_____________________________________________________________________ © THE HEFFERNAN GROUP 2012 Maths Methods CAS 3 & 4 Trial Exam 1 solutions

MATHS METHODS (CAS) 3 & 4 TRIAL EXAMINATION 1 SOLUTIONS 2012 Question 1

a. y = e2 x

x2 +3

22

222

)3(22)3(

+

−×+=

xxeex

dxdy xx

(quotient rule)

(1 mark) Avoid “further engagement” errors by leaving this fraction as it is. No like terms are obtained by expanding the brackets in the numerator.

b. f (x) = 2x tan x3!

"#$

%&

323

8

32141

32

32411

32

32

21

13

2

3tan2

3cos

13

23

tan23

sec3

2)('

rule)product (3

tan23

sec312)('

2

2

2

2

+π

=

+××π

=

+÷×π

=

+

⎟⎠

⎞⎜⎝

⎛×

π=

⎟⎠

⎞⎜⎝

⎛ π+

⎟⎠

⎞⎜⎝

⎛ π×

π=

⎟⎠

⎞⎜⎝

⎛ π+⎟

⎠

⎞⎜⎝

⎛ ππ=π

⎟⎠

⎞⎜⎝

⎛+⎟

⎠

⎞⎜⎝

⎛×=

f

xxxxf

(1 mark)

(1 mark)

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©THE HEFFERNAN GROUP 2012 Maths Methods CAS 3 & 4 Trial Exam 1 solutions

2

Question 2 a. sin(3x −1) dx∫

= −13cos(3x −1)

(1 mark) Note, because we are looking for ‘an’ antiderivative, the constant of antidifferentiation c does not have to be included because “an” antiderivative could be the case where c = 0 . Marks would not be lost though if c was included.

b. 12x +1

dx = 12

22x +1

dx0

2

∫0

2

∫

0)1(log since )5(log21

1log2114log

21

12log21 2

0

==

−+=

⎥⎦

⎤⎢⎣

⎡+=

ee

ee

e x

(1 mark) Question 3 a. f (x) = loge(x), x > 0

)()(log1)(log

0,1log1

1

ufu

u

uuu

f

e

e

e

−=

−=

=

>⎟⎠

⎞⎜⎝

⎛=⎟

⎠

⎞⎜⎝

⎛

−

as required. (1 mark)

b. Given ,, and )3()2()(2 +∈+= Rvuvfvfuf

6 so

,but 6 So,

6

)6(log)(log

)3(log)2(log)(log2

22

22

vu

Rvuvu

vu

vu

vvu

ee

eee

=

∈±=

=

=

+=

+

(1 mark)

(1 mark)

(1 mark)

©THE HEFFERNAN GROUP 2012 Maths Methods CAS 3 & 4 Trial Exam 1 solutions

3

Question 4

a. g :R→ R, g(x) = 3+ sin 2x3

"

#$

%

&'

π=

×π=

÷π=

=π

=

3232

322

32 where2period n

n

(1 mark)

The amplitude of the function is 1 and the graph of y = sin 2x3

!

"#

$

%& is translated 3 units

up to become the graph of ⎟⎠

⎞⎜⎝

⎛+=

32sin3 xy .

The range of g is therefore [2, 4] . (1 mark)

b. cos(3x) = 32, x ∈ R

1832

623

,26

,26

3

π±

π=

π±π=

∈π+π

π+π

−=

nx

nx

Znnnx

(1 mark) – correct base angles (1 mark) – correct use of πn2

(1 mark) – correct answer Question 5

x 0 1 2 3

Pr(X = x) 0.2 0.1 0.4 p

a. 0.2+ 0.1+ 0.4+ p =1

p = 0.3 (1 mark)

b. 4.01.02.01.02.0)31Pr(++

+=<≤ XX

(1 mark)

=0.30.7

=37

(1 mark) c. E(X) = 0×0.2+1×0.1+ 2×0.4+3×0.3 (1 mark)

= 0.1+ 0.8+ 0.9=1.8

(1 mark)

S

T

A

C

©THE HEFFERNAN GROUP 2012 Maths Methods CAS 3 & 4 Trial Exam 1 solutions

4

Question 6

)1(21)(,: −−=→ xexfRRf Let )1(21 −−= xey Swap x and y for inverse. )1(21 −−= yex Rearrange

)1(log211

1)1(log21

)1(2)1(log1

1)1(2

)1(2

xy

yx

yxex

ex

e

e

e

y

y

−+=

−=−

−=−

=−

−=−−

−

So )1(log211)(1 xxf e −+=−

(1 mark)

To find the domain of f −1 , Method 1

ff rd =−1

Do a quick sketch of y = f (x) . So )1,( so and )1,( 1 −∞=−∞= −ff dr

(1 mark)

Method 2

)1,( So1

1defined be ofunction t log for the 01

),1(log211)( Since

1

1

−∞=

<

−>−

>−

−+=

−

−

f

e

dxxx

xxf

(1 mark) Note that to find the inverse function f −1 you must give the domain and the rule.

