VCE Mathematical Methods (CAS) Units 3 & 4 - Cyberchalky!

Trial Examination 2010

VCE Mathematical Methods (CAS) Units 3 & 4 Written Examination 1

Suggested Solutions

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

Question 1 a.

y-intercept (x = 0) 2 π f ( 0 ) = sin  – --- – 1  2 2

= ( –1 ) – 1 A1

=0 b.

x-intercepts (y = 0) 2 π sin  2x – --- – 1 = 0  2 2 π sin  2x – --- = 1  2

π sin  2x – --- = ± 1  2

x ∈ [ 0, 2π ] ∴2x ∈ [ 0, 4π ] 7π ∴2x – π --- ∈ – π ---, ------2 2 2

π sin  2x – --- = – 1  2 π 3π 7π π 2x – --- = – ---, -------, ------2 2 2 2

M1

OR sin  2x – π --- = 1  2

M1

π 5π --- = ---, ------2x – π 2 2 2 2x = π, 3π

2x = 0, 2π, 4π

π 3π x = ---, ------2 2

x = 0, π, 2π 3π π x = 0, ---, π, -------, 2π 2 2

A1

Question 2 a.

1 2 2 –1 2 y = x log e  ---  = x log e ( x ) = – x log e ( x ) ,  x 2 1 thus dy ------ = – 2x ⋅ log e ( x ) – x ⋅ --- = – 2xlog e ( x ) – x x dx

A1

1 2 –1 dy (Alternatively, using the product rule directly gives ------ = 2xlog e  ---  + x  ------ , which is    x x dx equivalent to the answer above.)

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2


b.

dy For stationary points we require ------ = 0 . dx 1 – x ( 1 + 2log e ( x ) ) = 0 gives x = 0, log e ( x ) = – --2

M1 – 1---

2 --- ⇒ x = e . Clearly x > 0 so a stationary point exists if log e ( x ) = – 1 2 1 2    – --2-  1   Now y =  e  × log e  ------1-  – --   e 2

1 –1 = e × --2 1= ----2e 1

 – --2- 1  Thus the stationary point occurs at  e , ------ . 2e 

A1 1 – --2

Sign of the derivative test: Use suitable values such as for x < e , choose x = e

–1

and for

– 1--2

x > e , choose x = 1 . –1 –1 –1 –1 ------ = – e × [ 1 + 2log e e ] = – e × ( 1 – 2 ) > 0 x = e , dy dx

x = 1,

M1

dy ------ = – 1 × [ 1 + 2log e 1 ] = – 1 × 1 < 0 dx 1

 – --2- 1  ∴ a maximum turning point occurs at  e , ------ . 2e 

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.

b.

2

x = (y – 2) + 4

inverse

x – 4 = (y – 2)

A1 M1

2

y–2=± x–4 y = 2 – x – 4 (as y ∈ ( – ∞, 2 ) i.e. domain of f ) –1

∴ f (x) = 2 – x – 4


A1


3


c.

–1

rule of f ( f ( x ) ) = x –1

domain of f ( f ( x ) ) = domain f = ( – ∞, 2 ) ∴ graph is:

A1

y (2, 2)

2

0

x 2

Question 4 a.

y 2.0

f ( x ) = tan x

π P  ---, 1 4

1.0

g ( x ) = 2 cos x

O

π --4

π --2

x

correct identification of f and g A1 coordinates of P A1 b.

sin ( x -) = – tan ( x ) . ----------------Given y = log e ( cos ( x ) ) , then dy ------ = – dx cos ( x ) Thus

∫

tan ( x ) dx = – log e ( cos ( x ) ) .


M1

A1


4


c.

