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程序代写案例-MTH1020
时间：2022-06-21
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 1 of 24
Question 1: Complex Numbers
(a) In Cartesian form,
8i+ 4
i− 1 = a− bi
where a = ? and b = ?
[1 Marks]
Answer. We calculate
8i+ 4
i− 1 =
(4 + 8i)(−1− i)
(−1 + i)(−1− i) =
−4 + 8− 4i− 8i
(−1)2 + 12 =
4− 12i
2
= 2− 6i.
Hence a = 2 and b = 6.
(b) When
cis
(pi
2
)
cis
(−pi
4
)
is expressed in polar form, it can be written as
r cis
(
ppi
q
)
where r > 0 and p
q
is a fraction between 0 and 1 in simplest form.
Then p = ? and q = ?
[2 Marks]
Answer. We have
cis
(pi
2
)
cis
(−pi
4
)
= cis
(pi
2
− pi
4
)
= cis
(pi
4
)
.
This is in the desired form r cis
(
ppi
q
)
with r = 1, p = 1 and q = 4.
(c) For which integers n is in a real number?
[2 Marks]
Answer. We have i0 = 1, i1 = i, i2 = −1, i3 = −i, i4 = 1 and then the powers in form a
sequence which repeats every 4 elements. In particular, in only depends on the remainder
when n is divided by 4. We observe that i0 = 1 and i2 = −1 are real, while i1 = i and
i3 = −i are not. So in is real precisely when n leaves a remainder of 0 or 2 when divided by
4. These are just the even numbers. So the answer is: All even n.
Page: 1 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 2 of 24
(d) Again let
z = 2eipi/3.
Find the cube roots of z.
One of these cube roots has negative imaginary part; denote this particular cube root by w.
Then we can write
Arg(w) =
−ppi
q
,
where p, q are positive integers and p
q
, is a fraction in simplest form.
Then p = ? and q = ?
[2 Marks]
Answer. Suppose w satisfies w3 = z. Write w in polar form as w = r cis θ. Then w3 =
r3 cis(3θ) in polar form, while z = 2eipi/3 = 2 cis(pi/3). Comparing magnitude and argument
we then obtain
r3 = 2 and 3θ =
pi
3
+ 2kpi
for some integer k. So r = 3
√
2 and
θ =
pi
9
+
2pik
3
for some integer k. We may then take θ = −5pi
9
, pi
9
, 7pi
9
to obtain the three cube roots. Hence
the cube roots of z are given by
3
√
2 cis
(−5pi
9
)
,
3
√
2 cis
(pi
9
)
,
3
√
2 cis
(
7pi
9
)
.
Of these, the first has argument between −pi and 0, hence lies below the real axis, i.e. has
negative imaginary part. The second and third have arguments between 0 and pi, hence lie
above the real axis, i.e. have positive imaginary part. So the cube root w we are looking for
is the first one, w = 3
√
2 cis
(−5pi
9
)
, and we obtain p = 5 and q = 9.
Page: 2 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 3 of 24
Question 2: Vectors
(a) Let
b = 〈−1, 1, 3〉.
Then what is |b| ?
[1 Marks]
Answer. |b| = √(−1)2 + 12 + 32 = √11.
(b) Let
a = 〈−5,−15, 5〉 and b = 〈−1, 1, 3〉.
Calculate the dot product a · b, and determine whether or not a and b are perpendicular.
What is a · b ?
Are a and b perpendicular or not?
[2 Marks]
Answer. We calculate
a · b = 〈−5,−15, 5〉 · 〈−1, 1, 3〉 = (−5)(−1) + (−15) · 1 + 5 · 3 = 5− 15 + 15 = 5.
As this dot product is nonzero, a and b are not perpendicular.
(c) Let
a = 〈−5,−15, 5〉 and b = 〈−1, 1, 3〉.
The scalar projection of a in the direction of b can be written in simplest form as
compba =
m√
n
where m and n are positive integers.
Then m = ? and n = ?
[2 Marks]
Answer. We calculate
compba =
a · b
|b| =
5√
11
so m = 5 and n = 11.
(d) Let
a = 〈−5,−15, 5〉 and b = 〈−1, 1, 3〉.
Page: 3 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 4 of 24
The vector projection of a in the direction of b can be written as
projba =
〈
p
q
,
r
s
,
t
u
〉
where p
q
, r
s
, t
u
are fractions in simplest form.
Then t = ? and u = ?
[2 Marks]
Answer.
projba =
a · b
|b|2 b =
5
11
〈−1, 1, 3〉 = 〈−5
11
,
5
11
,
15
11
〉.
Hence t = 15 and u = 11.
Page: 4 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 5 of 24
Question 3: Functions
(a) Indicate whether the following statement is true or false, along with the best reasoning
from the choices below.
If f : R→ R is a continuous function, then for all real numbers x and y,
f(xy) = f(x)f(y).
[1 Marks]
Answer. This is not true. Several possible counterexamples are offered.
The function f(x) = x2 satisfies the equation since then f(xy) = (xy)2 = x2y2 = f(x)f(y).
The function f(x) = x satisfies the equation since then f(xy) = xy = f(x)f(y).
The function f(x) = 1 satisfies the equation since then f(xy) = 1 = 1 · 1 = f(x)f(y).
However the function f(x) = x + 1 does not satisfy the equation, since then f(xy) = xy + 1
while f(x)f(y) = (x+ 1)(y + 1) = xy + x+ y + 1 and in general xy + 1 6= xy + x+ y + 1.
Hence the statement is false, and the function f(x) = x+ 1 is a counterexample.
(b) Indicate whether the following statement is true or false, along with the best reasoning
from the choices below.
The function f : R→ R defined by f(x) = x3 − x is injective.
[2 Marks]
Answer. The function f(x) here is not injective. One way to see this is by plotting the graph
and finding where its critical points are. Another way is to find two distinct numbers x, y
such that f(x) = f(y). We can calculate f(−1) = f(0) = f(1) = 0. Since also f(2) = 6,
of the possibilities offered, the only one which demonstrates f is injective is the equality
f(−1) = f(1).
We can also rule out the other possibilities. As f is a function, its graph passes the vertical line
test. And f is indeed an odd function, but not all odd functions are injective (as we just
found). And this f is not increasing (one can check it is decreasing on a certain interval).
(c) Which of the following functions is continuous, but not differentiable?
1. f : (0,∞)→ R, f(x) = ln x.
2. f : R→ R, f(x) = e−x2 .
Page: 5 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 6 of 24
3. f : R→ R, f(x) = |x|.
4. f : R→ R, f(x) =
{
x2 x ≥ 0
−x2 x < 0
5. f : (0,∞)→ R, f(x) = 1/x.
6. f : R→ R, f(x) =
{
x− 1 x ≥ 0
−x+ 1 x < 0
[2 Marks]
Answer. It is straightforward to differentiate lnx, e−x
2
, and 1/x; these are all differentiable.
The hybrid function given by x2 for x ≥ 0 and −x2 for x < 0 is continuous and differentiable:
the functions x2 and −x2 are themselves differentiable; and they join at x = 0 taking the
same value 0 with the same gradient 0.
The hybrid function given by x− 1 for x ≥ 0 and −x+ 1 for x < 0 is not continuous. We have
f(0) = −1 but as x→ 0− we have f(x)→ 1.
Finally, the function |x|, which is given by x for x ≥ 0 and −x for x < 0, is continuous but
not differentiable. It is continuous since th two linear functions x and −x join at x = 0, but
they have different gradients and hence f(x) is not differentiable at x = 0.
Page: 6 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 7 of 24
Question 4: Differentiation
(a) Given f(x) = ln(sin x) esinx, then f ′(x) can be written as
f ′(x) = A(x)eB(x)
(
1
C(x)
+ ln(sinx)
)
,
where A(x), B(x) and C(x) are functions of x.
Then A(x) = ?, B(x) = ?, and C(x) = ?.
