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MATH1021 Calculus of One Variable
Solutions to MAIN EXAM 2020 - Extended Answer Questions
1. (a) Let z = 1 + i
√
3 and w = −√3− i.
(i) Calculate the modulus and principal argument of z and w.
Solution: For x = 1 and y =
√
3, one has
|z| =
√
x2 + y2 =
√
4 = 2
and
Arg(z) = sin−1
y
|z| =
pi
3
.
For a = −√3 and b = −1, one has
|w| =
√
a2 + b2 = 2
and
Arg(w) = −pi + sin−1 |b||w| = −pi +
pi
6
= −5pi
6
.
(ii) Write down the polar exponential form of z and w.
Solution: By (i), we have
z = |z| eiArg(z) = 2 eipi3 .
and
w = |w| eiArg(w) = 2 e−i 5pi6 .
(iii) Calculate the polar exponential form of z
w
.
Solution: By using the polar exponential form of z and w, we have that
z
w
=
2
2
ei(Arg(z)−Arg(w)) = ei
7pi
6 .
(iv) Determine the principal argument of z
w
.
Solution: Since arg( z
w
) = 7pi
6
6∈ (−pi, pi], we subtract 2pi and find Arg( z
w
) =
−5pi
6
.
(b) (i) Calculate cos3(2θ) by using the binomial theorem and the complex form of
cos θ.
Solution: Since cos θ = (eiθ + e−iθ)/2, we apply the binomial theorem with
n = 3:
(x+ y)3 = x3 + 3x2y + 3xy2 + y3
to x = ei2θ and y = e−i2θ. Then,
cos3(2θ) =
(
ei2θ + e−i2θ
2
)3
=
1
4
cos(6θ) +
3
4
cos(2θ).
(ii) Find
∫
cos3(2θ) dθ.
Solution: ∫
cos3(2θ) dθ =
1
4
∫
cos(6θ) dθ +
3
4
∫
cos(2θ) dθ
=
1
24
sin(6θ) +
3
8
sin(2θ) + C.
Copyright c© 2023 The University of Sydney 1
2. (a) (i) Calculate the following limits or show that they do not exist.
(A)
lim
x→−2+
x+ 2
|x+ 2|
Solution:
lim
x→−2+
x+ 2
|x+ 2| = limx→−2+
x+ 2
x+ 2
= lim
x→−2+
1 = 1.
(B)
lim
x→0
x3 sin(
3pi
x
)
Solution: Using the fact that | sin(3pi
x
)| ≤ 1 we have |x3 sin(3pi
x
)| ≤
|x3| so that
−|x3| ≤ x3 sin(3pi
x
) ≤ |x3|.
As limx→0±|x3| = 0, by using the squeeze theorem, we find that
limx→0 x3 sin(3pix ) = 0.
(C)
lim
x→+∞
(1 + x)
2
x
Solution: We set y = (1 + x)
2
x so that ln(y) = 2
x
ln(1 + x). Consequently
we have
lim
x→∞
ln(y) = 2 lim
x→∞
ln(1 + x)
x
L’Hoˆp
= 2 lim
x→∞
1
1 + x
= 0,
which implies that y → 1 as x→∞.
(ii) Find the 5th order Taylor polynomial P5(x) for the function cos(x) about
x = 0.
Solution:
P5(x) = 1− x
2
2!
+
x4
4!
(iii) Use the Taylor polynomial that you have found in part (ii) to approximate
the integral
∫ 1
0
cos(x3)dx. (You do not need to calculate the error in this
approximation.)
Solution: The 5-th Taylor polynomial for cos(x3) is
1− x
6
2!
+
x12
4!
∫
cos(x3)dx =
(
x− 1
2!
(
x7
7
) +
1
4!
(
x13
13
)
)∣∣∣1
0
= 1− 1
14
+
1
24
· 1
13
≈ 0.932.
3. (a) Given the function f(x) =
x2 + x− 1
x3
2
(i) Find the natural domain and vertical asymptotes, if any. Justify your an-
swers.
Solution: The natural domain of f is R \ {0} and we have the limits
lim
x→0−
f(x) = lim
x→0−
x2 + x− 1
x3
= −∞ and lim
x→0+
f(x) = lim
x→0+
x2 + x− 1
x3
=∞
so x = 0 is a vertical asymptote.
(ii) Find horizontal asymptotes, if any. Justify your answers.
Solution: Dividing numerator and denominator of f(x) by x3 yields the
limits
lim
x→±∞
f(x) = lim
x→±∞
x2 + x− 1
x3
= lim
x→±∞
1
x
+ 1
x2
− 1
x3
1
=
0 + 0 + 0
1
= 0
so y = 0 is a horizontal asymptote.
(iii) Calculate the first derivative f ′(x).
Solution: Using the quotient rule and simplifying we obtain
f ′(x) =
−x2 − 2x+ 3
x4
.
(iv) Find the critical points and intervals of increase/decrease of f .
Solution: The critical points are x = 0 (derivative does not exist) and the
solutions of −x2 − 2x+ 3 = 0, which are
x1 = −3
and
x2 = 1
The sign diagram for f ′(x) looks as follows:
x (−∞, x1) x1 (x1, 0) 0 (0, x2) x2 (x2,∞)
f ′(x) − 0 + undefined + 0 −
f(x) ↘ min ↗ undefined ↗ max ↘
(v) Calculate the second derivative f ′′(x).
Solution: Again, using the quotient rule and simplifying we obtain
f ′′(x) =
2(x2 + 3x− 6)
x5
.
(vi) Find the points of inflection and intervals of concavity of f .
Solution: The points of inflection are the solutions of −x2 − 2x + 3 = 0,
which are
x1 = −331/2/2− 3/2 ≈ −4.3723
and
x2 = 33
1/2/2− 3/2 ≈ 1.3723.
We also have that f ′′(x) is undefined when x = 0 so we obtain three points
where there may be a change in concavity.
The sign diagram for f ′′(x) looks as follows:
3
x (−∞, x1) x1 (x1, 0) 0 (0, x2) x2 (x2,∞)
f ′′(x) − 0 + undefined − 0 +
f(x) ∩ inflection ∪ undefined ∩ inflection ∪
(vii) Find the global maximum and global minimum of f(x) on the interval [3, 6].
Solution: Since there are no critical points of f on [3, 6], the global extrema
occur at the end points of the interval:
y1 = f(3) ≈ 0.4074
and
y2 = f(6) ≈ 0.1898
Therefore:
Global Maximum is y1
and
Global Minimum is y2