MULTIPLE CHOICE QUESTIONS
1
Functions
1.1
Domain and range of a function
s 9
Question 1 (L.O.1). Find the domain of function y = ln 3 +
.
x


9
9
x⩽−
x⩽−
9
9


2
2
C x⩽−
A − ⩽x<0
B
D
2
2
x>0
x⩾0
E x>0
Question 2 (L.O.1). Find the range of the function

 9x2 − 3, if x > 0
f (x) =

21x3 , if x ⩽ 0
A (0, +∞)
B (−∞, +∞)
D (−3, +∞)
C (−∞, 0)
E (−3, 0)
Question 3 (L.O.2). A family plans to organize a party at a restaurant. The overall service cost for the event
(venue, decoration, etc.) is 4 million Vietnamese dong. The restaurant requires a minimum of 2 tables, with
each table accommodating 8 people. The cost per guest is 258 thousand Vietnamese dong. The family plans
to spend a maximum of 36.6 million Vietnamese dong. If we denote x as the number of invited guests,
C(x) (in million Vietnamese dong) represents the cost for the party. Find the domain D and the range R of
C.
A D = N ∩ [8; 142], R = [2.064; 36.6]
B D = [16; 142], R = [8.128; 36.6]
C D = N ∩ [16; 142], R = [8.128; 36.6]
D D = N ∩ [0; 142], R = [0; 36.6]
E D = [16; 36.6], R = [4.016; 142]
1.2
Description of a function
Question 4 (L.O.2). A rectangular storage container with an open top has a volume of 14m3 . The length
of its base is quadruple its width. Material for the base costs 11 dollars per square meter; material for the
sides costs 8 dollars per square meter. Express the cost of materials as a function of the width w of the
base.
A 44w2 +
280
w
B 44w2 + 280
C 32w2 +
1120
w
D 32w2 +
385
w
E 44w2 +
140
w
Question 5 (L.O.2). To create an animated image appealing to players in a game, the game creator started
by creating a rectangular image with dimensions of 29 × 71 pixels. Then, each side is increased at a rate of
15 pixels per second. Determine the area of the image after t seconds.
A 2059
D 225t2
EXAM
B 225t2 + 1500t + 2059
C 225t2 + 1065t
E 225t2 + 435t
Page 1/1 — Question code: 1234
Question 6 (L.O.2). The amount of adrenaline in the body changes rapidly. Let’s assume that initially the
body has 18 mg of adrenaline, create a function describing the adrenaline level A = f (t) in the body after t
(minutes), given that A increases 0.39 mg per minute.
A 18t + 0.39
1.3
C 0.39 − 18t
B 7.02t
E 18 − 0.39t
D 0.39t + 18
Inverse functions
Question 7 (L.O.1). Let f (x) = x5 + 3x3 + 6x − 5. What is the value of f −1 (63).
A 1
2
B 0
D 2
C 6
E 3
The limit of a function, continuous functions
2.1
The limit of a function
e3.19x − cos(x)
lim
x→0
x
Question 8 (L.O.1). Find m > 0 such that
A 1.6481
2
.m +
ln(1 + 6.04x)
lim
x→0
x
C 1.2978
D 1.3338
6x
6x + m2
= e26.55m .
Question 9 (L.O.1). Find m > 1 such that lim
x→∞
6x − 1
C 27.3814
A 26.5123
B 26.5475
D 26.8591
12
Question 10 (L.O.1). Find lim 1 + 2x2 e7x 9x .
A e−7/9
B 1.2026
.m = 18.62.
E 2.3985
E 26.4543
x→0
C −e2/9
B 1
E e2/9
D 0
Question 11 (L.O.1). Find S = 4a + 8b where a and b are the two real constants such that
2
lim x ea/x + + b = 15.
x→+∞
x
A 43
B 40
C 48
D 44
e4ax − 1 − 9x
·
x→0
x + 13x2
9
B 4a − 9, ∀a ̸=
4
E 4a − 9, ∀a ∈ R
E 42
Question 12 (L.O.1). Evalulate I = lim
9
4
D 4a − 9, ∀a ̸= 0
A 4a, ∀a ̸=
2.2
C 4a, ∀a ∈ R
Continuity
Question 13 (L.O.1). Find real values of a such that f (x) =
at x = 13.
2
A
13
B −
24
13
C −
37
13


