Post on 16-May-2020
2018 Practice Exam 2A Letter
STUDENT NUMBER
MATHEMATICAL METHODS
Written examination 2
Reading time: 15 minutes Writing time: 2 hours
WORKED SOLUTIONS
Structure of book
Section Number of questions
Number of questions to be answered
Number of marks
A 20 20 20 B 5 5 60 Total 80
• Students are permitted to bring into the examination room: pens, pencils, highlighters, erasers, sharpeners, rulers, a protractor, set squares, aids for curve sketching, one bound reference, one approved technology (calculator or software) and, if desired, one scientific calculator. Calculator memory DOES NOT need to be cleared. For approved computer-based CAS, full functionality may be used.
• Students are NOT permitted to bring into the examination room: blank sheets of paper and/or correction fluid/tape.
Materials supplied • Question and answer booklet of 25 pages. • Formula sheet. • Working space is provided throughout the book. Instructions • Write your student number in the space provided above on this page. • Unless otherwise indicated, the diagrams in this book are not drawn to scale. • All written responses must be in English. At the end of the examination • You may keep the formula sheet.
Students are NOT permitted to bring mobile phones and/or any other unauthorised electronic devices into the examination room.
© TRIUMPH TUTORING 2017
2 2018 MATHMETH EXAM 2A - TRIUMPH TUTORING
WORKED SOLUTIONS
2018 Exam 2A - Section A
Question 1
Period of tangent function =
⇡
b
P =
⇡
3⇡
P =
1
3
Answer: A
Video Solution: http://bit.ly/2BkYVKj
Question 2
Original = P (2, �1)
Up = P (2, 2)
Left = P (�2, 2)
Reflected = P (�2, �2)
) Final = P (�2, �2)
Answer: E
Video Solution: http://bit.ly/2Apm15C
Question 3
If f 0(�1) = 0, the gradient is 0 at x = �1.
If f 0(x) > 0 for x 2 R\{1}, the gradient is
positive everywhere else. This means you have
a positive cubic which has a stationary point of
inflection at x = �1.
Answer: C
Video Solution: http://bit.ly/2keR65L
Question 4
f(x) = 3e
2x+1
g(x) = log
e
⇣x3
⌘� 1
g(f(x)) = log
e
✓1
3
(3e
2x+1
)
◆� 1
g(f(x)) = log
e
(e
2x+1
)� 1
g(f(x)) = log
e
(e)� 1
g(f(x)) = (2x+ 1)� 1
g(f(x)) = 2x
Answer: B
Video Solution: http://bit.ly/2zRH41m
Question 5
1.9 2.4 2.9 3.4 3.9 4.4 4.9
a
b
X
Z0 1 2 3-1-2-3
From diagram, a = b.
Pr(�1 < Z < 3) = Pr(2.9 < X < 4.9)
Pr(�1 < Z < 3) = Pr(�3 < Z < 1)
Pr(�1 < Z < 3) = Pr(1.9 < X < 3.9)
Answer: A
Video Solution: http://bit.ly/2ApdAan
3 2018 MATHMETH EXAM 2A - TRIUMPH TUTORING
Question 6
f(x) is the derivative of chosen graph (A).
A has a turning point at x = 0 meaning f(x)has an x-intercept at x = 0.
The gradient of A is negative until x = 0, then
positive from x > 0.
Answer: A
Video Solution: http://bit.ly/2kaDwQM
Question 7
g(x) = 3 log
e
⇣x2
⌘
g(4x) = log
e
(a)
g(4x) = 3 log
e
✓4x
2
◆
g(4x) = 3 log
e
(2x)
g(4x) = log
e
((2x)3)
g(4x) = log
e
(8x3
)
) a = 8x3
Answer: D
Video Solution: http://bit.ly/2Byy26C
Question 8
If graphs intersect, let them equal each other:
2ax� 3 = 2x2 � 4x
2x2 � 4x� 2ax+ 3 = 0
2x2
+ (�4� 2a)x+ 3 = 0
For two points of intersection, � > 0.
) b2 � 4ac > 0
(�4� 2a)2 � (4⇥ 2⇥ 3) > 0
16 + 16a+ 4a2 � 24 > 0
4a2 + 16a� 8 > 0
4(a2 + 4a� 2) > 0
a2 + 4a� 2 > 0
Use CAS: a < �2�p6 or a > �2 +
p6
Answer: D
Video Solution: http://bit.ly/2ArtYaF
Question 9
If the normal has a gradient of �1
2
, the tangent
has a gradient of 2.