[3 Marks]
Answer. Differentiating using the product and then chain rules, we obtain
f ′(x) =
d
dx
(ln(sinx)) esinx + ln(sinx)
d
dx
(
esinx
)
=
1
sinx
cosx esinx + ln(sinx) esinx cosx
= cosx esinx
(
1
sinx
+ ln(sinx)
)
.
This has the desired form with A(x) = cos x, B(x) = sin x and C(x) = sinx.
(b) Given y = Cos−1(x2), then dy
dx
can be written as
dy
dx
=
−AxB√
1− xC ,
where A,B,C are integer constants given by
A = ?, B = ?, and C = ?.
[3 Marks]
Answer. Since d
dx
Cos−1x = −1√
1−x2 , by the chain rule we obtain
dy
dx
= − 1√
1− (x2)2 2x =
−2x√
1− x4 .
This has the desired form with A = 2, B = 1 and C = 4.
(c) Given
f(x) = 2e−3(x−5)
2
,
then f ′(0) can be written as
f ′(0) = Ae−B,
where A and B are positive integers.
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MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 8 of 24
Then A = ? and B = ?
[2 Marks]
Answer. Differentiating using the chain rule, we obtain
f ′(x) = 2e−3(x−5)
2 d
dx
(−3(x− 5)2) = 2e−3(x−5)2 − 6(x− 5)
= −12(x− 5)e−3(x−5)2
so that f ′(0) = −12 · (−5)e−3(−5)2 = 60e−75. Hence A = 60 and B = 75.
(d) Given
cos(xy) = y,
find dy
dx
. This derivative can be written in the form
dy
dx
=
−y sin(xy)
A(x)B(xy) + 1
,
where A(x) is a polynomial function of x, and B(xy) is a trigonometric function of xy.
Then A(x) = ? and B(xy) = ?
[3 Marks]
Answer. Using implicit differentiation, we obtain
− sin(xy) d
dx
(xy) =
dy
dx
− sin(xy)
(
y + x
dy
dx
)
=
dy
dx
−y sin(xy) = x sin(xy)dy
dx
+
dy
dx
−y sin(xy) = dy
dx
(x sin(xy) + 1)
and hence
dy
dx
=
−y sin(xy)
x sin(xy) + 1
.
This has the desired form, with A(x) = x and B(xy) = sin(xy).
(e) Given
y = x5x
3+7x,
Page: 8 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 9 of 24
find dy
dx
. This derivative can be written uniquely in the form
dy
dx
= x5x
3+7x
(
(Ax2 +B) lnx+ Cx2 + 7
)
,
where A,B,C,D are positive integers.
Then A = ?, B = ?, and C = ?.
[3 Marks]
Answer. We apply logarithmic differentiation. From ln y = (5x3 + 7x) lnx we implicitly
differentiate and obtain
1
y
dy
dx
= (15x2 + 7) lnx+ (5x3 + 7x)
1
x
dy
dx
= y
(
(15x2 + 7) lnx+ 5x2 + 7
)
.
Substituting y = x5x
3+7x gives the desired form for dy
dx
with A = 15, B = 7 and C = 5.
Page: 9 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 10 of 24
Question 5: Investigating behaviour of functions
(a) Consider the function given by
g(x) =
|x+ 3|
(x− 1)(x− 2)
for x 6= 1, 2.
As a piecewise defined function, g(x) can be written as
g(x) =
{
Ax+B
(x−1)(x−2) x ≥ −C( and x 6= 1, 2),
−Ax−B
(x−1)(x−2) x < −C
where A,B,C are positive integers given by
A = ?, B = ?, C = ?
[2 Marks]
Answer. When x ≥ −3 we have x+3 ≥ 0 so |x+3| = x+3. When x < −3 we have x+3 < 0
so |x + 3| = −x − 3. Thus when x ≥ −3 we have g(x) = x+3
(x−1)(x−2) , and when x < −3 we
have g(x) = −x−3
(x+1)(x−2) . This gives g(x) in the required form, with A = 1, B = 3 and C = 3.
(b) Consider the function
g(x) =
|x+ 3|
(x− 1)(x− 2) .
Find the limits
lim
x→∞
g(x), lim
x→1−
g(x), and lim
x→1+
g(x).
The limit limx→∞ g(x) is ?
The limit limx→1− g(x) is ?
The limit limx→1+ g(x) is ?
[6 Marks]
Answer. When x is large positive, we have |x+ 3| = x+ 3 and so
g(x) =
x+ 3
(x− 1)(x− 2) =
x+ 3
x2 − 3x+ 2 =
1 + 3
x
x− 3 + 2
x
.
In the last expression we see that as x→∞ the numerator tends to 1 and the denominator
to ∞, so g(x)→ 0.
Page: 10 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 11 of 24
When x is close to 1 we have |x + 3| = x + 3 and hence g(x) = x+3
(x−1)(x−2) . As x approaches 1
from either side, the numerator approaches 4.
As x→ 1− we have (x− 1)→ 0− and (x− 2)→ −1 so the denominator (x− 1)(x− 2)→ 0+.
Hence the quotient g(x) has numerator approaching 4 and denominator approaching 0 from
above, so g(x)→∞.
Ax x→ 1+ we have (x− 1)→ 0+ and (x− 2)→ −1 so the denominator (x− 1)(x− 2)→ 0−.
Hence the quotient g(x) has numerator approaching 4 and denominator approaching 0 from
below, so g(x)→ −∞.
(c) Consider the function
g(x) =
|x+ 3|
(x− 1)(x− 2) .
For what values of x is g(x) continuous?
[2 Marks]
Answer. The numerator |x + 3| is a composition of the continuous functions x + 3 and |x|,
so is continuous. The denominator is a polynomial function, hence continuous. Hence g(x)
is continuous everywhere it is defined, i.e. for all x except when the denominator is zero.
Thus g is continuous for all x ∈ R \ {1, 2}.
(d) Consider the function
g(x) =
|x+ 3|
(x− 1)(x− 2) .
For what values of x is g(x) differentiable?
[2 Marks]
Answer. As discussed in a previous question, when x > −3 we have g(x) = x+3
(x−1)(x−2) , which
is a rational function, hence differentiable everywhere it is defined. Thus for x > −3, g(x)
is differentiable for all x except 1 and 2.
Similarly, when x < −3 we have g(x) = −x−3
(x−1)(x−2) , which is differentiable everywhere it is
defined. Thus g(x) is differentiable for all x < −3.
The remaining value to consider is x = −3. For x ≥ −3 we have g(x) = x+3
x2−3x+2 so that
g′(x) = 1(x
2−3x+2)−(x+3)(2x−3)
(x2−3x+2)2 =
−x2−6x+11
(x2−3x+2)2 , and as x→ −3+ we then have g′(x)→ −9+18+11(9+9+2)2 =
20
202
= 1
20
. On the other hand, for x < −3 we have g(x) the negative of the above, so
g′(x) = x
2+6x−11
(x2−3x+2)2 and as x→ −3− we then have g′(x)→ −1/20. So g(x) is not differentiable
at x = −3.
We conclude that g(x) is differentiable for x ∈ R \ {−3, 1, 2}.
Page: 11 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 12 of 24
(e) For the function
f : R→ R, f(x) = x2e−x,
the interval on which f(x) is increasing is of the form [a, b], where
a = ? and b = ?
[4 Marks]
Answer. Differentiating, we obtain f(x) = 2xe−x + x2(−e−x) = (2x − x2)e−x. To find
where f is increasing, we find where f ′(x) > 0. Since e−x > 0 for all x, we must find where
2x − x2 > 0. This is a quadratic function and by plotting the quadratic or analysing the
factorisation 2x−x2 = x(2−x), we find that f ′(x) ≥ 0 for x ∈ [0, 2]. Thus a = 0 and b = 2.
(f) The function
f : R→ R, f(x) = x2e−x
has one local maximum. It occurs when
x = ?
and at this value of x, the local maximum value of f is Ae−B
where A = ? and B = ?
[4 Marks]
Answer. We have f ′(x) = (2x − x2)e−x. If f ′(x) = 0 then, since e−x is never 0, we have
2x− x2 = 0 so x(2− x) = 0 and x = 0 or 2.