ax − 2,
√
 arctan( x − 13),
D −
50
13

3

 5 arctan x − 31
x−3
Question 14 (L.O.1). Find real values of a such that f (x) =


a,
at x = 3.
5π
3π
5π
A −
B
C
2
2
2
7π
D a does not exist
E
2
EXAM
if x ⩽ 13
is continuous
if x > 13
E −
if x ̸= 3
11
13
is continuous
if x = 3
Page 2/1 — Question code: 1234
2.3
Asymptotes
Question 15 (L.O.1). Find all real numbers a so that the line of equation x = 0 is a vertical asymptote of
ln(7x + a − 8)
the curve y =
·
x+9
C 6
A 11
B 7
D 8
E 12
9x
− 5.
1 − e−x
C y = 9x − 5
Question 16 (L.O.1). Find all asymptotes of the graph of the function y = 4x + √
A x = 0, y = 13x − 5
B y = 13x − 5
D x = 0, y = 4x − 5
E y = 4x − 5
Question 17 (L.O.2). A plastic molding company states that if they use x% of the company’s machines, the
total monthly cost for operating these machines is given by the function:
C(x) =
19x2 − 57x − 60610
4x2 − 100x − 7656
(hundreds of dollars).
The company employs an alternating maintenance mode to utilize close to 58% of the machines (optimal
capacity).What is the total cost that the company has to incur for operating this number of machines if the
machine usage reaches the optimal capacity?
A 5.8984
3
B 5.9918
C 5.7198
D 6.6802
E 5.7032
Derivatives
3.1
Tangents
Question 18 (L.O.1). A point M (a, b) is on the graph of the function f (x) = 1.55ex − x. Given that the
tangent line to the graph at M is parallel to the line y = 2.47x + 2.82. Find the value of a.
A 0.8059
C 1.7944
B 0.4466
D 1.4552
E 1.2394
Question 19 (L.O.1). Find ALL points on the graph of f (x) = 4x3 − 192x where the tangent line is parallel
to the x−axis.
A (4; 512)
B (4; −512) and (−4; 512)
D (4; 512) and (−4; −512)
C (4; −512)
E (−4; 512)
Question 20 (L.O.1). Find ALL points on the graph of f (x) =
9
where the tangent line is perpendicular to
x
the line y = 16x − 4.
A (−12; 3/4)
C (12; −3/4) and (−12; 3/4)
B (12; 3/4)
D (−12; −3/4)
E (12; 3/4) and (−12; −3/4)
√
Question 21 (L.O.1). Find the real value of a such that the tangent line to the graph of the function y = 3 x
21
at x = a has y−intercept .
2
C 49
A 46
B 52
D 44
E 48
3.2
Tangent to paramtric curve
Question 22 (L.O.1). Let the curve be given by paramtric equations

 x(t) = (3 − 4t) e3t
 y(t) = 3t2 − 5t − 4 e3t
EXAM
Page 3/1 — Question code: 1234
37
Find the set of all values of t such that the tangent at M (x(t), y(t)) has the slope of − ?
31 38
503
410
131
148
A − ;3
B −
C −
D −
E
; −1
;7
;4
;1
93
93
93
93
93
3.3
Differentiable functions
x2 − 11
is not differentiable.
x2 − 25
A x ̸= 5
B x = −5
D x = ±5
E x ̸= ±5
C x=5
p
Question 24 (L.O.2). Determine ALL values of x such that f (x) = 3 (x2 − 25)2 is not differentiable.
Question 23 (L.O.2). Determine ALL values of x such that f (x) =
A x = ±5
B x ̸= ±5
C x ̸= 5
D x = −5
E x=5
Question 25 (L.O.2). Find S = a + b such that function f is differentiable at x = 0, if