) f 0(1) = 2.
Let y = ae
u
, where u = x2
:
dy
dx=
dy
du⇥ du
dx
dy
dx= ae
u ⇥ 2x
dy
dx= 2axe
x
2
dy
dx= 2 at x = 1
2 = 2a(1)e(1)2
2 = 2ea
a =
1
e
Answer: E
Video Solution: http://bit.ly/2Aq5aPX
Question 10
f(x) = 2x3
+ 6x2
+ d
If it has three distinct solutions, the two turning
points are on either side of the x-axis.
We need to find where the turning points are,
so f 0(x) = 0.
f 0(x) = 6x2
+ 12x = 0
0 = 6x(x+ 2)
x = 0, x = �2
Let’s pretend d = 0 for a second.
4 2018 MATHMETH EXAM 2A - TRIUMPH TUTORING
f(x) = 2x3
+ 6x2
f(0) = 0
f(�2) = 2(�2)
3
+ 6(�2)
2
f(�2) = 8
Turning points at (0, 0), (�2, 8)
In order for there to be three x-intercepts, the
graph needs to be moved down by at most 8
units. Therefore, since we’re adding d, d must
be between �8 and 0.
(�2, 8)
(0, 0)x
y
Answer: E
Video Solution: http://bit.ly/2jxIuTF
Question 11
Drawing a Venn Diagram is the easiest way to
visualise this.
A B
x 1
5
x
Since Pr(A [ B) =
8
9
and
Pr(A \ B0) = Pr(A0 \ B),
x+
1
5
+ x =
8
9
2x =
8
9
� 1
5
2x =
31
45
x =
31
90
Pr(A | B) =
Pr(A \ B)
Pr(B)
Pr(A | B) =
✓1
5
◆
✓1
5
+
31
90
◆
Pr(A | B) =
1
5
a49
90
a
Pr(A | B) =
18
49
Answer: D
Video Solution: http://bit.ly/2zQHQeS
Question 12
f(x) = a cos(b(x+ c)) + d
Period =
2⇡
b
5⇡
3
=
2⇡
b
b =6
5
Midpoint of range =
�4 + 3
2
Midpoint of range = �1
2
Distance between midpoint and max:
= 3�✓�1
2
◆
=
7
2
) Amplitude =
7
2
Horizontal translation =
⇡
3
5 2018 MATHMETH EXAM 2A - TRIUMPH TUTORING
) f(x) =7
2
cos
✓6
5
⇣x� ⇡
3
⌘◆� 1
2
Answer: D
Video Solution: http://bit.ly/2kbJCAe
Question 13
Approximation:
Area = A1
+ A2
+ A3
+ A4
Area =
✓1⇥ 1
2
2
◆+
✓1⇥ 2
2
2
◆
Area =+
✓1⇥ 3
2
2
◆+
✓1⇥ 4
2
2
◆
Area =
1
2
+
4
2
+
9
2
+
16
2
Area = 15
Exact: Area =
5Z
0
x2
2
dx
Area =
125
6
(CAS)
% difference =
⇣1� approx
exact
⌘⇥ 100
% difference =
✓1� 15
125
6
◆⇥ 100
% difference =
✓1� 18
25
◆⇥ 100
% difference = 28 % ⇡ 30 %
Answer: D
Video Solution: http://bit.ly/2AfyFUq
Question 14
Confidence Interval = (45%, 60%)
Confidence Interval = (0.45, 0.60)
CI =
p̂� z
rp̂(1� p̂)
n, p̂+ z
rp̂(1� p̂)
n
!
) 0.45 = p̂� z
rp̂(1� p̂)
n
p̂ = 0.525 (halfway between 0.45 and 0.60)
z = 1.96 (known for 95% CI)
) 0.45 = 0.525� 1.96
r0.525(1� 0.525)
n
Solve for n in CAS:
n = 170
Answer: C
Video Solution: http://bit.ly/2katlvB
Question 15
2x� 1
3� x=
2x� 6 + 5
3� x
2x� 1
3� x=
�2(3� x) + 5
3� x
2x� 1
3� x=
�2(3� x)
(3� x)+
5
3� x
2x� 1
3� x= �2 +
5
3� x
) Asymptotes at x = 3 and y = �2
Alternatively, sketch this on your CAS and use
that to find asymptotes.
