2016 MATHEMATICAL METHODS UNIT 3 & 4

2016 MATHEMATICAL METHODS 3 & 4 EXAM 1 SOLUTIONS

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2016

MATHEMATICAL METHODS

UNIT 3 & 4

Trial examination 1

SOLUTIONS



Question 1 (4 marks)

a.

f x( ) = x2 −5e2x

Use the quotient rule.

′f x( ) = e2x 2x( )− x2 −5( ) 2e2x( )e2x( )2 1 method mark – applying quotient rule

= 2xe2x − 2x2e2x +10e2x

e4x 1 answer mark

=−2e2x x2 − x −5( )

e4x

=−2 x2 − x −5( )

e2x

b. f x( ) = sin 3x( )( )2

Use the chain rule Let u = sin 3x( )

dudx

= 3cos 3x( )

and

and y = u2

dydu

= 2u

= 2sin 3x( )

dydx

= dudx

× dydu

= 3cos 3x( )× 2sin 3x)( )

∴ ′f x( ) = 6cos 3x( )sin 3x( ) 1 answer mark

′f π( ) = 6cos 3π( )sin 3π( )

= 6 −1( ) 0( ) = 0 1 answer mark




a.

1

2x − 3( )2 dx∫ = 2x − 3( )−2dx∫

= − 1

22x − 3( )−1

+ c

1 method mark

= − 1

2(2x − 3)+ c 1 answer mark

b.

f (x)+ 2( )0

4

∫ dx = f (x)0

4

∫ dx + 20

4

∫ dx

= f (x)

0

4

∫ dx + 2x⎡⎣ ⎤⎦0

4 1 method mark

= 6+ 2 4( )− 2 0( )

= 14 1 answer mark Question 3 (4 marks) a. ran f = 0,∞( )

dom g = R 1 mark correct domain and range

0,∞( )⊂ R

∴ran f ⊂ dom g 1 method mark

∴g f (x)( ) exists

b. g f (x)( ) = g 1− x( )

= e 1−x domain = (−∞,1] 1 answer mark (correct rule and domain)

c.

g f 0( )( ) = e . 1 answer mark



Question 4 (3 marks) ex − 3e− x = 2 ex − 3e− x − 2 = 0 Multiply both sides by ex

e2x − 3e0 − 2ex = 0

ex( )2

− 2ex − 3= 0 Let a = ex

a2 − 2a − 3= 0 1 method mark – expressing as a quadratic

a − 3( ) a +1( ) = 0

a = 3 or a = −1

ex = 3 or ex = −1(not possible) 1 method mark

∴ x = ln 3( ) 1 answer mark


2log2 x −1( ) = −2+ log2 2x( )

log2 x −1( )2− log2 2x( ) = −2

log2

x −1( )2

2x

⎛

⎝⎜⎜

⎞

⎠⎟⎟= −2 1 method mark

2−2 =

x −1( )2

2x 1 method mark – writing in exponential form

14=

x −1( )2

2x

x = 2 x −1( )2

x = 2 x2 − 2x +1( )

x = 2x2 − 4x + 2 2x2 −5x + 2 = 0 1 method mark – writing as a quadratic

2x −1( ) x − 2( ) = 0

x = 1

2 or x = 2

The solution is x = 2 as from the original equation

x −1> 0x >1

1 answer mark must include justification



Question 6 (4 marks) a. amplitude = 3 1 answer mark

period = 2π

2

= π 1 answer mark b.

1 correct shape and period 1 correct endpoints



Question 7 (4 marks) a.

Pr bus departs on time( )= Pr T( )

= Pr G∩T( ) + Pr B∩T( )

= 0.7( ) 0.8( ) + 0.3( ) 0.6( ) 1 method mark

= 0.56+ 0.18 = 0.74 1 answer mark

b. Pr weather is good / bus is on time( )

= Pr(G |T )

=Pr G∩T( )

Pr T( )

=

0.7( ) 0.8( )0.74

= 28

37

1 method mark – applying conditional probability 1 answer mark




a. Pr X >12( ) = Pr Z > 12−8

2⎛⎝⎜

⎞⎠⎟

where Z = X −µ

σ

= Pr Z > 2( )

= Pr Z < −2( )

= 0.02 1 answer mark

b.

Pr X < 8 | X <12( ) = Pr X < 8∩ X <12( )

Pr X <12( )

=

Pr X < 8( )Pr X <12( )

=Pr Z < 8−8

2⎛⎝⎜

⎞⎠⎟

Pr Z < 12−82

⎛⎝⎜

⎞⎠⎟

=

Pr Z < 0( )Pr Z < 2( )

= 0.5

1− Pr Z < −2( )

= 0.5

1− 0.02

= 5

98

1 method mark – converting to standard form

1 answer mark



Question 9 (5 marks) a. p̂ = 0.6 1 mark

b. e = 1.96 p̂(1− p̂)n

= 1.96 0.6(0.4)n

1 method mark

= 1.96 0.24

n

= 1.96 24100n

= 1.96 625n

1 answer mark

c. If the sample size is n2

e = 1.96 6

25 n2

⎛⎝⎜

⎞⎠⎟

= 1.96 12n25 1 method mark

1.96 1225n

1.96 625n

= 1225n

× 25n6

= 2 1 answer mark

If the number of people in the sample were halved e is multiplied by 2 .




a. dydx

= 3

3ax2 = 3

x2 = 1a

x = 1a

, x ≥ 0 1 method mark

y = a 1

a⎛⎝⎜

⎞⎠⎟3

= 1a

Coordinates of A are 1a, 1a

⎛⎝⎜

⎞⎠⎟

1 answer mark

b. 11a

ax3( )0

1a∫ dx = 2

8 1 method mark

a ax4

4⎡

⎣⎢

⎤

⎦⎥0

1a= 28

a a4× 1a2

⎛⎝⎜

⎞⎠⎟ =

28

14 a

⎛⎝⎜

⎞⎠⎟= 28

4 a = 82

1 method mark

a = 2 1 answer mark

2016 MATHEMATICAL METHODS UNIT 3 & 4

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