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ZERAKI ACHIEVERS’ EXAMINATIONS (2023)
Term 1 - 2023
MATHEMATICS (121/2)
PAPER 2
FORM FOUR (4)
Time: 2 ½ Hours
MARKING SCHEME
NO. WORKING MARKS REMARKS
Maximum Area
¿ 10.5 × 14.5=152.
Actual Area
¿ 10 × 14 = 140
Absolute error in area ¿ 152.25− 140 =12.
% error =
× 100
M
M
A
Maximum A.E. in the area
Total 3
2 x
2
+ 2 y
2
+ 8 x − 10 = 0
= x
2
+ 4 x + y
2
x
2
+ 4 x + 4 + y
2
( x + 2 )
2
+( y − 0 )
2
2
Centre (− 2 , 0 ) and radius
3 units
B
B
B
B
Completing the square in x
Centre and radius obtained
Cartesian plane
Circle drawn
Total 4
tan 60
0
1 −cos 30
0
√
√
√
(
√
)
(
√ 3
)(
√ 3
)
B
M
Identifying tan 60
0
and cos
0
in terms of surds
Rationalizing the
denominator
√
√
(
√
)
× 4
¿ 4 √ 3 + 6 → 6 + 4 √ 3
a = 6 , b = 4
and
c = 3
A
Total 3
NO. WORKING MARKS REMARKS
log
3
2
−log
3
1
2
+log
3
log
3
81 −log
3
12 +log
3
log
3
(
81 × 972
)
=log
3
log
3
8
= 8 log
3
M
M
A
Total 3
Tn
2 m
√
L − A
2
3 k
T
2
n
2
4 m
2
L − A
2
3 k
3 k T
2
n
2
= L 4 m
2
− A
2
4 m
2
3 k T
2
n
2
+ A
2
4 m
2
= L 4 m
2
A
2
4 m
2
= L 4 m
2
− 3 k T
2
n
2
A
2
L 4 m
2
− 3 k T
2
n
2
4 m
2
A = ±
√
L 4 m
2
− 3 k T
2
n
2
4 m
2
M
M
A
Removal of square root/squaring
both sides
Term in A on one side
Total 3
- Binomial coefficients 1, 7, 21, 35, 35, …
( 1 + 2 x )
7
7
+ 7 × 2 x + 21 × ( 2 x )
2
+ 35 × ( 2 x )
3
+ 35 × ( 2 x )
4
( 1 + 2 x )
7
= 1 + 14 x + 84 x
2
+ 280 x
3
+ 560 x
4
1 + 2 x
7
7
→ 1 + 2 x =0.
2 x =0.98− 1 → 2 x =−0.02 → x =−0.
7
2
3
4
Total 3
θ =tan
− 1
(
)
0
Obtuse angle
¿ 180
0
0
0
A
B
Total 3
NO. WORKING MARKS REMARKS
A. s. f =
= 4 x −
{
2 ( x − 1 )
}
6 = 4 x −( 2 x − 2 )
6 = 4 x − 2 x + 2
6 − 2 = 4 x − 2 x
2 x = 4 → x = 2
M
M
A
Equating Asf to determinant of
matrix
Collection of like terms
Total 3
P = a Q
n
→ log P =log a + n log Q
log a = 1 → a =log
− 1
n =
B
M1, A
Total 3
5 s + 3 m = 2816
3 s + 5 m = 3360
(
)(
s
m
)
(
)
det ¿ 5 × 5 − 3 × 3 = 16
(
)
(
)
(
)
(
)(
s
m
)
(
)
(
)
(
)(
s
m
)
(
)
(
)
set = sh.
table = sh.
}
M
M
M
A
Matrix equation
Inverse
Pre-multiplication
Both
Total 4
× r
2
r =
√
346.5 × 7
=10.5 cm
Area of sector
×
× 10.5 × 10.
