KAMSSA 2022 MATH APPLIED
PAPER 2(P425/2)
SECTION A:40 MARKS
20.5−10
1 E(x)=0.4x25=10 Ƶ =
3.873
𝜎𝑥=√0.4 𝑥 0.6 𝑥 25 = 2.711
=√15 OR 3.873 (3d.p) ∴ p (Ƶ < 2.711) =0.5+0.4966
P ( ≤ 20 )= p (x ≤ 20.5) =0.9966
2 8
2 𝐹̃ m𝑎̃ = 500 ( ) =( )
−3 −12
1000 1000 8
=( )N Power developed =( ).( )
−1500 −1500 −12
From v = 𝑢̃ + 𝑎̃t = (26,000 watts )
0 2
= ( ) + ( )x4
0 −3
3(a)
X 10 15 20
Y 2.9 _ -0.1
𝑦−2.9 −0.1−2.9
=
15−10 20−10
Y = 1.4
3(b)
X 20 30 _
y - 0.1 - 2.9 - 3.2
𝑥−30 30−20
=
−3.2−−2.9 −2.9−−0.1
X = 31.07 or 31.1
4)
A
120cm
20N
α Ɵ
B Q C
50cm
For all forces drawn
120
tan 𝜃 =
50
𝜃 = 67.4°
∝ = 22.6°
̃ =∝ +−20 cos 22.6°
∓𝑅
0 −20 sin 22.6°
But 𝜃 + −20 cos 22.6° = 0
𝜃 = 18.462𝑵
5)
speed 20 - 30 30 - 40 40 - 60 60 - 80 80 - 100
f 2 7 20 16 5
F 2 9 29 45 50
40
i) 40th percentile is the x 50 =20thvalue
100
50𝑥40
−9
100
=40 +( ) 𝑥 20
20
20−9
=40 +( )x 20
20
=40 + 11
= 51
b) Number of vehicles whose speed > 45
60−45
16 + 5 + x 20
20
= 56
6) Range (max) = 0.12 x 1000
=120m
𝑢2
Range (max) = when 𝜃 = 45°
𝑔
𝑢2
120 =
9.8
U = 34.2929m/s
Or = 14√6𝑚/𝑠
b) from
𝑣 2 = 𝑢2 sin 2 𝜃 – 2gH
𝑂2 = 1176sin 2
45° - 2 x 9.8H
588𝑚
∴𝐻= or 30m
2 𝑥 9.8
7) By simple interoal arithmetic
2.55 𝑥 4.15
Min value = 6.225 – 3.15 - ( )
4.5
= 0.7233 (4dip).
2.45 𝑥 4.05
Max value= 6.235 – 3.05 - ( )
5.5
= 1.381
∴ 𝑀𝑎𝑥 𝑝0𝑠𝑠𝑖𝑏𝑙𝑒 𝑒𝑟𝑟𝑜𝑟 = 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑒𝑟𝑟𝑜𝑟
1.381 – 0.7238
=
2
= 0.329 (3𝑠. 𝑓 𝑠 )
8)
6
a) +2y +y = 1
7
1 − 6⁄ =3y
7
1
Y= or 0.0476.
21
6
b) 3y + P(M∩N) =
7
6 3
P(M∩ 𝑁} = -
7 21
15 5
= or
21 7
𝑂𝑅 0.7143
SECTION B : 60 MARKS
9a)
500
500
Lion =L
5N
A
10ms-1
900 -(VT)
A
−𝑉0
𝑇 α
Ɵ (VT)
1500
10ms-1
L
6
sin 𝜃 =
10
𝜃 = 36.9°
→ ∝= 180° - 90° − 36.9° = 53.1°
∴ 𝐷𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑟 𝑏𝑒𝑎𝑟𝑖𝑛𝑔𝑠 𝑖𝑠
180° - 53.1° = 126.9°
b) Closest or best distance occurs at A.
A
900
T
T
β 100m
L
𝛽 = 50° − 36.9° = 13.1°
∴ 𝑙𝑒𝑎𝑠𝑡 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
AT = 100sin 13. 1° = 22.6651m (4𝑑. 𝑝)
Time to reach the closest distance A.
𝐴𝐿
t=
| 𝐿𝑉𝑇 |
𝑉𝑇
| 𝐿 ~ | = √102 − 62
= 8𝑀𝑆 −1
AL = 100cos 13.1° = 97.3976m (4𝑑𝑝)
97.3976
Time t =
8
= 12.1747 seconds
4−0
10a) ℎ= = 0.8
5
X 0 0.8 1.6 2.4 3.2 4.0
𝑓(𝑥) 1 5.7995 33.6347 195.0662 1131.2954 6561
4 1
∴ ∫0 32𝑥 dx = 𝑥 0.8 (6562 + 2𝑥1365.7958)
2
= 3717.44
b) let U = 32𝑋
In U = 2x in 3
1 𝑑𝑢
= 2 in 3
𝑈 𝑑𝑥
𝑑𝑢
dx =
2 𝑖𝑛 3𝑥4
𝑑𝑢 𝑢 32𝑥 4
∫ 𝑈 . 2𝑢 𝑖𝑛 3 = |
2 𝑖𝑛 3 2 𝑖𝑛 3 0
38 − 1
=
2 𝑖𝑛 3 2 𝑖𝑛 3
= 2985.58
c)|𝑒𝑟𝑟𝑜𝑟 | = |2985.58 − 3717.44|
= 731.86
∴ 𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑒𝑟𝑟𝑜𝑟
731 . 86
=
2985.58
= 0.245
= 0.25(2𝑑. 𝑝)
Comment: By increasing sub interrals or ordinates or reducing the width of strips.