To see which of these is the local maximum, we consider the sign of f ′(x). As in the previous
question, e−x is always positive, so the sign of f ′(x) is the same as that of 2x − x2. Thus
f ′(x) is negative when x ∈ (−∞, 0), positive for x ∈ (0, 2) and negative for x ∈ (2,∞).
Hence x = 0 is a local minimum, and x = 2 is the desired local maximum.
The local maximum value of f is then f(x) = 22e−2 = 4e−2, so A = 4 and B = 2.
(g) For the function
f : R→ R, f(x) = x2e−x,
calculate the limits of f(x) as x approaches positive and negative infinity.
limx→∞ f(x) = ?
limx→−∞ f(x) = ?
[4 Marks]
Page: 12 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 13 of 24
Answer. Writing f(x) = x
2
ex
, as x → ∞ we have x2 → ∞ and ex → ∞. Recalling that
exponential functions grow much faster than polynomials, we conclude that limx→∞ f(X) =
0.
As x → −∞ we have x2 → ∞ and e−x → ∞. Thus f(x) = x2e−x → ∞ and we have
limx→−∞ f(x) =∞.
(h) For the function
f : R→ R, f(x) = x2e−x,
find the intervals on which the graph of y = f(x) is concave up and down.
The interval on which the graph is concave [up? — down?] is of the form [a, b].
The interval on which the graph is concave [up? — down?] is of the form (−∞, a]∪ [b,∞), for
the same a and b.
For this a and b, their sum is a+ b = ?
[4 Marks]
Answer. We have previously calculated f ′(x) = (2x − x2)e−x. Differentiating again yields
f ′′(x) = (2−2x)e−x+(2x−x2)(−e−x) = (2−4x+x2)e−x. As e−x is always positive, the sign
of f ′′(x) is the same as the sign of 2− 4x+x2. We plot or analyse this quadratic to find our
where it is positive and negative. Completing the square we have x2− 4x+ 2 = (x− 2)2− 2,
so x2 − 4x + 2 = 0 when (x− 2)2 = 2 i.e. x = 2±√2. We then have that x2 − 4x + 2 ≥ 0
for x ∈ (∞, 2 − √2] ∪ [2 + √2,∞) and x2 − 4x + 2 ≤ 0 for 2 ∈ [2 − √2,−2 + √2]. Thus
f ′′(x) ≥ 0 and f ′′(x) ≤ 0 on the same intervals.
So f is concave down on the interval [a, b] and concave up on the interval (−∞, a] ∪ [b,∞)
where a = 2−√2 and b = 2 +√2. So a+ b = 4.
Page: 13 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 14 of 24
Question 6: Integration
(a) Suppose that f is continuous on the interval [a, b]. Then the integral
∫ b
a
(1 + 2f(x)) dx
is necessarily equal to which of the following expressions?
Select one or more:
1.
∫ b
a
1 dx+ 2
∫ b
a
f(x) dx
2.
∫ b
a
1 dx+
∫ 2b
2a
f(x) dx
3. 1 +
∫ b
a
2f(x) dx
4. b− a+ 2 ∫ b
a
f(x) dx
5.
∫ 2b
2a
(
1
2
+ f(x)
)
dx
6.
∫ 2b
2a
(
1
2
+ f
(
x
2
))
dx
7.
∫ b−1
a−1 (1 + 2f(x+ 1)) dx
∫ b−1
a−1 (1 + 2f(x− 1)) dx
[3 Marks]
Answer. Using standard properties of integrals we have∫ b
a
(1 + 2f(x)) dx =
∫ b
a
1 dx+
∫ b
a
2f(x) dx =
∫ b
a
1 dx+ 2
∫ b
a
f(x) dx
so (a) is equal to the given integral.
In part (b) the second term 2
∫ b
a
f(x) dx is replaced with
∫ 2b
2a
f(x) dx, which is in general not
equal. So (b) is not equal to the given integral.
In part (c) the first term
∫ b
a
1 dx of (a) is replaced with a 1, which is in general not equal. So
(c) is not equal to the given integral.
In part (d) we use
∫ b
a
1 dx = b− a to observe it is equal to the integral in (a), so (d) is equal to
the given integral.
For parts (e) and (f) we consider the graph of f(x/2), which is obtained from that of f(x) by
a dilation of factor 2 in the x-direction from the y-axis. The integral of this function from
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MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 15 of 24
x = 2a from x = 2b gives twice the signed area of f(x) from x = a to x = b. Thus,∫ b
a
2f(x) dx =
∫ 2b
2a
f(x/2) dx.
Combined with
∫ b
a
1 dx = b−a and ∫ 2b
2a
1
2
dx = 1
2
(2b−2a) = b−a we conclude that (f) is equal
to the given integral. Alternatively, we could consider the substitution u = 2x. However, in
part (e) the f(x/2) is replaced with an f(x), and is in general not equal.
For part (g) we consider the function f(x + 1) whose graph is obtained from that of f(x) by
translating one unit to the left. The integral of this function from x = a − 1 to x = b − 1
gives the same signed area as that of f(x) from x = a to x = b. Alternatively we could
consider the substitution u = x− 1, using x = u + 1 and du = dx. Either way, (g) is equal
to the given integral.
In (h) the f(x+ 1) in part (g) is replaced with f(x− 1), hence gives a different result. So (h)
is not equal to the given integral.
We conclude that (a), (d), (f) and (g) are necessarily equal to the given integral, but the others
are not.
(b) Indicate whether the following statement is true or false, along with the best reasoning
from the choices below.
If f : R→ R is a continuous function, and ∫ b
a
f(x) dx 6= 0, then a 6= b .
1. True, because when a = b, it must be true that
∫ b
a
f(x) dx 6= 0.
2. True, because when a 6= b, the integral ∫ b
a
f(x) dx describes a signed area which must be
nonzero.
3. True, because if
∫ b
a
f(x) dx 6= 0, then the integral describes a signed area which is nonzero,
and this can only happen when a 6= b.
4. False, because
∫ b
a
f(x) dx = F (b)− F (a), where F (x) is an antiderivative of f(x), and if
this is nonzero then a 6= b.
5. False, because if
∫ b
a
f(x) dx were equal to zero, then necessarily a = b.
6. False, because although it is true when f is differentiable, it is not true when f is not
differentiable.
[1 Marks]
Page: 15 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 16 of 24
Answer. The statement is true. The reasoning in (c) is good: if the integral is nonzero then
the corresponding signed area is nonzero, and that can only happen if a 6= b.
However the reasoning in (a) and (b) is not good. When a = b we have
∫ b
a
f(x) dx is zero, not
onzero. And when a 6= b, in general the integral ∫ b
a
f(x) dx might be zero or nonzero.
(c) Calculate the following integral.∫ (
x 3
√
x+
1√
x
)
dx
This integral can be expressed as
A
B
xC/D +
E
F
xG/H + c
where A,B,C,D,E, F,G,H are positive integers, c is any constant, and A
B
, C
D
, E
F
, G
H
are
fractions in simplest form with C
D
> G
H
.
Then A = ?, B = ?, C = ? and D = ?
[4 Marks]
Answer. The integrand can be written as x4/3 + x−1/2 and hence the integral is given by
3
7
x7/3 + 2x1/2 + c. Thus A = 3, B = 7, C = 7 and D = 3.
(d) Calculate the following integral. ∫ 2
0
x4ex
5
dx
The result can be written as
eA − 1
B
,
where A and B are positive integers.
Then A = ? and B = ?
[4 Marks]
Answer. Substitute u = x5, so du = 5x4dx. The terminals x = 0, 2 correspond to u = 0, 32.
So the integral becomes ∫ 32
0
1
5
eu du =
[
eu
5
]32
0
=
e32 − 1
5
.
Hence A = 32 and B = 5.
Page: 16 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 17 of 24
(e) Calculate the following integral. ∫
(ln(x)− 5)2
x
dx
The result can be written uniquely in the form
(ln(x)− A)B
C
+ c
where A,B,C are positive integers and c is a constant.
Then A = ?, B = ? and C = ?
