121/2 MATHEMATICS PAPER 2 – MARKING SCHEME

NO WORKING MARKS COMMENTS

1 Log 559.3 = 2.7476
10 tan 30 = 10 x 0.0524
= 0.524

 2.7476 M1 Working out the

+ 0.5240 denominators to
3.2716 get 3.2716

0.7493 cos2 16.3350

3.2716
No. Std. form Log
0.7493 7.493 x 10-1 1.8747
Cos216.335 2(1.982) 1.9642 + M1 For attempt to
1.8389 add and subtract
3.2716 3.2716 x 100 0.5148 - correctly.
1.3241
-1 + 0.3241
2 M1 Division by 2
= 1.6621

= 10-1 x 4.593

= 0.4593 A1 CAO
04
2 ¼x2 + 1/9
¼ x2 + k + 1/9
a + 2ab + b2
a2 = ¼ x b2 = 1/9
2ab = k
2 x ½ x 1/3 = k
M1
k = 1/3 x A1
02
3
x6, x5 -9 , x4 a 2, x3 -a 3, x2 -a 4 , x -a 5, -a 6
x2 x2 x2 x2 x2 x2
1.x – 6.x a +15x a – 2x a 6 +15x a - 6x a 5 + 1. a 6
6 5 4 2 3 2 4

x2 x2 x2 x2 x2 x2
6 5 4 2 3 3 2 4 5 6 B1 For items and
x – 6ax + 15x a – 20x a + 15x a – 6xa + a coefficients
x2 x4 x6 x8 x10 x12 combined correctly
x6 – 6ax3 + 15a2 20a3 + 15a4 - 6a5 x a6
x3 x6 xa x12
15a2 = 1215
Equating to 1215
a2 = 81 M1
a = +9
C.A.O (+9)
A1
03

1
4

1.75r 24.8cm

θ = 2πr = c
360
1r = 57.2960
1.75 – (1.75 x 57.2960)
= 100.2680
100.26 x D x 22 = 24.8 M1 For substitution
560 7
0.8754D = 24.8
D = 24.8
0.874
D = 28.33cm For diameter
A1
03
5

AL2 = AE2 + EL2

= 52 + (1.5)2
= 27.5 M1
= 5.240

AM = 5.2442 + 42

= 27.5 + 16
M1
= 43.5

= 6.5955

Cos θ = 4
6.596
= 0.6064
θ = cos-1 0.6064 = 52.670 A1
04
2
6.
(0.0054667)½ = (54.667 x 10-4)½
= 7.3939 x 10-2 M1 Use of table, to
= 0.073939 get square root.

= 3
0.043279

= 3 x 0.2305 x 100
M1 Getting
= (69.15)2 reciprocals and
squaring it using
= (6.915 x 101)2 log tables.

= 47.82 x 102

= 4782

 0.073939 + 4782

= 4782.073989 Giving answer to

A1 5 significant
= 4782.1 figure.
03
7 A = P(1 + r)n Period = 4
100 Rate = 4
A = 1250000 ( 1 + 4.5)8
100 M1
= 1250000 (1.045)8
= 1777625.766
M1
Half withdrawn
Remaining 888812.883
A = 888812.883 (1 + 45)10 M1
100

= 888812.883 (1.045)10
= 1380299.229

Interest in phase one

= (1777625.766

Interest in phase two

1380299.229 – 888812.883
= 491486.346

Total interest
(527625.776 + 4914863.436)
=1019112.112
≈ 1019112 A1
03
3
8 16sm2 x + 4 cos x = 10
8sm2 x + 2 cos x = 5
8(1 – cos2x) + 2 cosx = 5
8 – 8 cos2x = 2 cos x = 5
-8 cos 2x + 2 cos x + 3 = 0
8 cos2x – 2 cosx – 3 = 0
Let cos x = y
M1 Accept use of sin
8y2 – 4y – 3 = 0

y=2± (-2)2 – 4 x 8x – 3
16

= 2± 4 + 96
16
= 2 ± 10
16
Cos x = 12 => cos x = 0.75 M1
16
and cos x = -8 => cos x = -1
16 2
Cos x = -0.5 – found in the 2nd and 3rd quadrant.

