1. No category

Leaving Certificate Higher Level Maths Solutions
SEC 2013 Sample Paper 1
1. (a)
(i)
____________
__
__
_____
r = |− 1 + √3 i| = √(− 1)2 + (√3 )2 = √1 + 3 = 2
__
p
2p
= __ but q is in the second quadrant. q = ___
1 3
3
2p
2p
r(cos q + i sin q) = 2 cos ___ + 2np + isin ___2np
3
3
√3
−1 ___
q = tan
[ (
(ii)
[ (
[( (
)
)
(
)]
)]
(
(
2p
2p
z2 = 2 cos ___ + 2np + isin ___2np
3
3
1
_
2p
2p
z = 2 cos ___ + 2np + isin ___ + 2np 2
3
3
__
p
p
z = √2 cos __ + np + isin __ + np
3
3
__
__
__ 1
__ 1
3
3
√
√
for n = 0: z = √2 __ + i___ : for n = 1: z = √2 − __ + i___
2
2
2
2
[ (
)
(i)
z4
))]
)]
(
(
(b)
)
)
(
)
Im(z)
z2
z1
z3
(ii)
2. (a)
(i)
Re(z)
1
k = __
2
To Prove:
n(n + 1)
1 + 2 + 3 + 4 ... n = _______
2
1(1
+
1)
2
For n = 1
1 = _______ = __ = 1.
∴ true for n = 1
2
2
2(2 + 1) 6
For n = 2
1 + 2 = _______ = __ = 3.
∴ true for n = 2
2
2
Assume true for n = k
k(k + 1)
⇒ 1 + 2 + 3 ... k = _______
2
Now for n = k + 1
(k + 1)(k + 2)
Is 1 + 2 + 3 ... k + (k + 1) = ___________
2
k(k + 1)
We assumed that 1 + 2 + 3 ... k = _______
2
k(k
+
1)
(k
+
1)(k
+
2)
So is _______ + (k + 1) = ___________?
2
2
k(k + 1) + 2(k + 1) ___________
(k + 1)(k + 2)
_______________
=
2
2
2
(k
+
1)(k
+
2)
k + k + 2k + 2 ___________
____________
=
2
2
2
(k
+
1)(k
+
2)
k + 3k + 2 ___________
_________
=
2
2
1
(k + 1)(k + 2)
(k + 1)(k + 2) ___________
___________
=
which is true.
2
2
∴ It is true for n = k + 1
∴ It is true for n = k
∴ It is true all values of n, where n ∈ N
(b)
|4x − 1| < 1
− 1 < 4x − 1 < 1
0 < 4x < 2
1
0 < x < __
2
∞
∑ (4x − 1)r
r=2
Writing out the series, starting at r = 2 we get
(4x − 1)2 + (4x − 1)3 + (4x − 1)4 + (4x − 1)5 + ...
This is a GP with a = (4x − 1)2 and r(common ratio) = (4x − 1)
a
S∞ = _____
1−r
(4x − 1)2
S∞ = __________
1 − (4x − 1)
(4x − 1)2
________
S∞ =
−4x + 2
3. (a)
f : x |→ x3 + (1 − k2)x + k
If x = −k is a root then x3 + (1 − k2)x + k = 0
(−k)3 + (1 − k2)(−k) + k = 0
−k3 − k + k3 + k = 0
0=0
⇒ −k is a root of f
(b)
x = −k is a root of f ⇒ x + k is a factor of x3 + (1 − k2)x + k
x2 − kx + 1
x + k | x3 + (1 − k2)x + k
x3 + kx2
−kx2 + (1 − k2)x + k
−kx2 − k2x
x+k
x+k
⇒ x2 − kx + 1 = 0
2
_______
−b ± √b2 − 4ac
x = _____________
2a
__________
k ± √k2 − 4(1)(1)
______________
x=
2(1)
_____
_____
k+
−4
k − √k2 − 4
x = __________
or
x = __________
2
2
2
One real root ⇒ b − 4ac < 0
√k2
k2 − 4 < 0
(k + 2)(k − 2) < 0
⇒−2<k<2
4. (a)
2x + 8y − 3z = − 1
2x + 8y − 3z = − 1
2x − 3y + 2z = 2
_______________
2x + y + z = 5
_______________
11y − 5z = − 3
7y − 4z = − 6
11y − 5z = − 3
7y − 4z = − 6
____________
44y − 20z = − 12
35y
− 20z = − 30
______________
9y
= 18
y=2
7(2) − 4z = − 6
z=5
2x + 8(2) − 3(5) = − 1
2x + 1 = − 1
x=−1
(b)
(i)
y
y = f(x)
D
A
B
y = g(x)
x
C
y= −x+3
y=x−3
y= −x+3
y=x−3
y__________
=2
y________
=2
y__________
=0
x________
=0
2=−x+3
2=x−3
x=3
y=3
x=1
x=5
A(1,2)
B(5,2)
C(3,0)
D(0,3)
3
(ii)
Solution from diagram: 1 < x < 5
Algebraic solution: x − 3 < 2
⇒x<5
and
x − 3 < −2
x>1
∴1<x<5
5. (a)
Height = h cm
Length = 20 − 2h cm
30 − 2h
Width = _______ = (15 − h) cm
2
(b)
Volume = h(20 − 2h)(15 − h)
= (20h − 2h2)(15 − h)
= 3,000h − 20h2 − 30h2 + 2h3
Volume = 2h3 − 50h2 + 300h
(c)
Square base ⇒ 20 − 2h = 15 − h
h=5
Volume = 2h3 − 50h2 + 300h
When h = 5
Volume = 2(5)3 − 50(5)2 + 300(5)
Volume = 250 − 1,250 + 1,500
Volume = 500 cm3, which is the correct volume
(d)
2h3 − 50h2 + 300h = 500
2h3 − 50h2 + 300h − 500 = 0
h = 5 is a solution ∴ h − 5 is a factor
2h2 − 40h + 100
h − 5 | 2h3 − 50h2 + 300h − 500
3
2
2h − 10h
_________
− 40h2 + 300h − 500
2
− 40h + 200h
_________________
+ 100h − 500
+
100h − 500
____________
⇒ 2h2 − 40h + 100 = 0
_______
_______________
−b ± √b2 − 4ac 40 ± √1,600 − 4(2)(100)
h = _____________ = ____________________
2a
2(2)
____
40 ± √800
h = _________
4
h = 17.1 (impossible as 17.1 + 17.1 >22)
h = 2.9 cm
4
(e)
Capacity
800
600
550 cm3
400
200
h
5
10
15 17.1 cm
20
–200
If the capacity of the box is increased by 10%, it would give a volume of 550 cm3.
If a line is drawn on the graph to give a volume of 550 cm3 (see line on graph),
the corresponding height will be ≈ 17.1 cm.
This is impossible as stated solution to part (d).
6. Initial Exploration
m×n
2
3
4
5
6
2
3
4
5
6
Interior pieces only if m > 2 and n > 2
Interior
Interior pieces
Edge
Edge pieces
3×3
1×1
1
8
3×3−1
4×4
2×2
4
12
4×4−4
5×5
3×3
9
16
5×5−9
6×6
4×4
16
20
6 × 6 − 16
m×n
(m – 2) × (n – 2)
mn – 2m – 2n + 4
5
m × n – Interior pieces = 2m + 2n – 4
6. (a)
Interior
mn – 2m – 2n + 4
Edges
2m + 2n – 4
3×3
9–6–6+4=1
6+6–4=8
3×4
12 – 6 – 8 + 4 = 2
6 + 8 – 4 = 10
4×3
12 – 8 – 6 + 4 = 2
8 + 6 – 4 = 10
4×4
16 – 8 – 8 + 4 = 4
8 + 8 – 4 = 12
Interior pieces = mn – Edge pieces for m > 2, n > 2, m ≤ 4 and n ≤ 4
(b)
If Edge pieces = Interior pieces
mn − 2m − 2n + 4 = 2m + 2n − 4
∴ mn − 4m − 4n + 8 = 0
⇒ m(n − 4) = 4n − 8
4n − 8
∴ m = ______
n−4
Divide 4n − 8 by n − 4
4
n − 4 | 4n − 8
4n
− 16
_______
+8
8
∴ (4n − 8) ÷ (n − 4) = 4 + _____
n−4
(c)
(d)
From part (a) we know that m and n must be greater than 4.
8
From part (b) we know that m = 4 + _____
n−4
As we can only have whole number values of m, then n − 4 must be a divisor of 8.
n−4=8
for
n = 12
n−4=4
for
n=8
n−4=2
for
n=6
n−4=1
for
n=5
8
8
m = 4 + _____ = 4 + __ = 5
n–4
8
8
8
m = 4 + _____ = 4 + __ = 6
n–4
4
8
8
m = 4 + _____ = 4 + __ = 8
n–4
2
8
8
m = 4 + _____ = 4 + __ = 12
n–4
1
Interior pieces < Edge pieces
(n − 2)(m − 2) < 2m + 2n − 4
nm − 2m − 2n + 4 < 2m + 2n − 4
mn − 4m < 4n − 8
m(n − 4) < 4n − 8
4n − 8
8
m < ______ i.e. m < 4 + _____
n−4
(n − 4)
6
For n = 5 and m < 12 ⇒ m ≤ 11
For n = 12 and m < 5 ⇒ m ≤ 4
For n = 6 and m < 8 ⇒ m ≤ 7
For n = 8 and m < 6 ⇒ m ≤ 5
NOTE:
For n = 7 and m < 6.66
8
For n > 12, m < 4 + _____ < 5 ⇒ m ≤ 4
n−4
For jigsaws with one side greater than 12 pieces and the other side less than 5 pieces
Interior pieces < Edge pieces
In part (a), if one of the sides is 4 or less, then the interior pieces will be less than the edge
pieces.