 2x2 + 7x + 5, x ⩽ 0
f (x) =

ln(ax + b), x > 0
A 7e5
3.4
B 8
C 8e5
D 9
E 6e5
The chain rule
Question 26 (L.O.1). Let f (x) and g(x) be functions satisfying the conditions: f (−4) = f ′ (−4) = 9 and
g(9) = g ′ (9) = −4. Consider the function h(x) = (f ◦g◦f )(x) = f (g(f (x))). Find the value of h′ (−4).
A −328
B −325
C −322
D −324
E −329
Question 27 (L.O.2). An importer of Rwandan coffee estimates that local consumers will buy approxi4346
pounds of the coffee per week when the price is p dollars per pound. It is estimated
mately D(p) =
p2
that t weeks from now, the price of Rwandan coffee will be p(t) = 0.02t2 + 0.08t + 9 dollars per pound. At
what rate will the weekly demand for the coffee be changing with respect to time 6 weeks from now?
A −3.1978
B −2.3627
C −3.4126
D −3.2408
E −2.621
Question 28 (L.O.2). When electric blenders are sold for p dollars apiece, local consumers will buy approx7928
imately D(p) =
blenders per month. It is estimated that t months from now, the price of the blenders
p√
will be p(t) = 0.03 t3 + 4 dollars. Compute the rate at which the monthly demand for the blenders will be
changing with respect to time 16 months from now.
A −40.9061
3.5
B −40.7186
C −40.9638
D −40.9057
Derivatives of inverse function
Question 29 (L.O.1). If f (x) = 2x + 2 sinh (x − 2), find (f −1 )′ (4).
9
19
13
15
A
B −
C
D −
4
4
4
4
3.6
E −41.3092
E
1
4
Undetermined Coefficients
Question 30 (L.O.2). Let y = Ax3 + Bx + C, where A, B, C ∈ R, satisfy the equation y ′′′ + 2y ′′ − 5y ′ + 9y =
27x3 − 45x2 + 54x + 89. Find the value of S = A + B + C.
A 11
EXAM
B 9
C 16
D 15
E 14
Page 4/1 — Question code: 1234
Question 31 (L.O.2). Let y = Ax cos x + Bx sin x, where A, B ∈ R, satisfy the equation y ′′ + y = −4 sin x +
10 cos x. Find the value of S = A + B.
A 6
B 9
C 8
D 5
E 7
Question 32 ([L.O.2). Let y = Axp + Bx + C, where A, B, C, p ∈ R, satisfy the equation y ′′ =
y ′ (1) = −3.0, y(1) = 5.0. Find the value of S = (A + B + C).p.
A 11
3.7
B 7
C 9
D 8
30
and
x7
E 12
Rates of change
Question 33 (L.O.2). The management fee for running services in an area is calculated by the formula
f (x) = 0.0692x2 + 0.0902x (in dollars), where x is the number of homes being present in the area. Suppose that at the moment when 63 homes are present in this area, the number of homes is increasing at
a rate of 5 homes per week. What is the rate of change (in dollars/week) in the management fee at this
moment?
A 44.047
B 44.5156
D 43.2984
C 43.0986
E 44.6353
25
· If the abscissa of
x
the particle increases by 0.3234 units per second, how does its ordinate vary as it passes through the point
Question 34 (L.O.2). A particle is moving along a hyperbolic trajectory given by y =
(5; 5)?
A Increases 0.0971 units per second
B Increases 0.8569 units per second
C Decreases 0.4064 units per second
D Decreases 0.1947 units per second
E Decreases 0.3234 units per second
3.8
Related Rates
Question 35 (L.O.2). Recall that volume of a right cylinder is V = π.r2 .h, where r is the radius of the base
and h is the height. Given a right cylinder with the radius of the base of 9 cm and the height of 5 cm. The
radius of the base is increasing at a rate of 0.8 cm/s, and its height is increasing at a rate of 0.21 cm/s. How
fast is the volume of cylinder increasing?
A 278.9661
B 280.2004
C 279.6332
D 279.1033
E 279.8283
Question 36 (L.O.2). Water pours into a conical tank of height 10(m) and radius 4(m) at a rate of 7(m3 /min).
4
r
10
h
At what rate is the water level rising when the level is 5(m) high?
A 0.5570
EXAM
B −0.3594
C 0.1141
D 0.0975
E 0.7804
Page 5/1 — Question code: 1234
Question 37 (L.O.2). A leaky water tank is in the shape of an inverted right circular cone with depth
of 11(m) and top radius 3(m). When the water in the tank is 3(m) deep, it is leaking out at a rate of
1/6(m3 /min).
3
r
11
h
How fast is the water level in the tank dropping at that time?
A 0.0792
B 0.3362
C −0.6292
D 0.1284
E −0.5812
Question 38 (L.O.2). Boyle’s law states that when a sample of gas is compressed at a constant temperature,
the pressure P and volume V satisfy equation P V = C = const. Suppose that at a certain instant the volume
is 647cm3 , the pressure is 184kP A, and the pressure is increasing at a rate of 28kP A/min. At what rate is
