Post on 23-May-2022
Solutions to CSEC Maths P2 May 2019
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Question 1a part (i)
Calculate the exact value of
πππ
β πππ
π
= (9
4β
8
5) Γ· 3
= (5(9) β 4(8)
20 Γ·
3
1)
= (45 β 32
20) Γ
1
3
=13
20 Γ
1
3
=13
60
Question 1a part (ii)
Calculate the exact value of
2.14 sin 75π
= 2.07 to 2 decimal places
Question 1b part (i)
Data Given
Item Amounts Allocated
Rent $π₯
Food $629
Other living expenses $2π₯
Savings $1,750
Total $4,320
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Calculate the annual take-home pay
π ππ‘π ππ πππ¦ = $4,320 πππ ππππ‘πππβπ‘ (ππ£πππ¦ 2 π€ππππ )
πππ π¦πππ = 52 π€ππππ
β΄ πβπππ πππ 52
2= 26 ππππ‘πππβπ‘π πππ π¦πππ
26 Γ $4,320 = $112,320 πππ π¦πππ
Question 1b part (ii)
Calculate the amount of money spent on βRentβ
Required to find rent
π₯ + 629+ 2π₯ + 1,750 = 4,320
3π₯ + 2,379 = 4,320
3π₯ = 1,941
π₯ = 1, 941
3
π₯ = $647
Question 1b part (iii)
Calculate how long it will take to pay off Tuition
ππ’ππ‘πππ πππ π‘ = $150,000
πππ£ππππ πππ ππππ‘πππβπ‘ = $1,750
πππ‘ππ π ππ£ππππ πππ π¦πππ = $1,750 Γ 26
= $45,500
# of years to pay off = $150,000
$45,500 β π.ππ ππ π π .π.
OR
The question says that she saves the same amount of money each MONTH. There are two
fortnights every month and 12 months per year. Therefore, there are 24 fortnights to consider.
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πππ£ππππ πππ π¦πππ = $1,750 Γ 24
= $42,000
# of years to pay off = $150,000
$42,000 β π.ππ ππ π π .π.
Either way she must work at least 4 years QED
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Question 2a part (i)
Simplify 3π2 Γ 4π5
= 12π2+5
= 12π7
Question 2a part (ii)
Simplify 3π₯
4π¦3 Γ·
21π₯2
20π¦2
= 3π₯
4π¦3 Γ 20π¦2
21π₯2
= 60π₯π¦2
84π₯2π¦3
=5π₯π¦2
7π₯2π¦3
= 5
7π₯π¦
Question 2b part (ii)
Solve 3
7π₯β1 +
1
π₯ = 0
3π₯ + 1 (7π₯ β 1)
π₯(7π₯ β 1 ) = 0
If a fraction = 0, then the numerator is 0.
3π₯ + 7π₯ β 1 = 0
10π₯ β 1 = 0
10π₯ = 1
π₯ =1
10
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Question 2c part (i)
Simplify (π₯ Γ 2)2 = π¦
(2π₯)2 = π¦
π¦ = 4π₯2
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Question 2c part (ii)
Calculate the values of x
π¦ = π₯ ---------- equation 1
π¦ = 4π₯2 ---------- equation 2
Let equation 1 = equation 2.
π₯ = 4π₯2
0 = 4π₯2 β π₯
0 = π₯(4π₯ β 1)
πβπππππππ, πππ‘βππ π₯ = 0 ππ 4π₯ β 1 = 0.
πΌπ 4π₯ β 1 = 0
πβππ 4π₯ = 1
π₯ = 1
4
β΄ π₯ = 0 πππ1
4
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Question 3a
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Question 3b part (i)
Question 3b part (ii)
The single transformation that maps π«πππ onto π«πππ is a 180o clockwise or anti-clockwise
rotation about the origin OR a reflection in the origin.