7. (a)
f ′(x) = 3x2 + 2x
f (x) = x3 + x2 − 4
f (xn)
xn + 1 = xn − _____
f ′(xn)
3
x1 = __
2
f (x1)
x2 = x1 − _____
f ′(x1)
( )
(2)
( _32 )3 + ( _32 )2 − 4 __4
3 ____________
_
x2 = 2 −
=
3 2
3
3
3( _2 ) + 2( _2 )
3
f _2
3 ____
_
x2 = 2 − 3
f′ _
(b)
(i)
5
dx (t + 2)(2) − (2t − 1)(1) ______
2t − 1
=
x = _____ ⇒ ___ = __________________
2
t+2
dt
(t + 2)
(t + 2)2
t
y = ____
t+2
dy (t + 2)(1) − (t)(1) ______
2
=
⇒ ___ = ______________
2
dt
(t + 2)
(t + 2)2
dy ___
dy ___
dt
___
=
×
dx
dt
dx
) (
2
)
dy ______
(t + 2)
2
2
___
=
× ______ = __
2
dx
(c)
(
(t + 2)
5
5
(ii)
The graph is a straight line.
(i)
f (x) = (1 + x)loge(1 + x)
1
f ′(x) = (1 + x) _____ + loge(1 + x)(1) = 0
1+x
1 + loge(1 + x) = 0
(
)
loge(1 + x) = − 1
⇒ 1 + x = e−1
7
f(x) = 0 at turning point
1
∴ x = __
e−1
1−e
⇒ x = _____
e
As y = (1 + x) loge(1 + x)
[ (
)]
[ (
1
1
__
⇒ y = 1 + __
e − 1 loge 1 + e − 1
)]
( )
1
1
__
⇒ y = __
e loge e
1
y = __
e (−1)
1
y = − __
e
(
1 − e __
1
∴ Turning point at _____
e , −e
(ii)
f ′(x) = 1 + loge(1 + x)
1
f ′′(x) = _____
1+x
1 − e __
1
1
f ′′(x) = _______
=e
at the point _____
e ,−e
1−e
1 + ____
e
1 − e __
1
As e > 0 _____
e , − e is a minimum point
(
)
(
8. (a)
)
)
∫(sin 2x + e4x)dx
∫sin 2x dx + ∫e4x dx
1 4x
1
__
− __
2 cos 2x + 4 e + c
(b)
(i)
cos x + sin x
y = ___________
cos x − sin x
(cos x − sin x)(−sin x + cos x) − (cos x + sin x)(−sin x − cos x)
dy __________________________________________________
___
=
(cos x − sin x)2
dx
2
2
2
2
(−cos x sin x + cos x + sin x − cos x sin x) − (−cos x sin x − cos x − sin x − cos x sin x)
dy ________________________________________________________________________
___
=
(cos x − sin x)2
(−2 cos x sin x + 1) − (−2 cos x sin x − 1)
dy _________________________________
___
=
dx
(cos x − sin x)2
dy _____________
2
___
=
dx (cos x − sin x)2
dx
(ii)
dy _____________
2
___
=
dx (cos x − sin x)2
dy
Is ___ = 1 + y2 ?
dx
cos x + sin x
y = ___________
cos x − sin x
dy
cos x + sin x
Assuming it is true ⇒ ___ = 1 + ___________
dx
cos x − sin x
2
(
dy
cos x + sin x + 2 cos x sin x
___
= 1 + ________________________
dx
2
(cos x − sin x)2
8
)
2
dy
1 + 2 cos x sin x
___
= 1 + ______________
(cos x − sin x)2
dx
2
dy ____________________________
(cos x − sin x) + 1 + 2 cos x sin x
___
=
dx
(cos x − sin x)2
dy _______________________________________
cos2 x + sin2 x − 2 cos x sin x + 1 + 2 cos x sin x
___
=
dx
(cos x − sin x)2
dy _____________
2
___
=
dx (cos x − sin x)2
dy
∴___ = 1 + y2
dx
b
(c)
(i)
cos x
dx
∫a _______
1 + sin x
du
du
⇒ ___ = cos x
⇒ dx = _____
cos
x
dx
b
b
b
b
du
cos x
cos x _____
1
__
=
I = ________ dx = _____
du
=
ln
u
=
ln(1
+
sin
x)
|
u cos x a u
a 1 + sin x
a
a
Let u = 1 + sin x
∫
∫
∫
I = ln(1 + sin b) − ln(1 + sin a)
(
1 + sin b
I = ln ________
1 + sin a
)
b
(ii)
sin x
∫a ________
1 + cos x dx
Let u = 1 + cos x
b
du
⇒ ___ = −sin x
dx
du
⇒ dx = _______
− sin x
b
∫
b
∫
∫
b
du
sin x
sin x ______
1
_____
__
=
−
J = ________
dx
=
du
=
ln
u
=
−
ln(1
+
cos
x)
|
u
u − sin x
a 1 + cos x
a
a
a
J = − ln(1 + cos b) + ln(1 + cos a)
1 + cos a
J = ln ________
1 + cos b
(
(iii)
)
p
p
p
a + b = __ ⇒ a = __ − b and b = __ − a
2
2
2
1 + sin b
I = ln ________
1 + sin a
p
1 + sin( _2 − a )
1 + cos a
____________
I = ln
= ln ________ = J
p
_
1 + cos b
1 + sin( 2 − b )
(
(
)
)
(
)
9
Leaving Certificate Higher Level Maths Solutions
SEC 2013 Sample Paper 2
1. (a)
1 − (0.383 + 0.575 + 0.004)
= 0.038
E(X ) = 13(0.383) + 14(0.575) + 15(0.038) + 16(0.004)
= 13.663
(b)
E(X) is the excepted values of the distribution.
It is the mean age of all second-year students in Irish schools.
(c)
Probability of success: p = 0.575
Probability of failure: q = 0.425
10
( p)6 (q)4
6
10
(0.575)6 (0.425)4
6
= 0.248
( )
( )
2. (a)
(b)
Stratified sampling: When subpopulations vary considerably, it is a good idea to sample
each subpopulation (stratum) independently. Stratification is the process of grouping
members of the population into relatively homogeneous subgroups before sampling. The
strata should be mutually exclusive: every element in the population must be assigned
to only one stratum. The strata should also be collectively exhaustive: no population
element can be excluded. Then random sampling is applied within each stratum. This
often improves the representativeness of the sample by reducing sampling error.
Cluster sampling: Cluster sampling is a technique used when ‘natural’ groupings are
evident in a population. It is often used in marketing research. In this technique, the
total population is divided into these groups (or clusters) and a sample of the groups
is selected. Then the required information is collected from the elements within each
selected group. A common motivation for cluster sampling is to reduce the average cost
per interview. Given a fixed budget, this can allow an increased sample size. Assuming a
fixed sample size, the technique gives more accurate results when most of the variation in
the population is within the groups, not between them.
(i)
1__
= ___
√n
1
_____
= ______
√1,111
= 0.03
= 3%
(ii)
234
_____
× 100 = 21%
1,111
21% ± Margin of error = 21% ± 3% ⇒ 18% → 24%
23% is within this interval so we can accept the party’s claim.
10
3. (b)
3
Slope l1 = __
4
4
∴ Slope l2 = – __
3
(4k – 2, 3k + 1) is on l2
y – y1 = m(x – x1)
4
y – (3k + 1) = – __ (x – (4k – 2))
3
4
y – 3k – 1 = – __ (x – 4k + 2)
3
3y – 9k – 3 = – 4x + 16k – 8
4x + 3y – 25k + 5 = 0
(c)
(3,11) is on 4x + 3y – 25k + 5 = 0
4(3) + 3(11) – 25k + 5 = 0
12 + 33 – 25k + 5 = 0
50 – 25k = 0
k=2
k=2
(d)
4x + 3y – 25k + 5 = 0
4x + 3y – 25(2) + 5 = 0
4x + 3y – 50 + 5 = 0
4x + 3y – 45 = 0
l1
3x – 4y + 10 = 0
l2
4x + 3y – 45 = 0
9x – 12y + 30 = 0
16x + 12y – 180 = 0
25x
– 150 = 0
x=6
18 – 4y + 10 = 0
4y = 28
y=7
(6,7)
4. Centre (−g, −f ) is on the line x + 2y – 6 = 0
– g –2f – 6 = 0 ……… Equation 1
x- and y-axes are tangents ⇒ g2 = c and f 2 = c
⇒ g2 = f 2
⇒ g = ±f
11
g = f ……… Equation 2 and g = –f ……… Equation 3
– f – 2f – 6 = 0
f – 2f – 6 = 0
– 3f – 6 = 0
–f – 6 = 0
f = –2
f = –6
g = –2
g=6
As g2 = c and f 2 = c
c=4
c = 36
The general equation of a circle is x2 + y2 + 2gx + 2fy + c = 0
5.
x2 + y2 + 2(–2)x + 2(–2)y + 4 = 0
x2 + y2 + 2(6)x + 2(–6)y + 36 = 0
x2 + y2 – 4x – 4y + 4 = 0
x2 + y2 + 12x – 12y + 36 = 0
(a)
3 y
y = 3 sin 2x
2
y = sin x
1
P
x
2y = 1
–1
y = sin 2x
–2
–3
(b)
sin 2x = 0.5
p 11p 13p 17p
p
2x = __ ⇒ x = ___, ____, ____, ____
6
12 12 12 12
17p 1
P ____, __
12 2
(
6A.
)
Proof by contradiction: A proof by contradiction is a proof in which an assumption is made.
Then using valid arguments, a statement is arrived at which is clearly false: so the orginal
assumption must have been false.
Example: When proving that a tangent is perpendicular to the radius of a circle, we assume
that it is not perpendicular and, using logical geometric arguments, we prove that this is false,
so our orginal assumption is therefore not true. Therefore, it can be concluded that the tangent
is perpendicular to the radius of a circle.
A
Example 1. We can use a proof by contradiction to prove
the converse of Theorem 7.
Theorem 7 states: In a ΔABC, suppose that |AC| > |AB|.
Then |∠ABC| > |∠ACB|. In other words, the angle opposite
the greater of two sides is greater than the angle opposite
the lesser side.