the volume decreasing at this instant?
A 98.1919
3.9
B 99.3825
C 98.4565
D 97.6455
E 98.362
D −299dx2
E −294dx2
Differentials
Question 39 (L.O.1). Find d2 y(0) if y = cos6 (7x).
A −292dx2
3.10
B −289dx2
C −298dx2
Applications of derivatives in physics
Question 40 (L.O.1). A particle moving on the x−axis has position s(t) = t3 − 39t2 + 432t + 12 (m) after an elapsed time of t seconds. What is the total distance travelled by the particle during the first 31
seconds?
A 6699
B 6707
C 6704
D 6709
E 6702
Question 41 (L.O.1). A particle moving on the x−axis has position s(t) = t3 − 45t2 + 600t + 9 (m) after
an elapsed time of t seconds. What is the displacement travelled by the particle during the first 37 seconds?
A 11243
3.11
B 11250
C 11248
D 11251
E 11253
Applications of derivatives in Economics
Question 42 (L.O.2). For Luce Landscaping, the total revenue from the yard maintenance of x homes is
given by R(x) = 1868x − 4x2 (dollars) and the total cost is given by C(x) = 2938 + 18x (dollars). Suppose
that Luce is adding 19 homes per day at the moment when the 458th customer is signed. At that moment,
what is the rate of change of total profit P in dollars per day, if the total profit P (x) = R(x) − C(x)?
A −34469
EXAM
B −34465
C −34471
D −34466
E −34463
Page 6/1 — Question code: 1234
3.12
Linear Approximations
Question 43 (L.O.1). Use the linear approximation formula to find an approximate value of f (x) at x =
12.17, given that f (12) = 4 and f ′ (12) = 0.74.
A 3.9803
B 4.4361
C 4.1258
D 3.8998
E 3.8271
Question 44 (L.O.2). A spherical balloon is inflated so that its radius increases from 20 cm to 20.44 cm in
4 seconds. By approximately how much has its volume increased (using linear approximation) (in cm3 /s)?
A 552.2614
B 552.9203
C 553.2891
D 552.9905
E 552.0326
Question 45 (L.O.2). A company estimates that volume sales S(x) of a type of commodity depends on the
amount of money x (in thousands dollars) spent on advertising, that is given by:
S(x) = −0.0079x3 + 0.412x2 + 2.8147x + 338.
Use the differential (also known as: the linear approximation formula) to estimate the change in the total sales of this commodity if the advertising expenditures are increased from 41 to 41.7583 (thousands
dollars).
A −2.4577
B −1.7600
C −2.4103
D −2.4215
E −1.9167
Question 46 (L.O.2). A ball of ice melts so that its radius decreases from 30 cm to 29.1647 cm. By approximately how much does the volume of the ball decrease?
A −9447.0725
B −9446.1664
C −9446.7456
D −9447.01
E −9447.2685
Question 47 (L.O.2). A water reservoir has a base shaped like a rectangle combined with half of a circle,
as illustrated in the picture.
x
4x
If we can measure x = 5 ± 0.0599(m). Approximate the error in the area of the reservoir’s base using
differential formula.
A 6.2392
B 5.6269
C 6.6628
D 6.6852
E 6.1596
Question 48 (L.O.1). Let f be an even function and f is differentiable in R. Given that f ′ (1) = 2.81,
f (1) = −3. Use differential formula to approximate f (−0.22).
A −6.0772
EXAM
B −5.1918
C −5.673
D −6.0849
E −6.0731
Page 7/1 — Question code: 1234
3.13
Monotonicity
ln(3x + 4)2
, then which statement is always true?
3x + 4
4 e 4 e
A The function y is decreasing on − − , − +
3 3 3 3 4 4 e
4 e 4
and − , − +
B The function y is increasing on − − , −
3 3 3
3 3 3
4 e 4 e
C The function y is increasing on − − , − +
3 3 3 3
Question 49 (L.O.1). If y =
D The function y is increasing on R
4 e
4 e
E The function y is increasing on − − , 0 ∪ 0, − +
3 3
3 3
tan x − 3
Question 50 (L.O.1). Find all real numbers m such that y =
is increasing on the interval
tan x − m
π
.
0;
4
A 1⩽m<3
B m>3
C m⩾0
D m ⩽ 0 or 1 ⩽ m < 3
3.14
E m⩽0
Local extreme values
Question 51 (L.O.1). Studying the local extreme values of the function f (x) = arctan
3x − 9
x2 + 72
, which
statement is always true?
A f has local minimum when x = −6, and no local maximum
B f has local minimum when x = −6, and local maximum when x = 12
C f has local maximum when x = −6, and local minimum when x = 12
D f has no local minimum, and local maximum when x = 12
E f has no local minimum, and no local maximum
3.15
Local extreme values of the parametric curve
Question 52 (L.O.1). Let y = y(x) be the parametric curve defined by

