Mat1330 sample final 1 of 2
.pdfSchool
University of Ottawa **We aren't endorsed by this school
Course
MAT 1330
Subject
Mathematics
Date
Nov 1, 2023
Type
Other
Pages
20
Uploaded by MegaStar16226 on coursehero.com
Winter 2023
Instructeur:
Punisher
sample final review
Instructions
◦
This is a
Closed Book Exam
of 80 minutes.
◦
The only calculators allowed are the ones approved by Faculty:
Texas Instruments TI-30, Texas Instruments TI-34, Casio fx-260, Casio fx-300
.
◦
There are
16 pages
including 16 for rough work.
◦
The points are indicated for each question.
◦
Read carefully each question.
◦
Questions 1-4 are
multiple choice
. Write your answer (choice) in the box. No work required.
◦
Questions 5-8 are
long answer
. You must show full and all work.
◦
May use the backs of pages and page 16 is for rough work.
Do not detach the last page with
rough work.
◦
Cellular phones, unauthorized electronic devices or course notes are not allowed during
this exam. Phones and devices must be turned off and put away in your bag. Do not
keep them in your possession, such as in your pockets. If caught with such a device or
document, the following may occur: you will be asked to immediately leave the exam
and academic fraud allegations will be filed, which may result in you obtaining a 0 (zero)
for the exam.
†
By signing below, you acknowledge that you have ensured that you are complying with
the above statement.
Last name:
Student ID:
First name:
†
Signature:
Goood Luck!
page 1 of
16
MC
1
2
3
4
5
6
7
8
Maximum
2
2
2
2
2
2
2
2
Note
9
10
11
12
13
14
15
16
Total
bonus!
3
3
4
7
7
3
7
3
50+3
Multiple choice
Questions 1 to 8 are
multiple choice
and each is worth 2 points. No need for justification in Multiple
choice.
Question 1.
The domain of
f
(
x
) =
r
x
x
2
−
4
is:
A.
(
−∞
,
−
2)
∪
(0
,
1]
B.
(
−∞
,
−
2]
∪
[0
,
1]
C.
(
−
2
,
0]
∪
(2
,
∞
)
D.
(
−∞
,
−
2)
E.
[
−
2
,
0)
∪
[2
,
∞
)
F.
(2
,
∞
).
Answer:
Answer:
C
Note that the denominator MUST not be zero:
x
2
−
4
̸
= 0 or (
x
−
2)(
x
+ 2)
̸
= 0, so
x
̸
= 2 and
x
̸
=
−
2. NEXT:
Case 1.
x
≥
0 and
x
2
−
4 = (
x
−
2)(
x
+ 2)
>
0, hence
x
∈
[0
,
∞
) and
x
∈
(
−∞
,
−
2)
∪
(2
,
∞
). INTERSECT (see
word AND) and we get:
x
∈
(2
,
∞
).
Case 2.
x
≤
0 and
x
2
−
4 = (
x
−
2)(
x
+ 2)
<
0, hence
x
∈
[
−∞
,
0] and
x
∈
(
−
2
,
2). INTERSECT and we get:
x
∈
(
−
2
,
0].
Union of the two cases gives C. Happy? HS stuff as u can see.
Question 2.
The table gives info on
f
(
x
),
g
(
x
),
f
′
(
x
) and
g
′
(
x
) for some
x
. Consider
h
(
x
) = (
f
◦
g
)(
x
).
x
1
2
3
4
f
(
x
)
1
4
2
3
g
(
x
)
3
2
4
1
f
′
(
x
)
2
3
1
4
g
′
(
x
)
4
4
3
2
Find the correct answer:
A.
h
(2) = 1 ,
h
′
(2) = 12.
B.
h
(2) = 8 ,
h
′
(2) = 9
C.
h
(2) = 1 ,
h
′
(2) = 6
D.
h
(2) = 8 ,
h
′
(2) = 6.
E.
h
(2) = 8 ,
h
′
(2) = 1.
F.
h
(2) = 4,
h
′
(2) = 12.
Answer:
Chain rule says:
h
′
(2) =
f
′
(
g
(2))
g
′
(2) =
f
′
(2)
×
4 = 3
×
4 = 12. Note
h
(2) =
f
(
g
(2)) =
f
(2) = 4, hiha SO:
Answer:
F
That s the full solution. But if u do not know the chain rule, no way we can finish this exercise...
Question 3.
Find the derivative of
f
(
x
) = ln
2
x
sin(2
x
)
(
x
+ 1)
5
e
x
2
:
A.
ln(2) + 2 cot(2
x
)
B.
2
x
+ 2 cot(2
x
)
−
1
(
x
+ 1)
5
−
2
x
C.
ln(2) +
2
cos(2
x
)
D.
ln(2)
−
2 cot(2
x
) + 2
x
−
5
x
+ 1
E.
ln(2) + 2 cot(2
x
)
−
2
x
−
5
x
+ 1
F.
2
x
−
2
cos(2
x
)
−
5
x
+ 1
−
2
x
Answer:
Answer:
What U need here (except an infinite love for Cal 1) is LOG LAWS: WE MUST rewrite:
f
(
x
) =
ln(2
x
) + ln(sin(2
x
))
−
ln(
x
+ 1)
5
−
ln(
e
x
2
), WHICH it is in fact giving:
f
′
(
x
) = ln(2) +
cos(2
x
)
sin(2
x
)
2
−
5
x
+ 1
−
2
x
.
DONE
E
page 3 of
16
