1. No category

Leaving Certificate Higher Level Maths Solutions
Sample Paper 1, no. 4
1. (a)
2
T2 = __
3
2
ar = __
3
2
a = __
3r
2
∴ __ = 3 − 3r
3r
2 = 9r − 9r2
S∞ = 3
a
_____
=3
1−r
a = 3 − 3r
9r2 − 9r + 2 = 0
(b)
(3r − 1) (3r − 2) = 0
1
2
r = __
or
r = __
3
3
1
r = __
3
2
=2
⇒ a = ____
1
3 _3
( )
2
r = __
3
2
⇒ a = ____
=1
2
3 _3
( )
a(1 − r5)
S5 = ________
1–r
( ( ))
1 5
1 1 − _3
_________
=
1
1−_
3
= 2.61
a(1
− r5)
_______
S5 =
1−r
( ( ))
2 5
2 1 − _3
_________
=
2
1−_
3
= 2.99
2. (a)
p(x) = 2x3 − x2 − 16x + 15
p(1) = 2(1)3 − 12 − 16(1) + 15
= 2 − 1 − 16 + 15
=0
∴ x − 1 is a factor.
1
2x2 + x − 15
x − 1 ) 2x3 − x2 − 16x + 15
–
2x3 +– 2x2
x2 − 16x
x 2 +–
x
15x + 15
+
–
15x +– 15
(x − 1) (2x 2 + x − 15)
(x − 1) (2x − 5) (x + 3)
(b)
p′(x) = 6x 2 − 2x − 16
6x 2 − 2x − 16 = 0 at turning points
3x2 − x − 8 = 0
______________
−(−1) ± √(−1)2 − 4(3)(−8)
x = ______________________
2(3)
___
1
±
97
√
x = _______
6
x = 1.81
or
x = −1.47
y = 2(1.81)3 − (1.81)2 − 16(1.81) + 15
x = 1.81
y = −5.38
(1.81, −5.38)
x = −1.47
y = 2(−1.47)3 − (−1.47)2 − 16(−1.47) + 15
y = 30.01
(−1.47, 30.01)
(c)
y
(–1.47,30.01)
30
25
20
15
10
5
0
(–3,0)
–3 –2.5
3
2
–2 –1.5 –1
2x – x – 16x + 15
–0.5
–5
(1,0)
0
0.5
1
1.5
(2.5,0)
2
(1.81,–5.38)
2
2.5
x
3
3.5
3. z = x + iy
(a)
(b)
_
z = x − iy
3
1 + __
__
=2+i
z
z
()
( )
__ 1
_ 3
_
zz __z + zz ___z = zz (2 + i)
_
_
_
z + 3z = zz (2 + i)
x − iy + 3(x + iy) = (x + iy) (x − iy) (2 + i)
x − iy + 3x + 3iy = (x2 + y2) (2 + i)
4x + 2iy = 2x2 + 2y2 + x2i + y2i
∴ 2x2 + 2y2 = 4x
x2 + y2 = 2y
x2 = −y2 + 2y
_______
_______
x = √−y2 + 2y
_______
2( √−y2 + 2y ) + 2y2 = 4√−y2 + 2y
−2y2
2
+ 4y +
2y2
_______
= 4√−y2 + 2y
_______
4y = 4√−y2 + 2y
_______
y = √−y2 + 2y
y2 = −y2 + 2y
2y2 − 2y = 0
2y (y − 1) = 0
y=0
or y = 1
If y = 0 ⇒ x = 0
________
__
If y = 1 ⇒ x = √−(1)2 + 2 = √1 = 1
∴ Answer: 1 + i
(c)
Im
5i
4i
3i
2i
z
i
Re
–3
–2
–1
1
–i
2
–
z
–2i
3
4. (a)
A−B=t
t2
1
A − B = 2 + ____ − 1 + ____
t+1
t+1
2
t
1
= 2 + ____ − 1 − ____
t+1
t+1
2
t
1
= 1 + ____ − _____
t+1 t+1
t +1 + t2 − 1
= __________
t+1
2
t + t t(t + 1)
= _____ = ______ = t
t + 1 (t + 1)
(
)
∴ Rate of flow of A is greater than rate of flow of B by t litres/s.
(b)
In 4 minutes A will be greater than B by 4t
= 4(4) = 16 litres.
5. (a)
−60,000
m = _______ = −4,000
15
(b)
The slope represents the decrease in the salvage value of the tractor each year.
(c)
S = −4,000n + 60,000
(d)
−4,000n + 60,000 ≥ 0
−4,000n ≥ −60,000
n ≤ 15
∴ Should not use any value of n > 15, as the salvage value will fall below 0.
(e)
S = 60,000 (1 − 0.20)14
S = €2,638.83
(f)
S = 60,000 (1 − 0.20)n
n
( )
8
S = 60,000 ___
10
as n → ∞
n
( 10 ) → 0
8
___
So the salvage value will approach 0 but won’t fall below zero. The rate of depreciation
will decrease as n → ∞ (n > 15 onwards).
4
6. (a)
32
x = ____
_____
100
2+x
100x = 64 + 32x
64
x = ___
68
x ≈ 0.941
(b)
Pattern 30, 31, 32 …
(i)
In 6 years ⇒ 35 = 243 hundred
= 24,300
7. (a)
(ii)
3n − 1
(iii)
No, the number of trees will not increase indefinitely. The number of trees will
increase for some time but will then plateau. Reasons for this could include
competition for water, light and/or nutrients.
