107 or 0.2893 – 0.1861
12
M1
= 0.1032 or 0.1033
A1
(ii)
P(10 ≤ X ≤ 14) = 0.9393 – 0.4437 or 0.5563 – 0.0607
B1B1
= 0.4956
B1
(iii) Prob = 0.8 × 0.8 × 0.2
M1A1
= 0.128
A1
(b) The week’s pay Y is given by
Y = 100 + 50X si
B1
E(Y) = 100 + 50 × 50 × 0.2 = 600
M1A1
M1A1
SD = 50 × 50 × 0.2 × 0.8 = 141
[FT on their sensible Y]
8
(a)
(b)
(c)
P(0.25 ≤ X ≤ 0.75) = F(0.75) – F(0.25)
M1
3
4
3
4
A1
= 4 × 0.75 − 3 × 0.75 − 4 × 0.25 + 3 × 0.25
= 0.6875
A1
F(0.6) = 0.4752
B1
Since F(med) = 0.5, it follows that the median is greater than 0.6.
[FT on their F(0.6)]
M1A1
2
3
M1A1
f(x) = F ′( x) = 12( x − x )
1
(d)
E ( X ) = 12 x( x 2 − x 3 )dx
M1A1
0
x4 x5
−
= 12
4
5
= 0.6
1
A1
0
A1
17
A/AS level Mathematics - S2 June 2008
Mark Scheme x=
1
32.6
36.78
, y=
5
6
0.25 2 0.25 2
+
5
6
The 95% confidence limits are
6.52 – 6.13 ± 1.96 × 0.1514
giving [0.093, 0.687]
SE of difference of means =
2
(a)
(b)
3
E(X) = 4, Var(X) = 2.4
Using Var(X) = E ( X 2 ) − [ E ( X )] 2
E ( X 2 ) = 2.4 + 16
= 18.4
E(Y) = 9, Var(Y) = 6.3
Similarly,
E (Y 2 ) = 6.3 + 81 = 87.3
E(U)=E(X)E(Y) = 36
2 2
E ( X Y ) = E ( X 2 ) E (Y 2 ) = 1606.32
Var(U) = E ( X 2Y 2 ) − [ E ( XY )]2
= 1606.32 − 36 2 = 310.32
49 − 50
= − 0 .5
2
Prob = 0.3085
(ii) Number weighing less than 49 g is B(6,0.3085) si
6
Required prob =
× 0.3085 3 × 0.6915 3 = 0.194
3
[Award M1A0 if combinatorial term omitted]
(b)
X-3Y is N(50-3 × 18, 2 2 + 9 × 1.2 2 ), ie N(- 4, 16.96)
4
z=
= (±)0.97
16.96
Prob = 0.166
(a) (i)
z=
18
B1
M1A1
m1A1
A1
B1
M1
A1
B1
B1
B1
B1
M1
A1
M1A1
A1
B1
M1A1
M1B1B1
A1
A1
4
5
48.32
(= 6.04)
8
0.05
SE of X =
(= 0.01767…)
8
[Award M1 only if the 8 appears in the denominator]
99% conf limits are
6.04 ± 2.576 × 0.01767..
[M1 correct form, A1 2.576]
giving [5.99,6.09]
(b) We solve
2.576 × 2 × 0.05
= 0.04
n
n = 41.467
giving n = 42
[FT on their n]
(a)
x=
B1
M1
m1A1
A1
M1
A1
B1
H 0 : p = 0.6 versus H 1 : p < 0.6
B1
Under H 0 , X is B(20,0.6) si
B1
and Y (No not germ) is B(20,0.4) (si)
B1
p-value = P(X ≤ 9 H 0 ) = P(Y ≥ 11 H 0 )
M1
= 0.1275
A1
(c)
X is now B(200,0.6) which is approx N(120,48)
B1
101.5 − 120
z=
M1A1A1
48
[M1A1A0 if no continuity correction]
= - 2.67
A1
p-value = 0.00379
A1
Strong evidence to support Malcolm (or germination prob less than 0.6). B1
[No c/c gives z = - 2.74, p = 0.00307, wrong c/c gives z = - 2.81, p = 0.00248]
(a)
(b)
19
6
(a)(i)
(ii)
(b)
AC = X 2 + X 2 = 2 X
(si)
We require P(√2X > 8) = P( X > 8 / 2 )
6 −8/ 2
=
= 0.172 (3 - 2v2)
6−4
X2
Area =
2
X2
< 10) = P(X < √20)
We require P(
2
20 − 4
=
= 0.236 (v5 – 2)
6−4
EITHER
f ( x) = 0.5 for 4 ≤ x ≤ 6
B1
M1A1
A1
B1
M1A1
A1
B1
6
Expected area = 0.5 x 2 × 0.5dx
M1
4
x3
= 0.25
3
=
6
A1
4
38
3
A1
OR
E ( X 2 ) = Var ( X ) + ( E ( X )) 2
22
76
+ 5 2 (= )
12
3
2
38
X
Expected area = E
=
2
3
=
7
(a)
H 0 : µ = 2.5 versus H1 : µ ≠ 2.5
(b)(i) Under H0 S is Po(15)
P(S ≤ 8) = 0.0374
P(S ≥ 23) = 0.0327
Significance level = 0.0701
(ii) If µ = 2, S is Po(12)
We require P(9 ≤ X ≤ 22) = 0.9970 – 0.1550
= 0.842
(c) Under H0, S is now Po(250) ≈ N(250,250)
269.5 − 250
z=
250
= 1.23
[Award M1A1A0 if the c/c is incorrect or omitted]
Prob from tables = 0.1093
p-value = 0.2186
Accept H0 (context not required but accept if correct)).
