MATH 100—Midterm Examination
version 1
Date: October 28, 2016 Time: 90 minutes
Surname: Given name(s):
(Please, print!)
ID#: Email: @ualberta.ca
Please, check your section/instructor!
Section Instructor X
EA1 V. Yaskin
EB1 E. Leonard
EC1 G. Tokarsky
ED1 V. Troitsky
EE1 V. Yaskin
EG1 S. Graves
Instructions
1. Place your U of A Student ID card on your table or desk.
2. The exam has 11 pages including this cover page. Please ensure that you have all
pages, and write your name at the top of each page.
3. The exam will be marked out of 60 points. There are 8 questions. The points for each
question are indicated beside the question number.
4. Answers for questions 1 through 7 must be accompanied by adequate justification.
Please note that this exam will be marked electronically. Anything written
on the back of the pages will not be marked.
5. For multiple-choice questions you are not required to show your work.
6. This is a closed book exam. No books, notes, calculators, cell phones or other
electronic aids are allowed!
,
3
1. [7 pts] Find the domain of f (x) = ln 2
− |x − 1| + |x − 2| − |x − 3| . Express the
solution using interval notation.
Solution.
Since the logarithmic function is only defined for positive values of the argument, we
conclude that
3
2
− |x − 1| + |x + 2| − |x − 3| > 0.
To solve this inequality, we consider several cases.
• Case x ≥ 3. In this case, |x − 1| = x − 1, |x − 2| = x − 2, and |x − 3| = x − 3,
hence the inequality becomes 23 − (x − 1) + (x + 2) − (x − 3) > 0. Solving this, we
get x < 27 . Thus, the answer for this case is [3, 72 ).
• Case 2 ≤ x < 3. In this case, |x−1| = x−1, |x−2| = x−2, and |x−3| = −(x−3),
hence the inequality becomes 32 − (x − 1) + (x + 2) + (x − 3) > 0. Solving this, we
get x > 25 . Thus, the answer for this case is ( 52 , 3).
• Case 1 ≤ x < 2. In this case, |x − 1| = x − 1, |x − 2| = −(x − 2), and
|x − 3| = −(x − 3), hence the inequality becomes 23 − (x − 1) − (x + 2) + (x − 3) > 0.
Solving this, we get x < 23 . Thus, the answer for this case is [1, 32 ).
• Case x < 1. In this case, |x − 1| = −(x − 1), |x − 2| = −(x − 2), and |x − 3| =
−(x − 3), hence the inequality becomes 32 + (x − 1) − (x + 2) + (x − 3) > 0. Solving
this, we get x > 21 . Thus, the answer for this case is ( 12 , 1).
Combining (taking the union of) the answers in the four cases, we obtain the final
answer:
( 21 , 32 ) ∪ ( 52 , 72 ).
version 1
Date: October 28, 2016 Time: 90 minutes
Surname: Given name(s):
(Please, print!)
ID#: Email: @ualberta.ca
Please, check your section/instructor!
Section Instructor X
EA1 V. Yaskin
EB1 E. Leonard
EC1 G. Tokarsky
ED1 V. Troitsky
EE1 V. Yaskin
EG1 S. Graves
Instructions
1. Place your U of A Student ID card on your table or desk.
2. The exam has 11 pages including this cover page. Please ensure that you have all
pages, and write your name at the top of each page.
3. The exam will be marked out of 60 points. There are 8 questions. The points for each
question are indicated beside the question number.
4. Answers for questions 1 through 7 must be accompanied by adequate justification.
Please note that this exam will be marked electronically. Anything written
on the back of the pages will not be marked.
5. For multiple-choice questions you are not required to show your work.
6. This is a closed book exam. No books, notes, calculators, cell phones or other
electronic aids are allowed!
,
3
1. [7 pts] Find the domain of f (x) = ln 2
− |x − 1| + |x − 2| − |x − 3| . Express the
solution using interval notation.
Solution.
Since the logarithmic function is only defined for positive values of the argument, we
conclude that
3
2
− |x − 1| + |x + 2| − |x − 3| > 0.
To solve this inequality, we consider several cases.
• Case x ≥ 3. In this case, |x − 1| = x − 1, |x − 2| = x − 2, and |x − 3| = x − 3,
hence the inequality becomes 23 − (x − 1) + (x + 2) − (x − 3) > 0. Solving this, we
get x < 27 . Thus, the answer for this case is [3, 72 ).
• Case 2 ≤ x < 3. In this case, |x−1| = x−1, |x−2| = x−2, and |x−3| = −(x−3),
hence the inequality becomes 32 − (x − 1) + (x + 2) + (x − 3) > 0. Solving this, we
get x > 25 . Thus, the answer for this case is ( 52 , 3).
• Case 1 ≤ x < 2. In this case, |x − 1| = x − 1, |x − 2| = −(x − 2), and
|x − 3| = −(x − 3), hence the inequality becomes 23 − (x − 1) − (x + 2) + (x − 3) > 0.
Solving this, we get x < 23 . Thus, the answer for this case is [1, 32 ).
• Case x < 1. In this case, |x − 1| = −(x − 1), |x − 2| = −(x − 2), and |x − 3| =
−(x − 3), hence the inequality becomes 32 + (x − 1) − (x + 2) + (x − 3) > 0. Solving
this, we get x > 21 . Thus, the answer for this case is ( 12 , 1).
Combining (taking the union of) the answers in the four cases, we obtain the final
answer:
( 21 , 32 ) ∪ ( 52 , 72 ).