Question 3b part (iii)
The 2 x 2 matrix that maps ΞABC onto ΞLMN is
[β1 00 β1
]
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Question 4a part (i)
Express P varies inversely as the square of V
π β 1
π2
β΄ π = π
π2
Question 4a part (ii)
Calculate the value of π when π = 1, given that V = 3 when P = 4.
4 = π
32
4 = π
9
4 Γ9 = π
β΄ π = 36
If P = 1, then
1 = 36
π2
π2 = 36
π = β36
π = 6
Question 4b (i)
Calculate the value of x for β7 < 3π₯ + 5 β€ 7
πΆπππ ππππ β 7 < 3π₯ + 5
β7β 5 < 3π₯
β12 < 3π₯
β4 < π₯
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πΆπππ ππππ 3π₯ + 5 β€ 7
3π₯ β€ 7β 5
3π₯ β€ 2
π₯ β€2
3
Hence, β4 < π₯ β€2
3.
Question 4b (ii)
Draw the graph of β4 < π₯ β€2
3
Question 4c (i)
Considering,
π₯
3+
π¦
7= 1
Find the coordinates of the y-intercept
State the equation in the form π¦ = ππ₯ + π,
And find the value of c.
Multiplying by the LCM of the denominators gives,
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7π₯ + 3π¦ = 21
3π¦ = β7π₯ + 21
π¦ = β7
3π₯ +
21
3
π¦ = β7
3π₯ + 7
β΄ πΆππππππππ‘ππ ππ π πππ (0,7).
Question 4c (ii)
State the gradient of the line
The gradient of the line is given by βmβ.
πΊπππππππ‘ = β7
3
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Question 5a (i)
Determine the Lower Class Boundary of the βdata set
Lower class boundary of 21 - 30 is 20.5 litres.
Question 5a (ii)
Calculate the class width of the data set
Class width = Upper Class Boundary β Lower Class Boundary
= 30.5β 20.5
= 10 vehicles
Question 5b
Calculate the range of the data set
Range = Highest Value β Lowest Value
= 101 β 59
= 42 vehicles
Question 5c
P (need > 50.5 litres to fill tank) = ππ’ππππ ππ π·ππ ππππ ππ’π‘πππππ
πππ‘ππ ππ’ππππ ππ ππ’π‘πππππ
= 150β129
150
=21
150
=7
50
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Volume of petrol (litres)
Fre
qu
ency
Question 5d
Bryonβs estimate is wrong because the median would be given at the 75th vehicle (half the
cumulative frequency) which would be found in the interval 31 - 40.
Question 5e
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Question 6a (i)
0.5 Γ 25,000 = 12,500 cm
= 0.125 km
Question 6a (ii)
An area of 1cm2 on the map is represented by a 1cm Γ 1cm square on the map. The actual area
represented in real like by this would be 25,000 Γ 25,000 = 625,000,000 ππ2.
Alternatively, since 25,000cm is one quarter of a km, then we can express this area as such:
0.25 Γ 0.25 = 0.0625km2
An area of 2.25 ππ2 is 2.25 times the area of 1 ππ2.
β΄ The area in real life will be 2.25 times the 0.0625 ππ2.
2.25 Γ 0.0625 = 0.140625 ππ2
Question 6b (i)
Volume = ππ2β
= Ο Γ (3π
2)2
Γ4
= Ο Γ 9π2
4 Γ 4
= 9d2 Ο cm3
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Question 6b (ii)
Calculate the height of Jar Y
Recall: V = Οr2h
β΄ β = π
ππ2
=9π2π
π(π
2)2
=9π2π
1Γ·
π2π
4
=9π2π
1Γ
4
π2π
=36cm
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Question 7 a (i)
Calculate the value of π1
ππ = 3π2 β 2
πΌπ π = 1,
πβππ π1 = 3(1) β 2
π1 = 3β 2
= 1
Question 7 a (ii)
Calculate the value of π3
ππ = 3π2 β 2
πΌπ π = 3,
πβππ π3 = 3(3)2 β 2
π3 = 27 β 2
= 25
Question 7 a (iii)
Calculate the value of n given that ππ = 145
145 = 3π2 β 2
3π2 = 147
π2 = 49
π = 7
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Question 7 b (i)
1, 1,2,3,5,8,13,21
U(1) = 1
U(2)= 1
U(3)=2
U(4)=3
U(5)=5
U(6)=8
U(7)=13
U(8)=21
Each term is the sum of the two terms that came before it (from the 3rd term onwards).
