MATHS METHODS 3 & 4 TRIAL EXAMINATION 1 SOLUTIONS 2009

Question 1

a.

(1 mark)

b. .Now

Since does not exist.(1 mark)

Question 2

a.

(1 mark) – use of product rule(1 mark) – correct derivative

b.

When ,

_____________________________________________________________________© THE HEFFERNAN GROUP 2009 Maths Methods 3 & 4 Trial Exam 1 solutions

(1 mark) use of quotient rule

(1 mark) substituting x = 0

P.O . B o x 11 8 0S u r re y H ill s N o rt h V I C 3 1 2 7

P h o n e 0 3 9 8 3 6 5 0 2 1F a x 0 3 9 8 3 6 5 0 2 5

in fo @ th e h eff er n a n g ro u p .c o m .a uw w w .th e h ef fer n a n g ro u p .c o m .a u

T H E

G R O U PH E F F E R N A N

(1 mark) – correct answer

Question 3

(1 mark) – for

(1 mark) – for remaining 3 correct answersQuestion 4

a. i. Method 1 – using a probability table or Karnaugh map.

This is what is given. This is what we can work out.

(1 mark)

Method 2 – using a Venn Diagram

(1 mark)

Method 3 –using Addition rule

©THE HEFFERNAN GROUP 2009 Maths Methods 3 & 4 Trial Exam 1 solutions

2

B

0 .2 5

0 .4 5

0 .1 5

0 .1 5

A

S

T

A

C

A A’

B 0.15 0.4

B’

0.3 1

A A’

B 0.15 0.25 0.4

B’ 0.15 0.45 0.6

0.3 0.7 1

(1 mark)

ii.

b. If A and B are mutually exclusive then . (1 mark)

Question 5

a.

(1 mark) – correct shape including cusps at (1 mark) correct labelling of intercepts and turning point

b. (1 mark)

c. (1 mark)

Question 6

a. (1 mark)

©THE HEFFERNAN GROUP 2009 Maths Methods 3 & 4 Trial Exam 1 solutions

3

x

y

1 3 5

5

4)4,3(

)( xfy

(1 mark)

©THE HEFFERNAN GROUP 2009 Maths Methods 3 & 4 Trial Exam 1 solutions

4

(1 mark)

b.

(1 mark)Question 7

a. From the diagram,

(1 mark)

b.

Again from the diagram,

by symmetry

(1 mark)

©THE HEFFERNAN GROUP 2009 Maths Methods 3 & 4 Trial Exam 1 solutions

5

(1 mark)

(1 mark)

5 1 0 2 01 5 2 5 3 0 3 510 2 3-3 -2 -1Z

X

Question 8

a.

(1 mark) – correct shape of graph(1 mark) – correct asymptotes

b.

Let

Swap x and y for inverse

Rearrange

So (1 mark) – correct rule

So (1 mark) – correct domain

©THE HEFFERNAN GROUP 2009 Maths Methods 3 & 4 Trial Exam 1 solutions

6

x1

2

y

1y

2x

)( xhy

Question 9

a.

The gradient of the tangent to f at .

Therefore the gradient of the normal to f at is 2. (1 mark)

The equation of the normal through with gradient of 2 is

(1 mark)b. The normal crosses the x-axis when .

Method 1

(1 mark) (1 mark)

©THE HEFFERNAN GROUP 2009 Maths Methods 3 & 4 Trial Exam 1 solutions

7

(1 mark)

Method 2

x2

y

)( xfy

2

3

Question 10

Method 1

Given

(1 mark)

(1 mark) correct antiderivative of (1 mark) correct answer

Method 2

Given

(1 mark) correct antiderivative of (1 mark) correct answer

©THE HEFFERNAN GROUP 2009 Maths Methods 3 & 4 Trial Exam 1 solutions

8

(1 mark)

Question 11

a.

(1 mark)

b.

(1 mark)

c.

Max/min occur when .

(1 mark)d. Method 1 – using part b.

This is the equation of an inverted parabola which has a local maximum and hence there will be a maximum rather than a minimum value to be found.

(1 mark) for reference to inverted parabolaMethod 2 – using a sign diagram

At .

At

From the sign diagram we see that we have a maximum surface area.(1 mark)

Total 40 marks

©THE HEFFERNAN GROUP 2009 Maths Methods 3 & 4 Trial Exam 1 solutions

9

(1 mark)

from part a.

(1 mark)

2 01 0

4

1 0 0