x = 2t + 3t2
dx
___
= 2 + 6t
dt
dy _____
−4t
−1
___
= ___
=
dt 2 + 6t 2
y = 1 − 2t2
dy
___
= − 4t
dt
−8t = − 2 − 6t
−2t = −2
t=1
If t = 1, x = 2(1) + 3(1)2 = 5
y = 1 − 2 (1)2 = −1
(5, −1)
∴ y − y1 = m(x − x1)
1
y + 1 = − __(x − 5)
2
2y + 2 = − x + 5
x+y−3=0
(b)
y2 = 2xy − 4
[
]
dy
dy
2y ___ = 2 y + x ___
dx
dx
dy
dy
2y ___ = 2y + 2x ___
dx
dx
dy
dy
2y ___ − 2x ___ = 2y
dx
dx
dy _______
y
2y
u
___
__
= _____
=
y
−
x=v
dx 2y − 2x
u
x
v
y
du
dv
v___ + u___
dx
dx
dy
y(1)+ x(1)___
dx
dy
y + x___
dx
5
2
dy
dy
dv
___
= v___ − u___
dx2
dx
dx
(
)
dy
dy
__
(y − x)__
dx − y dx − 1
_________________
=
(y − x)2
dy
dy
__
(y − x)__
dx − y dx + y
_______________
=
(y − x)2
=
dy
__
dx (y − x− y) + y
_____________
(y − x)2
dy
__
dx (− x) + y
= __________
(y −x)2
y
____
y − x (− x) + y
____________
=
(
)
(y − x)2
−xy
____
y−x +y
_______
=
(y − x)2
2
−xy + y − xy
_________
(y − x)
= _________
(y − x)2
y2 − 2xy
= _______
(y − x)3
(c)
1+x
f(x) = _____
2−x
(i)
Vertical ⇒ 2 − x = 0
2=x
1+x
Horizontal ⇒ lim _____
x→∞ 2 − x
_1 + 1
x
_____
= lim 2
x→∞ _
x−1
= −1
y = −1
(ii)
y
x=2
5
4
3
1+x
f(x) = ——
2–x
2
1
0
0 1
–8 –7 –6 –5 –4 –3 –2 –1
–1
y = –1
–2
x
2 3
4
5
–3
–4
–5
–6
6
6
7
8
9 10 11 12 13 14 15 16
(iii)
du
dv
__
v__
dx − u dx
________
f ′(x) =
v2
(2
−
x) (1) − (1 + x) (− 1)
f ′(x) = _____________________
(2 − x)2
2−x+1+x
f ′(x) = ___________
(2 − x)2
3
f ′(x) = _______2
(2 − x)
f ′(x) decreasing ⇒ f ′(x) < 0
3
_______
<0
(2 − x)2
⇒3<0
False
∴ f (x) is never decreasing.
−6
f ″(x) = −6(2 − x)−3 = _______3 ≠ 0
(2 − x)
as −6 ≠ 0
∴ No points of inflexion
8. (a)
(i)
∫
p
_
[
−cos 3q
_______
0 sin 3q dq =
2
]
p
_
2
3
0
3p
__
−cos 2
−cos 3(0)
= _______ − ________
3
3
0
1
= __ + __
3 3
1
= __
3
[
4 ______
(ii)
∫0 √9 + 4x dx
Let u = 9 + 4x.
1 25 _1
= __∫9 u2 du
4
[]
3
u_2
1 __
__
=
4 _3
[
du = 4dx
1
__ du = dx
4
x = 0, u = 9 + 4(0)
25
2 9
___
__
1 2√25 3 2√9 3
= __ ______ − _____
4
3
3
[
]
=9
]
x = 4, u = 9 + (4)(4)
u = 25
] [ ]
1 250 54 1 196 49
= __ ____ − ___ = __ ____ = ___
4 3
3
4 3
3
7
1
(iii)
∫0
x(x − 1)3 dx
Let u = x – 1.
0
= ∫−1 (u + 1)u3 du
du = dx
u +1 = x
0
= ∫−1 u4 + u3 du
x = 0 ⇒ u = –1
u5 u4 0
= __ + __
5 4 −1
(−1)4 __
(−1)5 _____
05 __
04 _____
1 1
1
__
= + −
−
= − __ = − ___
5 4
5
5 4
4
20
[
]
x=1⇒u=0
p
_
(b)
(i)
∫0 5 sin2 8x dx
1 1
= ∫0 5( __ − __ cos 16x ) dx
2 2
8
p
_
8
p
_
2 5
5
= ∫0 __ − __ cos 16x dx
2 2
p
_
5
5
= __x − ___ sin 16x 8
2
32
0
5
5 p
5
5
= __ __ − ___ sin 2p − __ (0) + ___ sin (0)
2 8 32
2
32
5p
= ___
16
[
]
( )
1
(ii)
∫−2
5 − 4x − x2
dx
___________
_________
√5 − 4x − x2
5 − (x2 + 4x)
1
dx
__________
= ∫−2 ___________
√9 − (x + 2)2
let u = x + 2
5 − (x2 + 4x + 4 − 4)
5 + 4 − (x2 + 4x + 4)
9 − (x + 2)2
du = dx
x = −2 ⇒ u = 0
x=1⇒u=3
3
du
______
⇒ ∫0 _______
√32 − u2
u 3
= sin−1 __
30
= sin−1 1 − sin−1 0
p
= __
2
[
(c)
]
y
3
2
–4y2 + x = 0
(16,2)
1
0
–4
x
0
–2
2
4
6
8
10
12
14
16
–1
(4,–1)
–2
–3
x – 4y = 8
8
18
20
x = 4y2
x − 4y = 8
4y2 − 4y − 8 = 0
y2 − y − 2 = 0
(y + 1) (y − 2) = 0
y = −1
y=2
∴ y = −1
x = 4(−1)2
x=4
(4, −1)
y=2
x = 4(2)2
x = 4(4)
x = 16
(16, 2)
∴ A = Area under line − Area under curve
2
2
= ∫−1 4y + 8 dy − ∫−1 4y2 dy
[
2
4y
= ___ + 8y
] [ ]
2
3 2
4y
− ___
2
3 −1
−1
4(4)
4(2)3 ______
4(−1)2
4(−1)3
____
______
_____
=
+ 8(2) −
− 8 (−1) −
−
2
2
3
3
[
32 4
= 8 + 16 − 2 + 8 − ___ − __
3 3
36
= 30 − ___
3
= 30 − 12
= 18 units2
9
]
Leaving Certificate Higher Level Maths Solutions
Sample Paper 2, no. 4
1. (a)
y
5
c1
c2
4
3
Radius = 3 Radius = 2
(4,2)
(9,2)
2
1
0
x
0 1
–1
–1
2
3
4
5
6
7
8
9
10 11 12 13
–2
–3
Centre c1 = (4,2)
Radius c1 = 3
Distance between the centres = 5
Centre c2 = (9,2)
Radius c2 = 2
Radius c1 + Radius c2 = 5
Therefore, the circles touch externally.
