vce mathematical methods (cas) units 3 & 4 - … · vce mathematical methods (cas) units 3...

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Trial Examination 2010

VCE Mathematical Methods (CAS) Units 3 & 4

Written Examination 1

Suggested Solutions

VCE Mathematical Methods (CAS) Units 3 & 4 Trial Examination 1 Suggested Solutions

Copyright © 2010 Neap TEVMMU34EX1_SS_2010.FM 2

Question 1

a. y-intercept (x = 0)

A1

b. x-intercepts (y = 0)

M1

OR M1

A1

Question 2

a. ,

thus A1

(Alternatively, using the product rule directly gives , which is

equivalent to the answer above.)

f 0( ) sin2 π2---–

1–=

1–( )2 1– =

0 =

sin2 2x π2---–

1– 0=

sin2 2x π2---–

1=

2x π2---–

sin 1±=

x 0 2π,[ ]∈2x 0 4π,[ ]∈∴

2xπ2--- π

2---–

7π2

-------,∈–∴

2x π2---–

sin 1–=

2x π2---–

π2--- 3π

2------- 7π

2-------, ,–=

2x 0 2π 4π,,=

x 0 π 2π, ,=

2x π2---–

sin 1=

2x π2---–

π2--- 5π

2-------,=

2x π 3π,=

xπ2--- 3π

2-------,=

x 0π2--- π

3π2

------- 2π, , , ,=

y x2loge 1x---

x2loge x 1–( ) x2loge x( )–= = =

dydx------ 2x– loge x( ) x2 1

x---⋅–⋅ 2xloge x( ) x––= =

dydx------ 2xloge

1x---

x2 1–x

------ +=



b. For stationary points we require .

gives x = 0, M1

Clearly so a stationary point exists if .

Now

Thus the stationary point occurs at . A1

Sign of the derivative test: Use suitable values such as for , choose and for

, choose .

, M1

,

∴ a maximum turning point occurs at . A1

Question 3

a. f is a parabola with turning point (2, 4).

∴ maximum value of a is 2, so f is one-to-one. A1

b. inverse M1

A1

dydx------ 0=

x 1 2loge x( )+( )– 0= loge x( ) 12---–=

x 0> loge x( ) 12---– x⇒ e

12---–

= =

y e

12---–

2

loge× 1

e

12---–

-------

=

e 1– 12---× =

12e------ =

e

12---– 1

2e------,

x e

12---–

< x e 1–=

x e

12---–

> x 1=

x e 1–= dydx------ e 1– 1 2logee 1–+[ ]×– e 1– 1 2–( )× 0>–= =

x 1= dydx------ 1 1 2loge1+[ ]×– 1 1 0<×–= =

e

12---– 1

2e------,

x y 2–( )2 4+=

x 4– y 2–( )2=

y 2– x 4–±=

y 2 x 4– (as y ∞ 2 ) i.e. domain of f ),–(∈–=

f 1– x( )∴ 2 x 4––=



c. rule of

domain of

∴ graph is: A1

Question 4

a.

correct identification of f and g A1

coordinates of P A1

b. Given , then . M1

Thus . A1

f 1– f x( )( ) x=

f 1– f x( )( ) domain f ∞ 2 ),–(= =

y

x0

2

2

(2, 2)

y

xO

1.0

2.0

P

π2---π

4---

f x( ) xtan=

g x( ) 2 xcos=

π4--- 1,

y loge x( )cos( )= dydx------ x( )sin–

x( )cos------------------ x( )tan–= =

x( )tan xd∫ log– e x( )cos( )=



c. Area A1

M1

(or equivalent: or ) A1

Question 5

As events A and B are independent, . A1

Also, events will be independent so we know .

From the question, .