For 3rd quadrant

Acute angle = 600
Then x – 1800 = 600
x = 2400 A1
Tan 2400= 1.7321
B1
04
9 (a) P = KL + B K & B where K and B are constants
√C
500 = 16K + B ……….. (i)
4
800 = 25K + B ……… (ii)
M1
5 Attempt to find
K = 2000 = 32.79 the constants.
61
B = -98.56
P = 32.79L - 98.56
√C A1

(b) P = 2000 (81) – 98.56

61 √81 M1
= 1327.87 – 10.95
= 1,316.92
A1
04
4
10

1250
Wall F. Post
(H)
620
70m
(a) Tan 620 = H => H = 131.65m
B1
70
0
(b) Tan 25 = x
70
 x = 70 Tan 250
= 32.64
Therefore (wall) = 131.05 – 32.64 M1
= 99.03m A1
03
11 Log 135 = log 33x5
= 3 log 3 + log 5 M1
= (1.583) + 2.322
= 7.071 A1

02
12 Max. Q = 12.7 = 1.7887
7.1
Min. Q = 12.5 = 1.7123 M1 Both values
7.3
Actual = 12.6 = 1.75
7.2
A.E = ½ (1.7887 – 1.7123) M1

= 0.0383

% Error = 0.0382 x 100

1.75
= 2.1829 A1
03

5
13 H
10
F
8
A E B 8 8 H
B1 Sketch
2 2 2 B1  Labelling
B 8 C
G 10 F 2 G
8
F
10
G

Area of part A = 2 x 10 = 20
Area of part B = 2 x 2 = 4
M1
Area of part C = ½ (2 + 8) 8 x 2 = 80
Area of part D = 8 X 2 = 16
Area of part E = 2 x 8 = 16
A1
Total area = 136cm3
04
x+1 2x+1
14 9 – 54 = 3
(3 ) (32) – 54 = (32x) (31)
2x
M1
Let 32x = y
9y – 54 = 3y
M1
6y = 54 = y = 9
32x = 32 => 2x = 2
A1
Therefore x = 1
04

6
15
B(2, -1)

A(-6,5)

M -6+2, 5 -1
2 2 B1
= M (-2, 2)
MB = 2 – -2 = 4
-1 2 -3

MB = Radius = 42 + (-3)2
=5 M1
(x + 2) + (y – 2)2 = 5
2

x2 + y2 + 4x – 4y – 17 = 0 A1
03
16 y + 2x + 1 = 0
x2 + xy = -6
y = (-2 – 1)
x2 + x (-2x – 1) = -6 M1
-x2 – x + 6 = 0
(x – 2) (x + 3) = 0 M1
x = 2 when y = -5
x = -3 when y = 5 A1
04
17 (a) Tree diagram
4
/5 T ………….RBT
1
B /5 N ………...RBN
4
/15
5
/7 R ……….. RNR B1 Labelling
18
/21 R N
1 2
/15 /7 B2 Indicating all
N ……… RNN probs
3 4
/21 B /5 T ……….. NBT
1
/13
1
N /5 N ……….. RBN
12 5
/13 N /7 R ………. NNR
2
/7 N ………. NNN
(b) (i) P (RNR) or P (RNN) or (RNN) or P (NNR) or (NNN)
6 x 11 x 5 + 6 x 1 x 2 + 3 x 12 x 5 + 3 x 12 x 2
7 15 15 7 15 7 21 13 7 21 13 7 M1

= 2 + 60 + 60 + 23 = 170
49 637 637 637 637 M1 A1

(ii) P (RNR) or P(NNR)

= 6 x 11 x 5 + 3 x 12 x 5
7
 7 15 7 21 13 7 M1
= 86
637 A1
(iii) P (R’) = 1 – P (RNR + P (NNR)
= 1 – 86 M1
637
= 557 A1
637
10
18 Total tax = PAYE – Relief
= Ksh. 2172 + 1093 M1
= Ksh. 3265p.m
= Ksh. 39,180 p.a. A1

Tax (K£ p.a.) Taxable income (K£p.a.)

4512 x 2 (9024) 4,512 M1
4512 x 3 (13,536) 4,512 M1
4x = 16,620 (16,2620) 4,155 M1
13,179
Therefore taxable income = Ksh. 13,179 p.a
= Kshs. 21,965p.m. M1
Basic salary = Ksh. 21965 – (5000 + 2000) M1
= Ksh. 14,965p.m. A1
Total deductions = Ksh. 2172 + 320 + 4000
= Ksh. 6,492p.m.
Gross income = Ksh. 21965p.m.
Therefore Koech’s net pay = Ksh. 21965 – 6492 M1
= Kshs. 15,473p.m. A1
10
19
B1 Construction of pt
C
B1 Construction of
ΔABC
B1 Locating the
centre
B1 Bisecting <s
Radius of circle
B1 construction.
Construction of
B1 angles 300.
Centre of circle
B1
Locus L
B1
(d) radius = 4.6 + o.1

Area = 120 x 22 x 4.6 x 4.6 M1

360 7
= 22.17cm2 A1
10
20
8
 x -30 0 30 60 90
+ cos 2x 2.0 4.0 2 -2.0 -4.0 B2 All 6
3 sin (x+30) 0 1.5 2.6 3.0 2.6