(b)
y
7
6
c1
5
k
c2
4
3
Radius = 5
2
(4,2)
(6,2)
(9,2)
1
0
–1
0 1
–1
x
2
3
4
5
6
7
8
9
–2
–3
Centre of k = (6,2)
Radius of k = 5
Equation of k: (x − 6)2 + (y − 2)2 = 25
This simplifies to x2 + y2 − 12x − 4y + 15 = 0
10
10 11 12 13
2. (a)
y
7
6
(h,k)
5
45
4
3
2
1
0
0 1
–8 –7 –6 –5 –4 –3 –2 –1
–1
x
2
3
4
–2
___
Perpendicular distance from (h,k) to 2x − y + 5 = 0 is √45
___
2(h)______
− (k) + 5
___________
= −√45
√22 + 12
___
_____
2h − k + 5 = ( −√45 )( √4 + 1 )
__
__
2h − k + 5 = ( −3√5 )( √5 )
2h − k + 5 = −15
2h − k = −20
Equation 1
(−1,3) is a point on the line 2x − y + 5 = 0
Slope of the line 2x − y + 5 = 0 is 2
1
Slope of ⊥ line is − __
2
y2 − y1 _____
k−3
1
__
Slope of ⊥ line is ______
x2 − x1 = h + 1 = − 2
⇒ 2k − 6 = −h − 1
2k + h = 5
Equation 2
Solving between Equation 1 and Equation 2 gives:
2k + h = 5
−k
+ 2h = −20
____________
2k + h = 5
−2k
+ 4h = −40
_____________
5h = −35
h = −7
⇒k=6
11
(b)
y
7
6
(h,k)
5
45
4
3
2
1
0 C
0 1
–8 –7 –6 –5 –4 –3 –2 –1
–1
F
–2
b
x
2
3
4
m1 − m2
b = tan−1________
1 + m1m2
2 − _13
b = tan−1________
1 + (2)(_13 )
b=
_5
−1
_
tan 38
_
3
5
b = tan−1__
8
b = 32°
3. (a)
1__
Margin of error = ___
√n
1__
___
∴
= ±3%
√n
3
1__
∴ ___
= ± ____
100
√n
9
1 ______
∴ __
n = 10,000
10,000
n = ______
9
n ≈ 1,100
(b)
1__ _______
1
___
= _____ = ± 3%
(c)
The null hypothesis: There is no increase in support for the president.
√n
√1,100
Accept the null hypothesis.
Reason: The result of the second poll is within the margin of error.
4. (a)
(b)
a = 3,
b = 0.5
Range: [3, −3]
Period: 4p
12
(c)
3
1
3 cos – x
2
y
2
1
0
x
0
p
p
—
2
–1
3p
—
2
5p
—
2
2p
3p
4p
7p
—
2
–2
–3
(d)
a sin bx = 0 means where the graph intersects the x-axis.
The graph intersects the x-axis at x = 0, x = 2p and x = 4p .
5. (a)
k=2
(b)
A
Q
R
P
D
B
C
Centre of Enlargement
(c)
F
BD2 = BA2 + AD2
RQ2 = RF2 + FQ2
BD2 = 12 + 12
RQ2 = 22 + 22
BD = √2
RQ = √8 = 2√2
__
(d)
__
Area of circle on ABCD = p r2
__
Area of circle on FRPQ = p r2
__
___
2
= p2
2
= 2p units2
p
= __ units2
2
(e)
__
= p ( √2 )2
( )
√2
=p( )
2
√2
= p ___
2
__
p
(Area of the circle on FRPQ) ÷ (Area of circle on ABCD) = 2p ÷ __ = 4
2
2
k (Area of circle on ABCD) = Area of the circle on FRPQ
13
6A.
Diagram:
A
A'
B"
C"
B
B'
C'
C
Given: Δ ABC and Δ A′B′C′ which are equiangular,
where |∠A| = |∠A′|, |∠B| = |∠B′| and |∠C| = |∠C′|
|BC|
|CA|
|AB|
To Prove: _____ = _____ = _____
|A′B′| |B′C′| |C′A′|
Construction: Pick B′′ on [AB] with |AB′′| = |A′B′|, and C′′ on [AC] with |AC′′| = |A′C′|
Proof: ΔAB′′C′′ is congruent to ΔA′B′C′................(SAS)
∴ |∠AB′′C′′| = |∠ABC|
∴ B′′C′′ is parallel to BC ..........................corresponding angles
|A′B′| |AB′′|
∴ _____ = _____.........................................choice of B′′, C′′
|A′C′| |AC′′|
|AB|
= ____ ......................................................Theorem 12
|AC|
|AB|
|AC| _____
_____
=
.........................................rearrange
|A′C′| |A′B′|
|BC|
|AB|
Similarly, _____ = _____
|B′C′| |A′B′|
Q.E.D.
6B.