This is Fibonacciβs sequence.
U(9) = U (8) + U (7) = 21 + 13 = 34
U(10) = U(9) + U (8) = 34 + 21 = 55
Question 7b (ii)
The term in the sequence which is the sum of U(18) and U(19) is U(20).
Any term is the sum of the two terms that came immediately before it.
Question 7b (iii)
RTS U(20) β U(19) β U(17)
Since U(20) = U(19) + U(18), then
U(20) β U(19) = U(18)
Since U(19) = U(18) + U(17), then
U(19) β U(17) = U(18)
Hence, U(20) β U(19) = U(19) β U(17)
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Question 8a (i)
Given,
π(π₯) = 9
2π₯ + 1 πππ π(π₯) = π₯ β 3
Calculate the value of x which cannot be in the domain of π(π₯) is the value which makes the
denominator equal to 0.
If 2π₯ + 1 = 0, π‘βππ
2π₯ = β1
π₯ = β1
2
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Question 8a (ii)
a. πΉπππ ππ(π₯)
ππ(π₯) = π(π₯ β 3)
= 9
2(π₯β3)+1
= 9
2π₯β6+1
= 9
2π₯β5
b. πΉπππ πβ1(π₯)
πΏππ‘ π¦ = π(π₯)
π¦ = 9
2π₯+1
ππ€ππ‘πβ π₯ πππ π¦
π₯ = 9
2π¦+1
πππππ πππ π πππ π¦
π₯(2π¦ + 1) = 9
2π₯π¦ + π₯ = 9
2π₯π¦ = 9β π₯
π¦ = 9βπ₯
2π₯
β΄ πβ1(π₯) = 9βπ₯
2π₯
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Question 8b (i)
π ππ π‘βππ‘ ππππ ππ π΄π΅πΆπ· ππ π₯2 + 2π₯ = 4 = 0
π΄πππ ππ ππππ‘πππππ= π Γ π
Substituting π = 4+ 3π₯ πππ π = 2+ 3π₯ gives
π΄ = (4+ 3π₯)(2+ 3π₯)
π΄ = 4(2+ 3π₯) + 3π₯(2+ 3π₯)
π΄ = 8+ 12π₯ + 6π₯ + 9π₯2
π΄ = 9π₯2 + 18π₯ + 8
πΊππ£ππ π‘βππ‘ ππππ = 44,
β΄ πππ + πππ+ π = ππ
9π₯2 + 18π₯ + 8β 44 = 0
9π₯2 + 18π₯ β 36 = 0
π·ππ£πππ π‘βπππ’πβππ’π‘ ππ¦ 9
π₯2 + 2π₯ β 4 = 0 QED
Question 8b (ii)
Using the quadratic formula to calculate π₯
π₯ =β2Β±β((2)2β4(1)(β4))
2(1)
π₯ =β2Β±β(4+16)
2
π₯ = β2Β±β20
2
π = π.πππ ππ ππ π π .π.
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Question 8b (iii)
Perimeter of unshaded region is as follows:
π = πΊπ΄+ π΄π΅ + π΅πΆ + πΆπΈ + πΈπΉ + πΉπΊ
π = 3π₯ + 2 + 3π₯ + 4+ 3π₯ + 3π₯ + 4 + 2
π = 12π₯ + 12
π = 12(1.236) + 12
π = ππ.πππππ ππ¨ π π .π.