120 150 150 210 240 270

-2.0 2.0 -4.0 2.0 -2.0 -4 B1 At least 4
1.5 0 -1.5 -2.6 -3.0 -2.6

(c) (i) x = 240, 1380, 3520 + 0.5 B2 all

B1 2
(ii) 3 sin (x + 300) = y
3 sin (x + 30) = y -y = y x = 168 + 0.5 B1

(iii) 4 cos 2x = y 2x = 360

x = 1800
Therefore period = 1800 B1
10
21 (a) 4800 = 80 0

60 M1
90 – 65 = 250
80 – 25 = 550
90 – 55 = 350 M1
R = 350 ………… A1

(b) X (650N, 360E)

Y (650N, 1440W)
Angle difference
144 + 36 = 1800 M1
Small circle
Therefore 1800 x 600 cos 650
9
 = 4564.28nm

Then: (650N, 1440W)

(350N, 1440W)
Angle difference
65 – 35 = 300 M1
Current circle
30 x 60 = 1800nm

Total distance
4564.28 + 1800
= 6364.28nm A1

Direct route
Angle difference = 800
Along a great circle (Longitude)
A1
Therefore 80 x 60 = 4800nm
Route difference in distance
6364.28 – 4800 = 1564.28nm A1

(c) Route difference in distance = 1564.28nm

Angle difference = 1564.28 = 26.070 M1
60
 35 + 26.07 = 61.070
0 0

(61.070N, 1440W) A1

10
2
22 (a) y = x + 2
y = 10 – x2
x2 + 2 = 10 – x2
x2 + x2 = 10 -2
2x2 = 8
x2 = 4
x = +2
x=2
y2 = x + 2 y = 10 – 2
y= 4 + 2 or y = 10 – 4
y=6 y=6
Q(2, 6) …………………………….. A1
(b) h = 2 – 0 = 2 = ¼ or 0.25
8 8

For y = 10 – x2
x 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00
10 10 10 10 10 10 10 10 10 10 T2
-x2 0.00 -0.0625 -0.25 -0.5625 -1.00 -1.5625 -2.25 -3.0625 -4.00
y 10.00 9.9375 9.75 9.4375 9 8.4375 7.75 6.9375 6
9.96875 9.4375 9.59375 9.21875 8.71875 8.09375 7.34375 6.46875

= 0.25(9.96875+09.84375+9.59375+9.21875+8.71875+8.09375+7.34375+6.46875)
= 0.25 (69.25)
= 17.3125…………………. A1

10
 For y = x2 + 2
X 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 T1
x2 0.00 0.0625 0.25 0.5625 1.00 1.5625 2.25 3.0625 4.00
2 2 2 2 2 2 2 2 2 2
y 2.00 2.0625 2.25 2.5625 3.00 3.5625 4.25 5.0625 6.00
3.03125 2.15625 2.40625 2.78125 3.28125 3.90625 4.65625 5.53125

=0.25(3.08125+2.15625+2.40625+2.78125+3.28125 + 3.90625 + 4.65625+5.53125)

= 0.25 (27.75) A1
= 6.9375…………………..

Shaded region:
17.3125 – 6.9375 A1
= 10.375

(c) y = 10 – x2
2
(10x – x2)dx
0 2
M1
10x – x3
3 0
M1
3 3
10(2) – (2) – (10(0) – (0)
3 3
(20 – 2.6667) – 0 A1
17.3333

10
23 (a) p = krn
Log p = Log rn + log k
Log p = n logg r + log k B1
(b)
p 1.2 1.5 2.0 2.5 3.5 4.5
r 1.58 2.25 3.39 4.74 7.86 11.5
Log p 0.792 0.1761 0.3010 0.3979 0.5441 0.6532
Log r 0.1987 0.3522 0.5302 0.6758 0.8954 1.0607 T2

n = the gradient

n = 0.3522 – 0.1987 = 0.1535 = 1.58

0.1761 – 0.0792 0.0969 M1 A1
Log k = 0.9
No Log
k 0.9
7.943
k = 7.943 ……………….

(b) A1

A1

11
 Log p

(c) P = 7.943r1.58

10

12
24 Scale 1cm rep 100km

B3 For locating ABC

B1 For locating D

A
(a) AC => 12.5cm
12.5 x 100
= 1250km A1 For 1250km

S=D
T
200 = 1250
T
T = 1250 = 6.25 M1 For correct
200 substitution
= 6 hours 15 minutes A1

(b) 1490 B1

(c) 8.3cm
8.3 x 100 = 830km A1
10

13