[3 Marks]
Answer. Letting u = lnx we have du = dx
x
and the integral becomes∫
(u− 5)2 du = 1
3
(u− 5)3 + c = (ln(x)− 5)
3
3
+ c
which is in the desired form with A = 5, B = 3 and C = 3.
Page: 17 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 18 of 24
Question 7: Applications of integration
(a) Suppose f : R → R is an odd function, and consider its average value on the interval
[−1, 1]. Which of the following statements about this average value is true?
1. It must be 0, because f(−x) = f(x), so ∫ 1−1 f(x) dx = 0 and hence the average value is 0.
2. It must be 0, because f(−x) = −f(x), so ∫ 1−1 f(x) dx = 0 and hence the average value is
0.
3. We can’t say for sure what it is, but it must be positive, because f(−x) = f(x) and hence∫ 1
−1 f(x) dx > 0.
4. We can’t say for sure what it is, but it must be positive, because f(−x) = −f(x) and
hence
∫ 1
−1 f(x) dx > 0.
5. It could be anything: any real number is a possible average value of an odd function on
the interval [−1, 1].
6. It must be infinite: the fact that f is odd means that the average value on the interval
[−1, 1] goes to infinity.
7. It’s not defined because f being odd means that when we calculate the average value we
have division by zero.
[2 Marks]
Answer. The reasoning in (b) is good. When f is odd we have f(−x) = −f(x) and this
implies
∫ 1
−1 f(x) dx = 0. The reasoning in (a) has the wrong definition of odd function. The
other options are then clearly false.
(b) Consider the closed region A bounded by the curves
y = 1/x and y = −10x+ 11.
The area of the region A is given by an integral whose terminals a, b correspond to the values
of x where the curves meet. Let these two values of x be a and b, where a < b.
Then 1/a = ? and b = ?
The area of the region A can be written uniquely in the form
L
M
− ln(N),
Page: 18 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 19 of 24
where L,M,N are positive integers and L
M
is a fraction in simplest form.
Then L = ?, M = ? and N = ?
[2 Marks]
Answer. The curves meet where 1/x = −10x + 11, which simplifies to 10x2 − 11x + 1 = 0,
and this has solutions x = 11±9
20
, i.e. x = 1/10 and x = 1. Thus a = 1/10 (so 1/a = 10) and
b = 1.
By plotting the graphs or otherwise, we observe that for x between a and b, 1/x < −10x+ 11.
So the area A is given by the integral∫ 1
1/10
(
−10x+ 11− 1
x
)
dx =
[−5x2 + 11x− lnx]1
1/10
= [−5 + 11− 0]−
[−5
100
+
11
10
− ln
(
1
10
)]
= [6]−
[
110− 5
100
+ ln 10
]
= 6− 21
20
− ln 10 = 99
20
− ln 10.
Thus L = 99, M = 20 and N = 10.
(c) A bowl is tipped on its side. The bowl is then described by rotating the curve
y = 9
√
x
3
+ 1,
from x = 0 to x = 9, around the x-axis. Here x and y are measured in centimetres.
Show that the volume of the bowl, from x = 0 to x = h, is given in cubic centimetres by
V (h) = 81pih+
27
2
pih2.
[10 Marks]
Answer.
Page: 19 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 20 of 24
y = 9 x
3
+ 1
9
x
9
18
y
The desired volume is a volume of revolution.
V (h) =
∫ h
0
piy2 dx =
∫ h
0
pi
(
9
√
x
3
+ 1
)2
dx =
∫ h
0
pi81
(x
3
+ 1
)
dx
=
∫ h
0
(27pix+ 81pi) dx =
[
27
2
pix2 + 81pix
]h
0
=
27
2
pih2 + 81pih.
Page: 20 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 21 of 24
Question 8: Applications of differentiation
(a) The bowl from the previous question is now placed upright.
In the previous question, you showed that the bowl, when placed upright and filled to a height
of h centimetres above its base, holds a volume of V (h) cubic centimetres, where
V (h) = 81pih+
27
2
pih2.
Liquid is poured into the bowl steadily at the rate of 100 cubic centimetres per second.
When the liquid reaches a height of 3 cm above the base of the bowl, how fast is the height
of the liquid increasing?
The height is increasing at a rate of A
Bpi
centimetres per second, where A and B are positive
integers and A/B is a fraction in simplest form.
Then A = ? and B = ?
[6 Marks]
Answer. Let t be time, measured in seconds. We must find dh
dt
when h = 3. From the given
V (h) we have dV
dh
= 81pi + 27pih, and we are also given that dV
dt
= 100. Thus we have
dh
dt
=
dh
dV
dV
dt
=
1
81pi + 27pih
· 100 = 100
27pi(3 + h)
and when h = 3 we have dh
dt
= 100
27pi·6 =
50
81pi
. Thus A = 50 and B = 81.
(b) Suppose the function f(x) is concave up for all x. Then which of the functions below is
necessarily concave down for all x ?
1. 2f(x) + x2
2. 2f(x)− x2
3. −2f(x) + x2
4. −2f(x)− x2
5. 2f(x)
6. −2f(x)
7. xf(x)
8. −xf(x)
Page: 21 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 22 of 24
[3 Marks]
Answer. Assuming f is twice differentiable, if f is concave up then f ′′(x) ≥ 0. We then
consider the proposed functions; they are necessarily concave down if their second derivatives
are necessarily ≤ 0.
1. has second derivative 2f ′′(x) + 2 ≥ 0, not ≤ 0, as f ′′(x) ≥ 0 and 2 ≥ 0.
2. has second derivative 2f ′′(x)− 2, which does not have definite sign.
3. has second derivative −2f ′′(x) + 2, again indefinite sign.
4. has second derivative −2f ′′(x)− 2 ≤ 0.
5. has second derivative 2f ′′(x) ≥ 0.
6. has second derivative −2f ′′(x) ≤ 0.
7. has first derivative f(x) + xf ′(x) hence second derivative 2f ′(x) + xf ′′(x), which is indef-
inite in sign.
8. is negative of the previous, hence indefinite in sign.
Thus (d),(f) are correct.
(c) A box has length equal to its width. The three side lengths of the box (i.e. its length,
width, and height) sum to 1 metre. What is the maximum volume of the box? Give a full
solution, explaining your reasoning and showing all working.
[10 Marks]
Answer. Let the length of the box, in metres, be x.
Then the width of the box is also x metres.
Let the height of the box, in metres, be y.
Let V be the volume of the box, in cubic metres. So V = x2y.
As the lengths are positive and sum to 1 we have x, y ≥ 0.
We are given 2x+ y = 1 and we wish to maximise V .
From 2x+ y = 1 we obtain y = 1− 2x. Since y ≥ 0 this gives x ≤ 1/2.
Thus V can be expressed as a function of x as V (x) = x2(1 − 2x) = x2 − 2x3. We must find
the maximum of V (x) over x ∈ [0, 1/2].
Page: 22 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 23 of 24
First we find critical points of V (x).
V ′(x) = 2x− 6x2 = 2x(1− 3x)
Thus V ′(x) = 0 when x = 0 or when x = 1/3. Checking these values and endpoints, we
calculate V (0) = 0, V (1/3) = 1/27 and V (1/2) = 0.
Hence the maximum possible volume of the box is 1/27 cubic metres.
Page: 23 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 24 of 24
Question 9: Differential equations
(a) Solve the initial value problem
dy
dx
=
y2
x+ 3
with y(−2) = 42.
Then when x = 100, we have
y =
A
1− A lnB,
where A and B are positive integers.
Then A = ? and B = ?
[10 Marks]
Answer. Separating and integrating we obtain∫
1
y2
dy =
∫
1
x+ 3
dx− 1
y
= ln(x+ 3) + c.
Substituting x = −2, y = 42 gives −1/42 = c. So we have −1
y
= ln(x+ 3)− 1
42
and hence
y =
−1
ln(x+ 3)− 1
42
=
42
1− ln(x+ 3) .
Substituing x = 100 gives y = 42
1−ln(103) . So A = 42 and B = 103.