12
B
C
The converse of Theorem 7 states: If |∠ABC| > |∠ACB|, then |AC| > |AB|. In other words, the
side opposite the greater of two angles is greater than the side opposite the lesser angle.
In the proof of the converse, it is assumed that |∠ABC| > |∠ACB|.
It could happen that |AC| ≤ |AB|, the either
Case 1: |AC| = |AB|, then |∠ABC| = |∠ACB| (Isosceles triangle). This contradicts our
assumption.
Case 2: |AC| < |AB|, then |∠ABC| < |∠ACB| (Theorem 7). This contradicts our assumption.
This cannot happen. Thus we conclude |AC| > |AB|.
6B.
To Prove: |∠DOC| = |∠DEC|
Proof:
A
E
|∠OEC| + |∠ODC| = 180°
O
∴ DOEC is a cyclic quadrilateral.
(circle is shown in the diagram)
C
∴ |∠DOC| = |∠DEC| ……… They are
standing on the same arc DC.
B
D
Q.E.D.
(ii)
70
60
50
40
30
20
10
0
RUN
SWIM
CYCLE
16 20 24 28 32 36
Time (minutes)
45
40
35
30
25
20
15
10
5
0
Competitors
70
60
50
40
30
20
10
0
Competitors
(i)
Competitors
30 34 38 42 46 50 54 58
Time (minutes)
80
70
60
50
40
30
20
10
0
10 14 18 22 26 30
Time (minutes)
RUN
16 20 24 28 32 36
Time (minutes)
Mean, Median ≈ 25 minutes
(iii)
This distribution is approximately normal so
that 99.7% of the data is within ±3 standard
deviations from the mean. Therefore
(Max – Min) ≈ 30 – 10 = 20 ≈ ±3s.
20 ÷ 2 ≈ 3s. Therefore s ≈ 10 ÷ 3 ≈ 3.3 minutes
Note: The standard deviation can also be found
by getting the square root of the variance.
13
SWIM
Competitors
(a)
Competitors
7.
80
70
60
50
40
30
20
10
0
10 14 18 22 26 30
Time (minutes)
(iv)
There is no modal time because no two competitors finished the events in the same
time (since times are given to the nearest 1/1000 minute).
(b)
There is a moderately strong positive correlation between run and cycle times, i.e. those
with fast cycle times tend to have fast run times.
There is also a positive correlation between run and swim times but less strong than in the
case of run versus cycle times, i.e. those with fast swim times tend to have fast run times.
There is also a positive correlation between cycle and swim times but less strong than in the
case of run versus cycle times, i.e. those with fast swim times tend to have fast cycle times.
(c)
y = 0.53(17.6) + 15.2 = 24.528
y = 0.58(35.7) + 0.71 = 21.46
24.528 + 21.46
_____________
2
= 22.972 min
= 23.0 min (to one decimal place)
(d)
95% of the athletes took between ±2 standard deviations of the mean.
Between 88.1 + 2(10.3) and 88.1 − 2(10.3)
Between 108.7 minutes and 67.5 minutes
(e)
x–m
z = _____
s
100 – 88.1
z = _________
10.3
z = 1.16
0.8770
∴ 87.7% finished in less than 100 min
224 × 0.877 = 196.448
196 athletes
(f)
F, F, S, F, F/S
F = Failure P(F) = 0.0877 S = Success P(S) = 0.123
5!
There are __ = 5 ways of arranging F, F, S, F, F. There is only one success at the end, S.
4!
This gives a probability of 5(0.877)(0.877)(0.123)(0.877)(0.877) × (0.123) = 0.0447
5
This is really
(0.123)1(0.877)4 × (0.123) = 0.0447
1
()
14
8. (a)
(i)
B
5 cm
C
30 cm
22 cm
D
E
5 cm
4 cm
a
A
F
25
22
______
= __________
C
sin 60º
sin | ∠AFC |
25
22
______
= __________
0.8660
25 cm
sin | ∠AFC |
25(0.8660)
sin | ∠AFC | = _________
22
25(0.8660)
| ∠AFC | = sin–1 _________
22
| ∠AFC | = 79.77°
22 cm
∴ | ∠ACF | = 40.23°
60°
A
F
C
|DE|2 = 202 + 182 –2(20)(18) cos 40.23°
20 cm
40.23°
|DE| = 13.20 cm
18 cm
D
E
(ii)
22
sin a = ___
25
C
22
a = sin–1 ___
25
a = 62°
25 cm
a
A
22 cm
90°
F
15
r
9.
120°
r
2a
h
2a
90°
r
(2a)2 = r 2 + r2 – 2(r)(r) cos 120°
1
4a2 = 2r2 + 2r2 __
2
2
2
2
4a = 2r + r
(2a)2 = h2 + r2
4a2 = h2 + r2
___
h2 = 4a2 – r2
√
4a2
r = ___
3
2a
___
r = __
√3
2a
__
Volume of cylinder = p r 2h = p ___
√3
_______
h = √4a2 – r 2
__________
√ ( )
2a
__
h = 4a2 – ___
√3
____
( ) √3
2
8a2
___
________
√
h = 12a – 4a
√ 3
h=
√8a3
4a2
h = 4a2 – ___
3
__
4a2
√
2a 2
= p ___ ___ __
3 1 3
__
_________
2
8a 2
= p ___ __
3
3
____
2
__ __
___
__
__
8a √6
= p ______
3
2
_________
√3
8a 2 √3
= p ___ __ ___
3 √3 √3
3
2
9
__
( )
8√6
= ____ p a3
9
16
Leaving Certificate Higher Level Maths Solutions
Sample Paper 1, no.1
1. (a)
To prove: 2n ≥ n2 [n ∈ N, n ≥ 4]
Test n = 4: 24 ≥ 42
16 = 16
∴ p(4) is true.
Assume p(k) is true, that is:
2k ≥ k2 *
Test p(k + 1)
2k + 1 ≥ (k + 1)2
2.2k ≥ 2(k2) from *
≥ k2 + k2
≥ k2 + k . k
≥ k2 + 3k since k ≥ 4
≥ k2 + 2k + k
≥ k2 + 7k + 1 since
k≥4
= (k + 1)2
⇒ 2k + 1 ≥ (k + 1)2
∴ p(k + 1) is true whenever p(k) is true.
Since p(4) is true, then by induction p(n) is true for any positive integer n, n ≥ 4.
(b)
(i)
3ex − 7 + 2e−x = 0
2
3ex − 7 + __x = 0
e
3ex ex − 7ex + 2 = 0
3(ex)2 − 7ex + 2 = 0
let y = ex
3y2 − 7y + 2 = 0
(3y − 1)(y − 2) = 0
1
y = __ on y = 2
3
1
x
e = __
ex = 2
3
1
x = ln __
x = ln 2
3
()
(ii)
Quadratic.
17
2. (a)
1.8 = i(1 + i)12
12
____
12
____
√ 1.18 = 1 + i
√ 1.18 – 1 = i
0.01388843 = i
130,000(0.01388843)2 = €133,636.07
(b)
Amount owing after three months 130,000(1.01388843)3
= €135,492.06
t
i(1 + i)
A = P _________
(1 + i)t – 1
[
54
0.01388843(1.01388843)
= 135,492.06 ______________________
54
(1.01388843) – 1
]
= €3,583.11
(c)
3. (a)
(b)
3,583.11 × 54 = €193,487.94
y = 2x + 3
C
xy = 2
A
x2 + y2 = 6
D
y = ln x
B
x2 + y2 = 6
xy = 2
22
__
2
y +y =6
4 + y4 = 6y2
()
y4 − 6y2 + 4 = 0
(y2)2 − 6(y2) + 4 = 0
let m = y2
__
m = √5 + 3
__
y2 = √ 5 + 3
y = ±2.29
or
__
m = 3 − √5
__
y2 = 3 − √ 5
y = ± 0.87
y = 2.29
y = − 2.29
y = 0.87
y = − 0.87
x = 0.87
x = − 0.87
x = 2.30
x = −2.30
18
4. (a)
(b)
No real roots ⇒ b2 − 4ac < 0
x2 + kx + 8 − k = 0
y
b2 − 4ac ≥ 0
x
k2 − 4(1)(k − 8) ≥ 0
–8 –6 –4 –2
k2 − 32 + 4k ≥ 0
k2 + 4k − 32 ≥ 0
k2 + 4k − 32 = 0
(k − 8)(k + 4) = 0
k = 4 or k = –8
−8 ≥ k ≥ 4
5. (i)
C − Po
A = ______
Po
11,500 − 600
A = ___________
600
A = 18.16666667
(ii)
c
p(t) = ________
1 + Ae−kt
11,500
1,800 = __________________
1 + 18.16666667e−k(1)
11,500
18.16666667e−k = ______ − 1
1,800
−k
e = 0.29663608
3.371134085 = ek
ln(3.371134085) = k
1.215249211 = k
(iii)
c
P(t) = ________
1 + Ae−kt
11,500
P(3) = ___________________________
1 + 18.16666667 . e− (1.25249211)(3)
P(3) = 8,075
19
2
4
(iv)
c
P(t) = ________
1 + Ae−kt
11500
4000 = _______________________
1 + 18.16666667 e−1.25249211t
(
11,500
_____
4,000 − 1
___________
−1.25249211t = ln
18.16666667
)
−1.25249211t = −2.270979754
t = 1.81316891 years
t ≈ 1.8 years
6. (a)
(b)
7.5 75 3
r = ___ = ____ = __
10 100 4
3
Tn = ar n−1 = 10 __
4
()
n−1
Height (ft)
(c)
10
9
8
7
6
5
4
3
2
1
2
1
3
4
5
6
7
8
9
10
n
10, 7.5, 5.625, 4.21875, 3.164, 2.373, 1.780, 1.335, 1.001, 0.751
(d)
(e)
() ()
()
()
() () () ()
() () () ]
3
32
33
3
Dn = 10 + 2.10 __ + 2.10 __ + 2.10 __ + 2.10 __
4
4
4
4
2
3
4
3
3
3
3
Dn = 10 + 20 __ + 20 __ + 20 __ + 20 __
4
4
4
4
2
3
3
3
3
D∞ = 10 + 20 __ + 20 __ + 20 __ + ...
