Calculus I
2023 – 2024 S1
Áurea Quintino
Francisco Agostinho
Helena Almeida
Islam Elbaz
João Farinha
Pamela Pacciani
Rodrigo Duarte
Sofia Henriques
Regular Season Exam
•
Date: December 21, 2023
•
Duration: 2 hours
Instructions:
1. This exam is composed of 9 multiple choice questions and 3 open-ended questions.
2. Multiple choice questions should be answered on this platform (Moodle).
3. Open-ended questions should be answered on paper, using the booklet provided, which should be
handed in at the end of the exam.
4. Each multiple choice question admits a single correct answer.
5. For each multiple choice question, each correct answer, wrong answer and blank answer submitted is
classified with, respectively, 1, - 0.33 and 0. The minimum grade of this group is 0.
6. In open-ended questions, present all your computations and justify conveniently all your answers.
7. Do not answer more than one group on the same sheet of paper.
8. No written support or calculators are allowed.
9. You may freely navigate through all questions.
10. On the right side of each page and at the end of the last page, you may click on Finish attempt when you
are ready to submit all your answers. However, when you do this, you still do not submit your answers. You
may click on Return to attempt and continue changing your answers, or on Submit all and finish and submit
them.
11. When you finish your answers, click on Submit all and finish and, after that, ask the invigilator to
introduce the exit password.
12. If time runs out before you submit your answers, they will be automatically saved as they are. Before you
leave the room, do not forget to ask the invigilator to introduce the exit password.
13. Once you finish the exam and have the exit password introduced, please take pictures of your answer
booklet and, after that, hand it in. The pictures of the answer booklet are exclusively for consultation
by the student in the occasion of exam revision. Only the paper version of the answer booklet will be
assessed, so no student may leave the room without handing in their answer booklet.
14. Mobile phones should remain switched off, and placed on top of the table, for the duration of the
exam.
Calculus I 2023 – 2024 S1
Regular Exam
1.
2.
Calculus I 2023 – 2024 S1
Regular Exam
3.
4.
5.
Calculus I 2023 – 2024 S1
Regular Exam
6.
7.
8.
Calculus I 2023 – 2024 S1
Regular Exam
9.
10.
11.
Calculus I 2023 – 2024 S1
Regular Exam
12.
Calculus I 2023 – 2024 S1
Regular Exam
Solution Topics
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
a)
𝑎 ∈ ℝ arbitrarily fixed.
• Recognize that the continuity of 𝑓 at (𝑎, 𝑎), accumulation point of 𝐷𝑓 , is equivalent to
lim 𝑓(𝑥, 𝑦) = 𝑓(𝑎, 𝑎).
(𝑥,𝑦)→(𝑎,𝑎)
•
•
Observe that 𝑓(𝑎, 𝑎) = 0.
Verify that
lim 𝑓(𝑥, 𝑦) = 0.
(𝑥,𝑦)→(𝑎,𝑎)
𝑦≤𝑥
Calculus I 2023 – 2024 S1
Regular Exam
𝑎
•
Verify that, if 𝑎 ≠ 0, then
•
continuous at points (𝑎, 𝑎) with 𝑎 ∈ ℝ ∖ {0}.
Recognize that the direct computation of
lim
lim
(𝑥,𝑦)→(𝑎,𝑎)
𝑦>𝑥
𝑓(𝑥, 𝑦) = √2 ≠ 0, and conclude, in this way, that 𝑓 is not
(𝑥,𝑦)→(0,0)
𝑦>𝑥
𝑓(𝑥, 𝑦) leads to an indeterminate form,
so that the definition of limit should be used, indicating a sequence of correct upper bounds
and finding a valid relation between 𝜀 and 𝛿 (for example, 𝜀 = 𝛿), concluding, in this way,
that 𝑓 is continuous at (0,0).
•
b)
Verify that 𝑓𝑥′ (0,0) = 0 = 𝑓𝑦′ (0,0).