Answer: D
Video Solution: http://bit.ly/2BAxnSf
Question 16
Turning points at x = �3 and x =
1
2
, and
domain must not ‘go past’ turning points to
allow for a one-to-one inverse function. The
only option that works with [b, 1) is D. Sketch
the graph on CAS to visualise.
Answer: D
Video Solution: http://bit.ly/2AfVOG6
6 2018 MATHMETH EXAM 2A - TRIUMPH TUTORING
Question 17
We need two simultaneous equations to solve
for a and b.
[1] : Area under graph = 1
) 1 =
bZ
0
ax2 dx [1]
[2] : E(X) =
1
2
) 1
2
=
bZ
0
x · ax2 dx [2]
Using ‘Solve Systems of Equations’ in CAS:
a =
81
8
and b =2
3
Answer: C
Video Solution: http://bit.ly/2BvCxih
Question 18
B = Takes a black coin
W = Takes a white coin
B
B2
2+k
Wk
2+k
3
3+k
W
B3
2+k
Wk�1
2+k
k
3+k
Pr(BB) =
1
7
) 1
7
=
✓3
3 + k
◆⇥✓
2
2 + k
◆
Solve in CAS:
k = �9 or k = 4.
Since k > 0,
k = 4
Answer: C
Video Solution: http://bit.ly/2AIg0kZ
Question 19
Chain rule: y = e
2u
u =
Zlog
e
(x) + 1 dx
u = x loge
(x) + c (using our CAS)
Passes through (3, loge
(27)):
log
e
(27) = 3 log
e
(3) + c
log
e
(27) = log
e
(27) + c
c = 0
) u = x loge
(x)
When u = log
e
(4):
log
e
(4) = x loge
(x)
log
e
(4) = log
e
(xx
)
4 = xx
) x = 2
Using our chain rule:
dy
dx=
dy
du⇥ du
dxdy
dx= 2e
2u ⇥ (log
e
(x) + 1)
Method 1:
Substitute u = x loge
(x)
) dy
dx= 2e
2x log
e
(x) ⇥ (log
e
(x) + 1)
) dy
dx= 2e
log
e
(
x
2x) ⇥ (log
e
(x) + 1)
) dy
dx= 2x2x ⇥ (log
e
(x) + 1)
When x = 2:
dy
dx= 2(2)
2(2) ⇥ (log
e
(2) + 1)
7 2018 MATHMETH EXAM 2A - TRIUMPH TUTORING
dy
dx= 32⇥ (log
e
(2) + 1)
dy
dx= 32 log
e
(2) + 32
Answer: A
Method 2:
When u = log
e
(4), x = 2:
dy
dx= 2e
2 log
e
(4) ⇥ (log
e
(2) + 1)
dy
dx= 2e
log
e
(
4
2) ⇥ (log
e
(2) + 1)
dy
dx= 2(16)⇥ (log
e
(2) + 1)
dy
dx= 32 log
e
(2) + 32
Answer: A
Video Solution: http://bit.ly/2Aekjnb
Question 20
Volume = Length ⇥ Width ⇥ Height
where L = 10� 2x
where W = 6� 2x
where H = x
V = (10� 2x)(6� 2x)x
V = 4x3 � 32x2
+ 60x
To find where V is a maximum, V 0(x) = 0
V 0(x) = 12x2 � 64x+ 60
According to CAS:
V 0(x) = 0 when x = 1.2 or x = 4.1
Since 2x < 6, x < 3
) x = 1.2
Answer: B
Video Solution: http://bit.ly/2Al6H7O
8 2018 MATHMETH EXAM 2A - TRIUMPH TUTORING
WORKED SOLUTIONS
2018 Exam 2A - Section B
Question 1a
h(x) = a+ b sin(cx)
Since maximum height = 18 m,
We use a � b = 18, because b is negative
(according to graph).
At point B, h(x) = 2. Since the track is
smooth here, this must be the minimum of
h(x).
If minimum height = 2 m,
We use a + b = 2, because b is negative
(according to graph).
