2018 Practice Exam 2A Letter - ghsmethods12.weebly.comghsmethods12.weebly.com/uploads/2/4/9/6/24960942/2018-exam-2… · 3 2018 MATHMETH EXAM 2A - TRIUMPH TUTORING Question 6 f(x)

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


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.


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


) 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


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)


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


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


Question 1c

Period =

2⇡

c

Period =

2⇡⇡

12

Period = 24 metres ( 1st mark


Question 1d

15 = 10� 8 sin

⇣⇡x12

⌘

�5

8

= sin

⇣⇡x12

⌘


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


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


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


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


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


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


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


Question 2diii

Find y-value of turning point:

g(x) =49

4

metres ( 1st mark


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


Question 3ci

✓1

2

◆3

=

1

8

( 1st mark



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


Question 3ciii

Pr(G Thur |G0Fri) =

Pr(G Thur \ G0Fri)

Pr(G0Fri)

* 1st mark

Pr(G Thur | G0Fri) =

✓1

2

⇥ 1

2

◆

3

8


a1

4

a

3

8


2

3

( 2nd mark


Question 3di

Use Binomial Pdf in CAS:

n = 12, p =

7

20

, x = 0

) Pr(X = 0) = 0.0057 ( 1st mark


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


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


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


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


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


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


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


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


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


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


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


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


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


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



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


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


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


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2018 Practice Exam 2A Letter - ghsmethods12.weebly.comghsmethods12.weebly.com/uploads/2/4/9/6/24960942/2018-exam-2… · 3 2018 MATHMETH EXAM 2A - TRIUMPH TUTORING Question 6 f(x)