 x(t) = −5t ln(t + 4)
(t ⩾ 0).
 y(t) = 2t3 − 24t2 + 72t + 4
Which statement is always TRUE?
A y attains local minimum value when x = −10 ln(6)
B y attains local maximum value when x = −10 ln(6)
C y attains local minimum value when x = 6
D y attains local maximum value when x = −30 ln(10)
E y attains local maximum value when x = 2
3.16
Concavity and inflection point
Question 53 (L.O.1). If y = x3 e(12/11)x , then the number of point(s) of inflection of the graph of y is
A 4
EXAM
B 1
C 0
D 2
E 3
Page 8/1 — Question code: 1234
3.17
Global maximum and minimum values
Question 54 (L.O.2). Find the maximum value of

 e2−x + x2 + 9x + 5, if 0 ⩽ x ⩽ 2
f (x) =

x2 − 24x + 72, if 2 < x ⩽ 12
A 23
B 31
C 25
D 30
E 28
Question 55 (L.O.2). A company specializing in producing athletic uniforms for students stated that the
cost to produce x batches (0 < x ⩽ 264) is
C(x) = 0.3x2 + 12x + 4915.2, (hundreds dollars)
The average cost per batch when producing x batches is given by the formula Cave =
C(x)
· Find x such that
x
the average cost atains minimum value.
A 125
3.18
B 131
C 124
D 123
E 128
Maclaurin-Taylor approximations
Question 56 (L.O.1). Find the coefficient of x8 in the Maclaurin polynomial of degree 9 for the function
f (x) = (2x − 5) sin (2x).
614
299
A
B
315
315
C −16/315
D −
1591
315
E −
646
315
Question 57 (L.O.1). Write the Taylor polynomial of degree 3 for the function f (x) = 2x2 − 8x + 11 e5x
where x is near x0 = 2.
79 (x − 2)2 e10 145 (x − 2)3 e10
+
+ O (x − 2)4 ; x → 2
2
2
79 (x − 2)2 e10 145 (x − 2)3 e10
10
10
3e − 2 + 15 (x − 2) e +
+
+ x + O (x − 2)4 ; x → 2
2
2
2 10
3 10
79
(x
−
2)
e
145
(x
−
2)
e
3e10 + 15 (x − 2) e10 +
+
− (x − 2)3 + O (x − 2)4 ; x → 2
2
2
79 (x − 2)2 e10
145 (x − 2)3 e10
2
4
10
10
3e + 15 (x − 2) e +
− (x − 2) +
+ O (x − 2) ; x → 2
2
2
2 10
3 10
79 (x − 2) e
145 (x − 2) e
3e10 + 15 (x − 2) e10 +
+
+ O (x − 2)4 ; x → 2
2
2
A 3e10 + 1 + 15 (x − 2) e10 +
B
C
D
E
4
Mixed problems
[L.O.2] Using the following information, answer the questions from 58 to 59.
A family plans to organize a party at a restaurant. The overall service cost for the event (venue, decoration,
etc.) is 8 million Vietnamese dong. The restaurant requires a minimum of 2 tables, with each table
accommodating 11 people. The cost per guest is 250 thousand Vietnamese dong. The family plans to spend a
maximum of 28.25 million Vietnamese dong. If we denote x as the number of invited guests, C(x) (in million
Vietnamese dong) represents the cost for the party.
Question 58. Find the domain of C(x) is
A N ∩ [22; 81]
EXAM
B [0; 81]
C [22; 28.25]
D N ∩ [0; 22]
E N ∩ [8; 28.25]
Page 9/1 — Question code: 1234
Question 59. The range of C(x) is
A [22; 81]
B [13.5; 81]
C [13.5; 28.25]
D [8; 28.25]
E [5.5; 28.25]
[L.O.2] Using the following information, answer the questions from 60 to 61.
In a certain country, income tax is assessed as follows. There is no tax on income up to $12340. Any income
over $12340 is taxed at a rate of 10%, up to an income of $23446. Any income over $23446 is taxed at 13%.
Question 60. Express the tax T as a function of income I.


