Final examination semester 2 academic year 2019-2020 course name Calculus 1 giúp các bạn học sinh có thêm tài liệu ôn tập, luyện tập nhằm nắm vững được những kiến thức, kĩ năng cơ bản, đồng thời vận dụng kiến thức để giải các bài tập một cách thuận lợi.
Trang 1No.: BM1/QT-PĐBCL-RĐTV Page: 1
Question 1 (1 point) Given f x( )=x3+3, ( )g x =cos (2 )x
a) f x and ( )( ) g x are even, odd or neither?
b) Find (f g x and ()( ) g f)( )x
Question 2 (1.5 points)
a) Find the value of the constant m such that the following piecewise - defined function
is continuous everywhere
( )
3 1
x e
x
=
b) With m found in question a), find f '( )x ,∀ x
Question 3 (1 point)
Let y be an implicit function of x satisfying the equation:
x2+xy−y3 = 1
Find the tangent line to the graph of the equation at the point M(1; 1)−
Question 4 (1 point)
Find the relative extrema of ( ) 0.5
g x = x− e−
Question 5 (1 point)
Let ( ) 22
4
x
f x
x
+
=
− Find the average value of f on the interval [ ]0, 1
Question 6 (1 point)
Find the particular solution of the separable differential equation
2
ln
x
dx = e
satisfying the initial condition y( )0 = e
Question 7 (1.5 points)
Assume that the position at time t of an object moving along a line is given by
s t = t − t + t for t on [1, 4]
a) Find the initial velocity and acceleration for the object
b) When is the object advancing and retreating?
c) When is the object accelerating and decelerating?
HCMC UNIVERSITY OF TECHNOLOGY AND
EDUCATION
HIGH QUALITY TRAINING FACULTY
GROUP OF MATHEMATICS
-FINAL EXAM, SEMESTER 2, 2019-2020 Subject: Calculus 1
Course code: MATH141601E Number of pages: 02 pages
Duration: 90 minutes
Date of exam: 31/07/2020 Materials are allowed during the exam
Trang 2Question 8 (1 point)
The volume of a spherical balloon is increasing at a constant rate of
10 3
/
cm s At what rate is the radius of the balloon increasing when the
radius is 3 cm?
Question 9 (1 point)
A cylinder box (Figure 1) is constructed with the volume
3
24
V = πcm The cost of the material used for the bottom is $2/cm2, the
cost of the material used for the lateral side is $2.5/cm2 and the cost of the
material used for the top is $5/cm2 Find the dimension of the box r and h
that minimizes the total cost
Notice: Invigilators should not explain the questions on the exam papers
[G 2.1]: Present mathematical information using words, statements,
numbers, formulas, graphs and diagrams
1
[G 1.1, 1.3, 5.2]: Students are able to find basic limits and test the
continuity of a function Students are able to find derivative and
differential
2, 3, 4
[G 3.1, 5.4] : Apply important rules and theorems effectively, such
as the mean value Students are able to apply theory to evaluate
indefinite and definite integrals
5
[ELO 1.4, 5.4]: Students are able to solve basic differential
equations
6
[G 4.1] Identify and analyze the information given in formulas,
graphs and tables relating to (a) rectilinear motion; (b) linear; (c)
optimization and applications in physics; (d) Riemann sum and
integration
7
[G 5.3]: Students are able to use derivative to solve problems
relating to rates of change and optimization
8,9
July 24th, 2020
Head of group of mathematics
Trang 3No.: BM1/QT-PĐBCL-RĐTV Page: 3
Đáp án
Q1: (1 point)
a) Because of (f −x)≠ ±f x( ), f(x) is neither odd nor even (0.25pt)
Because of (g − =x) g x( ), g(x) is even (0.25pt) b) (f g x)( )=(cos(2 ))x 3+3;(g f)( )x =cos(2(x3+3)). (0.25pt+0,25pt)
Q2:(1,5 points) D = ℝ f
a) x≠0 : ( )f x is continuous with all x ≠ 0
( )
f x is continuous with all x ∈ℝ ⇔ f x( ) is continuous with at x = (0.25pt) 0
3
1
x
e
x
−
b)
2
0 : ( )
e x e
x
′
3
2 0
2
x x
f
x
→
− −
Q3: (1 point)
F x y =x +xy− y − =
Differentiate both sides of the equation (*) with respect to x, we get
2
2
2
3
x y
+
− (0.25pt+0,25pt)
At M(1; -1): '(1) 1
2
y = (0.25pt) The quation of the tangent line to the graph of the equation at the point M is :
d y= x− − ⇔ =y x− (0.25pt)
Q4 (1 point)
g
D = ℝ
We have '( ) 0.5x(2 0.5 )
g(x) is increasing on (−∞, 4) and is decreasing on (4,+∞) (0.25pt)
So, g(x) has a relative maximum at x = 4, f(4) = 2e-2 (0.25pt)
Q5 (1 point)
The average value of f on [0, 1] is
1
2 0
2 4
x
x
+
=
−
Trang 4AV = 1 2
2
x
2
2 0
4
x
x
−
1 1
1
1
2
x
x
π
−
−
3
So AV = − +π
(0.25pt)
Q6 (1 point):
Then we have
2
2
x e
−
Because of y(0)= , e 1
2
C = Then
x e y
−
− (0.25pt)
Q7(1.5 points):
a)
2 ( ) '( ) 6 30 36 (1) 12
v t =s t = t − t+ ⇒v =
(0.25pt)
( ) '( ) 12 30 (1) 18
b) The object is advancing ⇔v t( )> ⇔ ∈0 t [1, 2)or t∈(3, 4] (0.25pt)
The object is retreating ⇔v t( )< ⇔ ∈0 t (2,3)
(0.25pt) c) The object is accelerating ⇔a t( )> ⇔ ∈0 t (2.5, 4]
(0.25pt) The object is decelerating ⇔a t( )< ⇔ ∈0 t [1, 2.5) (0.25pt)
Q8 ( 1 point)
3
4
3
2
R
dt = dR dt = π dt
(0.25pt)
2
3 10 4 3 dR
R
dt
π
5
( / ) 18
dR
cm s
So the radius is increasing at the rate of 5 ( / )
18π cm s (0.25pt)
Trang 5No.: BM1/QT-PĐBCL-RĐTV Page: 5
Q9 (1 point)
2
2
24 24
r
(0.25pt)
r
π
(0.25pt)
3 2
7
r
π π
(0.25pt)
3
240
r
π π
= + > ∀ >
(0.25pt)