Area =

∫

π --4

( 2 cos ( x ) – tan ( x ) ) dx

A1

0

= [ 2 sin ( x ) +

π --4 log e ( cos ( x ) ) ] 0

=

π π 2 sin  ---  + log e  cos  ---   – [ 2 sin 0 + log e cos 0 ]   4   4

=

1 1 2 × ------- + log e  -------  – [ 2 × 0 + log e ( 1 ) ]  2 2

1 1 = 1 – log e ( 2 ) (or equivalent: 1 + log e  -------  or 1 – --- log e ( 2 ) )  2 2

M1

A1

Question 5 3 1 As events A and B are independent, Pr ( A B ) = Pr ( A ) = --- ⇒ Pr ( A ′) = --- . 4 4

A1

Also, events A ′ and B ′ will be independent so we know Pr ( A ′ ∩ B ′) = Pr ( A ′) × Pr ( B ′) . From the question, Pr ( B ) = Pr ( A ′ ∩ B ′) . 1 Thus Pr ( B ) = --- ( 1 – Pr ( B ) ) , giving 4 Pr ( B ) = ( 1 – Pr ( B ) ) ⇒ 5 Pr ( B ) = 1 4

M1

Pr ( B ) = 1--5

A1

Question 6 x

e – 4 = 5e

–x x

Multiplying by e gives: x 2

M1

x

( e ) – 4e = 5 x 2

x

( e ) – 4e – 5 = 0 x

x

(e – 5 )(e + 1 ) = 0 x

x

e = 5 or –1, e ≠ – 1

A1

x = log e 5



5


Question 7 Pr ( 3 ≤ T ≤ 6 ) We require Pr ( T ≤ 6 T ≥ 3 ) = -------------------------------Pr ( T ≥ 3 )

∫

A1

6

–t 1 6 2 ------ + --- dt 36 6 –9 3 – –t t --------- + --- –  --------- + --- --------- + --50 5    100 5 100 5 100 5 3 = ------------------------------- = -------------------------3- = -----------------------------------------------------1 – Pr ( T < 3 ) 51 49 1 – ----------------100 100 – 27 3 --------- + --33- 100 100 5- = ------------× --------- = 33 = -----------------100 49 49 49 --------100

M1

A1

Question 8 2

f(a) = a + k f ′( x ) = 2x f ′( a ) = 2a

M1 2

Equation of tangent y – ( a + k ) = 2a ( x – a ) 2

A1

y = 2ax – a + k 2

Hence to pass through (0, 0): – a + k = 0 k=a

2

∴k ≥ 0

A1

Question 9 a.

X ′ = TX + B x' = 2 0 2 + – 4 y' 0 –1 3 2 = 4 + –4 –3 2 = 0 –1 ∴ image is (0, –1).


A1


6


x ′ = 2 0 x + –4 y′ 0 –1 y 2

b.

x ′ – –4 = 2 0 x y′ 2 0 –1 y 1 1 ------ – 1 0 x ′ + 4 = ------ – 1 0 2 0 x –2 0 2 y ′ – 2 –2 0 2 0 –1 y = 1 0 x 0 1 y 1 ∴ x = ------ – 1 0 x ′ + 4 y –2 0 2 y ′ – 2

M1

′+ 4 ∴x = x------------2

M1

y = – y′+ 2 x′+ 4 ∴y = 2x – 4 becomes – y ′ + 2 = 2  ------------- – 4 2 y′= 2 – x′

A1

∴ required image is y = 2 – x . Alternatively, a more elegant solution which avoids the need to find an inverse matrix is shown below: x TX + B = 2 0 + –4 0 – 1 2x + 4 2 =

∴X ′ =

2x + –4 – 2x + 4 2

M1

2x – 4 – 2x + 6

⇒ x ′ = 2x – 4

Μ1

y ′ = – 2x + 6 Adding gives x ′ + y ′ = 2 .

c.

∴ required image is x + y = 2 .

A1

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.



7


Question 10 a.

Approximate change = V ( x + h ) – V ( x )

M1

= hV ′( x ) 3

V = 6x , so V ′( x ) = 18x

2

x = 3, h = 0.02: hV ′( x ) = 0.02V ′( 3 ) = 0.02 × 18 × 3

2

= 3.24 = 3.24 cm b.

3

The gradient function at x = 3 is positive and increasing. Therefore the approximation of the change is less than the exact change.



A1 A1

8