Student
Number
Semester 1 Assessment, 2018
School Mathematics and Statistics
MAST10005 Calculus 1
Writing time: 3 hours
Reading time: 15 minutes
This is NOT an open book exam
This paper consists of 15 pages (including this page)
Authorised Materials
• Mobile phones, smart watches and internet or communication devices are forbidden.
• Calculators, tablet devices or computers must not be used.
• No handwritten or print materials may be brought into the exam venue.
Instructions to Students
• You must NOT remove this question paper at the conclusion of the examination.
• All questions may be attempted. All answers should be appropriately justified.
• Number the questions and question parts clearly. Start each question on a new page.
• Use the left pages for rough working. Write material that you wish to be marked on the
right pages only.
• There is a limited formula sheet provided on the back of this cover page.
• This exam paper consists of 15 pages in total and there are 11 questions.
• The total number of marks available is 102.
Instructions to Invigilators
• Students must NOT remove this question paper at the conclusion of the examination.
• Each candidate should be issued with an examination booklet, and with further booklets
as needed.
This paper may be held in the Baillieu Library
MAST10005 Semester 1, 2018
Formula sheet
Pythagorean identity
cos2(x) + sin2(x) = 1
Compound angle formulae
sin(x+ y) = sin(x) cos(y) + cos(x) sin(y)
cos(x+ y) = cos(x) cos y − sin(x) sin(y)
tan(x+ y) =
tan(x) + tan(y)
1− tan(x) tan(y)
Antiderivatives from inverse trigonometric functions∫
1√
a2 − x2 dx = arcsin
(x
a
)
+ C∫ −1√
a2 − x2 dx = arccos
(x
a
)
+ C∫
1
a2 + x2
dx =
1
a
arctan
(x
a
)
+ C
where a > 0 is constant, and C is an arbitrary constant of integration.
Complex exponential formulae
eiθ = cos(θ) + i sin(θ)
cos(θ) =
1
2
(
eiθ + e−iθ
)
sin(θ) =
1
2i
(
eiθ − e−iθ)
Vector projections
• The vector projection of v onto u is v‖ = (uˆ · v)uˆ.
• The vector component of v perpendicular to u is v⊥ = v − v‖.
Complex roots
The n-th roots of w = seiφ are s
1
n ei(
1
n
(φ+2kpi)) for k = 0, 1, . . . , n− 1.
Changes in speed
Provided r′(t) 6= 0, the speed function ‖r′(t)‖ is decreasing when r′(t) · r′′(t) < 0 and
increasing when r′(t) · r′′(t) > 0.
Page 2 of 15 pages
MAST10005 Semester 1, 2018
Question 1 (6 marks) Write each of the following expressions in cartesian form a + ib, where
a, b ∈ R:
(a) (3 + 2i)(4− 2i)
(b)
(3 + 2i)(4− 2i)
2− 3i
(c)
2
1 + 3i
+
∣∣∣∣(2 + 2i)3i(4− 3i)(1− i)(6 + 8i)
∣∣∣∣ .
Solution:
(a) (3 + 2i)(4− 2i) = (3 + 2i)(4 + 2i)
= 12 + 6i+ 8i− 4
= 8 + 14i 1A
(b)
(3 + 2i)(4− 2i)
2− 3i =
8 + 14i
2− 3i
=
8 + 14i
2− 3i ×
2 + 3i
2 + 3i
1M Multiply by conjugate
=
16 + 28i+ 24i− 42
13
=
52i− 26
13
= 4i− 2 1A
(c)
2
1 + 3i
+
∣∣∣∣(2 + 2i)3i(4− 3i)(1− i)(6 + 8i)
∣∣∣∣ = 21 + 3i +
∣∣∣∣2(1 + i)3i(4− 3i)(1− i)2(3 + 4i)
∣∣∣∣
=
2
1 + 3i
+
∣∣∣∣3i1
∣∣∣∣ 1M Use | · | properties
=
2
1 + 3i
+ 3
=
2
1 + 3i
× 1− 3i
1− 3i + 3 1M Multiply by conjugate
=
1− 3i
5
+ 3
=
16− 3i
5
=
16
5
− 3
5
i 1A First form is OK
Page 3 of 15 pages
MAST10005 Semester 1, 2018
Question 2 (7 marks) Let u = (1, 0,−1) and v = (5, 2,−1).