0,
I ⩽ 12340


A T =
0.1, 12340 < I ⩽ 23446



0.13,
I > 23446


0,
I ⩽ 12340


B T =
C T =
D T =
0.1I, 12340 < I ⩽ 23446



0.1I,
I > 23446


0,
I ⩽ 12340


0.13I, 12340 < I ⩽ 23446



0.13I,
I > 23446


0,
I ⩽ 12340


0.1I, 12340 < I ⩽ 23446



0.13I,
I > 23446


0,
I ⩽ 12340


E T =
0.13I, 12340 < I ⩽ 23446



0.1I,
I > 23446
Question 61. How much tax (in dollars) is assessed on an income of $27148.
A 3529.6
B 3529.5
C 3528.7
D 3529.7
E 3529.2
[L.O.2] Using the following information, answer the questions from 62 to 64.
A water filter equipment manufacturer has a fixed monthly cost of 340 million Vietnamese dong, and the
variable cost to produce x devices is C(x) = −0.005x2 + 6.0x, (0 ⩽ x ⩽ 800) (million Vietnamese dong). The
selling price per device is p(x) = −0.002x + 8.6, (0 ⩽ x ⩽ 800) (million Vietnamese dong).
Question 62. The company’s revenue (million Vietnamese dong) when selling 144 devices in a month
is
A 1197.9280
B 1193.9280
C 1199.9280
D 1196.9280
E 1192.9280
Question 63. The company’s profit (million Vietnamese dong) when selling 144 devices in a month is
A 99.6080
B 94.6080
C 91.6080
D 96.6080
E 98.6080
Question 64. To achieve the profit of 2300 million Vietnamese dong in a month, how many devices does
the company need to sell in that month?
A 605
B 604
C 601
D 600
E 602
[L.O.2] Using the following information, answer the questions from 65 to 67.
EXAM
Page 10/1 — Question code: 1234
All pyramids with a square base and a height of h = 4 have a surface area given by
S(v) = 0.75v + 2
p
0.28125v 2 + 12v,
where v is the volume of the pyramid. If x is the length of each side of the base then we can consider the volume
dv
(7) = 18.67.
of a pyramid as a function v(x) of x. In addition, suppose that v(7) = 65.33 and
dx
Question 65. Compute the slope of the slant asymptote of the graph of function S(v) with respect to
v.
A 1.1961
B 1.0757
C 2.3389
D 2.7592
E 1.8107
Question 66. Compute the rate of change of the surface area of the pyramid with respect to the length x of
each side when x = 7.
A 34.1516
B 34.2931
C 34.1693
D 34.4337
E 35.2992
Question 67. Approximate the change in surface area of a pyramid when the length of each side of the base
increases from x = 7 to x = 7.75.
A 25.8357
5
B 25.1149
C 25.2789
D 25.8253
E 26.8070
Integration
5.1
Indefinite integrals
Question 68 (L.O.1). Find the antiderivative F (x) of the function f (x) = 8 sin x + 4 cos x which satisfies
π the condition F
= 12.
2
C −8 cos x + 4 sin x − 8
A −8 cos x + 4 sin x + 8
B −8 cos x − 4 sin x + 8
D 8 cos x − 4 sin x − 8
E 8 cos x + 4 sin x + 8
Question 69 (L.O.1). Find the antiderivative F (x) of the function f (x) = 8ex + 4x which satisfies the
condition F (0) = 14.
A 8ex + 2x2 − 6
B −8ex + x2 + 6
D 8ex + 2x2 + 6
C −8ex + 2x2 − 6
E ex + 2x2 + 6
Question 70 (L.O.1). Suppose that f ′′ (x) = −18x − 100 sin(5x) + 9e3x and f ′ (0) = 23, f (0) = −379.
Evaluate f (2).
A −2.2641
5.2
B −2.7473
C −2.0485
D −3.6915
E −3.5396
Definite integrals
Z
Question 71 (L.O.1). Let
π
2
Z
f (x) dx = 9.51. Evaluate I =
0
π
2
[6.49f (x) + 2.63 sin x] dx.
