ACK SCHOOLS FORM FOUR JOINT EXAMINATIONSKenya Certificate of Secondary Education (K.C.S.E)

121/2MATHSPAPER 2MARKING SCHEME

1. No s.f. Log45.3 4.53 x 101 1.66510.0067 6.97x 10-3 3.8432

1.50830.534 5.34 x 10-1 1.7275

1. 7808 3

0.845 8.45 x 10-1 1. 9269

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All logs correct

Operations + &

Division by three

Answer even if in standard form04

2.a)16+6.15(−1

2x)+15.14(−1

2x)

2

+20.13(−12x)

3

=1-3x+ 154x

2

−54x

3

b) -12x= -0.02

x= 0.04

= 1-3(0.04 )+154

(0.04 )2+54

(0.04 )3

= 1.11392

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Attempt to expand simplification.

043.

550,000 = 800000(1− r100 )

4

0.9106=1-r

100r=8.942%

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034. A.S.F= Determinant of the matrix

6010=x( x+3 )-12

6=x2+3x-12(x2−3 x )+(6 x−18 )=0

( x+6 ) ( x−3 )=0X= -6 or 3X=3

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035. √3

1−cos300=√3

1−√32

= 2√3– (2+√3 )2√3−(2+√3)

= 2√3(2+√3)4−3

= 4√3+ 6

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Accept Equivalent

03

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6

OD = OA + 23AB

= ( 44

−6)+( 23 )(10 −4

4 −412 −6 )

= ( 44

−6)+( 23 )( 6

018)

= ( 44

−6)+( 40

12)8i + 4j + 6kCoordinates of D are; (8, 4, 6)

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Expression of OD

Simplification

037

Z X2

√Y Z =K X

2

√YZ1 = ¿¿

= 1.44 K X2

0.9√Y = 1.6Z

= % change (1.6 z−zz

¿ x100%

= 60%

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038 a)

12−1

5= 3

10

Required time = 103

= 313 or 3 hours 20 mins

b)12−1

5−1

6= 4

30

Required time = 303

= 7 hrs. 30mins or 7 ½

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049 ( 2

3x+20)+5

6x+100=90

96x+30=90

96x=600

X= 400

=Tan ( x+20 )

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=Tan 6003

10 Log x+log y p=logQ pxLog x + p log y=px log QLog x=px log Q- p log yLog x= p( x logQ−log y )

log xx logQ−log y= p

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. 0311

Midpoint ( 2±22, 4+6

2 )(0,5 )

Length=√ (0−2 )2 (5−4 )2

Radius √4+1 =5units( x−0 )2+ ( y−5 )2=(√5 )2

X2+y2-10y+20=0

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A103

12 P(R) bag A 12× 4

12= 16

P(R) bag B 12× 3

10= 3

2016+ 3

10= 38

120

= 1960

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0313 a) a, a+d, a+5d

2+d2

=2+5d2+d

4+4d+d2=4+10d4d+d2=10dd(4+d)=10d4+d=10d=6

b) r= (6+2)×12

8×12

=4

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0314 12x-8-6x+3=6-5x

6 66x-5=6-5x11x=11x=1

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0215 4.05 2.15

Limits 4.1 2.2 4.15 2.25Maximum area 4015x2.25=9.3375 B1 For three areas

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Actual area 4.1x2.2=9.02Minimum area 4.05x2.15=8.7075

A.e = 9.3375−8.7075

2 = 0.315

%error=(0.3159.02

×100)%

= 3.492%

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Accept equivalent

0316 a) 4y=3 sin

25x

y=34 sin

25x

Amplitude = 34 or 0.75

b) period = 360

25

= 360×5

2 = 9000

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0317 a) Frequency of modal class = 15

x d f t ft Ft2

9 -9 4 -3 -12 3612 -6 7 -2 -14 2815 -3 11 -1 -11 1118 0 15 0 0 021 3 8 1 8 824 6 5 2 10 2027 9 3 3 9 27

∑ f=53 ∑ ft=−10∑ ft2=130

i)Mean(x) = 18+ (−1053 )x3

= 18- 3053

= 17.434

ii) Variance =s2=32[13053

−(−1053 )

2] =9[2.4538−(−0.1887 )2 ] =21.7548iii) Standard deviation = √variance = √21.7548 = 4.664

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For x valuesforft values

Ft2 values

1018 a) sin<CDE =

4.58

CDE= sin-1 4.58

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= 34.230

b)i AC= √102+82

√164 = 12.81 ii) DE=82-4.52=6.61cm AE= √6.612+102

=√143.75 = 11.99

Tan CAE = 4.5

11.99CAE = 20.570

c)i MB = √102+42

= √116 = 10.77

ii) Sin CBM = 4

10.77 CBM = 21.80

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1019 a) Shortest distance

i) in Kilometres

= θ

360x2πR

30+30 = 600

60360x 22

7x 6370

= 6673.3ii) in nautical mile (nm) 60x 66 = 3600nm

b) 3600600 = 6hrs

c) Time difference at Q = 180x4= 720hrs720

6 = 12hrsDistance via parallel latitude= 60x1800cos 60= 5400nm

Time of travel = 5400600 = 9hrs

Time at P =1000hrsTime at Q = 100+1200+900 3100-2400 = 0700hrs Tuesday

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1020 (a)i) AC=AO+0C

=-p+q =q-p ii) OB=OA+AB

=p+34q

iii) BC=BO+OC

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= -p-34q+q

= 14q−P

b(i) OX=h(P+34q¿

= hp + 34hq

OX=OA+AX =p+t(q-p) = (1-t)p+tq

hp+34hq=(1−t ) p+ tq

34hq=tq

34h=t

hp=1-t

h=1-34h→1 3

4h=1

h=47

34× 4

7=t

t=37

ii) OB=BX77

: −37

7 :-3

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10

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21 1 2 3 4 5 6

1 2 3 4 5 6 72 3 4 5 6 7 83 4 5 6 7 8 94 5 6 7 8 9 10

a) i) P(6) = 424 =

16

ii) P(Total of odd) = 1224=

12

iii) P(6 or 9) = 14

P(6) = 4

24

P(9) = 2

24

424 +

224

624 = 14

b) P( win) 14

P( loss) 34

P(wins atleast once) P(WW) + P(WL)+P(LW)

(14× 1

4 ) +(14× 3

4 ) +(34× 1

4 )

116 +

316+

316 =

716

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1022

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CUBE

TETRAHEDRON

a) Gross tax = 5512 + 1162 = 6672

1st 9680 x 10

100 = 968

2nd 9120 x 15

100 = 1368

3rd 9120 x 20

100 = 1824

4th 9120 x 25

100 = 2280

5th 780 x 30

100 = 234

37820 6674

Basic salary = 37320 – 15220 = Ksh. 22600

b) Net payTotal deductions = 320 + 200 + 7500 + 7500 = 13532Net pay = Gross – deduction

= 37820 – 13532 = 24288

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1023 a)

72000n−5

−7200n

72000n−72000n+36000n (n−5 )36000n (n−5 )

b)36000n (n−5 )=1200

n2−5n−300=0(n-20)(n+15)=0n=20

c) Original contribution72000

20 =3600

Final contribution72000

15 = 4800

( 4800−36003600

×100)%

=33.3%

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1024 X 0 15 30 45 60 75 90 105 120 135 150 165

3sinx 0 0.78 1.50 2.12 2.60 2.90 3.00 2.90 2.60 2.12 1.5 0.784cos(2x + -10)

3.94 3.76 2.57 0.69 -1.37 -3.06 -3.94 -3.76 -2.57 -0.69 +1.37 3.06

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