Mark scheme 4024/2 – Paper 2 November 2001
1
(a)
152 − 82
M1
12.7 m
(b)
152 + 112
A1
± 2 × 15 × 11 cos 55° soi
M1
dep
M1
(dep on first M1)
A1
further
A1
Correct formula, simplification and square root taken soi
156(.7…) or 535(.2…) soi
12.5 m
(c)
2
(a)
ˆB =
cos AD
8
15
oe
M1
ADˆ B = 57. 8° soi
ˆB ) √
67.2° or (125° – their AD
A1
p = 17
B3
q = 36
A1
r = 125 (B1 each)
s = 178 or their (p + q + r) v
(b)
x = 3n + 2 oe
y = (n + 1)2 oe
t = n 3 + n 2 + 5n + 3
3
oe
B1
z = n 3 oe
(B1 each)
or their (x + y + z) v
B3
B18
(a)
20°
B1
(b)
105°
B1
(c)
55° or 180 – their (a) – their (b) v
B1
(d)
55°
B2
(allow B1 for any indication ECˆ B = 75 ° )
(e)
30° or 85 – their (d) v or 50 – their (a) v
(allow B1 for indication BAˆ C = BEˆ C or ABˆ E = ACˆ E )
http://www.xtremepapers.net
B2
Mark scheme 4024/2 – Paper 2 November 2001
4
Final answer t =
(a)
mv − mu
oe
F
B2
(allow M1 for reaching m(v – u) or mv – mu = Ft
(b)(i)
(ii)
oe)
43 = c + 50d and 49 = c + 80d oe
1
c = 33 and d = 0.2 or
5
B1
B3
(allow B2 for one correct www
or M1 for correct elimination of one variable)
(iii)
40 = their c + m × their d
35
(iv)
5
(a) (i)
M1
soi
A1
cao
Statement such as “Length of unloaded string”
B1
Fig (π × 242 × 125)
M1
soi
A2
226 litres cao
(allow A1 for 226 000 seen)
(ii)
Fig {2π × 24 × 125 + (2 ×) π × 242 } soi
M1
2
(b)
Division by 100 soi or works in metres
M1
2.07 m2 cao
A1
fig
48 000
soi
150 × 20
M1
16 cm
(c)
(Drop =) fig
A1
20 000
π × 24 2
or fig
Divides their drop by 2.5 or
4.42 cm/h cao
20 000
20 × 150
20 000
2.5
M1
soi
indep
M1
A1
Mark scheme 4024/2 – Paper 2 November 2001
6
(a)
Triangle A correctly plotted
B1
(b)
Triangle B with vertices at (10, 1), (12, 1) and (10, -2)
B1
(c)
Triangle C with vertices at (-3, 1), (-3, -1) and (-6, 1)
B2
(allow B1 for 2 vertices correct or vertices at (9, 7), (12, 7) and (9, 9)
(d)
Triangle D with vertices at (1, 4), (1, 2) and (-2, 4)
B2
(allow B1 for 2 vertices correct or vertices at (0, –3), (0, –5), (–3, –3)
(e)
Shear
M1
 1 2

Factor 2 or with matrix 
 0 1
A1
[In (b), (c), (d) allow √ from earlier mistakes]
7
200 (x + 4) − 200 x
oe
x( x + 4 )
(a)
Final answer
(b)(i)
(ii)
M1
800
oe
x (x + 4)
A1
200
x
B1
200
or (their (i) – 5) seen
x+4
B1
 200 
200
their 
=
 – their
x+4
 x 
±5
or (x + 4) [their (i) – 5] = 200 oe
Correct derivation of x2 + 4x – 160 = 0
(c)
AG
− 4 ± 656
2
M1
A1
B2
(allow B1 for –4 and 2 and B1 for
656 or 25.6…)
10.81 and –14.81 (B1 each)
B2
(allow B1 for both 10.8(….) and –14.8(….) seen)
(d)
40
120
+
(dep on (their 10.81) > 0)
their 10.81 4 + their 10.81
M1
cao (but accept 11 800 cm3 )
A1
11.8 litres
Mark scheme 4024/2 – Paper 2 November 2001
8
(a)
Vertical lines at t = 60, 80, 90, 95, 100, 110 and 130
B1
Rectangles completed, with heights in ratio 1 : 5 : 14 : 20 : 12 : 2
B2
(allow B1 for 4 or 5 correct heights)
(b)
Interval 95 < t ≤ 100 indicated
B1
(c)
98
3
oe or 98.2 or better
16
B3
(allow B2 for 98 or
or allow M1 for
7855
seen
80
4t1 + 10 t2 + ...
where 60 ≤ t1 ≤ 80 etc
4 + 10 + ...
and dep M1 for t 1 = 70, t 2 = 85, t 3 = 92.5, t 4 = 97.5, t 5 = 105, t 6 = 120)
(d)
7
oe
40
B1
(e) (i)
7
790
B2
cao
(allow B1 for equivalents or for
(ii)
8×7
7
or for
)
80 × 79
800
4
oe
395
(allow B1 for
B2
2× 4×8
2
or for answer
oe)
80 × 79
395
Mark scheme 4024/2 – Paper 2 November 2001
9
(a)
p = 29(.0)
B1
(b)
All 8 points plotted v
P2
(allow P1 for at least 6 correct plots v)
(c) (i)
(ii)
(d)
Smooth curve through all plots of which at least 6 v
C1
1.28 to 1.33
B1
13.7 to 13.99
B1
Tangent drawn at x = 1.5 and using
(e)
changein y
change in x
M1
-9(.0) to –12(.0)
A1
Ruled straight line through (0, 25) and (4, 13)
B2
(allow B1 for short ruled line or good freehand line)
(f) (i)
(ii)
1.2(0) to 1.25 and 3.2(0) to 3.28
B1
7x3 – 25x2 + 25 = 0 oe
B1
(accept a = 7, b = –25, c = 0, d = 25)
10
1
of whole angle at O
5
(a)
Reasonable explanation, such as
(b)
½ 1.52 sin 72° or other complete method for area of triangle
5 × their (area of triangle)
M1
dep
2
(c) (i)
B1
M1
5.35 cm
A1
Reference to angle at centre of circle property, (or other method)
B1
M1
(ii)
2 × 1.5 cos 18° or other complete method
2.85 cm convincingly obtained
A1
AG
M1
(iii)
½ (their 2.85)2 sin 36° or other complete method
A1
2
2.39 cm
M1
(iv)
36
π (their 2.85)2
360
A1
0.162 to 0.166 cm2
B1
(v)
2
6.15 to 6.19 cm
or [their (b) + 5 × their (c)(iv)] v
Mark scheme 4024/2 – Paper 2 November 2001
11
(a) (i)
b–a
B1
(ii)
2
(b - a ) oe
3
(iii)
1
2
a + b oe or a + their (ii) v
3
3
B1
(iv)
5
b oe
3
B1
2
(their (i) ) v
3
B1
cao
CD = CO + OD or CD = CB + BD or better seen
(b)
b–
1
a convincingly obtained
3
(c)
5
5
b− a
3
9
(d)(i)
Shows ED =
(ii)
(e)
or
AG
M1
A1
B1
5
5
CD (accept k = )
3
3
5
length of CD
3
E, C and D lie on a straight line
Length of ED =
oe
B1
B1
B1
AE
or a correct method seen
OE
M1
4
5
A1
oe