500𝑚
11 (a) x = = 5m OR 8 x 103 ug
100
OR 2 X 103 ug
U=0
From 𝑉 2 = 𝑈 2 + 2ax
𝑉 2 = 𝑂2 + 2 X 10 X 5 = 100
10𝑚
∴𝑉=
𝑠
OR P.E = K.E =8 𝑋 103 X 10 X 5
1
= 𝑋 8 𝑋 103 𝑋 𝑈 2
2
=𝑉 2 = 100
=V = 10m/s
d) Initial total mom = 8 x 103 x 10 = 8 x 104 ng m/s
let 𝑉1 be the common speed
final total mom = ( 8 + 2 ) x 103 x 𝑉1
But 104 𝑉1 = 8 X 104
∴ 𝑉1= 8m/s
c)
R1=106N
W = (8 + 2) x 103 𝑥 10 = 105 𝑁
By Newton’s 2nd law downwards
105 - 106 = ( 8 + 2) x 103 𝑎
a = −90 𝑚/𝑠 2
Let h be distance penetrated.
From 𝑉 2 = 42 + 2ax
𝑂0 = 82 + 2 𝑥 ( −90) 𝑥 ℎ
64
h= m
180
32
or m
90
16
or m
45
or 0.3556m.
12a) P( x > 200) = P( Ƶ > Ƶ1 ) = 0.63
Ƶ1 = −0.332
200−𝑁
⇒ −0.332 = … … … … … (𝑖)
𝜎
P( X < 250) = p ( Ƶ <Ƶ2 ) = 054
Ƶ2 = 0.101 ,0R = 0.10, OR = 0.11
250−𝑁
⇒ 0.101 = … … … … . . (𝑖𝑖)
𝜎
i – ii
− 0.433𝜎
= -50
𝜎
= 115.4734 OR 113.1738
Put 𝜎 𝑖𝑛 (𝑖) 𝜑 = 200 + 0.332 x 115.4734
= 238.3372
OR = 237.5504
b) P( x > 195)
195−238.3372
Ƶ=
115.4734
= - 0.375
P( Ƶ > −0.375 ) = 0.5 + 0. 1462 (or = 0.465)
= 0.6462 or 0.6465
∴ % 𝑎𝑔𝑒 = 0.6462 𝑥 100
= 64.62
13a)
P
R
Ɵ
---
------ Mg cos 300
-
------
--
-
---
µR Mg
----
300
Resolve parallel to the plane
𝑚𝑔
Pcos 𝜃 𝜇ℛ + mgsin 30° =𝜇ℛ + ……………….(i)
2
√3
Resolve ∴ 𝑅 + 𝑃 sin 𝜃 = mgcos 30° = mg…….(2)
2
√3
R= mg - Psin 𝜃.
2
√3 𝑚𝑔
From (1) and (2) Pcos 𝜃 = 𝜇( 𝑚𝑔 − 𝑃 sin 𝜃) +
2 2
sin 𝜆
For equilibrium , 𝑡 𝑎𝑛𝑑 𝜆 =
cos 𝜆
sin 𝜆 √3 𝑚𝑔
Pcos 𝜃 = ( 𝑚𝑔 − sin 𝜃) +
cos 𝜆 2 2
√3 𝑚𝑔
Pcos 𝜃 cos 𝜆 = mg sin 𝜆 - P sin 𝜃 sin 𝜆 + cos 𝜆
2 2
𝑚𝑔
P(cos 𝜃 sin 𝜆 + sin 𝜃 sin 𝜆) = (√3 sin 𝜆 + cos 𝜆)
2
For 𝑃𝑚𝑖𝑛 , cos(𝜃 − 𝜆) = + 1
𝑚𝑔
∴ 𝑃𝑚𝑖𝑛 = (√3 sin 𝜆 + cos 𝜆)
2
13b)
P1
R1 µR1
Ɵ
mg
300
For all forces
𝜇ℛ1 = mgsin 30° + 𝑃1 cos 𝜃
𝑚𝑔
= + 𝑃1 cos 𝜃 ……………….(1)
2
ℛ1 = 𝑃1 sin 𝜃 + mgcos 30°
√3
= 𝑃1 sin 𝜃 + mg ………………….(2)
2
From (1) and (2)
√3 𝑚𝑔
𝜇 (𝑃1 sin 𝜃 + 𝑚𝑔) = 𝑥 𝑃1 cos 𝜃
2 2
sin 𝜆
For equilibrium,𝜇 = tan 𝜆 = .