4
4
4
a
S∞ = ____
1–r
[
= 10 +
[ ]
3
3
20 _4
20 _4
_____
_____
= 1
=
3
_
1–_
( )
4
( )
4
3 4
= 20 __ __
4 1
( )( )
= 10 + 60
= 70 feet
20
4
____
(f )
√
2.10
t1 = ____
32_______
3
2.10 __
4
t2 = 2. _______
32
________
32
2.10 __
4
t3 = 2. _______
32 _______
________
________
3
32
33
__
__
____
2.10
2.10
2.10 __
4
4
4
2.10
t∞ = ____ + 2. _______ + 2. _______ 2. _______ + ...
32 ________
32
32
32
________
________
3
32
33
__
__
___
2(10)
2(10)
2(10) __
4
4
4
20
= ___ + 2 _______ + 2 ________ + 2 ________ + ...
32
32
32
32
√
√
√
()
()
√
[√
√
()
√
√
20
20 3
3
3
=
√32 + 2√32 [ √4 + √4 + √4
3
√
4
20
20
+2
+
=
√32 √32 1 − 3
√4
___
___
()
___
__
__
___
__
__
()
()
__
2
3
__ + ...
__
___
___
___
___
]
√
√
()
()
[ ]
__
______
__
__
=11.011 seconds
7. (a)
y = 3x3 − 2x2 + 12x − 2
y ′ = 9x2 − 4x + 12
x0 = 0.5
7
3 _8
f (x0)
_____
____
x1 = x0 −
⇒ 0.5 − 1
f ′(x )
12_
0
4
= 0.183673469
= 0.18
0.112696
x2 = 0.18 − ________ = 0.17976
11.5716
= 0.1798
0.110381664
x3 = 0.1798 − ___________
11.57175236
= 0.17026111
= 0.17 to two significant figures
(b)
(i)
__
Let f (x) = √x
_____
f (x + h) = √x + h
_____
_____
__
__ __________
x + h + √x
√_____
√x .
__
f (x + h) − f(x) = √x + h −
√x + h + √x
h
_____
= __________
__
√x + h + √x
21
]
f (x + h) − f(x) __
1
1 _____h
____________
_____
= __________
= · __________
__
__
h √x + h + √x √x + h + √x
f(x + h) − f(x)
1
_____
lim ___________ = lim __________
__
h
h→0
h→0 √x + h + √x
1__
= ____
2√x
h
_____
(ii)
√
( )
( )
1
y = _____
3−x
1 _1
y = _____ 2
3−x
dy __
1 1 − _1
___
= _____ 2 · − 1 (3 − x)−2 · − 1
dx 2 3 − x
1
1 _____
1
= __ ______
· _______2
2 _____
(3 − x)
1
3−x
√
(c)
(i)
u v
y = x2 ex
dy
___
= x2 · ex + 2xex
dx
= ex(x2 + 2x)
(ii)
ex(x2 + 2x) = 0
ex = 0
x2 + 2x = 0
or
no solutions
x(x + 2) = 0
x=0
or
y=0
(
4
Turning points: (0,0), –2, __2
e
u
(iii)
x = −2
4
y = __2
e
)
v
dy
___
= ex(x2 + 2x)
dx
d2y
___
= ex(2x + 2) + ex(x2 + 2x)
dx2
= ex(2x + 2 + x2 + 2x)
= ex(x2 + 4x + 2)
2
|
dy
|
dx (
dy
___
dx2
2 > 0 ∴ Local minimum at (0,0)
(0,0)
2
___
4
2 −2, __
e2
)
= e−2((−2)2 + 4(−2) + 2)
= − 0. 2706 … < 0
4
∴ Local maximum at −2, __2
e
(
)
22
8.
(a)
(i)
∫e−2x dx
1
⇒ − __ e−2x + c
2
(ii)
∫2 cos 2x dx
(
)
1
2 __ sin 2x + c
2
sin 2x + c
(b)
(i)
r
Let y = __x
h
2
r
y2 = __2x2
h
Volume of cone ⇒ p
h
∫0 y2 dx
r2
= p Ι0:h:__2 x2dx
h
∫
h
r2
= __2 p 0 x2dx
h
[ ]
r2 x3 h
= __2 p __
3 0
h
2
h3
pr __
1
= ___
·
− 0 = __pr2h
2
3
3
h
(ii)
y = 2x + 3
y−3
_____
=x
2
y_____
−3 2
= x2
2
( )
6
p
(y − 3)
dx
∫4 _______
4
p
= __
4
2
6
∫4 y2 − 6y + 9 dx
[
3
p y
= __ __ − 3y2 + 9x
4 3
p
1
= __ 18 − 9__
4
3
13p
= ____
6
[
]|
6
4
]
23
2
(c)
_____
∫0 √4 − x2 dx
Let x = 2 sin q
dx = 2 cos q dq
α units x = 0
2 sin q = 0
sin q = 0
q=0
x=2
2 sin q = 2
sin q = 1
p
q = __
2
2
_____
p
_
__________
∫0 √4 − x2 dx = ∫0 √4 − 4 sin2 q 2 cos q dq
___________
= ∫0 √4 (1 −sin2 q) 2 cos q dq
_______
= ∫0 √4 cos2 q 2 cos q dq
= ∫0 2 cos q 2 cos q dq
= 4∫0 2 cos2 q dq
1 (1 + cos 2q) dq
= 4∫0 __
2
= 2∫0 (1 + cos 2q) dq
2
p
_
2
p
_
2
p
_
2
p
_
2
p
_
2
p
_
2
[
]
sin 2q
= 2 q + _____ dq
2
sin 2(0)
p sin p
p
= 2 __ + ____ − 0 − _______ = 2 __ = p
2
2
2
2
[
]
( )
24
Leaving Certificate Higher Level Maths Solutions
Sample Paper 2, no. 1
U
1. (a)
A
B
0.2
0.4
0.1
0.3
(b)
From the Venn diagram ( A ∪ B )′ = 0.3
(c)
P(B ∩ A) 0.4
P(B|A) = ________ = ___ = 0.6666
0.6
P(A)
(d)
Events are not independent because P(B|A) is not equal to P(B).
2. (a)
(i)
(ii)
Number of Favourable Outcomes __
1
___________________________
=
6
5
P(Success) = P(not getting a six) = __
6
1
__
P(Failure) = P(getting a six) =
6
Probability of not getting a six in the first two throws
Total Outcomes
= Failure, Success or
1 5
= __ × __
+
6 6
10
= ___
36
5
= ___
18
(iii)
Success, Failure
5 __
1
__
×
6 6
Method 1.
51
1st throw: Probability of Failure = __ = 0.833 > 0.5
6
5
5
52
2nd throw: Probability of Failure = __ × __ = __ = 0.6944 > 0.5
6
6
6
5
5
5
53
3rd throw: Probability of Failure = __ × __ × __ = __ = 0.5787 > 0.5
6
6
6
6
5
5
5
5
54
4th throw: Probability of Failure = __ × __ × __ × __ = __ = 0.4822 < 0.5
6
6
6
6
6
()
() () ()
() () () ()
() () () () ()
∴ Four throws will give a player a better than equal chance of starting the game.
1n 1
Method 2. Solving the inequality __ > __
6
2
log 2
1
1 __
n
__
>
>
2
⇒
n
log
6
>
log
2
⇒ n > _____
⇒
6
n
log 6
6 2
n > 0.3868 ∴ n = 4
()
25
3. (a)
3
k
2
1
0
–3 –2 –1 0 1
–1
2
3
4
5
6
7
8
9 10
–2
–3
l
–4
–5
(b)
Method 1. Solving simultaneously
Method 2. Putting in the point to both equations
x−y−5=0
3x + y + 5 = 0
2x
+ 3y − 6 = 0
______________
3x − 3y − 15 = 0
3(−2) + (1) + 5 = 0
2x
+ 3y − 6 = 0
______________
∴ (−2,1) is on 3x + y + 5 = 0
5x
x − 3y + 5 = 0
⇒ −6 + 6 = 0
− 21 = 0
1
21
⇒ x = ___ = 4__
5
5
1
4__ − y − 5 = 0
5
4
−y − __ = 0
5
4
⇒ y = − __
5
1 4
∴ The point 4_5 ,− _5 is the
(
−2 − 3(1) + 5 = 0
⇒ −5 + 5 = 0
∴ (−2,1) is on x − 3y + 5 = 0
(−2,1) is on both lies
∴ It is the point of intersection.
)
point of intersection.
(c)
3
k
2
1
0
–3 –2 –1 0 1
–1
A (3,0)
2
3
4
B (5,0)
5
6
7
8
9 10
1 4
P 4 5,– 5
(
–2
)
–3
l
–4
–5
1
Area of ABP = __|AB| × Height
2
4
1
= __ (2) × __
5
2
4
= __ units2
5
()
26
4.
(a)
1
Area of triangle = __ a × b × sin C
2
1
= __ (3)(3) sin 120°
2
= 4.5(0.8660)
= 3.90 units2
(b)
1
Area of sector OCRB = __ r 2q (q in radians)
2
2p
1
= __ (3)2 ___
2
3
= 3p
= 9.4248 sq. units
Area of region CRB = Area of sector OCRB − Area of triangle OBC
= 9.4248 − 3.90
= 5.52 sq. units
5.
6A.
(a)
Equation of circle with centre at A:
(x + 1)2 + (y − 1)2 = 36
Equation of circle with centre at B:
(x − 7)2 + (y − 1)2 = 9
(b)
Centre: (2,1),
Radius: 8
(c)
Equation of the third circle:
(x − 2)2 + (y − 1)2 = 64
A
Diagram:
D1
Dm = D
E1
I
E = Em
C = Em + n
Dm + n = B
Given: Δ ABC
To Prove: If a line l is parallel to BC and cuts [AB] in the ratio m : n, then it also cuts [AC] in
the same ratio.