(a)
S
c
2
U
E
3
1
R
T
In ΔTES and ΔTUR:
|∠1| = |∠2| .............................. isosceles triangle TSE
|∠1| = |∠3| .............................. isosceles triangle TUR
∴ |∠2| = |∠3|
In ΔTES and ΔTUR
|∠1| = |∠1|...................................common angle
|∠2| = |∠3|...................................from above
14
∴ ΔTES and ΔTUR are equiangular
|TR| |TS| 1
∴ ____ = ____ = __ as R is mid point of [TE]
|TE| |TU| 2
∴ U is mid point of [ST]
(b)
(i)
A
V
Z
W
B
In ΔAVB and ΔAZW:
|∠AVB| = |∠AZW| ..............................both 90°
|∠ABV| = |∠AWZ| ............................angles standing on the same arc
∴ ΔAVB and ΔAZW are equiangular
(ii)
As ΔAVB and ΔAZW are equiangular:
|AV| |AB|
⇒ ____ = ____
|AZ| |AW|
|AV| |AB|
∴ ____ = ____
|AZ| |AW|
6
4
⇒ ____ = __
|AZ| 5
∴ |AZ| = 3.333
7. (a)
(i)
3
–
5
1
–
6
5
–
6
Good Shot
Correct
Club
2
–
5
2
–
7
Bad Shot
Good Shot
Wrong
Club
5
–
7
Bad Shot
15
From the tree diagram, the probability of a good shot is:
5 2
or __ × __
6 5
6 7
5 2
3
1 __
__
+ __ × __
×
6 5
6 7
3 ___
10
___
+
30 42
71
____
210
3
1 __
__
×
(
(ii)
) (
)
The probability that she has chosen the correct club, given that she plays a good
shot, is:
P(Correct club) × P(Good shot)
P(Correct club/Good shot) = __________________________
P(Good shot)
3
1 __
__
×
5
6
= _____
71
____
210
71
3
= ___ ÷ ____
30 210
21
= ___
71
(iii)
The probability that she has chosen the wrong club, given that she plays a bad shot, is:
P(Wrong club) × P(Bad shot)
P(Wrong club/Bad shot) = ________________________
P(Bad shot)
5
5 __
__
×
7
6
= _______
71
1 − ____
210
25 139
= ___ ÷ ____
42 210
125
= ____
139
(b)
(i)
Mean is 73% of 130 = 94.9
Standard deviation is 6% of 130 = 7.8
x−µ
z = _____
σ
80 − 94.9
z = ________
7.8
z = −1.910
P(<80) = 1 − 0.9713
–3.5 –3 –2.5 –2 –1.5 –1–0.5 0 0.5 1 1.5 2 2.5 3 3.5
P(<80) = 0.0287
16
(ii)
Mean is 73% of 140 = 102.2
Standard deviation is 6% of 140 = 8.4
P(>a) = 1 − 0.02 = 0.98
∴ z = 2.06
x−µ
z = _____
σ
x − 102.2
2.06 = ________
8.4
x > 199.5
Answer: 120 bottles
8. (a)
B
E
76°
F
104°
D
38°
104°
38°
6m
A 1.5 m
h
38°
C
(b)
h
38°
2.5 m
C
h
tan 38° = ___
2.5
h = 1.953 m
= 1 m 95 cm
(c)
|∠DCE| = 14°
∴ |∠ACE| = 42°
In the triangle ACE:
|AE|
sin |∠ACE| = ____
|AC|
|AE|
sin 42° = ____
10
⇒ |AE| = 6.69 m
(d)
17
(e)
72°
xm
72°
54° 54°
5m
5m
5m
5m
5m
5
x
______
= ______
sin 72°
sin 54°
x = 4.253 m
Timber required:
16(5) + 10(4.253)
= 122.53 m
= 123 m
18
Leaving Certificate Higher Level Maths Solutions
Sample Paper 1, no. 5
1. (a)
2x2 − 2 ≤ 3x
y
2x2 − 3x − 2 ≤ 0
Solve 2x2 − 3x − 2 = 0
x
(2x + 1)(x − 2) = 0
1
x=2
x = − __
2
2x2 − 3x − 2 ≤ 0 when
(b)
(i)
–1 – –1
2
1
2
1
− __ ≤ x ≤ 2
2
2 + y = − 2x
y = −2x − 2
−2x2 + 8xy + 42 = y
−2x2 + 8x(−2x − 2) + 42 = − 2x − 2
−2x2 − 16x2 − 16x + 42 + 2x + 2 = 0
−18x2 −14x + 44 = 0
9x2 + 7x − 22 = 0
(9x − 11)(x + 2) = 0
11
x = ___ or x = − 2
9
11
y = −2 ___ − 2
9
22
= − ___ − 2
9
40
___
=−
9
( )
(ii)
(i)
=2
The line 2 + y = −2x and the curve −2x2 + 8xy + 42 = y intersect at two points:
40
11 ,− ___
___
and (−2,2).
9 9
_________
___ 14 ___ = _____________
__ __ 14 __ __
√63 − √28 √9 √7 − √4 √7
(
(c)
y = − 2(− 2) − 2
)
__
7
√__
__ 14 __ × ___
= _________
3√7 − 2√7
__
√7
14√7
= _______
21 − 14
__
14√7
= _____
7
__
= 2√7
∴ a = 2 and b = 7
(ii)
__
___
2√7 = √28
∴ a = 1 and b = 28 is another way of expressing the solution.
19
2. (a)
1st year: 150
2nd year: 150 + 18
8
3rd year: 168 + __(18) = 184
9
8 n–2
after n years.
Growth = 18 __
9
8 17 – 2
= 3.08 cm
17th year growth is 18 __
9
()
()
OR
Tree Height
(cm)
Growth (cm )
Tree Height
(cm)
Growth (cm )
(b)
yr 1 yr 2
150
168
18
yr 3 yr 4
184 198.2
yr 5
210.84
yr 6
222.07
yr 7
232.06
yr 8
240.94
16
12.64
11.23
9.99
8.88
14.2
yr 9 yr 10 yr 11 yr 12 yr 13 yr 14 yr 15 yr 16 yr 16
248.83 255.84 262.08 267.62 272.55 276.93 280.82 284.28 287.36
7.89
7.01
6.24
5.54
4.93
4.38
3.89
3.46
3.08
Height after 10 years
( ( )9 )
8
18 1 – _9
150 + __________
8
1–_
9
= 150 + 105.8768146
= 255.88 cm
OR
Read from table given in part (a) – no penalty for early rounding if answer from table is given.