(1 mark)

x

y

1=y

)(xfy =

©THE HEFFERNAN GROUP 2012 Maths Methods CAS 3 & 4 Trial Exam 1 solutions

5

Question 7 Method 1

⎥⎦

⎤⎢⎣

⎡

−+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡

31

4002

''

yx

yx

where )','( yx is the image point

3412340'102'

−=+=

−+=++=

yxyxyyxx

(1 mark)

43'

21'

43'21' Rearrange+

=−

=

=+=−

yyxx

yyxx

Substitute into x

y 1=

31'8'

1'83'

1'21

43'

21'1

43'

−−

=

−=+

−×=

+

−÷=

+

xy

xy

xy

xy

So the image equation is y = 8x −1

−3 (1 mark)

RE-READ THE QUESTION!

Since y = ax + b

+ c, 3 and 1,8 −=−== cba .

(1 mark) Method 2

⎥⎦

⎤⎢⎣

⎡

−+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡

31

4002

''

yx

yx

We have a dilation by a factor 2 from the y-axis

so xx

yx

y 2

2

1 becomes 1=== .

We have a dilation by a factor of 4 from the x-axis

so .8 so 24

becomes 2x

yx

yx

y === (1 mark)

We have a translation of 1 unit to the right and 3 units down

so 31

8 so 1

83 becomes 8−

−=

−=+=

xy

xy

xy

So the image equation is y = 8x −1

−3 (1 mark)

RE-READ THE QUESTION!

Since y = ax + b

+ c, 3 and 1,8 −=−== cba . (1 mark)

©THE HEFFERNAN GROUP 2012 Maths Methods CAS 3 & 4 Trial Exam 1 solutions

6

Question 8

Pr(A) = 13, Pr(B) = 1

2

a. If A and B are mutually exclusive then Pr(A∩B) = 0 Method 1 – Venn diagram

Note: 13+

12

=26+

36

=56

)'Pr( BA∪ is shaded.

(1 mark)

21

211)'Pr( So =−=∪ BA

(1 mark) Method 2 – Addition formula

21

31

21

31

)'Pr()'Pr()Pr()'Pr(

=

−+=

∩−+=∪ BABABA

(1 mark) b. If A and B are independent events then Pr(A)×Pr(B) = Pr(A∩B)

Method 1 – Venn diagram

So Pr(A∩B) = 13×

12

=16

(1 mark) )'Pr( BA∪ is shaded.

32

62

61

61)'Pr( So

=

++=∪ BA

(1 mark) Method 2 – Addition formula

32

61

63

62

61

21

31

)'Pr()'Pr()Pr()'Pr(

=

−+=

−+=

∩−+=∪ BABABA

(1 mark)

31

61

A

B

21

61

61

A

62

62

B

62

(1 mark)

(1 mark)

©THE HEFFERNAN GROUP 2012 Maths Methods CAS 3 & 4 Trial Exam 1 solutions

7

Question 9 Pr(June shops locally exactly once in the coming 3 weeks)= Pr(L,M,M )+Pr(M,L,M )+Pr(M,M,L)= 0.4×0.7×0.6+ 0.6×0.4×0.7+ 0.6×0.6×0.4= 0.168+ 0.168+ 0.144= 0.48

(1 mark) Question 10

y = x35 + c

dydx

=35x−25

At x =1 , dydx

=35×1

=35

The gradient of the tangent at x =1 is 35

so the gradient of the normal at x =1 is −53

.

(1 mark) The normal passes through the point )0,(a which is it’s x-intercept and through the point of

normalcy which occurs at 1=x . At ccyx +=+== 11 , 1 53

. So the point of normalcy is )1,1( c+ . The gradient of the normal is therefore given by

583

3853355

)1(3)1(511

35 So

11

101

12

12

+=

+=

+=+−

+=−−−

+=

−−

+=

−

−+=

−

−

ca

cacacaacacac

xxyy

(1 mark)

(1 mark)

(1 mark)

(1 mark)

©THE HEFFERNAN GROUP 2012 Maths Methods CAS 3 & 4 Trial Exam 1 solutions

8

Question 11 The graphs intersect when 2sin(x) = 2cos(x)

sin(x) = cos(x)sin(x)cos(x)

=1

tan(x) =1

x = π4

, 5π4

(1 mark)

[ ]

units square 24242

22

222

21

21

21

212

4sin

4cos

45sin

45cos2

)sin(2)cos(2

))cos(2)sin(2(

))()((area shaded

45

4

45

4

45

4

=

−×−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−−=

⎭⎬⎫

⎩⎨⎧

⎟⎟⎠

⎞⎜⎜⎝

⎛+−⎟⎟

⎠

⎞⎜⎜⎝

⎛−−−=

⎭⎬⎫

⎩⎨⎧

⎟⎟⎠

⎞⎜⎜⎝

⎛ π+⎟

⎠

⎞⎜⎝

⎛ π−⎟

⎠

⎞⎜⎝

⎛ π+

π−=

−−=

−=

−=

π

π

π

π

π

π

∫

∫

xx

dxxx

dxxgxf

(1 mark)

S

T

A

C

xO

y

4π

2π

23π

45π

)(xgy =

)(xfy =

(1 mark) for correct terminals of integration (1 mark) for correct integrand

(1 mark)