) We have two simultaneous equations:
a� b = 18 [1]
a+ b = 2 [2]
From [1], a = b+ 18
Sub into [2]:
(b+ 18) + b = 2
2b+ 18 = 2
b = �8 ( 1st mark
Since b = �8,
a = �8 + 18
a = 10 ( 2nd mark
We know that from A to B is 30 metres.
This is 1 and
1
4
periods of h(x). To find
what one period is:
P ⇥ 5
4
= 30
P = 30÷ 5
4
P = 24 metres
P =
2⇡
c
24 =
2⇡
c
c =⇡
12
( 3rd mark
a = 10, b = �8, c =⇡
12
Video Solution: http://bit.ly/2BwR3pL
Question 1b
Max height = 18 metres
18 = 10� 8 sin
⇣⇡x12
⌘
�1 = sin
⇣⇡x12
⌘
sin
�1
(�1) =
⇡x
12
3⇡
2
=
⇡x
12
) x = 18 metres at maximum height
* 1st mark
Video Solution: http://bit.ly/2BwR3pL
Question 1c
Period =
2⇡
c
Period =
2⇡⇡
12
Period = 24 metres ( 1st mark
Video Solution: http://bit.ly/2BwR3pL
Question 1d
15 = 10� 8 sin
⇣⇡x12
⌘
�5
8
= sin
⇣⇡x12
⌘
9 2018 MATHMETH EXAM 2A - TRIUMPH TUTORING
sin
�1
✓�5
8
◆=
⇡x
12
⇡x
12
= �0.675, 3.817, 5.608 ( 1st mark
x = �2.58, 14.58, 21.42
Since x > 0, first time h reaches 15 metres is
at:
x = 14.58 ( 2nd mark
Video Solution: http://bit.ly/2BwR3pL
Question 1e
Coaster is only above 15 metres high between
x = 14.58 and x = 21.42.
So, above 15 metres for (21.42 � 14.58)horizontal distance.
Horizontal distance travelled while above 15
metres = 6.84 metres ( 1st mark
Distance travelled below 15 metres:
= 30� 6.84
= 23.16 metres ( 2nd mark
Video Solution: http://bit.ly/2BwR3pL
Question 2a
Starts falling at turning point:
TP = (10, 14)
h(x) = a(x� 10)
2
+ 14 ( 1st markEmily is 30 metres away, so x-int = (30, 0)
0 = a(30� 10)
2
+ 14
�14 = a(400)
a =
�7
200
) h(x) = � 7
200
(x� 10)
2
+ 14 ( 2nd mark
Video Solution: http://bit.ly/2ApimF4
Question 2b
h(0) = � 7
200
(0� 10)
2
+ 14
h(0) =
✓� 7
200
⇥ 100
◆+ 14
h(0) = �7
2
+ 14
h(0) =21
2
metres ( 1st mark
Video Solution: http://bit.ly/2ApimF4
Question 2c
Draw triangle to figure out angle.
A
B✓
✓21
2
30
* 1st mark
✓ = angle of depression
Due to parallel lines, ✓ at A = ✓ at B.