Page: 24 of 24
Question 1: Complex Numbers
(a) In Cartesian form,
8i+ 4
i− 1 = a− bi
where a = ? and b = ?
[1 Marks]
Answer. We calculate
8i+ 4
i− 1 =
(4 + 8i)(−1− i)
(−1 + i)(−1− i) =
−4 + 8− 4i− 8i
(−1)2 + 12 =
4− 12i
2
= 2− 6i.
Hence a = 2 and b = 6.
(b) When
cis
(pi
2
)
cis
(−pi
4
)
is expressed in polar form, it can be written as
r cis
(
ppi
q
)
where r > 0 and p
q
is a fraction between 0 and 1 in simplest form.
Then p = ? and q = ?
[2 Marks]
Answer. We have
cis
(pi
2
)
cis
(−pi
4
)
= cis
(pi
2
− pi
4
)
= cis
(pi
4
)
.
This is in the desired form r cis
(
ppi
q
)
with r = 1, p = 1 and q = 4.
(c) For which integers n is in a real number?
[2 Marks]
Answer. We have i0 = 1, i1 = i, i2 = −1, i3 = −i, i4 = 1 and then the powers in form a
sequence which repeats every 4 elements. In particular, in only depends on the remainder
when n is divided by 4. We observe that i0 = 1 and i2 = −1 are real, while i1 = i and
i3 = −i are not. So in is real precisely when n leaves a remainder of 0 or 2 when divided by
4. These are just the even numbers. So the answer is: All even n.
Page: 1 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 2 of 24
(d) Again let
z = 2eipi/3.
Find the cube roots of z.
One of these cube roots has negative imaginary part; denote this particular cube root by w.
Then we can write
Arg(w) =
−ppi
q
,
where p, q are positive integers and p
q
, is a fraction in simplest form.
Then p = ? and q = ?
[2 Marks]
Answer. Suppose w satisfies w3 = z. Write w in polar form as w = r cis θ. Then w3 =
r3 cis(3θ) in polar form, while z = 2eipi/3 = 2 cis(pi/3). Comparing magnitude and argument
we then obtain
r3 = 2 and 3θ =
pi
3
+ 2kpi
for some integer k. So r = 3
√
2 and
θ =
pi
9
+
2pik
3
for some integer k. We may then take θ = −5pi
9
, pi
9
, 7pi
9
to obtain the three cube roots. Hence
the cube roots of z are given by
3
√
2 cis
(−5pi
9
)
,
3
√
2 cis
(pi
9
)
,
3
√
2 cis
(
7pi
9
)
.
Of these, the first has argument between −pi and 0, hence lies below the real axis, i.e. has
negative imaginary part. The second and third have arguments between 0 and pi, hence lie
above the real axis, i.e. have positive imaginary part. So the cube root w we are looking for
is the first one, w = 3
√
2 cis
(−5pi
9
)
, and we obtain p = 5 and q = 9.
Page: 2 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 3 of 24
Question 2: Vectors
(a) Let
b = 〈−1, 1, 3〉.
Then what is |b| ?
[1 Marks]
Answer. |b| = √(−1)2 + 12 + 32 = √11.
(b) Let
a = 〈−5,−15, 5〉 and b = 〈−1, 1, 3〉.
Calculate the dot product a · b, and determine whether or not a and b are perpendicular.
What is a · b ?
Are a and b perpendicular or not?
[2 Marks]
Answer. We calculate
a · b = 〈−5,−15, 5〉 · 〈−1, 1, 3〉 = (−5)(−1) + (−15) · 1 + 5 · 3 = 5− 15 + 15 = 5.
As this dot product is nonzero, a and b are not perpendicular.
(c) Let
a = 〈−5,−15, 5〉 and b = 〈−1, 1, 3〉.
The scalar projection of a in the direction of b can be written in simplest form as
compba =
m√
n
where m and n are positive integers.
Then m = ? and n = ?
[2 Marks]
Answer. We calculate
compba =
a · b
|b| =
5√
11
so m = 5 and n = 11.
(d) Let
a = 〈−5,−15, 5〉 and b = 〈−1, 1, 3〉.
Page: 3 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 4 of 24
The vector projection of a in the direction of b can be written as
projba =
〈
p
q
,
r
s
,
t
u
〉
where p
q
, r
s
, t
u
are fractions in simplest form.
Then t = ? and u = ?
[2 Marks]
Answer.
projba =
a · b
|b|2 b =
5
11
〈−1, 1, 3〉 = 〈−5
11
,
5
11
,
15
11
〉.
Hence t = 15 and u = 11.
Page: 4 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 5 of 24
Question 3: Functions
(a) Indicate whether the following statement is true or false, along with the best reasoning
from the choices below.
If f : R→ R is a continuous function, then for all real numbers x and y,
f(xy) = f(x)f(y).
[1 Marks]
Answer. This is not true. Several possible counterexamples are offered.
The function f(x) = x2 satisfies the equation since then f(xy) = (xy)2 = x2y2 = f(x)f(y).
The function f(x) = x satisfies the equation since then f(xy) = xy = f(x)f(y).
The function f(x) = 1 satisfies the equation since then f(xy) = 1 = 1 · 1 = f(x)f(y).
However the function f(x) = x + 1 does not satisfy the equation, since then f(xy) = xy + 1
while f(x)f(y) = (x+ 1)(y + 1) = xy + x+ y + 1 and in general xy + 1 6= xy + x+ y + 1.
Hence the statement is false, and the function f(x) = x+ 1 is a counterexample.
(b) Indicate whether the following statement is true or false, along with the best reasoning
from the choices below.
The function f : R→ R defined by f(x) = x3 − x is injective.
[2 Marks]
Answer. The function f(x) here is not injective. One way to see this is by plotting the graph
and finding where its critical points are. Another way is to find two distinct numbers x, y
such that f(x) = f(y). We can calculate f(−1) = f(0) = f(1) = 0. Since also f(2) = 6,
of the possibilities offered, the only one which demonstrates f is injective is the equality
f(−1) = f(1).
We can also rule out the other possibilities. As f is a function, its graph passes the vertical line
test. And f is indeed an odd function, but not all odd functions are injective (as we just
found). And this f is not increasing (one can check it is decreasing on a certain interval).
(c) Which of the following functions is continuous, but not differentiable?
1. f : (0,∞)→ R, f(x) = ln x.
2. f : R→ R, f(x) = e−x2 .
Page: 5 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 6 of 24
3. f : R→ R, f(x) = |x|.
4. f : R→ R, f(x) =
{
x2 x ≥ 0
−x2 x < 0
5. f : (0,∞)→ R, f(x) = 1/x.
6. f : R→ R, f(x) =
{
x− 1 x ≥ 0
−x+ 1 x < 0
[2 Marks]
Answer. It is straightforward to differentiate lnx, e−x
2
, and 1/x; these are all differentiable.
The hybrid function given by x2 for x ≥ 0 and −x2 for x < 0 is continuous and differentiable:
the functions x2 and −x2 are themselves differentiable; and they join at x = 0 taking the
same value 0 with the same gradient 0.
The hybrid function given by x− 1 for x ≥ 0 and −x+ 1 for x < 0 is not continuous. We have
f(0) = −1 but as x→ 0− we have f(x)→ 1.
Finally, the function |x|, which is given by x for x ≥ 0 and −x for x < 0, is continuous but
not differentiable. It is continuous since th two linear functions x and −x join at x = 0, but
they have different gradients and hence f(x) is not differentiable at x = 0.
Page: 6 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 7 of 24
Question 4: Differentiation
(a) Given f(x) = ln(sin x) esinx, then f ′(x) can be written as
f ′(x) = A(x)eB(x)
(
1
C(x)
+ ln(sinx)
)
,
where A(x), B(x) and C(x) are functions of x.
Then A(x) = ?, B(x) = ?, and C(x) = ?.
[3 Marks]
Answer. Differentiating using the product and then chain rules, we obtain
f ′(x) =
d
dx
(ln(sinx)) esinx + ln(sinx)
d
dx
(
esinx
)
=
1
sinx
cosx esinx + ln(sinx) esinx cosx
= cosx esinx
(
1
sinx
+ ln(sinx)
)
.