•
Verify that the approximation error is given by 𝑅(𝑥, 𝑦) = 𝑓(𝑥, 𝑦), so that 𝑓 is differentiable
at (0,0) if and only if
•
Verify that
•
Show that
lim
𝑓(𝑥,𝑦)
lim
(𝑥,𝑦)→(0,0) ||(𝑥,𝑦)||
𝑓(𝑥,𝑦)
(𝑥,𝑦)→(0,0) ||(𝑥,𝑦)||
𝑦≤𝑥
𝑓(𝑥,𝑦)
lim
(𝑥,𝑦)→(0,0) ||(𝑥,𝑦)||
𝑦>𝑥
= 0.
= 0.
is not zero: for example,
𝑓(𝑥, 𝑦)
𝑥|𝑦|
2𝑥 2 2
= lim
=
lim
= ≠ 0.
(𝑥,𝑦)→(0,0) ||(𝑥, 𝑦)||
(𝑥,𝑦)→(0,0) 𝑥 2 + 𝑦 2
𝑥→0+ 5𝑥 2
5
lim
𝑦=2𝑥
𝑥>0
•
𝑦=2𝑥
𝑥>0
Conclude that 𝑓 is not differentiable at (0,0).
11.
a)
•
2
State that lim+ 𝑓(𝑥) = lim+ 𝑥 (𝑥 ) can be computed using the Cauchy Rule.
𝑥→0
𝑥→0
(𝑥 2 )
2
ln(𝑥 (𝑥 ) )
lim 𝑥 2 ln 𝑥
lim
𝑥→0+
ln 𝑥
1
𝑥2
•
Write lim+ 𝑥
•
Apply the Cauchy Rule to the last limit, since it results in a indeterminate form of the
∞
type ∞ , and obtain 𝑒 0 = 1.
•
Compute lim− 0 𝑥
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the result is 1.
Conclude that we can define a continuity extension of 𝑓 to 𝑥 = 0 since 0 ∉ 𝐷𝑓 and the
limit of 𝑓 at 𝑥 = 0 is finite (both limits are equal to 1).
𝑥→0
= lim+ 𝑒
𝑥→0
𝑥
=𝑒
using the Fundamental Theorem of Calculus and conclude that
•
b)
•
.
2
∫ 𝑒 𝑡 𝑑𝑡
𝑥→0
=𝑒 𝑥→0+
𝑥
2
∫ 𝑒 𝑡 𝑑𝑡
Realise that when 𝑥 < 0, 𝑓 is defined by 𝑓(𝑥) = 0 𝑥
.
Calculus I 2023 – 2024 S1
Regular Exam
𝑥
2
0
2
•
Use the properties of the integrals to conclude that ∫0 𝑒 𝑡 𝑑𝑡 = − ∫𝑥 𝑒 𝑡 𝑑𝑡.
•
Since 𝑒 𝑡 > 0, ∀𝑡 ∈ ℝ and 𝑥 < 0, then ∫𝑥 𝑒 𝑡 𝑑𝑡 > 0, ∀𝑥 < 0.
•
Therefore, − ∫𝑥 𝑒 𝑡 𝑑𝑡 < 0, ∀𝑥 < 0.
•
Conclude that the proposition is true, since the numerator and the denominator of the
general expression of 𝑓 are both negative when 𝑥 < 0.
0
2
0
2
2
12.
a)
•
Represent the level curve of level 0 of 𝑓 as the union of the curve with equation 𝑥 = 0 with
the curve with equation 𝑥 = 𝑦 3 .
b)
• Recognize that 𝐷𝑔∘𝑓 = {(𝑥, 𝑦) ∈ 𝐷𝑓 : 𝑓(𝑥, 𝑦) ∈ 𝐷𝑔 }.
• Observe that 𝐷𝑓 = {(𝑥, 𝑦) ∈ ℝ2 : 𝑦 ≠ −𝑥} ∪ {(0,0)} and that 𝐷𝑔 = ℝ ∖ {0}.
•
Conclude that 𝐷𝑔∘𝑓 = {(𝑥, 𝑦) ∈ ℝ2 : 𝑦 ≠ −𝑥 ∧ 𝑥 ≠ 𝑦 3 ∧ 𝑥 ≠ 0}.