MASTER CLASS PROGRAM UNIT 4 — MATHEMATICAL METHODS SEMESTER TWO 2014

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MASTER CLASS PROGRAM
UNIT 4 — MATHEMATICAL METHODS
SEMESTER TWO 2014
WRITTEN EXAMINATION 2 — SOLUTIONS
SECTION 1 — MULTIPLE CHOICE QUESTIONS
QUESTION 1
Answer is D


1
2
1 
2 
Graph of f ( x) = −( x − 1)(2 x − 1) is positive for x values  − ∞,  ∪  ,1 .
2
QUESTION 2
Answer is A
Gradient of 2 y + 3 x − 6 = 0 is −
3
2
so perpendicular gradient is
2
3


Equation of line through point 1,−
Answer y =
3
3 2
 is y + = ( x − 1)
2
2 3
2
13
x−
3
6
 The School For Excellence 2014
Unit 4 Master Classes – Maths Methods – Exam 2
Page 1
QUESTION 3
Sketch f
−1
Answer is D
and f : (− ∞,3) → R, f ( x) = − x 2 + 6 x together, noting domain and range.
Answer f −1 : (− ∞,9) → R, f −1 ( x) = 3 − 9 − x
QUESTION 4
Answer is D
f ( x) = a log e (bx − c)
c

f ( x) = a log e b x − 
b

The function has a vertical asymptote at x =
QUESTION 5
c
c 
, therefore the domain is  , ∞  .
b
b 
Answer is D
2
2b 2
 x−a
Substitute f 
 into f ( x) = 2 giving y =
3x
3( x − a) 2
 b 
QUESTION 6
Answer is C
In the polynomial P( x) = ax 3 − bx + c solve simultaneously using
f (−2) = 0, f (−1) = 1, f (−3) = 2
QUESTION 7
Answer is B
Asymptotes for y = −
3
1
− 6 are x = , y = −6
3x − 1
3
Page 2
QUESTION 8
Answer is E
(
)
(
f / ( x) when f ( x) = e − x x 2 + 2 . Answer is f / ( x) = − e − x x 2 − 2 x + 2
QUESTION 9
)
Answer is B
g (x) when g / ( x) = e − x + cos(2 x) . Answer is − e − x +
1
sin(2 x)
2
Page 3
QUESTION 10
Answer is D
There are 3 solutions to the equation {x : 2 sin(3 x) = cos(3 x)} for x ∈ [− π ,0] based on angle
x=
1
1
tan −1 , altered for the given domain.
3
2
Answer x =
1 2π
1
1 π
1
1
1
tan −1 − π , x = tan −1 − , x = tan −1 −
2 3
2
2 3
3
3
3
QUESTION 11
If f ( x ) =
Answer is E
x and g ( x ) = 1 − x then g ( f ( x)) = 1 − x so g ( f (9)) = −2
QUESTION 12
Answer is C
 − 1 0    x   − 2 
T : R2 → R2 : 
    +    gives the equations
0 2   y  1  
− ( x − 2) = x new giving x = 2 − x new
2( y + 1) = y new
Substitute in y =
giving
y=
x giving
y new
−1
2
ynew
− 1 = 2 − xnew
2
Page 4
QUESTION 13
Answer is A
Cusp at point (1, 3) . For there to be 2 solutions, m must be greater than 3.
QUESTION 14
Answer is E
y = log e (2 x − 1) area bounded by the x-axis and the lines with equations x = 1 and x = 3 .
Answer is not a decimal approximation. Answer
5 log e 5 − 4
.
2
Page 5
QUESTION 15
Answer is C
 1
X  Bi ( n, p ) = Bi 12,  and Pr ( X ≤ 4) . Answer is 0.1938
 2
QUESTION 16
Answer is B
Median is at x = 1 .
QUESTION 17
Answer is D
For f ( x) = 5 x − 7 , Test D.
f ( x + y ) = f ( x) + f ( y ) + 7
LHS f ( x + y ) = 5( x + y ) − 7 = 5 x + 5 y − 7
RHS f ( x) + f ( y ) = 5 x − 7 + 5 y − 7 + 7 true.
QUESTION 18
Answer is A
N (μ , σ 2 ) = N (51, σ 2 ) and Pr ( X < 50 ) = 0.42
Giving Pr ( Z < −0.20189 ) = 0.42 Using z =
x−μ
σ
. Answer σ = 4.95
Page 6
QUESTION 19
Answer is D
1
2
Find the value of a such that
 π sin(2πx)dx = 0.2 .
Answer a = 0.35
a
QUESTION 20
Variance =
1
2
1
2
x
Answer is C
2
0
f ( x)dx − μ 2 where μ =  xf ( x)dx
0
 12