This has the desired form with A(x) = cos x, B(x) = sin x and C(x) = sinx.
(b) Given y = Cos−1(x2), then dy
dx
can be written as
dy
dx
=
−AxB√
1− xC ,
where A,B,C are integer constants given by
A = ?, B = ?, and C = ?.
[3 Marks]
Answer. Since d
dx
Cos−1x = −1√
1−x2 , by the chain rule we obtain
dy
dx
= − 1√
1− (x2)2 2x =
−2x√
1− x4 .
This has the desired form with A = 2, B = 1 and C = 4.
(c) Given
f(x) = 2e−3(x−5)
2
,
then f ′(0) can be written as
f ′(0) = Ae−B,
where A and B are positive integers.
Page: 7 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 8 of 24
Then A = ? and B = ?
[2 Marks]
Answer. Differentiating using the chain rule, we obtain
f ′(x) = 2e−3(x−5)
2 d
dx
(−3(x− 5)2) = 2e−3(x−5)2 − 6(x− 5)
= −12(x− 5)e−3(x−5)2
so that f ′(0) = −12 · (−5)e−3(−5)2 = 60e−75. Hence A = 60 and B = 75.
(d) Given
cos(xy) = y,
find dy
dx
. This derivative can be written in the form
dy
dx
=
−y sin(xy)
A(x)B(xy) + 1
,
where A(x) is a polynomial function of x, and B(xy) is a trigonometric function of xy.
Then A(x) = ? and B(xy) = ?
[3 Marks]
Answer. Using implicit differentiation, we obtain
− sin(xy) d
dx
(xy) =
dy
dx
− sin(xy)
(
y + x
dy
dx
)
=
dy
dx
−y sin(xy) = x sin(xy)dy
dx
+
dy
dx
−y sin(xy) = dy
dx
(x sin(xy) + 1)
and hence
dy
dx
=
−y sin(xy)
x sin(xy) + 1
.
This has the desired form, with A(x) = x and B(xy) = sin(xy).
(e) Given
y = x5x
3+7x,
Page: 8 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 9 of 24
find dy
dx
. This derivative can be written uniquely in the form
dy
dx
= x5x
3+7x
(
(Ax2 +B) lnx+ Cx2 + 7
)
,
where A,B,C,D are positive integers.
Then A = ?, B = ?, and C = ?.
[3 Marks]
Answer. We apply logarithmic differentiation. From ln y = (5x3 + 7x) lnx we implicitly
differentiate and obtain
1
y
dy
dx
= (15x2 + 7) lnx+ (5x3 + 7x)
1
x
dy
dx
= y
(
(15x2 + 7) lnx+ 5x2 + 7
)
.
Substituting y = x5x
3+7x gives the desired form for dy
dx
with A = 15, B = 7 and C = 5.
Page: 9 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 10 of 24
Question 5: Investigating behaviour of functions
(a) Consider the function given by
g(x) =
|x+ 3|
(x− 1)(x− 2)
for x 6= 1, 2.
As a piecewise defined function, g(x) can be written as
g(x) =
{
Ax+B
(x−1)(x−2) x ≥ −C( and x 6= 1, 2),
−Ax−B
(x−1)(x−2) x < −C
where A,B,C are positive integers given by
A = ?, B = ?, C = ?
[2 Marks]
Answer. When x ≥ −3 we have x+3 ≥ 0 so |x+3| = x+3. When x < −3 we have x+3 < 0
so |x + 3| = −x − 3. Thus when x ≥ −3 we have g(x) = x+3
(x−1)(x−2) , and when x < −3 we
have g(x) = −x−3
(x+1)(x−2) . This gives g(x) in the required form, with A = 1, B = 3 and C = 3.
(b) Consider the function
g(x) =
|x+ 3|
(x− 1)(x− 2) .
Find the limits
lim
x→∞
g(x), lim
x→1−
g(x), and lim
x→1+
g(x).
The limit limx→∞ g(x) is ?
The limit limx→1− g(x) is ?
The limit limx→1+ g(x) is ?
[6 Marks]
Answer. When x is large positive, we have |x+ 3| = x+ 3 and so
g(x) =
x+ 3
(x− 1)(x− 2) =
x+ 3
x2 − 3x+ 2 =
1 + 3
x
x− 3 + 2
x
.
In the last expression we see that as x→∞ the numerator tends to 1 and the denominator
to ∞, so g(x)→ 0.
Page: 10 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 11 of 24
When x is close to 1 we have |x + 3| = x + 3 and hence g(x) = x+3
(x−1)(x−2) . As x approaches 1
from either side, the numerator approaches 4.
As x→ 1− we have (x− 1)→ 0− and (x− 2)→ −1 so the denominator (x− 1)(x− 2)→ 0+.
Hence the quotient g(x) has numerator approaching 4 and denominator approaching 0 from
above, so g(x)→∞.
Ax x→ 1+ we have (x− 1)→ 0+ and (x− 2)→ −1 so the denominator (x− 1)(x− 2)→ 0−.
Hence the quotient g(x) has numerator approaching 4 and denominator approaching 0 from
below, so g(x)→ −∞.
(c) Consider the function
g(x) =
|x+ 3|
(x− 1)(x− 2) .
For what values of x is g(x) continuous?
[2 Marks]
Answer. The numerator |x + 3| is a composition of the continuous functions x + 3 and |x|,
so is continuous. The denominator is a polynomial function, hence continuous. Hence g(x)
is continuous everywhere it is defined, i.e. for all x except when the denominator is zero.
Thus g is continuous for all x ∈ R \ {1, 2}.
(d) Consider the function
g(x) =
|x+ 3|
(x− 1)(x− 2) .
For what values of x is g(x) differentiable?
[2 Marks]
Answer. As discussed in a previous question, when x > −3 we have g(x) = x+3
(x−1)(x−2) , which
is a rational function, hence differentiable everywhere it is defined. Thus for x > −3, g(x)
is differentiable for all x except 1 and 2.
Similarly, when x < −3 we have g(x) = −x−3
(x−1)(x−2) , which is differentiable everywhere it is
defined. Thus g(x) is differentiable for all x < −3.
The remaining value to consider is x = −3. For x ≥ −3 we have g(x) = x+3
x2−3x+2 so that
g′(x) = 1(x
2−3x+2)−(x+3)(2x−3)
(x2−3x+2)2 =
−x2−6x+11
(x2−3x+2)2 , and as x→ −3+ we then have g′(x)→ −9+18+11(9+9+2)2 =
20
202
= 1
20
. On the other hand, for x < −3 we have g(x) the negative of the above, so
g′(x) = x
2+6x−11
(x2−3x+2)2 and as x→ −3− we then have g′(x)→ −1/20. So g(x) is not differentiable
at x = −3.
We conclude that g(x) is differentiable for x ∈ R \ {−3, 1, 2}.
Page: 11 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 12 of 24
(e) For the function
f : R→ R, f(x) = x2e−x,
the interval on which f(x) is increasing is of the form [a, b], where
a = ? and b = ?
[4 Marks]
Answer. Differentiating, we obtain f(x) = 2xe−x + x2(−e−x) = (2x − x2)e−x. To find
where f is increasing, we find where f ′(x) > 0. Since e−x > 0 for all x, we must find where
2x − x2 > 0. This is a quadratic function and by plotting the quadratic or analysing the
factorisation 2x−x2 = x(2−x), we find that f ′(x) ≥ 0 for x ∈ [0, 2]. Thus a = 0 and b = 2.
(f) The function
f : R→ R, f(x) = x2e−x
has one local maximum. It occurs when
x = ?
and at this value of x, the local maximum value of f is Ae−B
where A = ? and B = ?
[4 Marks]
Answer. We have f ′(x) = (2x − x2)e−x. If f ′(x) = 0 then, since e−x is never 0, we have
2x− x2 = 0 so x(2− x) = 0 and x = 0 or 2.