(c)
Max height
= 150 + S∞
18
= 150 + _____8
1 – _9
= 150 + 162 = 312 cm
∴ The tree will never exceed 312 cm in height.
3. (a)
2
2
2
2
3
48
24
12
6
3
1
48 = 24 × 3
2
3
3
19
342
171
57
19
1
342 = 2 × 32 × 19
2 2400
2 1200
2 600
2 300
2 150
2 75
5 75
5 15
3 3
1 1
2,400 = 25 × 52 × 3
∴ H.C.F = 2 × 3 = 6
20
(b)
7n − 1 = 48
7n − 1 = 342
7n = 49
7n = 343
n=2
n = log7 343
49 = 72 − 1
n=3
73 − 1 = 343
2,400 = 7n − 1
2,401 = 7n
log7 2,401 = n
4=n
∴ 2,400 = 74 − 1
(c)
P(n): 7n − 1 is divisible by 6, for all n ∈N.
P(1): 71 − 1
= 6 which is divisible by 6
∴ P(1) is true.
Assume P(k) is true.
∴ 7k − 1 is divisible by 6.
P(k + 1): 7k + 1 − 1
= 7k · 7 − 1
= 7k · 7 − 7 + 7 − 1
= 7(7k − 1) + 6
Now 7k − 1 is divisible by 6.
∴ 7(7k − 1) is divisible by 6.
and 6 is divisible by 6.
∴ 7k + 1 − 1 is divisible by 6
∴ P(k + 1) is true if P(k) is true.
Hence, by the principle of mathematical induction P(n) is true for all n ∈ N.
4. (i)
2 + 3i
_____
5+i
2 + 3i 5 − i
= _____ ____
5+i 5−i
10
− 2i + 15i − 3i2
= _______________
25 − i2
10 + 13i + 3
= __________
25 + 1
13
+ 13i
= _______
26
1
1
= __ + __i
2 2
1
= __(1 + i)
2
(
)(
)
21
4
(ii)
2 + 3i
1
= __(1 + i) )
( _____
5 + i ) (2
4
14
= __ (1 + i)4
2
1
= ___((1 + i)2)2
16
1
= ___(1+ 2i + i2)2
16
1
___
= (2i)2
16
1
= ___(4i2)
16
1
= ___(−4)
16
1
= −__ ∈ R
4
(iii)
(a)
z = 1 + 2i is a root.
(1 + 2i)2 − (p − i)(1 + 2i) + q − i = 0
1 + 4i + 4i2 − (p + 2pi − i − 2i2) + q − i = 0
1 + 4i + 4i2 − p − 2pi + i + 2i2 + q − i = 0
−5 − p + q + 5i − 2pi − i = 0
−p+q=5
or −2pi + 4i = 0
−2p + 4 = 0
−2p = −4
p=2
∴ −2 + q = 5
q=7
(b)
z2 − (2 − i)z + 7 − i = 0
_______
−b ± √b2 − 4ac
z = _____________
2a
_________________
(2 − i) ± √(−(2 − i))2 − 4(1)(7i)
z = __________________________
2(1)
_________________
2 − i ± √4 − 4i + i2 − 28 + 4i
= ________________________
____ 2
2 − i ± √−25
= ___________
2
2 − i ± 5i
= ________
2
2 − i + 5i
2 − i − 5i
= ________ or ________
2
2
2 − 6i
2 + 4i
= _____
or _____
2
2
= 1 + 2i
or 1 − 3i
∴ The other root is 1 − 3i.
22
5. (a)
(b)
( )
15 9
s = 16 − 16 ___ = 7.04
16
The lift is expected to stop about seven times.
n
( )
16
12 = 17 − 17 ___
17
n
16
5 = 17 ___
17
( )
5
16
___
= ___
17 ( 17 )
5
16
ln ___ = ln( ___ )
17
17
n
n
( )
5
16
ln ___ = n ln ___
17
17
5
ln __
17
____
=n
16
ln __
17
20.186 = n
There were approximately 20 people in the lift.
(c)
(i)
(ii)
( )
1
A = b2 + 4 __bs
2
= b2 + 2bs
s2 = (1.5)2 + (2.5)2
s2 = 8.5
s = 2.915475947
A = (3)2 + 2(3)(2.915475947)
A = 26.49 m2
(iii)
• IF a = b, then the shape is a cuboid – he has been given instructions not to use a
cuboid.
• IF a = 0, then the shape becomes a pyramid – this has proven impractical.
(iv)
1, 9, 25, 49
12, 32, 52, 72
i.e. odd numbers squared
Tn = (2n − 1)2
23
(v)
(2n − 1)2 = 1,100
_____
2n − 1 = √1,100
_____
2n = √1,100 + 1
_____
n=
√1,100 + 1
__________
2
= 17.08
= 17
width = (17)(0.1 m)
= 1.7 m
6. (a)
(b)
100,000 × 0.15 = €15,000.
1.0675 = 1(1 + i)12
12
______
√1.0675 = 1 + i
1.00545813 = 1 + i
0.00545813 = i
0.55% = i
(c)
i(1 + i)t
A = P _________
(1 + i)t − 1
[
0.005458(1.005458)120
= 85,000 ___________________
(1.005458)120 − 1
]
= 85,000[0.01138005072]
= €967.30
(d)
Year 0:
−20,000
Year 1: _______
=
(1.06)1
25,000
Year 2: ______2 =
(1.06)
35,000
Year 3: ______3 =
(1.06)
50,000
Year 4: ______4 =
(1.06)
NPV =
−80,000
−18,867.92
22,249.91
29,386.67
39,604.68
−€7,626.66
As NPV < 0, it would not be worthwhile to invest in Patricia’s business.