Construction: Let l cut [AB] in D in the ratio m : n with natural numbers m, n.
Proof: We prove only the commensurable case.
Let l cut [AB] in D in the ratio m : n with natural numbers m, n. Thus, there are points
(see diagram)
D0 = A, D1, D2 , . . . , Dm − 1, Dm = D, Dm + 1 , . . . , Dm + n − 1, Dm + n = B,
equally spaced along [AB], i.e. the segments
[D0 D1], [D1 D2 ], . . . [Di Di + 1 ], . . . [Dm + n – 1 Dm + n]
have equal length.
27
Draw lines D1 E1, D2 E2, . . . parallel to BC with E1, E2, . . . on [AC].
Then all the segments
[AE1], [E1 E2], [E2 E3], . . . , [E m + n – 1C] have the same length, ...........................[Theorem 11]
and Em = E is the point where l cuts [AC] ....................................................[Axiom of Parallels]
Hence, E divides [AC] in the ratio m : n.
Q.E.D.
6B.
(a)
G
6
J
1
5
H
3
E
4
2
E
F
In the triangles HJE and GFE:
|∠1| = |∠2| .............................Given
|∠3| = |∠4| .............................Same angle
∴ |∠5| = |∠6| .....................Third angle in triangles
∴ The triangles HJE and GFE are equiangular.
∴ The triangles HJE and GFE are similar.
|EH| |JE|
∴ ____ = ____
|GE| |FE|
Rearranging gives |EH| : |EJ| = |EG| : |EF|
(b)
[JH] is parallel to [FR] ......................Given
|JE| |HE|
∴ ____ = ____
|ER| |EF|
|HE| |EF|
Rearranging gives ____ = ____
|EJ| |ER|
∴ |HE| : |EJ| = |EF| : |ER|
But
|EG|
|HE| ____
____
=
............................Part (a)
|EJ|
|FE|
|FE| |EG|
∴ ____ = ____
|ER| |FE|
|FE|2 = |EG| × |ER|
28
7. (a)
E
30°
90°
40°
20 m
S
(b)
Circumference (Length) of a circle = 2pr = 3.5
3.5
r = ___ = 0.557
2p
Diameter = 2(0.557) = 1.1 m
(c)
E
d
E
h
30°
90°
90°
s
d
20 m
S
h
40°
s
S
202 = d2 + s2
h
tan 30° = __
d
h
tan 40° = __s
√3
___
h
0.8391 = __s
__
h
= __
3 d __
d√3
h = ____
3
__
d√3
s(0.8391) = ____
3
s(0.8391) = d(0.5774)
s = d(0.6880)
202 = d2 + (0.6880d)2
400 = 1.4734d2
d2 = 271.28
d = 16.47
__
d√3
As h = ____
3
__
16.47√3
_______
h=
3
= 9.509
= 9.5 m
29
h = s(0.8391)
8. (a)
24
Time in Hours
22
20
18
16
14
12
10
8
6
4
2
Age in Years
0
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46
(b)
Estimation of correlation coefficient 0.7
(c)
Correlation coefficient 0.7365
Error 0.7 − 0.7365 = 0.0365
Percentage error:
100
Error
% Error = ______________ × ____ = 4.96%
1
Accurate Figure
0.0365 100
% Error = ______ × ____ = 4.96%
0.7365
1
(d)
The correlation coefficient seems to suggest a strong positive correlation between the
ages of the swimmers and the times for their swims. However, one would want to be
very careful in making general statements about the ages of the swimmers and their
corresponding swim times. Does the general statement above lead us to conclude that
the younger you are, the faster your swim time is?
(e)
(i)
Male: Mean 15.9 hours. Standard deviation 2.78
(ii)
Female: Mean 17.2 hours. Standard deviation 2.78
(f)
The male times on average for the swim seem to be faster. There seems to be little
variation in the times when comparing men with men and women with women.
There is a greater spread of times among the female swimmers compared to the male
swimmers.
The correlation between the ages and swim times in the females is stronger
(coefficient = 0.82) than the correlation between the ages and swim times in the males
(coefficient = 0.66), but this could be due to a number of factors such as having a
smaller sample of females.
30
Leaving Certificate Higher Level Maths Solutions
Sample Paper 1, no. 2
x3 + 2x2 + 4x + (a + 4)
1. (a)
x − 1| x4 + x3 + 2x2 + ax + b
– 4+ 3
x – x
2x3 + 2x2
– 2x3 +– 2x2
4x2 + ax
– 4x2 +– 4x
(a + 4)x + b
– (a + 4)x +– (a + 4)
b+a+4
Remainder is 7
∴b+a+4=7
b+a=3
(b)
(i)
a3 – a2b – ab2 + b3
= a2(a – b) – b2(a – b)
= (a – b)(a2 – b2)
= (a – b) (a – b) (a + b)
= (a – b)2(a + b)
(ii)
a3 + b3 ≥ a2b + ab2?
a3 – a2b – ab2 + b3 ≥ 0?
(a – b)2(a + b) ≥ 0?
(a – b)2 ≥ 0 as (any number)2 ≥ 0
If a, b ∈ N ⇒ a + b ≥ 0
(a – b)2(a + b) ≥ 0
a3 + b3 ≥ a2b + ab2
2. (a)
ar = 750
ar4 = −6
−6
a = ___
r4
31
ar = 750
−6r
____
= 750
r4
−6
___
= 750
r3
−6
____
= r3
750
750
−1
___
=r
∴ a = ____
= −3,750
−1
__
5
5
(b)
(c)
3. (a)
Tn =
arn – 1
1 n–1
= (–3,750) –__
5
Sum to ∞
a
_____
as
|r| < 1
1−r
−3,750
S∞ = ______
−1
1 − __
5
S∞ = −3,125
( )
log5(x − 2) = 1 − log5(x − 6)
log5(x − 2) + log5(x − 6) = 1
log5[(x − 2)(x − 6)] = 1
(x − 2)(x − 6) = 51
x2 − 6x − 2x + 12 − 5 = 0
x2 − 8x + 7 = 0
(x − 7)(x − 1) = 0
x=7
or
x=1
Reject as
x>6
2
(b)
3
_
− x2
2x __
_______
√x
2x2
3
_
x2
___ − __
1
_
x2
1
_
x2
3
_
2x2 − x1
4. (a)
z = cos q + i sin q
zn = cos nq + i sin nq
1 ______________
1
__
= (cos nq + i sin nq )−1
=
n
z cos nq + i sin nq
= cos (−nq ) + i sin (−nq )
= cos nq − i sin nq
1
= cos nq + i sin nq + cos nq − i sin nq
zn + __
zn
= 2 cos nq
32
(b)
64 cos6 q Expand using binomial or otherwise:
16
= z + __z = z6 + 6z4 + 15z2 + 20 + 15z−2 + 6z−4 + z−6
1
1
1
= z6 + __6 + 6 z4 + __
+ 15 z2 + __
+ 20
4
z
z
z2
64 cos6 q = 2 cos 6q + 12 cos 4q + 30 cos 2q + 20
(
)
(
) (
) (
)
32 cos6 q = cos 6q + 6 cos 4q + 15 cos 2q + 10
∴p=1
q=6
r = 15
s = 10
5. (a)
n = 3.24
p = 2.15
V = 35.28
Find T:
R = 0.0821
PV = nRT
PV
___
=T
nR
(2.15)(35.28)
____________
=T
(3.24)(0.0821)
285.1536067 = T (kelvin)
285.1536067 − 273 = T (celsius)
12.15°C = T
(b)
PV = nRT
nRT
V = ____
P
nM = m
m
D = __
V
m
n = ___
M
m
Substitute both into D = __:
V
nM
___
nRT
___
D=
P
PM
D = ____
RT
(c)
PM
D = ____
RT
M = 64.1
P = 1.226
T = 65°C (* Must be changed to Kelvin)
33
Tk = 65 + 273
= 338° K
(1.226)(64.1)
D = ____________ = 2.83196996 g/L
(0.0821)(338)
6. (a)
42 = x2 + x2 − 2(x)(x)cos 60°
1
16 = 2x2 − 2x2 __
2
2
2
16 = 2x − x
()
16 = x2
60°
4=x
x
[
1
∴ Area of hexagon = 6 __(4)(4) sin 60°
__
2
√3
___
=6 8
__2
= 24√3 cm2
[ ( )]
(b)
]
4
Radius = perpendicular height of each Δ
∴ 42 = h2 + 22
4
16 = h2 + 4
h
12 = h2
___
√12 = h
2
__
2√3 = h
__
r = 2√3 cm
(c)
x
Volume of cone = 105.5 cm3
1 2
__
p r h = 105.5
3
105.5 ________
105.5
h = _____
=
__
1 2
_
1
__
p
r
p
(2
3 )2
√
3
3
= 8.395423248 cm
∼
− 8.4 cm
(d)
3 cm
1 cm
4 cm
(
)
3+4
A = _____ (1)
2
7
A = __ (1) = 3.5 cm2
2
(3.5)(11) = 38.5 cm2
__
__
Hexagon ⇒ 24√3 × 2 = 48√3 cm2
34
Rectangular sides = 4 × 8.4
= 33.6 cm2 × 6
= 201.6 cm2
__
Total S.A. = 38.5 + 48√3 + 201.6
= 323.24 cm2
(e)
V = 155.5
1 2
__
p r h = 155.5
3
155.5 ________
155.5
__ = 12.37 cm
h = _____
=1
1
_p r 2 _p ( 2 3 )2
√
3
3
∴ Rectangular sides = 4 × 12.37
= 49.48 cm2 × 6
= 296.88 cm2
__
T.S.A. = 296.88 + 48√3 + 38.5
= 418.52 cm2
418.52 − 323.24 = 95.28 cm2
95.28
______
= 29.48% increase
323.24
7. (a)
u
2t
x = ____ = __
t+3 v
(t + 3)(2) − (2t)(1)
dx _______________
___
=
dt
(t + 3)2
2t + 6 − 2t
= _________
(t + 3)2
6
= ______2
(t + 3)
3t
u
y = ____ = __
t+3 v
(t + 3)(3) − (3t)(1)
dy _______________
___
=
dt
(t + 3)2
3t + 9 − 3t
= _________
(t + 3)2
9
= ______2
(t + 3)
9
dy _____
__
dy __
(t + 3)2 __
9
dt _____
___
= dx = 6 =
_____ 6
dx __
dt
(t + 3)2
35
(b)
u v
y2(x − 2) = 3x
dy
⇒ y2 + (2xy − 4y)___ = 3
dx
dy
(2xy − 4y) ___ = 3 − y2
dx
2
dy _________
3−y
___
=
dx
(2xy − 4y)
y2 (x − 2)
dv
du
u___ + v___
dx
dx
dy
y2(1) + (x − 2)(2y) ___
dx
dy
___
y2 + (2xy − 4y)
dx
3 − 32
At (3,3) ⇒ M = _____________
(2(3)(3) − 4(3))
−6
= ___
6
= −1
y − y1 = m(x − x1)
y − 3 = −1(x − 3)
y − 3 = −x + 3
x+y−6=0
(c)
1
y = __
x loge x
y = x−1 loge x
dy
du
dv
___
= v___ + u___
dx
dx
dx
1
= loge x(−1x−2) + (x−1) __
x
loge x __
1 1
+ x · __
= −_____
x
2
x
loge x __
1
+ 2
= −_____
2
x
x
−loge x + 1
= _________
x2
dy
At maximum/minimum, ___ = 0
dx
−log
x
+
1
e
⇒ _________
=0
x2
−loge x + 1 = 0
( )
1 = loge x
e=x
36
1
When x = e, y = __
e · logee
1
y = __
e·1
1
y = __
e
( )
1
y has max/min value at e, __
e .