Thus , giving M1

A1

Question 6

Multiplying by gives: M1

A1

2 x( )cos x( )tan–( ) xd0

π4---

∫ =

2 x( )sin loge x( )cos( )+[ ]0

π4---

=

2π4---

sin logeπ4---

cos + 2 0sin loge 0cos+[ ]– =

2 1

2------- loge

1

2-------

+× 2 0× loge 1( )+[ ]– =

1 loge 2( )–= 1 loge1

2-------

+ 1 12---– loge 2( )

Pr A B( ) Pr A( ) 34--- Pr A ′( )⇒ 1

4---= = =

A ′ and B ′ Pr A ′ B ′∩( ) Pr A ′( ) Pr× B ′( )=

Pr B( ) Pr A ′ B ′∩( )=

Pr B( ) 14--- 1 Pr B( )–( )= 4 Pr B( ) 1 Pr B( )–( ) 5 Pr B( )⇒ 1= =

Pr B( ) 15---=

ex 4– 5e x–=

ex

ex( )2

4ex– 5=

ex( )2

4ex– 5– 0=

ex 5–( ) ex 1+( ) 0=

ex 5 or –1, ex 1–≠=

x loge5=



Question 7

We require A1

M1

A1

Question 8

M1

Equation of tangent

A1

Hence to pass through (0, 0):

A1

Question 9

a.

∴ image is (0, –1). A1

Pr T 6 T 3≥≤( ) Pr 3 T 6≤ ≤( )Pr T 3≥( )

--------------------------------=

t–50------ 1

5---+ td

3

6

∫1 Pr T 3<( )–-------------------------------- =

t2–100--------- t

5---+

3

6

1 51100---------–

--------------------------

36–100--------- 6

5---+

9–100--------- 3

5---+

–

49100---------

------------------------------------------------------= =

27–100--------- 3

5---+

49100---------

------------------- = 33

100--------- 100

49---------× 33

49------= =

f a( ) a2 k+=

f ′ x( ) 2x=

f ′ a( ) 2a=

y a2 k+( )– 2a x a–( )=

y 2ax a2– k+=

a2– k+ 0=

k a2=

k 0≥∴

X ′ TX B+=

x'

y'

2 0

0 1–

2

3

4–

2+=

4

3–

4–

2+ =

0

1– =



b.

M1

M1

becomes

A1

∴ required image is .

Alternatively, a more elegant solution which avoids the need to find an inverse matrix is shown below:

M1

⇒ Μ1

Adding gives .

∴ required image is . A1

c. A dilation of factor 2 from the y-axis and a reflection in the y-axis A1

followed by translations of 4 units to the left and 2 upwards. A1

Note: the order of the first two transformations could be exchanged and similarly, the order of the translations is not significant, however the translations must follow the dilation and reflection.

x ′y ′

2 0

0 1–

x

y

4–

2+=

x ′y ′

4–

2– 2 0

0 1–

x

y=

12–

------ 1– 0

0 2

x ′ 4+

y ′ 2–

12–

------ 1– 0

0 2

2 0

0 1–

x

y=

1 0

0 1

x

y=

x

y

12–

------ 1– 0

0 2

x ′ 4+

y ′ 2–=∴

x∴ x ′ 4+2

-------------=

y y ′– 2+=

y∴ 2x 4–= y ′– 2+ 2x ′ 4+

2-------------

4–=

y ′ 2 x ′–=

y 2 x–=

TX B+ 2 0

0 1–

x

2x 4+

4–

2+=

2x

2x– 4+

4–

2+=

X ′∴ 2x 4–

2x– 6+=

x ′ 2x 4–=

y ′ 2x– 6+=

x ′ y ′+ 2=

x y+ 2=



Question 10

a. Approximate change M1

A1

b. The gradient function at is positive and increasing. Therefore the approximation of the change is less than the exact change. A1

V x h+( ) V x( )– =

hV ′ x( ) =

V 6x3, so V ′ x( ) 18x2==

x 3, h 0.02: hV ′ x( ) 0.02V ′ 3( )= = =

0.02 18 32××=

3.24=

3.24 cm3 =

x 3=

vce mathematical methods (cas) units 3 & 4 - … · vce mathematical methods (cas) units 3...

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