Nov 2001 P2

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

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

(dep on first M1)

A1

further

A1

ADˆ B = 57. 8° soi ˆB ) √ 67.2° or (125° – their AD

A1

p = 17

B3

q = 36

A1

r = 125 (B1 each)

x = 3n + 2 oe

y = (n + 1)2 oe

t = n 3 + n 2 + 5n + 3 3

M1

M1

s = 178 or their (p + q + r) v

(b)

dep

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 )

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