du
dv
__
v__
d2y ________
dx − u dx
___
=
dx2
v2
−1
x2( __
x ) − (−loge x + 1)(2x)
_____________________
=
x4
−x + 2xloge x − 2x
= _______________
x4
−e + 2e logee − 2e
When x = e ⇒ _______________
e4
−e + 2e − 2e
___________
e4
−e
= ___4
e
−1
= ___
< 0 ⇒ maximum value
e3
1
1
__
∴ y = __
x loge x has maximum value at e .
4
8. (a)
(i)
∫1 (x + √__x )2 dx
4
=
∫1 (x2 + x + 2x√__x ) dx
4
=
∫1 (x2 + x + 2x ) dx
3
_
2
[
3
2
[
]
5 4
_
2
x x 2x2
= __ + __ + ___
5
_
3 2
1
5
_
][
5
_
4 4 2(4) · 2
1 1 2(1) · 2
= __ + __ − _______ − __ + __ − _______
3
2
3 2
11 1
= 3___ − ___
15 30
2
5
3
2
3
2
5
7
= 3___ = 3.7
10
2
(ii)
x−1
∫1 ( _____
dx
x3 )
2
=
∫1 __xx3 − __x13 dx
2
=
∫1 __x12 − __x13 dx
2
=
2
∫1 x−2 − x−3 dx
37
]
2
[
]
1 2
−1 2
−1 2
= [−x + ] = [ + ] − [ + ]
2
4
1 1
x
x−1 x−2
= ___ − ___
−1 −2
__
__
2
1
2
___
__
___
__
1
=0−1
= −1
2
(b)
1
_____
∫0 _______
dx
2
(i)
√4 − x
x 2
= sin−1 __
20
0
2
__
= sin−1 − sin−1 __
2
2
[
]
= sin−1 1 − sin−1 0
p
= __ − 0
2
p
__
=
2
1
e
∫0 ______
dx
1 + e2x
(ii)
x
Let u = ex
du = ex dx
x = 0 ⇒ u = e0 = 1
x = 1 ⇒ u = e1 = e
e
⇒
1
∫1 _____
du
1 + u2
= [tan−1 u]1e
= tan−1 e − tan−1 1
p
= tan−1 e − __
4
2
(c)
∫0 sin(x − 2) · cos(x + 2)
1
__
2
2
∫0 sin(x − 2 + x + 2) + sin(x − 2 − x − 2)
2
∫0 sin 2x + sin(−4)
1
= __
2
2
1 −cos 2x
= __ _______ + sin(−4)x
2
2
0
−cos 4
1 ______
1 −cos 0
__
+ 2 sin(−4) − __ ______ + 0 · sin(−4)
=
2 2
2 2
1
1 1
= __ [1.840426801] − __ − __
2
2 2
1
= 0.9202134 + __
4
= 1.170213401
[
[
]
] [
[ ]
= 1.17 (to two decimal places)
38
]
Leaving Certificate Higher Level Maths Solutions
Sample Paper 2, no. 2
1. (a)
S
BP
C
0.16 0.11 0.21
0.52
(b)
P(BP ∩ C) = 0.11
(c)
P(BP) = 0.27
(d)
P(PB ∩ C) 0.11
P(BP|C) = P(BP|C) = _________ = ____ = 0.344
0.32
P(C)
2. (a)
Line
k
l
m
(b)
Slopes
3
– __
5
y-intercept
Equation
Intersection point
2
3x + 5y – 10 = 0
k ∩ m = (2,0)
2
4
– __
7
0
y = 2x
k ∩ p = (–2,3)
2
4x + 7y – 14 = 0
k ∩ w = (2,1)
(
)
(
)
p
∞
Undefined
None
x = –2
22
p ∩ m = –2,___
7
w
2
–3
2x – y – 3 = 0
10 20
l ∩ m = ___,___
13 13
k: 3x + 5y − 10 = 0
l: y = 2x
3x + 5(2x) − 10 = 0
3x + 10x − 10 = 0 ⇒ 13x = 10
10
⇒ x = ___:
13
10
20
y = 2 ___
⇒ y = ___
13
13
10 20
l ∩ k is ___,___
13 13
( )
(
)
(
)
10 20
The three points that form the triangle are (−2,3), (−2,−4) and ___,___ .
13 13
1
__
Area = base × height
2
10
1
= __(7) 2___
2
13
36
1
__
___
= (7)
2
13
126
= ____ units2
13
( )
( )
39
(c)
1
Slope = − __ (opposite slope of w)
2
Equation: y − y1 = m(x − x1)
1
y − (−3) = − __(x − (2))
2
1
y + 3 = − __(x − 2)
2
2y + 6 = − x + 2
Point (2,−3)
x + 2y + 4 = 0
3. (a)
Equation of altitude from D to
AB (slope of AB = −2):
1
Point (3,4) Slope = __ (opposite slope
2
to AB)
A
6
5
4
3
2
1
0
y
D
–7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7
–1
–2
–3
–4
–5
B
–6
–7
–8
Equation: y − y1 = m(x − x1)
1
y − 4 = __(x − 3)
2
2y − 8 = x − 3
x − 2y + 5 = 0
Equation of altitude from A to BD
(slope of BD = 3):
Point (–4,3)
1
Slope = – __ (opposite slope to BD)
3
Equation: y – y1 = m(x – x1)
1
y – 3 = –__(x + 4)
3
3y – 9 = –x – 4)
x + 3y – 5 = 0
Solving simultaneously:
x + 3y − 5 = 0
x − 2y + 5 = 0
5y − 10 = 0
⇒y=2
⇒ x = −1
Orthocentre (−1,2)
40
x
(b)
Distance from (0,0) to A( − 4,3):
_________________
Distance from (0,0)
to B(0,− 5):
Distance from (0,0) to D(3,4)
_________________
_________________
(x2 − x1 + (y2 − y1
√________________
2
2
√(x
2 − x1) + (y2 − y1)
_______________
= √16 + 9
= √0 + 25
= √9 + 16
= √25
= √25
= √25
=5
=5
=5
(x2 − x1 + (y2 − y1
√________________
)2
)2
− 0)2 + (3 − 0)2
√(−4______
___
)2
)2
− 0)2 + (−5 −0)2
√(0______
− 0)2 + (4 − 0)2
√(3______
___
___
∴ (0,0) is the circumcentre of the triangle ABD
4. (a)
Answer: Skewed to the right.
Reason: The mean is greater than the median.
(b)
(c)
Answer: Between €310 and €820.
Statistical description
Mean salary
Median salary
Lowest salary
Lower quartile
Interquartile range
Standard deviation
Range
(d)
Statistical description
Mean salary
Median salary
Lowest salary
Lower quartile
Interquartile range
Standard deviation
Range
5. (a)
Value in €
New value after
€30 increase
€630
€480
€310
€340
€510
€370
€1,000
€600
€450
€280
€310
€510
€370
€1,000
Value in €
€600
€450
€280
€310
€510
€370
€1,000
New value after
10% increase
€660
€495
€308
€341
€561
€407
€1,100
x−µ
z = _____
s
150 − 130 20
z = _________ = ___
8
8
z = 2.5
P(x > 150) = P(z > 2.5)
= 1 − P(z < 2.5)
= 1 − 0.9938
–3.5 –3 –2.5 –2 –1.5 –1 –0.5 0
= 0.0062
0.6% of 500 bars = 3.1
Answer: 3 bars
41
0.5
1
1.5
2
2.5
3
3.5
(b)
x−µ
z = _____
s
x−µ
z = _____
s
125 − 130
5
z = _________ = − __
8
8
z = − 0.625
140 − 130 10
z = _________ = ___
8
8
z = 1.25
P(z) = P( − 0.625 < z < 1.25)
= P(z < 1.25) − P(z < − 0.625)
= P(z < 1.25) − [1 − P(z < 0.625)]
= 0.8944 − [1 − 0.7324]
= 0.8944 − 0.2676
–3.5 –3 –2.5 –2 –1.5 –1 –0.5 0
= 0.6268
62.68% of 500 bars = 313.4
Answer: 313 bars
6A.
A standard construction of incentre of equilateral triangle.
30°
6B.
(a)
|∠ HGK| = |∠ KJI| ……Same arc
|∠ GKH| = |∠ JKI|…….Vertically opposite
Therefore, the triangles are equiangular.