(e)
F = P(1 + i)t
= 15,000(1 + 0.035)5
= 15,000(1.035)5
= €17,815.29
24
(f)
50,000 − 17,815.29 = €32,184.71
32,184.71 = A + A(1.02)1 + A(1.02)2 + ..…
… + A(1.02)59
= A(1 + (1.02)1 + (1.02)2 + .… + (1.02)59)
[
]
1 − 1.0259
= A 1 + _________
1 − 1.02
= A[1 + 110.8348426]
= A[111.8348426
32,184.71
∴ A = ___________
111.8348426
A = €287.79
7. (a)
dy __________________________________
(1 + sin 3q)(2 cos 2q) − (sin 2q)(3 cos 3q)
___
=
(1 + sin 3q)2
dx
p
p
p
p
1 + sin __2 2 cos __3 − sin __3 3 cos __2
p
__
_______________________________
at q = ⇒
6
p 2
1 + sin __
)(
(
) (
(
__
=
( ( )) ( )
2
√3
1
(2) 2 _2 − __
2 (3(0))
__________________
)(
)
)
(1 + 1)2
2−0
= _____
22
1
= __
2
(b)
3
(i) _____ = 3(5 − x)−1
5−x
dy
3
___
= −3(5 − x)−2(−1) = _______2
dx
(5 − x)
_____
(ii)
√x + 2 ( x + 2 )
4x
4x = _____
_____
_____
(1
1
_
2
1
_
(4x)
= _______
2
1
_
(x + 2)2
−1
__
)
___
(1
−1
__
)
√x + 2 _2(4x) 2 · 4 − √4x _2 (x + 2) 2 (1)
dy
_________________________________
___ =
1
_
dx
((x + 2)2 )2
___
_____
4x
√_____
2
___
− ______
√x + 2 · ___
4x
2
x
+2
√
√
= _________________
_____
(x + 2)
___
2√x___
+ 2 ______
√4x
______
− _____
2√x + 2
√4x
_____________
=
(x + 2)
_____
__
2√x +
2 ______
2_____
√x
______
__
2√x − 2√x + 2
_____________
=
x+2
25
_____
__
√x
2
√x +
_____
_____
__ − _____
x
√
x
+2
√
= ___________
(x + 2)
(x + 2) − x
_______
__ _____
√x √x + 2
= _______
x+2
2
1
_____
= _________
· _____
__
+2 x+2
2
= ___________
__ _____ 3
√x ( √x + 2 )
√x √x
(iii)
______
2x ( √3x + 1 )
3
1
_
= (2x)(3x + 1)3
−2
3 ______
dy
__
1
___
= 2x __(3x + 1) 3 (3) + √3x + 1 · 2
3
dx
3 ______
2x
__________
= 3 ______ 2 + 2√ 3x + 1
( √ 3x + 1 )
[
]
3
=
______ 3
2x + 2(√ 3x + 1 )
______________
3
_______
√3x + 12
2x + 2(3x + 1)
= ____________
______
( 3√3x + 1 )2
8x + 2
= _________
3 ______ 2
( √3x + 1 )
(c)
(i)
y3 + y2 − x4 = 1
dy
dy
3y 2 ___ + 2y___ − 4x3 = 0
dx
dx
dy 2
___
(3y + 2y) = 4x3
dx
dy _______
4x3
___
= 2
dx 3y + 2y
4(1)3
4
At (1,1), m = ___________
= __
2
3(1) + 2(1) 5
y − y1 = m(x − x1)
4
y − 1 = __(x − 1)
5
5y − 5 = 4x − 4
4x − 5y + 1 = 0
(ii)
f (x) = x3 + x2 − 1
f ′(x) = 3x2 + 2x
f (1) = 13 + 12 −1
f ′(1) = 3(1)2 + 2(1)
=1
=5
26
f (x1)
x2 = x1 − _____
f ′(x1)
1 4
x2 = 1 − __ = __
5 5
3
4
4
42
f __ = __ + __ − 1
5
5
5
19
= ____
125
() () ()
42
4
4 = 3 __
+ 2 __
f ′ __
5
5
5
88
= ___
25
() () ()
f (x2)
x3 = x2 − _____
f ′(x2)
19
___
125
4 ____
__
= − 88
__
5
25
8. (a)
333
= ____ = 0.757
440
(
0.6 ___
0.4
∫ ___
− 6 ) dx
4
x
x
= ∫0.6x−4 − 0.4x−6 dx
0.6x−3 0.4x−5
= ∫ _____ − _____ + c
–5
−3
0.08
−0.2 ____
= ____
+ 4 +c
x
x3
p
_
∫ 0 cos2 3q dq
6
(b)
(i)
p
_
=
∫ 0 __2(1 + cos 6q) dq
6
1
1 _p
= __ 6 1 + cos 6q dq
2 0
p
sin 6q _6
1
__
_____
= q+
2
6 0
sin 0
1 p sin p 1
= __ __ + ____ − __ 0 + ____
2 6
6
2
6
p
___
=
12
∫
[
[
]
] [
]
4
(ii)
∫2 5xex − 1 dx
15
5
__
= ∫3 eu du
2
2
let u = x2 – 1
du = 2x dx
1 du = x dx
__
2
x = 2, u = 22 − 1 = 3
5
= __[eu]315
2
5
= __[e15 – e3]
2
x = 4, u = 42 − 1 = 15
= 8,172, 493.22
27
(c)
x2 + y2 = 36
y
cuts x-axis at 6,−6
∴−6≤x≤6
y2 = 36 − x2
6
∫ –6 y2 dx
6
= p ∫–6 36 − x2 dx
x
Volume = p
[
x3
= p 36x − __
3
]
–6
6
–6
[
3
(−6)
63
= p 36(+6) − __ − 36(−6) + _____
3
3
= 288p units3
28
]
6
Leaving Certificate Higher Level Maths Solutions
Sample Paper 2, no. 5
1. (a)
(i)
–3 –2.5 –2 –1.5 –1 –0.5 0 0.5 1 1.5 2 2.5 3
–3 –2.5 –2 –1.5 –1 –0.5 0 0.5 1 1.5 2 2.5 3
P (≤ 1) = 0.8413
P (≤ −1) = 1− 0.8413 = 0.1587