(a) Calculate 2u− v.
(b) Calculate u · v. Is the angle between u and v less than pi2 or greater?
(c) Find a unit vector parallel to u.
(d) Find a unit vector perpendicular to u.
Solution:
(a) 2u− v = 2(1, 0,−1)− (5, 2,−1)
= (−3,−2,−1) 1A
(b) u · v = 5 + 1 = 6. 1M Use dot product
This implies cos(θ) > 0 so θ < pi2 . 1A
(c) uˆ = 1‖u‖u =
1√
2
(1, 0,−1). 1A
(d) If w = (w1, w2, w3) is perpendicular to u then
u ·w = 0⇒ w1 − w3 = 0⇒ w1 = w3. 1M Use dot product
A vector satisfying this condition is w = (1, 0, 1), 1M Find an example
so a unit vector is wˆ = 1‖w‖w =
1√
2
(1, 0, 1). 1A
Page 4 of 15 pages
MAST10005 Semester 1, 2018
Question 3 (8 marks) Consider the sets
A =
{
x ∈ R | e−x < x} ,
B = {x ∈ (0,∞) | log(x) + x > 0} .
(a) Prove that A ⊆ (0,∞), stating any properties of the exponential function that you use.
(b) Prove that A ⊆ B. Be careful to explain why it is valid to apply the logarithm function
in your proof.
(c) Prove that B ⊆ A. In view of part (b), what can you now conclude?
(d) Prove that (0,∞) 6⊆ A. You may assume that e ∈ (2, 3).
Solution:
(a) x ∈ A⇒ x > e−x and since exponentials are always positive, 1M Explain x > 0.
this gives x ∈ (0,∞).
(b) Let x ∈ A, so x ∈ (0,∞) = dom(log) by part (a). 1M Explain x ∈ dom(log).
Since log is order preserving 1M Explain this (OK to say increasing)
e−x < x⇒ −x < log(x)⇒ log(x) + x > 0, so x ∈ B. 1M
(c) Let x ∈ B, so log(x) + x > 0⇒ −x < log(x)⇒ x > e−x > 0, so x ∈ B.
We conclude that A = B. 1A
(d) Let x = 12 , 1A Any correct counterexample
so x ∈ (0,∞). 1M State this
Now e−x = 1√
e
and e < 4⇒ 1√
e
= e−x > 12 = x, so x /∈ A. 1M
Page 5 of 15 pages
MAST10005 Semester 1, 2018
Question 4 (11 marks)
(a) Write
(1 + i)9
in Cartesian form a+ ib, where a, b ∈ R.
(b) (i) Show that −i is a root of the polynomial
P (z) = 2z4 + 2iz3 − 6z − 6i,
where z ∈ C.
(ii) Using the above information (or otherwise) find all roots of the polynomial P (z).
Express your answers in complex exponential form.
(c) Sketch the set
{z ∈ C | Arg(z) ∈ [0, pi4 ] and |z − 2i| ≤ 1}.