To see which of these is the local maximum, we consider the sign of f ′(x). As in the previous
question, e−x is always positive, so the sign of f ′(x) is the same as that of 2x − x2. Thus
f ′(x) is negative when x ∈ (−∞, 0), positive for x ∈ (0, 2) and negative for x ∈ (2,∞).
Hence x = 0 is a local minimum, and x = 2 is the desired local maximum.
The local maximum value of f is then f(x) = 22e−2 = 4e−2, so A = 4 and B = 2.
(g) For the function
f : R→ R, f(x) = x2e−x,
calculate the limits of f(x) as x approaches positive and negative infinity.
limx→∞ f(x) = ?
limx→−∞ f(x) = ?
[4 Marks]
Page: 12 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 13 of 24
Answer. Writing f(x) = x
2
ex
, as x → ∞ we have x2 → ∞ and ex → ∞. Recalling that
exponential functions grow much faster than polynomials, we conclude that limx→∞ f(X) =
0.
As x → −∞ we have x2 → ∞ and e−x → ∞. Thus f(x) = x2e−x → ∞ and we have
limx→−∞ f(x) =∞.
(h) For the function
f : R→ R, f(x) = x2e−x,
find the intervals on which the graph of y = f(x) is concave up and down.
The interval on which the graph is concave [up? — down?] is of the form [a, b].
The interval on which the graph is concave [up? — down?] is of the form (−∞, a]∪ [b,∞), for
the same a and b.
For this a and b, their sum is a+ b = ?
[4 Marks]
Answer. We have previously calculated f ′(x) = (2x − x2)e−x. Differentiating again yields
f ′′(x) = (2−2x)e−x+(2x−x2)(−e−x) = (2−4x+x2)e−x. As e−x is always positive, the sign
of f ′′(x) is the same as the sign of 2− 4x+x2. We plot or analyse this quadratic to find our
where it is positive and negative. Completing the square we have x2− 4x+ 2 = (x− 2)2− 2,
so x2 − 4x + 2 = 0 when (x− 2)2 = 2 i.e. x = 2±√2. We then have that x2 − 4x + 2 ≥ 0
for x ∈ (∞, 2 − √2] ∪ [2 + √2,∞) and x2 − 4x + 2 ≤ 0 for 2 ∈ [2 − √2,−2 + √2]. Thus
f ′′(x) ≥ 0 and f ′′(x) ≤ 0 on the same intervals.
So f is concave down on the interval [a, b] and concave up on the interval (−∞, a] ∪ [b,∞)
where a = 2−√2 and b = 2 +√2. So a+ b = 4.
Page: 13 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 14 of 24
Question 6: Integration
(a) Suppose that f is continuous on the interval [a, b]. Then the integral
∫ b
a
(1 + 2f(x)) dx
is necessarily equal to which of the following expressions?
Select one or more:
1.
∫ b
a
1 dx+ 2
∫ b
a
f(x) dx
2.
∫ b
a
1 dx+
∫ 2b
2a
f(x) dx
3. 1 +
∫ b
a
2f(x) dx
4. b− a+ 2 ∫ b
a
f(x) dx
5.
∫ 2b
2a
(
1
2
+ f(x)
)
dx
6.
∫ 2b
2a
(
1
2
+ f
(
x
2
))
dx
7.
∫ b−1
a−1 (1 + 2f(x+ 1)) dx
∫ b−1
a−1 (1 + 2f(x− 1)) dx
[3 Marks]
Answer. Using standard properties of integrals we have∫ b
a
(1 + 2f(x)) dx =
∫ b
a
1 dx+
∫ b
a
2f(x) dx =
∫ b
a
1 dx+ 2
∫ b
a
f(x) dx
so (a) is equal to the given integral.
In part (b) the second term 2
∫ b
a
f(x) dx is replaced with
∫ 2b
2a
f(x) dx, which is in general not
equal. So (b) is not equal to the given integral.
In part (c) the first term
∫ b
a
1 dx of (a) is replaced with a 1, which is in general not equal. So
(c) is not equal to the given integral.
In part (d) we use
∫ b
a
1 dx = b− a to observe it is equal to the integral in (a), so (d) is equal to
the given integral.
For parts (e) and (f) we consider the graph of f(x/2), which is obtained from that of f(x) by
a dilation of factor 2 in the x-direction from the y-axis. The integral of this function from
Page: 14 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 15 of 24
x = 2a from x = 2b gives twice the signed area of f(x) from x = a to x = b. Thus,∫ b
a
2f(x) dx =
∫ 2b
2a
f(x/2) dx.
Combined with
∫ b
a
1 dx = b−a and ∫ 2b
2a
1
2
dx = 1
2
(2b−2a) = b−a we conclude that (f) is equal
to the given integral. Alternatively, we could consider the substitution u = 2x. However, in
part (e) the f(x/2) is replaced with an f(x), and is in general not equal.
For part (g) we consider the function f(x + 1) whose graph is obtained from that of f(x) by
translating one unit to the left. The integral of this function from x = a − 1 to x = b − 1
gives the same signed area as that of f(x) from x = a to x = b. Alternatively we could
consider the substitution u = x− 1, using x = u + 1 and du = dx. Either way, (g) is equal
to the given integral.
In (h) the f(x+ 1) in part (g) is replaced with f(x− 1), hence gives a different result. So (h)
is not equal to the given integral.
We conclude that (a), (d), (f) and (g) are necessarily equal to the given integral, but the others
are not.
(b) Indicate whether the following statement is true or false, along with the best reasoning
from the choices below.
If f : R→ R is a continuous function, and ∫ b
a
f(x) dx 6= 0, then a 6= b .
1. True, because when a = b, it must be true that
∫ b
a
f(x) dx 6= 0.
2. True, because when a 6= b, the integral ∫ b
a
f(x) dx describes a signed area which must be
nonzero.
3. True, because if
∫ b
a
f(x) dx 6= 0, then the integral describes a signed area which is nonzero,
and this can only happen when a 6= b.
4. False, because
∫ b
a
f(x) dx = F (b)− F (a), where F (x) is an antiderivative of f(x), and if
this is nonzero then a 6= b.
5. False, because if
∫ b
a
f(x) dx were equal to zero, then necessarily a = b.
6. False, because although it is true when f is differentiable, it is not true when f is not
differentiable.
[1 Marks]
Page: 15 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 16 of 24
Answer. The statement is true. The reasoning in (c) is good: if the integral is nonzero then
the corresponding signed area is nonzero, and that can only happen if a 6= b.
However the reasoning in (a) and (b) is not good. When a = b we have
∫ b
a
f(x) dx is zero, not
onzero. And when a 6= b, in general the integral ∫ b
a
f(x) dx might be zero or nonzero.
(c) Calculate the following integral.∫ (
x 3
√
x+
1√
x
)
dx
This integral can be expressed as
A
B
xC/D +
E
F
xG/H + c
where A,B,C,D,E, F,G,H are positive integers, c is any constant, and A
B
, C
D
, E
F
, G
H
are
fractions in simplest form with C
D
> G
H
.
Then A = ?, B = ?, C = ? and D = ?
[4 Marks]
Answer. The integrand can be written as x4/3 + x−1/2 and hence the integral is given by
3
7
x7/3 + 2x1/2 + c. Thus A = 3, B = 7, C = 7 and D = 3.
(d) Calculate the following integral. ∫ 2
0
x4ex
5
dx
The result can be written as
eA − 1
B
,
where A and B are positive integers.
Then A = ? and B = ?
[4 Marks]
Answer. Substitute u = x5, so du = 5x4dx. The terminals x = 0, 2 correspond to u = 0, 32.
So the integral becomes ∫ 32
0
1
5
eu du =
[
eu
5
]32
0
=
e32 − 1
5
.
Hence A = 32 and B = 5.
Page: 16 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 17 of 24
(e) Calculate the following integral. ∫
(ln(x)− 5)2
x
dx
The result can be written uniquely in the form
(ln(x)− A)B
C
+ c
where A,B,C are positive integers and c is a constant.
Then A = ?, B = ? and C = ?