–3 –2.5 –2 –1.5 –1 –0.5 0 0.5 1 1.5 2 2.5 3
P (–1 ≤ x ≤ 1) = 0.8413 – 0.1587 = 0.6826
Answer 0.6826
(ii)
x–μ
z = _____
σ
170 – 150
z = _________ = 1
20
190 – 150
z = _________ = 2
20
P (1 ≤ x ≤ 2)
–3 –2.5 –2 –1.5 –1 –0.5 0 0.5 1 1.5 2 2.5 3
–3 –2.5 –2 –1.5 –1 –0.5 0 0.5 1 1.5 2 2.5 3
P (≥ 1) = 1− 0.8413 = 0.1587
P (≤ 2) = 1− 0.9772 = 0.0228
–3 –2.5 –2 –1.5 –1 –0.5 0 0.5 1 1.5 2 2.5 3
P (1 ≤ x ≤ 2) = 0.1587 – 0.0228 = 0.1359
Answer 0.1359
U
2. (a)
M
P
5%
5% 20%
70%
29
(b)
(i)
P(P∩M) 5 1
5
P(P|M) = ________ = ___ = __ or from Venn diagram P(P|M) = ___ = 0.5
10 2
10
P(M)
(ii)
P(P∩M)
5
5 1
P(M|P) = ________ = ___ = __ or from Venn diagram P(M|P) = ___ = 0.2
25 5
25
P(P)
(iii)
P(NP∩NM) 70 14
P(NM|NP) = __________ = ___ = ___ or from Venn diagram
75 15
P(NP)
70 14
P(Not P|Not M) = ___ = ___
75 15
3. (a)
2x2 + 2y2 + 6x – 14y – 3 = 0
3
x2 + y2 + 3x – 7y – __ = 0
2
Centre = (–g, –f )
(–1.5, 3.5)
_________
_________________
___
Radius = √g2 + f 2 – c = √(1.5)2 + (3.5)2 + 1.5 = √16 = 4
(b)
(i)
Centre of circle x2 + y2 – 10x = 0 = (–g,–f )
_________
____________
(5, 0)
___
Radius = √g2 + f 2 – c = √(5)2 + (0)2 – 0 = √25 = 5
If line is a tangent, the perpendicular distance from centre to the line should equal
the radius.
Perpendicular distance from (5, 0) to 3x – 4y + 10 = 0 is:
ax1 + by1 + c
3(5) –______
4(0) + 10
___________
______
= _____________
=5
√32 + 42
√a2 + b2
|
15 ___
+ 10
_______
=5
√25
25
___
=5⇒5=5
5
(ii)
∴The line is a tangent to the circle.
If (2,4) is on the line and the circle, then (2,4) must be the point of contact.
Is (2,4) on 3x – 4y + 10 = 0?
3(2) – 4(4) + 10 = 0
6 – 16 + 10 = 0
16 = 16
∴ (2,4) is on the circle.
Is (2,4) on x2 + y2 – 10x = 0?
(2)2 + (4)2 – 10(2) = 0
4 + 16 – 20 = 0
20 = 20
∴ (2,4) is on the circle.
∴ As (2,4) is on both the line and the circle, it is the point of contact.
30
_________________
4. (a)
|AB| = √(x2 – x1)2 + (y2 –y1)2
________________
= √(0 – 2t)2 + (–t – 0)2
______
___
= √4t2 + t2 = √5t2
___
___
= √5t2 = √20
5t2 = 20
⇒ t2 = 4
⇒ t = ±2
(b)
6
5 45°
4
(4,3)
3
2
1
0
–2
0
–1
(3,0)
45°
1
2
3
4
x
5
6
7
8
9
–1
–2
m1 – m2
Tan 45° = ± ________
1 + m1m2
–6 – m2
1 = ± _________
1 + (–6)m2
–6 – m2
1 = + _________
1 + (–6)m2
–6 –m2
1 = – _________
1 + (–6)m2
1 – 6m2 = –6 – m2
5m2 = 7
7
m2 = __
5
1 – 6m2 = 6 + m2
7m2 = –5
–5
m2 = ___
7
Equations of lines
−5
Point (4,3), Slope = ___
7
y − y1 = m(x − x1)
7
Point (4,3), Slope = __
5
y − y1 = m(x − x1)
5y − 15 = 7x − 28
5
y − 3 = − __ (x − 4)
7
7y − 21 = − 5x + 20
7x − 5y − 13 = 0
5x + 7y − 41 = 0
7
y − 3 = __
5 (x − 4)
31
5.
(a)
Standard book proof
(b)
92 = 32 + 82 − 2(3)(8) cos ∠DAB
81 = 9 + 64 − 48 cos ∠DAB
81 − 73 = −48 cos ∠DAB
8
− ___ = cos ∠DAB
48
1
cos−1 − __
6 = ∠DAB
∠DAB = 99.594°
1
Area of triangle = __ a.b sin C
2
1
= __(3)(8) sin 99.594
2
= 12 sin 99.594
( )
= 11.83 cm2
6A.
(a)
Centroid: The point at which the medians (lines from vertices to the mid point
of opposite sides) meet.
Incentre: The point at which bisectors of the angles meet.
Orthocentre: The point at which the altitudes (lines from vertices drawn perpendicular
to the opposite sides) meet.
Circumcentre: The point at which the perpendicular bisectors of the sides meet.
(b)
D
Circumcentre
Incentre
Centroid
A
B
Orthocentre
(c)
Explaination: The circumcentre is outside because the triangle has an angle greater
than 90o.
(d)
Explanation: Concurrent means they are collinear (on the same line).
32
6B.