2
Answer π  x sin( 2πx)dx −  π  x sin( 2πx)dx 
 0

0


1
2
QUESTION 21
Answer is B
(
2
)
The point (0,3) is transformed to the point 3,3 + 2 giving the transformation
y = f ( x − 3) + 2 .
In the equation y = 3 − x 2 the new equation is y = 3 + 2 − ( x − 3) 2 .
QUESTION 22
Answer is D
Consider f ( x) =
x , using linear approximation f ( x + h) ≈ f ( x) + hf / ( x) , where h = 0.02
f (4 + 0.02) ≈ f (4) + 0.02 f / (4)
∴ f (4.02) ≈ 2 + 0.02 f / (4) where f / (4) =
Answer matches f (4.02) ≈ 2 + 0.02 ×
∴ f (4.02) ≈ 2 +
1
4
1
4
1 1
×
50 4
Page 7
SECTION 2 — EXTENDED ANSWER QUESTIONS
QUESTION 1
1
a.
(i)
f / ( x) =
3 2
3
x −1 =
x −1
2
2
3
x − 1 = 0 gives
2
2
4
x = ∴x =
3
9
From graph this is a minimum.
Answer x =
4
9
1M, 1A
4
9


(ii) Strictly increasing for x ∈  , ∞ 
b.
(i)
1A
y
1
x
3A
4 
4
 ,− 
 9 27 
(ii)
y = f (x)
4 4 
, 
 9 27 
Max at 
1A
Page 8
c.
(i)
4−0 4
=
4 −1 3
4
Equation of line y − 0 = ( x − 1)
3
4
Equation of line y = ( x − 1)
3
Gradient of line =
(ii) Let g / ( x) =
1M, 1A
3
4
x −1 =
2
3
 196 980 
,

 81 729 
Solve for x giving answer ( a, g(a ) ) = 
1M, 1A
Page 9
d.
(i)
y = g (a) gives y =
196
81
Area =
 729 − (x
980
980
729
)
x − x dx
1M, 1A
0
(ii) Area = 2.54
1A
Page 10
QUESTION 2
a.
tan(2 x) =
2 tan( x)
1 − tan 2 ( x)
π 
2 tan  
π 

 12 
tan  2 ×  =
 12  1 − tan 2  π 
 
 12 
1
π
2(2 − 3)
π 
∴ tan =
∴ tan   =
2
6
 6  1 − (2 − 3)
3
b.
Equation of the line is y = tan
π
12
2M
c
x
∴ y = (2 − 3 ) x
c.
d.
1M
Using simultaneous equations b = 2
So equation of the cubic curve y = −( x − 2 )( x + 1)
Gradient = 2 − 3 so perpendicular gradient = −
2
1A
1
2− 3
= −2 − 3
Also using the point (0,2 ) solving simultaneously giving p = ±
In domain p =
3 3 + 15
,b = 2
3
3 3 + 15
giving m = 3, n = 15
3
2M, 1A
Page 11
e.
Point shown is point p (1.50,3.13)
Point of intersection (1.94,0.52)
f.
3A
Vertical strips of width one half between x = 0 and x = 1.5 .
(i)
Area of the left endpoint rectangles between the cubic curve and the x-axis.
Area =
1
( y(0) + y(0.5) + y (1) ) = 75
2
16
Page 12
1M, 1A
(ii) Area of the left endpoint rectangles under the straight line.
1
( y(0) + y (0.5) + y(1) ) = 6 − 3 3
2
4
Area =
1M, 1A
(iii) Approximate area between the curves.
75 6 − 3 3
−
= 4.49
16
4
g.
1A
Area required for painting in yellow.
 − ( x − 2)( x + 1) − (2 − 3 )xdx = 5.48
1.93986
Area =
2
1M, 1A
0
h.
Graph is half a sin graph with horizontal but no vertical translation.
Period of graph =
2π
π
=4
2
Horizontal translation of =
2
π
So x-intercepts could be 0 +
2
and 2 +
π
2
π
Let total area be 4 square units.
2+
Area = m
2
π