Solution:
(a) 1 + i =
√
2ei
pi
4 ⇒ (1 + i)9 = 292 ei9pi4 1M
= 16
√
2ei
pi
4 = 16
√
2( 1√
2
+ i√
2
)
= 16 + 16i 1A
(b) (i) P (−i) = 2(−i)4 + 2i(−i)3 − 6(−i)− 6i 1M
= 2− 2 + 6i− 6i
= 0
(ii) P (z) = 2(z + i)(z3 − 3) 1A Long division
and z3 − 3 = 0⇒ z3 = 3 1M Observe this
So z ∈ {31/3ei2kpi/3, k = 0, 1, 2}
= {31/3ei0, 31/3ei2pi/3, 31/3e−i2pi/3} 1M Use root finding formula
= {e−ipi/2, 31/3ei0, 31/3ei2pi/3, 31/3e−i2pi/3} 2A 1 mark for mostly correct
(c) From the diagram, the intersection is clearly empty. 1M Explain (diagram or otherwise)
2A Empty set
Im
Re−1 1 2
2 i
3 i
i
Page 6 of 15 pages
MAST10005 Semester 1, 2018
Question 5 (8 marks) Rival energy companies Energy King and Discount Dan have power
stations located at K = (2, 10) and D = (0, 0) where the coordinates are measured in kilometres
from the origin. Each company has a direct (straight) power line supplying the town of Rotcev,
which is located at R = (10, 10).
Old farmer Banks, whose house is located at B = (5, 8), has finally decided he needs electricity
on his farm. He will need to build a private power line to connect up with either the Energy
King or Discount Dan power line.
(a) Which of the two power lines passes closest to Old farmer Banks’ house? You may use
the fact that 75 <
√
2 < 32 .
(b) Suppose that old farmer Banks has been quoted $1000
√
2 per kilometre to construct his
private power line, that Energy King charges a $1000 connection fee, but that Discount
Dan charges a $700 connection fee to connect to their power lines. Which is the cheapest
option for the farmer?
K R
D
B
Solution:
(a) Closest point is measured perpendicular to line.
Since Energy King line is horizontal, perpendicular measurement is vertical 10−8 = 2km.
1A
For Discount Dan line, we use vector projections. Letting u =
−−→
OR = (10, 10) and u =−−→
OB = (5, 8):
v‖ = (uˆ · v)uˆ =
(
13
2 ,
13
2
)
1A
v⊥ = v − v‖ =
(−32 , 32) 1A
so distance to line is ‖v⊥‖ = 3√2 . 1A
Now
√
2 < 32 <
3√
2
, so the Energy King line is closest. 1A
(b) Cost of connecting to Discount Dan line is 3√
2
1000
√
2 + 700 = 3700. 1A
Cost of connecting to Energy King line is
2000
√
2 + 1000 > 75 × 2000 + 1000 = 3800. 1A
Being a mean old bastard, farmer Banks connects to the Discount Dan line. 1A
Page 7 of 15 pages
MAST10005 Semester 1, 2018
Question 6 (7 marks) Suppose that the motion of two particles called Edward and Thomas are
given by the parametric curves r1 : R −→ R and r2 : R −→ R respectively, where:
r1(t) =
(
t2 − t3) i+ (2t− 3t2) j
r2(t) =
(
t3 − t2) i+ (3t2 − 2t) j
(a) Find the time and position where the particles collide.
(b) Find the speed of each of the particles at the time of collision.
Solution:
(a) r1(t) = r2(t)⇒ t3 − t2 = t2 − t3and 2t− 3t2 = 3t2 − 2t 2M
⇒ 2t3 = 2t2and 6t2 = 4t
⇒ t ∈ {0, 1}and t ∈ {0, 23}
⇒ t = 0 1A
So only collision is when t = 0 and occurs at r1(0) = 0. 1A
(b) 1M Calculate derivatives
r′1(t) = (2t− 3t2)i+ (2− 6t)j
so speed at collision is ‖r′1(0)‖ = ‖2j‖ = 2. 1A
r′2(t) = (3t2 − 2t)i+ (6t− 2)j
so speed at collision is ‖r′2(0)‖ = ‖ − 2j‖ = 2. 1A
Page 8 of 15 pages
MAST10005 Semester 1, 2018
Question 7 (8 marks) John and Daniel are on a skiing trip and plan to meet at the chairlift at
the top of a particular hill. The cross section of the hilltop is given by the curve C with implicit
equation
C : x2y3 + 3x+ y2 + log(y) = −1.