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Question 9a
i. π ππΉ β DBA
DBA = ABC β DBC
= πππ β πππ
= πππ
ABC is 90π because the angle in a semicircle is a right angle.
ii. π ππΉ β DAC
π«π¨πͺ = π«π©πͺ
= πππ
Angles in the same segment standing on the same arc are equal.
iii. RTF β BCE
β BCE = β DAB
= 46π + 28π
β πππ = πππ
The external angle of a cyclic quadrilateral is equal to the opposite internal angle.
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Question 9b
i. π ππΉ π‘βπ πππππ‘β ππ ππ
N. B. β PQS = β RSQ because of alternating or Z-angles.
sin πππ = ππ
ππ
β΄ sin πππ Γ ππ = ππ
π·πΊ = π¬π’π§πππ Γ π
= πππ
ii. π ππΉ π‘βπ πππππ‘β ππ ππ
ππ πππ ππ¦π‘βπππππs'Theorem,
ππ2 = ππ2 + ππ2
β΄ ππ2 β ππ2 = ππ2
ππ2 = 82 β 42
ππ = β64β 16
π·πΈ = βππ
= π.ππππ ππ ππ .π.
iii. π ππΉ π‘βπ ππππ ππ πππ π
PQRS is a trapezium
Area of trapezium =1
2 (ππ + π πβ) Γ ππ
=1
2 (6.93 + 8.54) Γ 4
= ππ.ππππ (πππππππ ππ π π .π. )
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*Using the sine rule to find RS:
π π
π πππ ππ=
ππ
π ππππ πββ
π π = 8
sin68π Γ sin 82π
π π = 8.54ππ π‘π 2 π.π
** To find β QRS:
Angle QRS = 180Β° β (82Β° + 30Β°)
= 180Β° β 112Β°
= 68Β°
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Question 10a (i)
a. Find the following product
(β1 34 β
) (π5) = (
(β1Γ π) + (3Γ 5)(4Γ π) + (βΓ 5)
)
= (βπ + 154π + 5β
)
b. Find the values of β and π
πΌπ (βπ + 154π + 5β
) = (00), π‘βππ
βπ + 15 = 0, πππ
π = ππ
β΄ 4(15) + 5β = 0
60 + 5β = 0
5β = β60
β = β60
5
π = βππ
Question 10a (ii)
(2 3β5 1
)(π₯π¦) = (
513
)
To solve for x and y multiply the inverse of
(2 3
β5 1) ππ¦ (
513
)
Therefore,
πΌππ£πππ π = 1
(2Γ1)β(3Γβ5) Γ (
1 β35 2
)
= 1
17(1 β35 2
)
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(π₯π¦) =
1
17(1 β35 2
)Γ (513
)
= 1
17((1Γ 5) + (β3Γ 13)(5Γ 5) + (2 Γ 13)
)
= 1
17(5 β 3925 + 26
)
=1
17(β3451
)
= (
π
ππΓ βππ
π
ππΓ ππ
)
= (βππ
)
Question 10b
ππ΄ββββ β = (90), ππ΅ββ ββ β = (
36), π΄π·ββ ββ β =
1
3π΄π΅, ππΈββββ β =
1
3ππ΄
i. RTF π΄π΅ββββ β π΄π΅ββββ β = π΄πββββ β + ππ΅ββ ββ β
π¨π©ββββββ = (βππ
)+ (ππ)
= (βππ
)
ii. RTF ππ·ββββββ ππ·ββββββ = ππ΄ββββ β + π΄π·ββ ββ β
ππ·ββββββ = (90)+
1
3π΄π΅ββββ β
ππ·ββββββ = (90)+
1
3(β66
)
πΆπ«ββββ ββ = (ππ)+ (
βππ
)
= (ππ)
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iii. RTF π΅πΈββββ β π΅πΈββββ β = π΅πββ ββ β + ππΈββββ β
π΅πΈββββ β = (β3β6
) +1
3(ππ΄ββββ β)
π΅πΈββββ β = (β3β6
) + 1
3(90)
π©π¬ββββββ = (βπβπ
)+ (ππ)
= (πβπ
)