tan(✓) =21
2
30
✓ = tan
�1
✓21
60
◆= 19.29� ( 2nd mark
Video Solution: http://bit.ly/2ApimF4
Question 2di
Create transformation matrix:
T
0
@
2
4x
y
3
5
1
A=
2
42
3
0
0
1
2
3
5
2
4x
y
3
5+
2
4 0
21
4
3
5
* 1st mark
x0=
2
3
x
y0 =1
2
y +21
4
10 2018 MATHMETH EXAM 2A - TRIUMPH TUTORING
x =
3
2
x0
y = 2y0 � 21
2
Sub x and y into h(x) for g(x):
2y � 21
2
=
�7
200
✓3
2
x� 10
◆2
+ 14
* 2nd mark
2y = � 7
200
✓3
2
x� 10
◆2
+
49
2
y = � 7
400
✓3
2
x� 10
◆2
+
49
4
) g(x) = � 7
400
✓3
2
x� 10
◆2
+
49
4
* 3rd mark
Video Solution: http://bit.ly/2ApimF4
Question 2dii
Need to find x-intercept of g(x)
x-int when g(x) = 0
0 = � 7
400
✓3
2
x� 10
◆2
+
49
4
�49
4
= � 7
400
✓3
2
x� 10
◆2
49
4
⇥ 400
7
=
✓3
2
x� 10
◆2
700 =
✓3
2
x� 10
◆2
x =
2
3
(±p700 + 10)
x =
20
3
(1�p7) and x =
20
3
(1 +
p7)
Since x > 0, x =
20
3
(1 +
p7)
x = 24.31 ( 1st mark(30� 24.31) = 5.69
Decreased by 5.69 metres ( 2nd mark
Video Solution: http://bit.ly/2ApimF4
Question 2diii
Find y-value of turning point:
g(x) =49
4
metres ( 1st mark
Video Solution: http://bit.ly/2ApimF4
Question 3a
E(X) = mean =
bZ
a
x · f(x) dx
E(X) =
3Z
1
x
✓1
9
✓x3
3
� 5x2
2
+ 6x
◆◆dx
) E(X) = 2.01 hours ( 1st mark
Video Solution: http://bit.ly/2ApUK31
Question 3b
0.43 =
3Z
n
1
9
✓x3
3
� 5x2
2
+ 6x
◆dx
0.43 =
1
9
x4
12
� 5x3
6
+ 3x2
�3
n
( 1st mark
3.87 =
3
4
12
� 5(3)
3
6
+ 3(3)
2 � n4
12
+
5n3
6
� 3n2
) 0 =
n4
12
� 5n3
6
+ 3n2 � 7.38
According to CAS:
n = �1.32 or 2.15
Since 1 n 3,
n = 2.15 hours ( 2nd mark
Video Solution: http://bit.ly/2ApUK31
Question 3ci
✓1
2
◆3
=
1
8
( 1st mark
Video Solution: http://bit.ly/2ApUK31
11 2018 MATHMETH EXAM 2A - TRIUMPH TUTORING
Question 3cii
G = Goes to the gym
G0= Doesn’t go to the gym
G
G
G1
2
G01
2
1
2
G0G
3
4
G01
4
1
2
FRI
THURWED
* 1st mark
Pr(G0) = Pr(GGG0
) + Pr(GG0G0)
Pr(G0) =
✓1
2
⇥ 1
2
◆+
✓1
2
⇥ 1
4
◆
Pr(G0) =
1
4
+
1
8
Pr(G0) =
3
8
( 2nd mark
Video Solution: http://bit.ly/2ApUK31
Question 3ciii
Pr(G Thur |G0Fri) =
Pr(G Thur \ G0Fri)
Pr(G0Fri)
* 1st mark
Pr(G Thur | G0Fri) =
✓1
2
⇥ 1
2
◆
3
8
Pr(G Thur | G0Fri) =
a1
4
a
3
8
Pr(G Thur | G0Fri) =
2
3
( 2nd mark
Video Solution: http://bit.ly/2ApUK31
Question 3di
Use Binomial Pdf in CAS:
n = 12, p =
7
20
, x = 0
) Pr(X = 0) = 0.0057 ( 1st mark
Video Solution: http://bit.ly/2ApUK31
Question 3dii
Use Binomial Cdf in CAS:
Pr(X > 5) = Bi ⇠ X
✓12,
7
20
◆( 1st mark
lower = 6, upper = 12, n = 12, p =
7
20
) Pr(X > 5) = 0.2127 ( 2nd mark
Video Solution: http://bit.ly/2ApUK31
Question 3ei
p̂ 3
10
= X ✓
3
10
⇥ 20
◆
p̂ 3
10
=X 6
p̂ � 1
10
= X �✓
1
10
⇥ 20
◆
p̂ � 1
10
=X � 2
Pr
�p̂ 3
10
| p̂ � 1
10
�= Pr(X 6 | X � 2)
Pr(X 6 | X � 2) =
Pr(2 X 6)
Pr(X � 2)
* 1st mark
Use Binomial Cdf on CAS to find
Pr(2 X 6):
lower = 2, upper = 6, n = 20, p =
7
20
) Pr(2 X 6) = 0.4145
12 2018 MATHMETH EXAM 2A - TRIUMPH TUTORING
Use Binomial Cdf on CAS to find
Pr(X � 2):
lower = 2, upper = 20, n = 20, p =
7
20
) Pr(X � 2) = 0.9979 ( 2nd mark
) Pr(X 6 | X � 2) =
0.4145
0.9979
) Pr(X 6 | X � 2) = 0.4154
Pr
✓p̂ 3
10
| p̂ � 1
10
◆= 0.4154
* 3rd mark
Video Solution: http://bit.ly/2ApUK31
Question 3eii
Confidence Interval =
p̂� z
rp̂(1� p̂)
n, p̂+ z
rp̂(1� p̂)
n
!