The top of the hill is quite slippery due to ice, so platforms made of long straight pieces of wood
have been installed on the hill at coordinates (−1, 1) and (−2, 1). The slope of each platform is
the same as the slope of the tangent line to C at each point.
John and Daniel will meet at the point where the platforms intersect. By considering the
tangent lines at each of the points, determine the meeting point.
Solution:
Implicit differentiation gives:
d
dx
(x2y3) +
d
dx
(3x) + y2 + log(y) =
d
dx
(1) 2M
⇒2xy3 + 3x2y2 dy
dx
+ 3 + 2y
dy
dx
+
1
y
dy
dx
= 0
⇒dy
dx
(3x2y2 + 2y +
1
y
) = −2xy3 − 3
⇒dy
dx
=
−2xy3 − 3
3x2y2 + 2y + 1y
1A
We want the tangent lines at (−1, 1) and (−2, 1).
Gradient at (−1, 1) is −16 and at (−2, 1) is 115 2A
so tangent lines are
y = −16(x− 5) and y = 115(x+ 17). 2A
The intersection point of these two lines is (−97 , 2221). 1A
Page 9 of 15 pages
MAST10005 Semester 1, 2018
Question 8 (5 marks)
(a) Explain why sin(x) is an increasing function on [−pi2 , pi2 ].
(b) What is the implied domain of the function with formula
f(x) =
[
3 arcsin
(x
2
)
− C
]1
2
where C ∈ (0, pi6 ).
Solution:
(a) sin′(x) = cos(x) > 0 for all x ∈ (−pi2 , pi2 ) so sin is increasing on [−pi2 , pi2 ]. 1A
(b) dom(f) =
{
x ∈ R
∣∣∣∣ x2 ∈ [−1, 1] and 3 arcsin(x2)− C ≥ 0
}
=
{
x ∈ [−2, 2]
∣∣∣∣ arcsin(x2) ≥ C3
}
1M
Using part (a) 1M Explain why sin can be applied to inequality
arcsin
(x
2
)
≥ C
3
⇒ x
2
≥ sin
(
C
3
)
1M
⇒ x ≥ 2 sin
(
C
3
)
so dom(f) = [2 sin
(
C
3
)
, 2]. 1A
Page 10 of 15 pages
MAST10005 Semester 1, 2018
Question 9 (16 marks) For the function defined by the rule f(x) =
e−x
x2 − 2x+ 1 find:
(a) The implied (or maximal) domain dom(f).
(b) The x and y intercepts.
(c) Any asymptotes of f .
[Hint: To determine what happens as x −→∞, try multiplying by e
x
ex
.]
(d) All intervals on which f is increasing or decreasing.
(e) The set of stationary points of f and the nature of each.
(f) All intervals on which f is concave up or down.
(g) Any inflection points of f .
(h) Using all the above information, sketch the graph of f . You may assume that f(x) −→∞
as x −→ −∞.
Solution:
(a) f : R \ {1}. Only point where the function is not defined is x = 1. 1A
(b) The function is never zero so no x-intercept.
When x = 0 we have y = 1, so intercept at (0, 1). 1A
(c) Vertical asymptote at x = 1. 1A
e−x
x2 − 2x+ 1 =
1
ex(x− 1)2 −→ 0 1M
as x −→∞, so horizontal asymptote (in positive direction) at y = 0. 1A
(d) f ′(x) = −e
−x(x+ 1)
(x− 1)3 , 1A
so f ′(x) > 0 (respectively < 0) when x+1x−1 < 0 (respectively > 0).
Hence f is increasing on (−1, 1) 1A
and decreasing on (−∞,−1) ∪ (1,∞) 1A
(e) When f ′(x) = 0⇒ x = −1.
So the only stationary point is (−1, e/4) and the point is a local minimum 1A
because f ′(x) goes from negative to positive at this point. 1M Explane why min
(f)
f ′′(x) =
e−x(x2 + 2x+ 3)
(x− 1)4 > 0 1A
for all x in domain so the function is concave up everywhere. 1A
(g) There are no inflection points since the concavity does not change. 1A
Page 11 of 15 pages
MAST10005 Semester 1, 2018
(h) 1A Turning point 2A Asymptotes
x = 1
x
y
−4 −3 −2 −1 1 2 3 4 5
1
2
3
4
Page 12 of 15 pages
MAST10005 Semester 1, 2018
Question 10 (14 marks) Evaluate the following integrals.