Therefore, the triangles are similar.
(b)
7.
(a)
As the triangles HGK and KJI are similar,
|GK| ____
|HK|
____
=
|KJ| |KI|
Therefore, |GK|.|KI| = |HK|.|KJ|
6
7
8
9
10
11
1
0
6
3
2
5
1
0
6
3
1
0
6
3
1
0
6
7
6
7
Key: 6|1 = 6.1
42
0.5
1
1.5
2
2.5
3
3.5
(b)
The distribution is skewed slightly to the right. Median 8.15
(c)
Mean: 8.02
(d)
Range [6.56 km, 9.48 km]
(e)
None of the following factors were taken into consideration:
Standard deviation: 1.46
1. Engine size
2. Makes and model
3. Type of driving (city or motorway)
4. Fuel type (diesel or petrol)
8. (a)
Volume of tank = pr2h
= 3.14(30)2(200)
= 565,200 cm3
= 565 litres
(b)
O
10
A
30
B
C
10 1
cos |∠COB| = ___ = __
30 3
−1
|∠COB| = cos 0.3
|∠COB| = 70.5°
|∠AOB| = 141°
Area of oil face = Area of sector − Area of ΔAOB
141p 1
1
= __ (30)(30) _____ − __
(30)(30) sin 141°
2
180 2
= 1,107.4 − 283.2
= 824.2 cm2
Volume of oil = 824.2 × 200
= 164,840 cm3
= 165 litres
43
(c)
S
P
20
R
30
O
20 2
cos |∠POR| = ___ = __
30 3
−1
|∠POR| = cos 0.66666
= 48.2°
|∠SOR| = 96.4°
Area of empty part = Area of sector − Area of ΔSOR
96.4p 1
1
= __(30)(30) _____ − __ (30)(30) sin 96.4°
2
180
2
= 757.12 − 447.2
= 309.92 cm2
Volume of empty part = 309.92 × 200 = 61,984 cm3
= 62 litres
Volume of oil = Volume of tank − Volume of empty part
= 565 − 62
= 503
Volume of extra oil = 503 − 165
= 338 litres
(d)
O
30
20
L
F
W
From part (b)
Area below |LW| = 309.92 cm2
O
25
30
+++++++++++++++++++++++++++++++++++++++++++
J
H
G
Area below |JH|
44
25
cos |∠GOH| = ___ = 0.833 ⇒ |∠GOH| = cos−10.833
30
|∠GOH| = 33.6°
|∠JOH| = 67.2°
Area below |JH| = Area of sector − Area of ΔJOH
67.2p 1
1
= __(30)(30) _____ − __ (30)(30) sin 67.2°
2
180
2
= 525.8 − 414.8
Area below |JH| = 111 cm2
Area of strip = 309.92 − 111 = 198 cm2
45
Leaving Certificate Higher Level Maths Solutions
Sample Paper 1, no. 3
1. (a)
f (m + 1) − f (m) = 34m + 4 + 24m + 6 −34m −24m + 2
= 34m(34 − 1) + 24m + 2(24 − 1)
= 80 · 34m + 15 · 24m + 2
= 80 · 34m − 1 · 3 + 15 · 24m + 2
= 240 · 34m − 1 + 15 · 24m + 2
= 15[16 · 34m − 1 + 24m + 2]
∴ f (m + 1) − f (m) is divisible by 15
(b)
f (x) = 34x + 24x + 2
Test for n = 1; 34(1) + 24(1) + 2 = 145, which is divisible by 5.
Assume f (k) = 34k + 24k + 2 is divisible by 5.
i.e. f (k) = 5λ
Test if f (k + 1) is divisible by 5.
From (a) f (k + 1) − f (k) is divisible by 15.
f (k + 1) − f (k) = 15m
f (k + 1) = f (k) + 15m
f (k + 1) = 5λ + 15m = 5(λ + 3m)
∴ f (k + 1) is divisible by 5. Using induction, f (x) is divisible by 5.
(c)
We are being asked if f (x) is divisible by 15.
If f (1) is not divisible by 15, then f (2) etc. is not divisible by 15.
f (1) = 145 which is not divisible by 15.
∴ f (x) is not divisible by 15 for positive integers.
2. (a)
w = −1
Im
3i
2i
i
Re
w
–2
–1
1
2
–i
–2i
–3i
46
(b)
arg(w) = p
|w| = 1
w = r (cos(q + 2np) + i sin(q + 2np))
= 1 (cos(p + 2np) + i sin(p + 2np))
= cos(p + 2np) + i sin(p + 2np)
(c)
z3 − w = 0
z3 + 1 = 0
z3 = − 1
z3 = cos(p + 2np) + i sin(p + 2np)
1
_
z = [cos(p + 2np) + i sin(p + 2np)]3
p + 2np
p + 2np
= cos _______ + i sin _______
3
3 __
p
p 1 i√3
n = 0 ⇒ z = cos __ + i sin__ = __ + ___
3
3 2
2
n = 1 ⇒ z = cos p + i sin p = − 1
__
3
i
√
5p
5p
1
n = 2 ⇒ z = cos___ + i sin___ = __ − ___
3
3 2
2
−1, w , −w 2 __
1 i√3
let w = __ + ___
2
2__
__
3
3
i
i
√
√
1
1
__
___
__
___
w2 = +
+
2
2 2 __ 2
1 i23 2i√3
= __ + ___ + ____
4 4
__4
1 3 i√3
= __ − __ + ___
4 4 __2
1 i√3
w 2 = − __ + ___
2 __2
i√ 3
1
−w 2 = __ − ___
2
2
∴ Roots are −1, w , −w 2
(
(
(d)
)
)(
(
)
)
w2 − w + 1 = 0
__
−w 2 + w − 1 = 0?
__
i√3 __
1 i√3
1 − ___
__
+ + ___ − 1 = 0?
2
2
2
2
1 − 1 = 0?
0=0ü
47
3. (a)
y
5
x
7.5
(b)
|2x − 5| ≥ 3
(2x − 5)2 ≥ 9
4x2 − 20x + 25 ≥ 9
4x2 − 20x + 16 ≥ 0
x2 − 5x + 4 ≥ 0
(x − 4)(x − 1) = 0
x=4
or
x=1
y
x
x
4. (a)
1
4
1
x
4
8x2 + 4x3 + 2x4 + …
x
a = 8x2 r = __
2
Tn = arn − 1
x n−1
= 8x2 __
2
8xn + 1
8x2 xn −1 _____
=
= _______
2n −1
2n − 1
n
+
1
8x
= ______
2n · 2−1
16xn + 1
= ______
2n
= 24 − n xn + 1
( )
(b)
The series converges when |r| < 1.
x
__
<1
2
x
−1 < __ < 1
2
− 2 < x < 2 ⇒ |x| < 2
| |
The series converges when |x| < 2.
48
(c)
8x2
a
S∞ = _____ = ______x
1−r 1−
2
=
=
( )
3 2
8 _2
_____
3
1 − _4
72
__
4
___
1
_
4
= 72
5. (a)
A cross-section is created when a 2D plane intersects a 3D shape.
The cross-sectional area is the area of the shape left on the plane, e.g. circle.
(b)
(c)
(150)(490)
sin α ≤ _________
1,32,000
α ≤ 33.8°
s(490)
________
≤1
1,32,000
s ≤ 269.3877551 MPa
6. (a)
(i)
A = 55,000(1 + 0.1275(4))
= 83,050
83,050
Monthly instalment = ______
4 × 12
= €1,730.21
(ii)
55,000 =
[ (
)
]
0.2 −12 × 4
x 1 − 1 + ___
12
______________
x = €1,673.67
0.2
___
12
Option 2 is the better option because monthly repayments are lower.
(iii)
1,673.67 × 48 = 80,336.16
80,336.16 = 55,000(1 + 4i)
1.460657455 = 1 + 4i
0.11516436 = i
∴ Rate = 11.52%
49
(b)
25,000 (1 − (1 + 0.1375)−n)
80,000 = _______________________
0.1375
11
___
= 1 − (1 + 0.1375)−n
25
14
___
= (1 + 0.1375)−n
25
14
log ___ = −n log(1.1375)
25
( )
n = 4.50054779
The money will last four full years.
(c)
Linear
(i)
(ii)
A
(
P
0
(iii)
Must have
P>O
Slope > O
)
n
A = P + (Pi)n
∴ Slope = Pi
A = (Pi)n + P
Therefore, for every increase of 1 in n, A increases by Pi.
7. (a)
1
f (x) = __
x
1
1
____
_
f (x + h) − f (x) _______
x+h − x
____________
=
h
h
−h
______
x(x + h) _______
−1
______
=
=
h
x(x + h)
f (x + h) − f (x)
−1
lim ____________ = _______
h
x(x + 0)
h→0
−1
= ___
x2
(b)
Let f (x) = u(x) + v(x)
f (x + h) = u(x + h) + v(x + h)
f (x + h) − f (x) = u(x + h) + v(x + h) − u(x) − v(x)
= u(x + h) − u(x) + v(x + h) − v(x)
50
u(x + h) − u(x) ____________
v(x + h) − v(x)
f (x + h) − f (x) ____________
____________
=
+
h
h
h
f____________
(x + h) − f (x)
u(x
+
h)
−
u(x)
v(x + h) − v(x)
lim
= lim____________ + lim ____________
h
h
h
h→0
h→0
h→0
dv
du
= ___ + ___
dx dx
(c)
f (x) = x3 − 6x2 + 18x + 5
(i)
f ′(x) = 3x2 −12x +18
= 3(x2 − 4x + 6)
= 3(x2 − 4x + 4 + 2)
= 3((x − 2)2 + 2)
= + (+ + + )
= + ∴ Increasing for all x.
(ii)
f ′(x) = 3x2 −12x + 18
f ″(x) = 6x − 12
6x − 12 = 0
6x = 12
x = 12
x = 2; f (x) = 25
(2, 25)
(iii)
f (x) = x3 − 6x2 + 18x + 5
We know that when x = 2, f (x) = 25.