[No c/c gives z = 1.26, prob = 0.1038, p-value = 0.2076;
incorrect c/c gives z = 1.30, prob = 0.0968, p-value = 01936]
20
M1
A1A1
A1
B1
B1
B1
B1
B1
B1
M1A1
A1
B1
M1A1A1
A1
A1
B1
B1
AS/A Mathematics - S3 – June 2008
Markscheme
1.
(a)
(b)
2
6 5 4 1
× × =
10 9 8 6
4 3 2 1
P(3£1) = × × =
10 9 8 30
6 5 4
1
P(2£2,1£1) = × × × 3 =
10 9 8
2
6 4 3
3
P(1£2,2£1) = × × × 3 =
10 9 8
10
[Do not accept sampling with replacement]
The sampling distribution is
Total £3
£4
£5
£6
Prob
1/30 3/10 1/2 1/6
[-1 each error]
[FT on their probabilities]
P(3£2)) =
1
3
1
1
× 3 + × 4 + × 5 + × 6 = 4 .8
30
10
2
6
6
4
× 2 + × 1 = 1.6 as
Mean value =
10
10
E (Total) =
630
= 0.525
1200
0.525 × 0.475
= 0.0144
(ii)
ESE =
1200
(b)
95% confidence limits are
0.525 ± 1.96 × 0.0144
giving [0.497,0.553]
(c)
No, because 0.6 is not in the confidence interval.
(a)(i)
pˆ =
B1
B1
B1
B1
M1A2
M1A1
A1
B1
M1A1
M1A1
A1
B1B1
Σx = 3528 ; Σx 2 = 1038840
B1
UE of µ = 294
B1
2
1038840 3528
−
M1
UE of σ 2 =
11
11 × 12
= 146.18
A1
H 0 : µ = 300 versus H 1 : µ < 300
B1
(b)
294 − 300
Test stat =
M1A1
146.18 / 12
= -1.72
A1
DF = 11
B1
(i) At the 5% significance level, critical value = 1.796
B1
Accept farmer’s claim with correct reasoning.
B1
(ii) At the 10% significance level, critical value = 1.363
B1
Reject farmer’s claim with correct reasoning.
B1
[FT in (b) from (a) ; FT conclusions from their critical values but not from p-values]
3
(a)
21
4
(a)
(b)
5
UE of µ = 1815/75 (= 24.2)
44213 1815 2
UE of Var(X) =
−
(= 3.92)
74
74 × 75
[Accept division by 75 giving 3.87]
3.92
= 0.229 (3.87 gives 0.227)
ESE =
75
Approx 90% confidence limits for µ are
24.2 ± 1.645 × 0.229
giving [23.8, 24.6] (cao)
No since the Central Limit Theorem ensures the approximate
normality of X .
(a)
H 0 : µ x = µ y versus H 1 : µ x ≠ µ y
(b)
x = 1506 / 60(= 25.1); y = 1530 / 60(= 25.5)
B1
M1A1
M1A1
M1A1
A1
B1
B1
B1B1
38124 1506 2
−
= 5.4813... (/60 gives 5.39)
M1A1
59
59 × 60
39327 1530 2
s y2 =
−
= 5.2881... (/60 gives 5.20)
A1
59
59 × 60
[Accept division by 60]
25.5 − 25.1
Test stat =
M1A1
5.4813.. 5.2881..
+
60
60
= (±)0.94
(0.95)
A1
Prob from tables = 0.1736
(0.1711)
A1
p-value = 0.347
(0.342)
B1
Insufficient evidence to doubt that the mean weights are equal.
B1
s x2 =
1/ 2
6
(a)
(1 + λx)dx
P(X > 0) =
M1
0
λx 2
= x+
2
(b)(i)
(ii)
1/ 2
=
0
1 λ
+
2 8
Y is B(n,1/2+λ/8)
1 λ
E(Y) = n +
2 8
1 λ
E (U ) = 8 × n +
÷n−4 = λ
2 8
Var(Y) = n
Var (U ) =
SE =
1 λ
+
2 8
1 λ
1 λ2
−
=n −
2 8
4 64
64
1 λ2
n
×
−
4 64
n2
16 − λ2
n
A1AG
M1
A1
M1A1
M1A1
M1A1
A1
22
7
(a)
(b)
S xy = 23538 − 270 × 517 / 6 = 273
S xx = 13900 − 2702 / 6 = 1750
273
b=
= 0.156
1750
517 − 270 × 0.156
a=
6
= 79.15
Estimated length = 79.15 + 0.156 × 60 (= 88.51)
1 15 2
+
(= 0.0815)
6 1750
The 99% confidence interval is
88.51 ± 2.576 × 0.0815
giving [88.3, 88.7]
[FT from (a)]
SE = 0.15
23
B1
B1
M1A1
M1
A1
B1
M1A1
M1A1
A1
WJEC
245 Western Avenue
Cardiff CF5 2YX
Tel No 029 2026 5000
Fax 029 2057 5994
E-mail: exams@wjec.co.uk
website: www.wjec.co.uk/exams.html