(i)
B
|∠CIB| = 71°...... Given
|∠CIG| = 90° .... Angle in a semi-circle
|∠BIG| = 90° − 71° = 19°
C
But |∠BIG| = |BCG| ...... Same arc [BG]
G
|∠BCG| = 19°
(ii)
71°
In the triangle CBH
I
|∠IHG| = 33°
|∠CBH| = 90°
33°
∴ |∠BCH| = 57°
But |∠BCG| = 19°
H
∴ |∠GCI| = 38°
(iii)
B
C
19°
76°
G
C
19°
38°
G
I
As the angle at the centre of the circle standing on IG (Diagram 1) is twice the angle
standing on the arc BG (Diagram 2), the ratio of the length of the arc IG to the arc
BG is 2 : 1.
7.
(a)
(b)
The pentagon is made up of three triangles so the missing angle is
3(180o) – (96o + 140o + 76o + 131o) = 97o
1
2
Area of triangle ABC = __
2 (8.82)(9.61) sin 97° = 42.064 m
1
2
Area of triangle DAE = __
2 (8.83)(8.8) sin 76° = 37.7 m
33
C
96°
8.8 m
8.82 m
140°
B
D
97°
8.83 m
9.61 m
76°
131°
E
8.8 m
A
To find the area of triangle ACD, we need |DA| and |CA|.
|DA|2 = 8.832 + 8.82 − 2(8.83)(8.8) cos 76° = 117.812
_______
|DA| = √117.812
= 10.854 m
|CA|2 = 8.822 + 9.612 − 2(8.82)(9.61) cos 97° = 190.803
_______
|CA| = √190.803
= 13.81 m
C
96°
8.8 m
8.82 m
140°
D
13.81 m
B
97°
8.83 m
10.854 m
9.61 m
76°
131°
A
8.8 m
Now we must find the angle ADC.
34
E
|AC|2 = |AD|2 + |DC|2 − 2|AD|.|DC| cos |∠ADC|
|13.81|2 = |10.845|2 + |8.8|2 − 2|10.845|.|8.8| cos|∠ADC|
190.716 = 117.614 + 77.44 − 190.872 cos|∠ADC|
cos |∠ADC| = − 0.0229
|∠ADC| = 91.31°
1
1
Area of triangle ACD = __|CD|.|DA| sin|∠ADC| = __ (8.8).(10.854) sin 91.31° = 47.745 m2
2
2
Total area = 42.064 + 37.7 + 47.745 = 127.509 m2 = 128 m2
8. (a)
Proof: A regular hexagon is made up of 6
equilateral triangles.
A
Join each of the vertices of ABCDEF to the centre
point O.
B
F
In the triangle AFO
O
|AF| = |AO| = |OF| ….. Radii of equal circles
∴ |∠FAO| = |∠AOF| = |∠OFA| = 60°
E
C
Similarly, |∠OAB| = |∠ABO| = |∠BOA| = 60°
D
∴|∠FAB| = 120°
Similarly, all the vertex angles of ABCDEF = 120°
∴ ABCDEF is a regular hexagon.
(b)
Proof: Join each of the vertices of FBCE
to the centre point O.
|OF| = |OB| = |OC| = |OE| ….. Radii of
equal circles
A
In the triangles OFE and OBC
P
F
all sides are equal ……. Equal radii
B
O
∴ |∠EFO| = |OBC| = 60°
Join A to F and join O to A.
E
|∠AFP| = |OFP| = 30°, as ΔAFP and
ΔOFP are congruent.
C
D
Similarly, |∠OBP| = 30°
∴ | ∠EFB| = 90° and |CBF| = 90°
∴ FBCE is a rectangle.
35
9. (a)
Sample Data
Sample Data
Sample Size, 10
Sample Size, 20
Single Sample
100
Clear Samples
Sample
Samples
Information
Single Sample
100
s/√n:
Sample
Samples
Data Set Size, N: 100
Mean, m: 164.72
Std. Dev., s: 10.2079
s/√n:
Std. Dev., s: 10.2079
3.282
s/√n:
Sample Means: 100
1.864
Sample Means: 100
Means, m: 165.286
Means, m: 164.853
Means, m: 164.735
Std. Dev., s: 2.96105
Std. Dev., s: 2.17909
Std. Dev., s: 1.85509
Sample Data
Sample Data
Sample Size, 40
100
Sample Data
Sample Size, 50
Single Sample
Clear Samples
Sample
Samples
Information
Sample Size, 60
Single Sample
100
Clear Samples
Sample
Samples
Information
Data Set Size, N: 100
Mean, m: 164.72
Std. Dev., s: 10.2079
s/√n:
Std. Dev., s: 10.2079
1.614
s/√n:
Sample Means: 100
1.444
Sample Means: 100
Means, m: 164.717
Means, m: 164.63
Std. Dev., s: 1.61937
Single Sample
100
Clear Samples
Sample
Samples
Information
Data Set Size, N: 100
Mean, m: 164.72
Std. Dev., s: 1.44511
Data Set Size, N: 100
Mean, m: 164.72
Std. Dev., s: 10.2079
s/√n:
1.317
Sample Means: 100
Means, m: 164.791
Std. Dev., s: 1.3297
The mean of the sample means approaches the mean of the population as the sample
size increases. The means seem to become equal for sample sizes of 30 or greater.
(ii)
σ__
≈s
If the sample size is 30 or over ___
√n
(i)
The distribution seems not to be normal.
(ii)
This distribution seems to be normal.
µ = 164.75 s = 10.23 n = 30 x = 160
x − µ ___________
164 − 164.75
= − 0.4016
z = _____
σ__ =
___
10.23
_____
___
√n
√30
–4
100
Clear Samples
Information
3.228
Sample Means: 100
(d)
Sample
Samples
Single Sample
Mean, m: 164.72
Std. Dev., s: 10.2079
(c)
Clear Samples
Data Set Size, N: 100
Mean, m: 164.72
(i)
Sample Size, 30
Information
Data Set Size, N: 100
(b)
Sample Data
–2
0
2
4
P(z < − 0.4016) = 1 − P(0.4016) = 0.344
36