2
π

sin  x − 1dx = 4
2

π
Gives m = π .
1M, 1A
Page 13
QUESTION 3
a.
Height = 10, so radius = 5
b.
h = 2r ∴ r =
c.
Given
1A
h
2
1A
dV
1
= − cm3 / sec
dt
2
Chain rule gives
dV dV dh
=
×
dt
dh dt
1 2
πr h
3
dV 25π
1 h3
dV π 2
∴V = π
so
= h At h = 5
=
dh 4
3 4
dh
4
Now V =
∴−
∴
1 25π dh
=
×
dt
2
4
dh
2
cm / sec
=−
dt
25π
2M, 1A
The cone was completely full with ‘Cut-the-Fat’ when Lurch began his evil deed.
d.
Volume of cone V =
1 2
πr h when height =10, radius = 5
3
250
1
∴V = π 250 =
π cm 3 as required
3
3
e.
1M
Cylinder with diameter of x cm and a height 12-x
To find maximum volume.
 x
V = πr 2 h with radius =  
2
2
 x
V = π   (12 − x )
2
∴
dV
 − 3 x( x − 8) 
= π
 = 0 gives stationary points x = 0, x = 8 .
dx
4


From graph local max at x = 8 so diameter = 8, height =12 – 8 = 4
3M, 1A
f.
250
1
π = 83 π cm3
3
3
2
Volume of contents of cylinder = πr h = π 4 2 × 4 = 64π cm3
Volume of contents of cone =
He will NOT be able to catch all contents in the bowl.
1A
Page 14
QUESTION 4
a.
Find a transition matrix, T.
0.6

T =


0.4
b.
(i)
0.6

7
T S0 = 


0.4
0.95




0.05
1  ← W
 
where S 0 =  
 
 
0  ← C
0.95




0.05
7
1A
1 
0.704 ← W
 


  = 


 

 


0 
0.296 ← C
So probability of cycling = 0.296
1M, 1A
(ii) In the long term probability of walking = 0.704
c.
(i)
1A
X  Bi ( n, p ) = Bi ( 7, 0.35 )
Pr( X ≥ 4) = 0.1998
1M, 1A
Page 15
(ii)
Bi (n, p ) = Bi (30,0.35)
Pr( X = 15 \ X ≤ 20) =
d.
Pr( X = 15)
0.0351
= 0.0351
==
Pr( X ≤ 20)
0.9998
1M, 1A
1
1
k + 0.1 + 0.2 + k + k + k + 2k + k =1
3
2
3
∴k =
25
1M
e.
(i)
1 3
1
Pr( X = 4) = ×
=
3 25 25
1A
1
1
0 × k + 0.1 + 2 × 0.2 + 3k + 4× k + 5× k + 6 × 2k + 7 k
3
2
18
Answer = E ( X ) =
= 3.6
5
1
Pr( X = 4)
2
25
==
(iii) Pr( X < 5 \ X > 3.6) =
=
1
3
9
Pr( X > 3.6)
23
+ +
25 50 25
(ii)
1M, 1A
1M, 1A
Page 16

MASTER CLASS PROGRAM UNIT 4 — MATHEMATICAL METHODS SEMESTER TWO 2014

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