Test x = 0; then f (x) = 5.
Test x = 5; then f (x) = 70.
Test x = −1; then f (x) = −20.
∴ There is a root between x = 0 and x = −1.
−1
Let xn = ___.
2
f (xn)
xn + 1 = xn − _____
f ′(xn)
−1 −5.625
x2 = ___ − ______
2
24.75
x2 = −0.2727
8. (a)
(i)
∫ 3 sin x dx
−3cos x + c
51
(ii)
2x +
__ 3 dx
∫______
√x
∫(2x + 3x ) dx
1
_
−1
__
2
2
3
1
_
_
2
__
· 2 x2 + 2 . 3x2 + c
3
1
_
4 _2
__
x + 6x2 + c
3
3
(b)
(i)
y2 = 9 − x2
= p∫y2 dx
= p∫9 − x2 dx
(
x3
= p 9x − __
3
)|
[(
3
−3
33
= p 9(3) − __
)] [ (
3
= 18p − [−18p]
3
(−3)
− p 9(−3) − _____
3
= 36p
1
(ii)
e
∫2 _______
dx
x
(e − 1)2
x
Let u = ex − 1.
du
___
= ex
dx
du = ex dx
1
= ∫__2 du
u
= ∫u−2 du
= − u−1
1 2
= − _____
x
e −11
|
[
−1
−1
− _____
= _____
2
1
e −1 e −1
]
= 0.425459064
1
_
(c)
(i)
1
∫0 ______
dx
1 + 9x2
3
1
_
1
∫0 ________
dx
2
1 + (3x)2
3
3x _1
1
3 . __ tan−1 ___ 3
1
1 0
−1
3 tan 1 − 3 tan−1 0
3p
___
−0
4
3p
___
4
|
52
)]
b
(ii)
∫0 1 + x
1
1
_____
dx = __
2
a
1
dx
∫0 _____
1+x
1
∫ _____
dx
1+x
Let u = 1 + x.
du = dx
∫ __1u du = ln u
= ln(1 + x)
1
ln(1 + x ) |0b = __ln(1 + x ) |0a
2
2[ln(1 + b) − ln 1] = ln(1 + a) − ln 1
2 ln(1 + b) − 2 ln 1 = ln(1 + a) − ln 1
ln(1 + b)2 − ln 1 = ln(1 + a)
ln(1 + b)2 = ln(1 + a)
(1 + b)2 = 1 + a
_____
b = ± √1 + a − 1
53
Leaving Certificate Higher Level Maths Solutions
Sample Paper 2, no. 3
1. (a)
2. (a)
Function
Range
Period
What is the Function?
A
[3,−3]
2p
3 cos x
B
[1,−1]
2p
sin x
C
[3,−3]
2p
3 sin x
D
[ 12, − 12 ]
p
1
__
sin 2x
__
__
2
The reaction times seem to be symmetric about 1.5 seconds. About 68% of all
reaction times are between 1.4 and 1.8 seconds. About 95% of all reaction times
are between 1.2 and 2 seconds. About 99.7% of all reaction times are between
1 and 2.2 seconds.
(b)
Percentage: 8.2%
(c)
Percentage: 24.1%
(d)
Interquartile range: Quartiles are 1.38 and 1.62; IQR is 0.24 seconds.
(e)
1.58 seconds or more
3. (a)
4
3
B = (10.5,2.5)
2
1
0
R
–5 –4 –3 –2 –1 0 1 2
–1
P
A = (–3,–2)
3
4
5
6
7
8
–2
–3
–4
(b)
P(1.5, −0.5 )
(c)
Using the distance formula √(x2 − x1)2 + (y2 − y1)2
R(6,1)
_________________
The distance |AP| = |PR| = |RB| = 4.47
54
9 10 11 12 13
4. (a)
sin(A + B) = sin A cos B + cos A sin B
__
( 5 )( 2√3 ) ( 5 )( 2√3 )
2
3 2√__
4 2__
= __ ____
+ __ ____
= 0.9518
(b)
tan A + tan B
tan(A + B) = ____________
1 − tan A tan B
3 ___
2
_
+ __
4
2
2
√
= ____________
3
2__
1− _ + ___
( 4 ) ( 2√2 )
= 3.10 (to two decimal places)
5. (a)
The line can be written as mx − y + c = 0
p
(−4, −4) is on y = mx + c
k
⇒ −4 = m(−4) + c
⇒ 4m − 4 = c
__
m(−4) − (−4) + c
_________
= ______________
= 2√2
√m2 + (−1)2
__ _________
m(−4) − (−4) + c = 2√2 ( √m2 + (−1)2 )
__
−4m + 4 + c = 2√2
(–4,0)
2√2
(–4, –4)
_________
( √m2 + (−1)2 )
But as c = 4m − 4:
__
_________
( √m2 + (−1)2 )
__ _________
0 = 2√2 ( √m2 + (−1)2 )
−4m + 4 + 4m − 4 = 2√2
0 = 4(2)(m2 + 1)
⇒ 8m2 + 8 = 0
⇒ m = ±1
As c = 4m − 4:
when m = 1
when m = −1
c=0
c = −8
y=x
y = −x −8
Equation of k: y = x
Equation of p: x + y − 8 = 0
(b)
Point: (−4, 0), Slope: −1
Point: (−4, 0), Slope: −1:
y − y1 = m(x − x1)
y − y1 = m(x − x1)
y − 0 = −1(x + 4)
y − 0 = 1(x + 4)
y = −x − 4
y=x−4
x+y+4=0
x−y+4=0
55
2√2
6A. Theorem 2 (Isosceles Triangles)
In an isosceles triangle the angles opposite the equal sides are equal.
Converse: If in a triangle two angles are equal then the triangle is isosceles.
Theorem 7
In ΔABC, suppose that |AC| > |AB|. Then |∠ABC| > |∠ACB|. In other words, the angle opposite
the greater of the two sides is greater than the angle opposite the lesser side.
Converse
If |∠ABC| > |∠ACB| then |AC| > |AB|. In other words, the side opposite the greater of two
angles is greater than the side opposite the lesser angle.
Theorem 13
If two triangles ABC and A′B′C′ are similar, then their sides are proportional, in order:
|BC| _____
|CA|
|AB| _____
_____
=
=
|A′B′| |B′C ′| |C ′A′|
Converse
|BC|
|CA|
|AB|
If _____ = _____ = _____, then the two triangles ABC and A′B′C ′ are similar.
|A′B′| |B′C′| |C ′A′|
6B. (a)
In the triangle QMR
|SW| is parallel to |QR|
∴ |QS| : |SM| = |RW| : |WM| ...........Theorem
But |QS| = |SM| ................................Given
∴ |RW| = |WM|
∴ W is the mid-point of RM
(b) In the triangle PQM
|PQ| is parallel to |SV|
∴ |QS| : |SM| = |PV| : |VM| ......... Theorem
But |QS| : |SM| = |RW| : |WM|........Part(a)
∴ |PV| : |VM| |RW| : |WM|
∴ in the triangle PRM
|WV| is parallel to |RP| .................Theorem
(c)
|VW| is parallel to |PR|.............Part(b)
∴ |∠RPS| = |∠SWV| and |∠PRV| = |∠SVW|
∴ The triangle VSW and the triangle SPR are equiangular.
∴ The triangle VSW and the triangle SPR are similar.
∴ |PS| : |SW| = |RS| : |SV|
Rearranging: |PS| : |RS| = |SW| : |SV|
56
But |RS| = |PQ|
∴ |PS| : |PQ| = |SW| : |SV|
But |PS| : |PQ| = 2 : 1
∴ |SW| : |SV| = 2 : 1
∴ |SW| = 2|SV|
(d)
7. (a)
(b)
Value of ratio:
2:1
0
1
2
3
4
5
6
7
9 8 8 7 7 7 5 5 1 8
2 2 1 0 9
10
Key: 3|5 = 35
100
5
0
5
6
5
3
80
7
60
Class A
40
Class B
20
5
5
0
5
0
6
3
4
0
Each block represents 1 student
Class A: Bimodal; median = 43.3
Class B: Negatively skewed; median = 65
(c)
Class B
Reason: Higher median value; all above 80% except for two outliers, 0 and 6.
(d)
(i)
2
Probability of passing is __
3
1
∴ Probability of not passing is __
3
1 30
∴ Probability of 30 students not passing = __
3
2
Probability of passing is __
3
1
∴ Probability of not passing is __
3
30 2 15 __
1
∴ Probability of 15 students passing = ___ __
15 3
3
= 0.0247
()
(ii)
( )( ) ( )
57
15
8. (a)
D
C
O
9 cm
A
12 cm
18 cm
B
A
|AC|2 = 122 + 92
|AO|2 = 182 + 7.52
________
__________
|AC| = √144 + 81
|AO| = √324 + 56.25
____
______
= √380.25 = 19.5 cm
= √225 = 15 cm
(b)
7.5 cm
Using the Cos Rule:
122 = 19.52 + 19.52 − 2(19.5)(19.5) cos|∠AOB|
O
144 = 380.25 + 380.25 − 760.5 cos|∠AOB|
19.5 cm
= 760.5 − 760.5 cos|∠AOB|
19.5 cm
cos |∠AOB| = 0.8107
A
|∠AOB| = cos−1 (0.8107)
B
12 cm
= 35.84°
(c)
Using the Cos Rule:
152 = 19.52 + 19.52 − 2(19.5)(19.5) cos|∠AOC|
O
225 = 380.25 + 380.25 − 760.5 cos|∠AOC|
= 760.5 − 760.5 cos|∠AOC|
19.5 cm
19.5 cm
cos |∠AOC| = 0.7041
|∠AOC| = cos−1 (0.7041)
A
= 45.24°
(d)
C
15 cm
4.5
tan |∠O| = ___
18
tan |∠O| = 0.25
O
|∠O| = 14.04°
18 cm
∴ |∠ROS| = 2(14.04°)
= 28.08°
R
58
4.5 cm
Midpoint of |RS|