[3 Marks]
Answer. Letting u = lnx we have du = dx
x
and the integral becomes∫
(u− 5)2 du = 1
3
(u− 5)3 + c = (ln(x)− 5)
3
3
+ c
which is in the desired form with A = 5, B = 3 and C = 3.
Page: 17 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 18 of 24
Question 7: Applications of integration
(a) Suppose f : R → R is an odd function, and consider its average value on the interval
[−1, 1]. Which of the following statements about this average value is true?
1. It must be 0, because f(−x) = f(x), so ∫ 1−1 f(x) dx = 0 and hence the average value is 0.
2. It must be 0, because f(−x) = −f(x), so ∫ 1−1 f(x) dx = 0 and hence the average value is
0.
3. We can’t say for sure what it is, but it must be positive, because f(−x) = f(x) and hence∫ 1
−1 f(x) dx > 0.
4. We can’t say for sure what it is, but it must be positive, because f(−x) = −f(x) and
hence
∫ 1
−1 f(x) dx > 0.
5. It could be anything: any real number is a possible average value of an odd function on
the interval [−1, 1].
6. It must be infinite: the fact that f is odd means that the average value on the interval
[−1, 1] goes to infinity.
7. It’s not defined because f being odd means that when we calculate the average value we
have division by zero.
[2 Marks]
Answer. The reasoning in (b) is good. When f is odd we have f(−x) = −f(x) and this
implies
∫ 1
−1 f(x) dx = 0. The reasoning in (a) has the wrong definition of odd function. The
other options are then clearly false.
(b) Consider the closed region A bounded by the curves
y = 1/x and y = −10x+ 11.
The area of the region A is given by an integral whose terminals a, b correspond to the values
of x where the curves meet. Let these two values of x be a and b, where a < b.
Then 1/a = ? and b = ?
The area of the region A can be written uniquely in the form
L
M
− ln(N),
Page: 18 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 19 of 24
where L,M,N are positive integers and L
M
is a fraction in simplest form.
Then L = ?, M = ? and N = ?
[2 Marks]
Answer. The curves meet where 1/x = −10x + 11, which simplifies to 10x2 − 11x + 1 = 0,
and this has solutions x = 11±9
20
, i.e. x = 1/10 and x = 1. Thus a = 1/10 (so 1/a = 10) and
b = 1.
By plotting the graphs or otherwise, we observe that for x between a and b, 1/x < −10x+ 11.
So the area A is given by the integral∫ 1
1/10
(
−10x+ 11− 1
x
)
dx =
[−5x2 + 11x− lnx]1
1/10
= [−5 + 11− 0]−
[−5
100
+
11
10
− ln
(
1
10
)]
= [6]−
[
110− 5
100
+ ln 10
]
= 6− 21
20
− ln 10 = 99
20
− ln 10.
Thus L = 99, M = 20 and N = 10.
(c) A bowl is tipped on its side. The bowl is then described by rotating the curve
y = 9
√
x
3
+ 1,
from x = 0 to x = 9, around the x-axis. Here x and y are measured in centimetres.
Show that the volume of the bowl, from x = 0 to x = h, is given in cubic centimetres by
V (h) = 81pih+
27
2
pih2.
[10 Marks]
Answer.
Page: 19 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 20 of 24
y = 9 x
3
+ 1
9
x
9
18
y
The desired volume is a volume of revolution.
V (h) =
∫ h
0
piy2 dx =
∫ h
0
pi
(
9
√
x
3
+ 1
)2
dx =
∫ h
0
pi81
(x
3
+ 1
)
dx
=
∫ h
0
(27pix+ 81pi) dx =
[
27
2
pix2 + 81pix
]h
0
=
27
2
pih2 + 81pih.
Page: 20 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 21 of 24
Question 8: Applications of differentiation
(a) The bowl from the previous question is now placed upright.
In the previous question, you showed that the bowl, when placed upright and filled to a height
of h centimetres above its base, holds a volume of V (h) cubic centimetres, where
V (h) = 81pih+
27
2
pih2.
Liquid is poured into the bowl steadily at the rate of 100 cubic centimetres per second.
When the liquid reaches a height of 3 cm above the base of the bowl, how fast is the height
of the liquid increasing?
The height is increasing at a rate of A
Bpi
centimetres per second, where A and B are positive
integers and A/B is a fraction in simplest form.
Then A = ? and B = ?
[6 Marks]
Answer. Let t be time, measured in seconds. We must find dh
dt
when h = 3. From the given
V (h) we have dV
dh
= 81pi + 27pih, and we are also given that dV
dt
= 100. Thus we have
dh
dt
=
dh
dV
dV
dt
=
1
81pi + 27pih
· 100 = 100
27pi(3 + h)
and when h = 3 we have dh
dt
= 100
27pi·6 =
50
81pi
. Thus A = 50 and B = 81.
(b) Suppose the function f(x) is concave up for all x. Then which of the functions below is
necessarily concave down for all x ?
1. 2f(x) + x2
2. 2f(x)− x2
3. −2f(x) + x2
4. −2f(x)− x2
5. 2f(x)
6. −2f(x)
7. xf(x)
8. −xf(x)
Page: 21 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 22 of 24
[3 Marks]
Answer. Assuming f is twice differentiable, if f is concave up then f ′′(x) ≥ 0. We then
consider the proposed functions; they are necessarily concave down if their second derivatives
are necessarily ≤ 0.
1. has second derivative 2f ′′(x) + 2 ≥ 0, not ≤ 0, as f ′′(x) ≥ 0 and 2 ≥ 0.
2. has second derivative 2f ′′(x)− 2, which does not have definite sign.
3. has second derivative −2f ′′(x) + 2, again indefinite sign.
4. has second derivative −2f ′′(x)− 2 ≤ 0.
5. has second derivative 2f ′′(x) ≥ 0.
6. has second derivative −2f ′′(x) ≤ 0.
7. has first derivative f(x) + xf ′(x) hence second derivative 2f ′(x) + xf ′′(x), which is indef-
inite in sign.
8. is negative of the previous, hence indefinite in sign.
Thus (d),(f) are correct.
(c) A box has length equal to its width. The three side lengths of the box (i.e. its length,
width, and height) sum to 1 metre. What is the maximum volume of the box? Give a full
solution, explaining your reasoning and showing all working.
[10 Marks]
Answer. Let the length of the box, in metres, be x.
Then the width of the box is also x metres.
Let the height of the box, in metres, be y.
Let V be the volume of the box, in cubic metres. So V = x2y.
As the lengths are positive and sum to 1 we have x, y ≥ 0.
We are given 2x+ y = 1 and we wish to maximise V .
From 2x+ y = 1 we obtain y = 1− 2x. Since y ≥ 0 this gives x ≤ 1/2.
Thus V can be expressed as a function of x as V (x) = x2(1 − 2x) = x2 − 2x3. We must find
the maximum of V (x) over x ∈ [0, 1/2].
Page: 22 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 23 of 24
First we find critical points of V (x).
V ′(x) = 2x− 6x2 = 2x(1− 3x)
Thus V ′(x) = 0 when x = 0 or when x = 1/3. Checking these values and endpoints, we
calculate V (0) = 0, V (1/3) = 1/27 and V (1/2) = 0.
Hence the maximum possible volume of the box is 1/27 cubic metres.
Page: 23 of 24
MTH1020 Final Exam — Solutions, Semester 1, 2021 Page: 24 of 24
Question 9: Differential equations
(a) Solve the initial value problem
dy
dx
=
y2
x+ 3
with y(−2) = 42.
Then when x = 100, we have
y =
A
1− A lnB,
where A and B are positive integers.
Then A = ? and B = ?
[10 Marks]
Answer. Separating and integrating we obtain∫
1
y2
dy =
∫
1
x+ 3
dx− 1
y
= ln(x+ 3) + c.
Substituting x = −2, y = 42 gives −1/42 = c. So we have −1
y
= ln(x+ 3)− 1
42
and hence
y =
−1
ln(x+ 3)− 1
42
=
42
1− ln(x+ 3) .
Substituing x = 100 gives y = 42
1−ln(103) . So A = 42 and B = 103.
Page: 24 of 24