CI =
0
BBB@6
20
� 1.96
vuut6
20
✓1� 6
20
◆
20
,6
20
+ 1.96
vuut6
20
✓1� 6
20
◆
20
1
CCCA
CI = (0.099, 0.501) ( 1st mark
Video Solution: http://bit.ly/2ApUK31
Question 4ai
For points of intersection, let h(x) = g(x).
x2
=
x2
2
+ 2x+ t
0 = �x2
2
+ 2x+ t ( 1st mark
To solve, use quadratic formula, where:
a = �1
2
, b = 2, c = t
x =
�b±pb2 � 4ac
2a
x =
�2±q(2)
2 ��4⇥�1
2
⇥ t�
2
��1
2
�
x =
�2±p4 + 2t
�1
x = 2±p4 + 2t ( 2nd mark
g(2 +p4 + 2t) =
�2 +
p4 + 2t
�2
g(2 +p4 + 2t) = 4 + 4
p4 + 2t+ 4 + 2t
g(2 +p4 + 2t) = 2
�4 + 2
p4 + 2t+ t
�
g(2�p4 + 2t) =
�2�
p4 + 2t
�2
g(2�p4 + 2t) = 4� 4
p4 + 2t+ 4 + 2t
g(2�p4 + 2t) = 2
�4� 2
p4 + 2t+ t
�
) A =
�2�
p4 + 2t, 2
�4� 2
p4 + 2t+ t
��
)B =
�2 +
p4 + 2t, 2
�4 + 2
p4 + 2t+ t
��
* 3rd mark
Video Solution: http://bit.ly/2Bn9mgq
13 2018 MATHMETH EXAM 2A - TRIUMPH TUTORING
Question 4aii
Area =
bZ
a
h(x)� g(x) dx
Area =
2+
p4+2tZ
2�p4+2t
x2
2
+ 2x+ t� x2 dx
Area =
�x3
6
+ x2
+ tx
�2+
p4+2t
2�p4+2t
* 1st mark
Area =
��2 +
p4 + 2t
�3
6
+
�2 +
p4 + 2t
�2
+ t�2 +
p4 + 2t
�!
�
��2�
p4 + 2t
�3
6
+
�2�
p4 + 2t
�2
+ t�2�
p4 + 2t
�!
A =
4
p2 (t+ 2)
32
3
units
2
* 2nd mark
Video Solution: http://bit.ly/2Bn9mgq
Question 4b
At t = 2, Area = k
) k =
4
p2 (2 + 2)
32
3
k =
4
p2 (8)
3
k =
32
p2
3
units
2 ( 1st mark
Video Solution: http://bit.ly/2Bn9mgq
Question 4c
A =
4
p2 (t+ 2)
32
3
dA
dt=
4
p2
3
⇥ 3
2
(t+ 2)
12 ( 1st mark
dA
dt= 2
p2⇥
pt+ 2
dA
dt= 2
p4 + 2t ( 2nd mark
Video Solution: http://bit.ly/2Bn9mgq
Question 4d
log
e
(2) = 2
p2 + 4t
log
e
(2)
2
=
p4 + 2t
4 + 2t =(log
e
(2))
2
4
2t =(log
e
(2))
2
4
� 4
t =(log
e
(2))
2
8
� 2 ( 1st mark
Video Solution: http://bit.ly/2Bn9mgq
Question 5a
f(x) = ke
x � 2
f(0) < 0
) 0 > ke
0 � 2
0 > k � 2
k < 2
Since k > 0 (because the graph is a positive
exponential),
0 < k < 2 ( 1st mark
Video Solution: http://bit.ly/2iqdYi2
14 2018 MATHMETH EXAM 2A - TRIUMPH TUTORING
Question 5bi
P lies at (p, f(p))
Since f(x) = e
x � 2 (after letting k = 1)
f(p) = e
p � 2
) P at (p, e
p � 2) ( 1st mark
Length formula:
L =
q(x
2
� x1
)
2
+ (y2
� y1
)
2
L =
q(p� 0)
2
+ (e
p � 2� 0)
2
L =
pp2 + (e
p � 2)
2
L =
pp2 + e
2p � 4e
p
+ 4 ( 2nd mark
Video Solution: http://bit.ly/2iqdYi2
Question 5bii
If OP is a minimum, L is a minimum. When
L is a minimum,
dL
dp= 0.