cos 𝜆
sin 𝜆 √3 𝑚𝑔
(𝑃1 sin 𝜃 + 𝑚𝑔) = + 𝑃1 cos 𝜃
cos 𝜆 2 2
√3 𝑚𝑔
𝑃1 (sin 𝜃 sin 𝜆 + 𝑚𝑔 sin 𝜆) = cos 𝜆 + 𝑃1 cos 𝜃 cos 𝜆
2 2
𝑚𝑔
(√3 sin 7 − cos 𝜆) = 𝑃1 (cos 𝜃 cos 𝜆 − sin 𝜃 sin 𝜆)
2
=𝑃1 cos(𝜃 + 𝜆)
For 𝑃𝑚𝑖𝑛 cos(𝜃 + 𝜆) = + 1
NO 14a
14b)
x 82 78 86 72 91 80 95 72 89 74
y 75 80 93 65 87 71 98 68 84 77
ℛ𝑥 5 7 4 9.5 2 6 1 9.5 3 8
ℛ𝑦 7 5 2 10 3 8 1 9 4 6
𝐷2 4 4 4 0.25 1 4 0 0.25 1 4
∑ 𝑑2 =
22.5
6 𝑥 22.5
𝑃 = 1−
10(102 −1)
= 0.86
Comment : At 1% level of significance there is reasonable for much significant of math’s on
Economics
15 a) 𝑓(0) = 0+0-1 = -1
𝑓(1) = 1 +2 -1 = 2 OR
X 0 1
𝑓(𝑋) - 2
1
Since 𝑓 (0) x 𝑓 (1) < 0
or a change in sign of 𝑓(x), there in a real root between x=0 and x= 1
b)
𝑥 0 𝑥0 1
𝑓(𝑥) -1 0 2
𝑥0 −0 1−0
=
0− − 1 2− − 1
1
𝑥0 = ≈ 0.3
3
c) 𝑓 1(𝑥) = 3𝑥 2 + 2 , (𝑁𝑜𝑡 𝑓 1(𝑥𝑛 ) = 3𝑥𝑛2 + 2)
(0.3)3 + (2 𝑥 𝑜.3)−1
𝑋1 = 0.3 - = 0.464
3𝑥 (0.3)2 +1
(0.464)3 +2 (0.464)2 −2
𝑋2 = 0.464 -
3 𝑥 (0.464)2 +2
= 0.453
(0.453)3 +2 𝑋 0.453−1
𝑋3 = 0.453 -
3𝑋(0.453)2 +2
= 0.453
=|0.453 − 0.453|
∴ 𝑅𝑜𝑜𝑡 = 0. 45
16a) for 𝑓(𝑋 ) < 0 , 𝑓 (𝑋 ) = 0
For 0 ≤ 𝑥 ≤ 1
𝑋
𝑓 (𝑋 ) = 0 + ∫0 𝐾𝑥 (1 − 𝑋 2 ) 𝑑𝑋
(1 − 𝑋 2 ) 2 | 𝑋
−𝑘
=
4 0
−𝐾 𝐾
𝑓 (𝑋 ) = (1 − 𝑋 2 )2
4 4
𝐾 𝐾
𝑓 (1) = 0 + =
4 4
0 , 𝑥 < 0 0𝑟 𝑥 ≤ 0
𝐾 2 2 𝐾
∴ 𝑓 (𝑋 ) = | 4 − 4 (1 − 𝑋 ) ; 0 ≤ 𝑋 ≤ 1
𝐾
; 𝑥 ≥ 1 𝑜𝑟 𝑥 ≥ 1
4
b (i)
𝐾
=1
4
𝐾=4
𝐾 𝐾 𝑚 1
(ii) − (1 − 𝑋 2 )2 | =
4 4 𝑜 2
4 4 4 4 1
(ii) − (1 − 𝑚2 )2 + − =
4 4 4 4 2
1
1 − -(1 − 𝑚2 )2 =0
2
−𝑚4 + 2𝑚2 − 1 = 0
1
+𝑚2 − 2𝑚2 + =0
2
√16−8 4 ±2√2
𝑚2 = 4 ± =
4 4
√2
M = √1 − =√0.2929
2
= 0.5412
c) 𝑚𝑒𝑎𝑛 𝑜𝑓 𝑥
1
E(X) = ∫0 𝑥 . 𝑘 𝑥 (1 − 𝑋 2 ) 𝑑x
1
=∫0 𝑘 (𝑥 2 − 𝑥 4 )𝑑 x
𝑋3 𝑋5 1
=𝐾 ( − )|
3 5 0
1 1
=4 ( − ) = 0
3 5
8
= OR 0.5333
15
END
1
Area of 1st triangle = x 30 x 45
2
= 675 𝑠𝑞 . 𝑢𝑛𝑖𝑡𝑠
1
Area of 2nd triangle = x 60 x 25
2
= 750 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠
1
Area of a trapezium = ℎ (𝑎 + 𝑏 )
2
1
= x 60 (45 + 65)
2
= 3150 𝑠𝑞 . 𝑢𝑛𝑖𝑡𝑠
Total are = 675 + 750 + 3150
= 4575𝑠𝑞 . 𝑢𝑛𝑖𝑡𝑠
END