(a)
∫
x3 + 4x2 + 7x+ 6
x2 + 3x+ 4
dx.
(b)
∫ 2pi
3
pi
6
cos(x)
sin(x)
dx.
(c)
∫
(x+ 2)
(x− 1)1/2 dx.
(d)
∫
2x+ 3
(x− 2)2(4− x) dx.
Solution:
(a)
∫
x3 + 4x2 + 7x+ 6
x2 + 3x+ 4
dx =
∫
x+ 1 +
2
x2 + 3x+ 4
dx 1M Long division
= x2/2 + x+
∫
2
(x+ 3/2)2 + 7/4
dx 1M Complete the square
= x2/2 + x+
4√
7
arctan
(
2x+ 3√
7
)
1A
(b)
∫ 2pi
3
pi
6
cos(x)
sin(x)
dx =
∫ √3
2
1
2
1
u
du where u = sin(x) and dudx = cos(x). 1M Substitution
Hence terminals are sin(pi6 ) =
1
2 and sin(
2pi
3 ) =
√
3
2 .
∫ √3
2
1
2
1
u
du = log(u)
∣∣∣∣
√
3
2
1
2
1M
= log(
√
3
2
)− log(1
2
) = log(
√
3) 1A
(c) Let u = x− 1. Then dudx = 1 and x = u+ 1. 1M Linear substitution
∫
(x+ 2)
(x− 1)1/2 dx =
∫
u+ 3
u1/2
du 2M
=
∫
u1/2 +
3
u1/2
du
=
2
3
u3/2 + 6u1/2 + C
=
2
3
(x− 1)3/2 + 6(x− 1)1/2 + C 1A
(d) Partial fractions gives
2x+ 3
(x− 2)2(4− x) =
7
2(x− 2)2 +
11
4(x− 2) +
11
4(4− x) 2M Partial fractions
so ∫
2x+ 3
(x− 2)2(4− x) dx =
∫
7
2(x− 2)2 +
11
4(x− 2) +
11
4(4− x) dx
= − 7
2(x− 2) +
11
4
log(|x− 2|)− 11
4
log(|4− x|) + C 2A
Page 13 of 15 pages
MAST10005 Semester 1, 2018
Question 11 (12 marks) Consider the separable differential equation
dy
dx
=
1
y2
√
4− x, x ∈ (−1, 2)
(a) Does this equation have any constant solutions. Explain your answer.
(b) Solve this equation given the initial value y = 1 when x = 1.
Solution:
(a) No. 1A
There are no y values for which the RHS equals 0. 1M Explain
(b) 3M Overall
dy
dx
=
1
y2
√
4− x ⇒ y
2 dy
dx
=
1√
4− x 1M Separate variables
⇒
∫
y2
dy
dx
dx =
∫
1√
4− x dx
⇒
∫
y2 dy =
∫
1√
4− x dx [Substitution] 1M Clear use
⇒ 1
3
y3 =
∫
u−
1
2 (−du
dx
) dx [u = 4− x]
⇒ 1
3
y3 = −
∫
u−
1
2 du [Substitution] 1M Clear use
⇒ 1
3
y3 = −2u12 + C
⇒ y3 = −6√4− x+B
⇒ y = 3
√
−6√4− x+B 1A General solution
Initial values give
1 =
3
√
−6
√
3 +B ⇒ 1 = −6
√
3 +B ⇒ B = 1 + 6
√
3. 2M
Hence
y =
3
√
1 + 6
√
3− 6√4− x. 1A
Page 14 of 15 pages
MAST10005 Semester 1, 2018
End of Exam—Total Available Marks = 102
Page 15 of 15 pages