L =
pp2 + e
2p � 4e
p
+ 4
Let L =
pu, where u = p2 + e
2p � 4e
p
+ 4
dL
dp=
dL
du⇥ du
dp
dL
dp=
1
2
pu⇥ (2p+ 2e
2p � 4e
p
)
dL
dp=
p+ e
2p � 2e
p
pp2 + e
2p � 4e
p
+ 4
( 1st mark
dL
dp= 0 at minimum:
0 =
p+ e
2p � 2e
p
pp2 + e
2p � 4e
p
+ 4
Solve for p in CAS:
p = 0.52 ( 2nd mark
f(0.52) = e
0.52 � 2
f(0.52) = �0.31
) P lies at (0.52, �0.31) ( 3rd mark
Video Solution: http://bit.ly/2iqdYi2
Question 5biii
L =
pp2 + e
2p � 4e
p
+ 4
L(0.52) =q
(0.52)2 + e
2(0.52) � 4e
0.52
+ 4
L = 0.609 units ( 1st mark
Video Solution: http://bit.ly/2iqdYi2
Question 5ci
P occurs at (0.52, �0.31)
f(x) = e
x � 2
f 0(x) = e
x
f 0(0.52) = e
0.52
f 0(0.52) = 1.69 ( 1st mark
For tangent: y � y1
= m(x� x1
)
y � (�0.31) = 1.69(x� 0.52)
y = 1.69x� 0.89� 0.31
) g(x) = 1.69x� 1.20 ( 2nd mark
Video Solution: http://bit.ly/2iqdYi2
15 2018 MATHMETH EXAM 2A - TRIUMPH TUTORING
Question 5cii
Area =
log
e
(2)Z
0
f(x)� g(x) dx
A =
log
e
(2)Z
0
(e
x � 2)� (1.69x� 1.20) dx
A =
log
e
(2)Z
0
e
x � 1.69x� 0.80 dx
A = [e
x � 0.84x2 � 0.8x]loge
(2)
0
( 1st mark
A =
⇥e
log
e
(2) � 0.84 (loge
(2))
2 � 0.8 (loge
(2))
⇤
�⇥e
0 � 0.84 (0)2 � 0.8 (0)⇤
A = 0.04 units
2 ( 2nd mark
Video Solution: http://bit.ly/2iqdYi2
Question 5di
P = (0.52, �0.31)
Dilation by a factor of 0.43 from the y-axis
(multiply x-value by 0.43):
New point = (0.52⇥ 0.43, �0.31)
New point = (0.22, �0.31) ( 1st mark
Translation of 0.19 in positive direction of
y-axis:
Q = (0.22, �0.31 + 0.19)
Q = (0.22, �0.12) ( 2nd mark
Video Solution: http://bit.ly/2iqdYi2
Question 5dii
Length of OQ:
L =
q(x
2
� x1
)
2
+ (y2
� y1
)
2
L =
qx2
+ (ke
x � 2)
2
According to CAS:
dL
dx=
k2
e
2x � 2ke
x
+ xpk2
e
2x � 4ke
x
+ x2
+ 4
dL
dx= 0 at minimum length of OQ.
Since Q = (0.22, �0.12),
x = 0.22 at
dL
dx= 0
) 0 =
k2
e
2(0.22) � 2ke
(0.22)
+ (0.22)qk2
e
2(0.22) � 4ke
(0.22)
+ (0.22)2 + 4
* 1st mark
Multiply both sides by denominator:
0 = 1.56k2 � 2.50k + 0.22
According to CAS:
k = 0.09 or k = 1.51 ( 2nd mark
) a = 0.09 or a = 1.51
Since a > 1,
a = 1.51 ( 3rd mark
Video Solution: http://bit.ly/2iqdYi2