M
M
A
Expression for area of circle
¿ 69.3 m
2
Total 3
Amount = 750,000 – 225, ¿ 525,
Amount to pay = 15×55,000=825,
Let the rate of interest be
r %
per month
(
r
)
15
(
r
)
15
r
15
√
r
100 + r =103.
r =103.059− 100 =3.059%
M
M
A
Total 4
NO. WORKING MARKS REMARKS
- Total mass
¿ 20 × 200 = 4000 kg
New mass
× 20 × 200 = 3625 kg
¿ 375 kg
M
M
A
Total 3
- (a) Monthly basic pay
Let the taxable income be A
Gross Tax = 4312 + 1062 = 5374
9680 ×
9120 ×
9120 ×
( A − 27920 ) ×
A − 27920 =
A = 27920 + 4856 =32,
Basic Salary
(b) Net salary
M1, A
M
M
A
M1, A
M
M
A
Total 10
NO. WORKING MARKS REMARKS
(a)
6
2
2
2
− 2 × 11 × 7 cos C
2
2
2
=− 154 cos C
cos C =
C =cos
− 1
(
)
C =29.
0
(b) Radius of the circle
sin 29.
0
= 2 R
R =
2 sin 29.
0
R =6.1 cm
(c) Shaded area
AOB
¿ 2 × 29.
0
0
Area of sector
× 3.142 × 6.1 × 6.
Area of AOB
× 6.1 × 6.1sin 59
0
Shaded area
¿ 3.2 cm
2
M
M
A
M
A
B
M
M
M
A
Application of cosine rule
Attempt to get C
Equating sine rule to radius
of circumcircle
Angle at the centre
Area of sector
Area of triangle
Total 10
NO. WORKING MARKS REMARKS
- (a) (i) Geometric series
= 2 = common ratio
(ii)
16384 = 2 × 2
n − 1
n − 1
13
n − 1
13 = n − 1 → n = 14
(iii) Sum of first 14 terms
S
14
( 2
14
)
S
14
(b) Last term of the sequence
n
{ 2 × 3 +( n − 1 ) 4 }
1640 = 4 n
2
+ 2 n
4 n
2
+ 2 n − 1640 = 0 → 2 n
2
+ n − 820 = 0
( n − 20 ) ( n +20.5)= 0
n =−20.
Hence
n = 20
( 3 + l )
820 = 10 ( 3 + l )
82 = 3 + l
l = 82 − 3 = 79
B
M
M
A
M
A
M
A
M
A
Total 10
Total 10
NO. WORKING MARKS REMARKS
- (a) Total rate of flow in litres
¿ 120 + 150 = 270 litres / minute
Time taken
¿ 70 minutes
hours
(b) (i) Part of tank filled after 25 minutes
¿ 270 × 25
¿ 6750 litres
Time taken to fill the remaining part
¿ 48.6 minutes
Time taken to fill the tank
¿ 25 +48.6 minutes
¿ 73.6 minutes ≅ 74 minutes = 1 hr 14 minutes
11.15 a. m .+ 1 hr 14 minutes
¿ 12.29 p. m.
M
M
A
M
M1, M
A
M
M
A
Total 10
− 2 a + 2 b = 2 − 2 c + 2 d =− 2
4 a = 0 → a = 0 4 c = 4 → c = 1
b = 1 d = 0
Hence U =
(
)
(ii) Reflection along
y = x
or
y − x = 0
B
Total 10
NO. WORKING MARKS REMARKS
- (a) Tree
diagram
(b) (i) Exactly 2 students pass
¿ ( 0.8 × 0.6 × 0.8) +( 0.8 × 0.4 × 0.2) +( 0.2 × 0.6 × 0.2)
(ii) At most 2 students pass
¿ 1 −{ P
(
PJL ∧ P
'
J
'
L
'
) }
{
( 0.8 × 0.6 × 0.2) × ( 0.2 × 0.4 × 0.8)
}
{ 0.096+0.
}
(iii) Only 1 passes
0.8 × 0.4 × 0.
0.2 × 0.6 × 0.
0.2 × 0.4 × 0.
(iv) At least 1 passes
P (all fail)
¿ 1 −( 0.2 × 0.4 × 0.8)
Total 10
NO. WORKING MARKS REMARKS
- (a) (i)
SR = SO +¿
p +
q
(ii) QS = QO + OS
¿− q +
p
(iii)
PT = PS + ST
p +
(− QS )=
p +
(
q −
p
)
p +
q −
p =
p −
p +
q
p +
q =
q −
p
(iv)
TR = TQ + QR
QR −
q
(
− q +
p
)
q =
q −
q −
p
q −
p
(b) Collinearity between P, T and R
PT =
q −
p ∧ TR =
q −
p
⇒ TR = 3
(
q −
p
)
⇒ TR = 3 PT
Hence
PT /¿ TR
Point T is common hence P, T and R are collinear
B
B
M
M
A
M
A
B
B
B
Total 10