0
A 63.8129
B 64.3499
C 64.0289
D 64.5836
E 64.7637
Z 6.4
Z 6.4
Z 6.4 h
Question 72 (L.O.1). Let
f (x) dx = 6.88 and
g(x) dx = −2.47. Evaluate I =
18.86x +
−2.5
−2.5
−2.5
i
5.55f (x) − 7.76g(x) dx.
A 384.0257
EXAM
B 384.9296
C 385.2162
D 384.6665
E 385.0722
Page 11/1 — Question code: 1234
Question 73 (L.O.2). Let g(x) = f −1 (x) be the inverse function of function y = f (x) = ex/2 +5.34. Evaluate
Z 11.01
g(x)dx.
6.34
A 9.6536
5.3
B 10.3370
C 10.9418
D 10.7950
E 9.4292
Riemann sum
Question 74 (L.O.2). Using the midpoint rule with 5 equal interval on [3; 11] to estimate the following
Z 11
f (x)dx where f (x) is given by the following table
integral I =
3
x
3
3.8 4.6 5.4 6.2 7.0 7.8 8.6 9.4 10.2
f (x) 6.8 4.3 6.6 5.0 3.4 5.8 5.1 5.5 4.2
A 41.8
5.4
B 41.88
C 42.24
5.8
11
6.0
D 41.6
E 42.26
The fundamental theorem of calculus
Z
x
Question 75 (L.O.2). Find the maximum and minimum values of f (x) =
(t2 − 9t − 22)dt on the interval
−9
[−9, 9].
A fmax ≈ 433.8311; fmin ≈ −0.1873
B fmax ≈ 433.4807; fmin ≈ 0.638
C fmax ≈ 432.9731; fmin ≈ 0.1885
D fmax ≈ 433.7281; fmin ≈ 0.0568
E fmax ≈ 432.8333; fmin ≈ 0
6
Applications of integration
6.1
The Area problems
1
; y = 0; x = 0;
Question 76 (L.O.2). Find the area of the region bounded by the graphs of y = √
9 16 − x2
and x = 4.
A 0.5049
B 0.3040
C 0.2790
D 0.2973
E 0.1745
Question 77 (L.O.2). Find the area of the region bounded by the graphs of y = e2.53x ; y = e4.75x and
x = 3.03.
A 374034.3816
6.2
B 374034.2241
C 374034.3701
D 374034.9554
E 374034.1488
The Volume of the solid of revolution
Question 78 (L.O.2). Let the region D be bounded by the curve y =
p
15x2 + 2, x−axis and the lines x = 0,
x = 5. Evaluate the volume V of the solid, revolving the region D about x−axis.
A 1995.0119
B 1995.0890
C 1994.9113
D 1994.7434
E 1993.9313
Question 79 (L.O.2). Find the volume of the solid obtained by rotating the region bounded by the curves
x = 0, y = −16, y = x2 − 8x about the y−axis.
A 134.0465
EXAM
B 134.0413
C 133.9908
D 133.6074
E 134.2850
Page 12/1 — Question code: 1234
6.3
The arc length
Question 80 (L.O.2). Find the length of the arc y = 3 ln x, where 6 ⩽ x ⩽ 9.
A 3.2399
B 2.8595
C 3.9077
D 2.5529
E 3.1169
Question 81 (L.O.2). A steady wind blows a kite due west. The kite’s height above the ground from hori√
3
zontal position x = 0 to x = a > 0 meters is given by y =
3 + 2x . Find the real number a such that the
distance travelled by the kite is 20.66 meters.
A 3.6074
B 3.2822
C 2.5416
D 2.8492
E 3.5988
Question 82 (L.O.2). Find the length of the arc y = P (x), where 3 ⩽ x ⩽ 9. Here P (x) is a polynomial of
degree 2 such that P (3) = 62, P (4) = 96, P (9) = 386.
C 323.3469
D 324.0597
E 324.9452
Z xp
Question 83 (L.O.2). Find the length of the arc y =
5t2 + 7dt, where 2 ⩽ x ⩽ 8.
A 323.5810
B 323.6091
A 69.4857
B 69.7190
2
6.4
C 69.7552
D 69.0281
E 68.6696
The Area of the surface of revolution
√
Question 84 (L.O.2). Find the area of the surface of revolution obtained by rotating the curve y = 2 x + 6,
where 1 ⩽ x ⩽ 8 about the x−axis.
A 502.2993
B 502.5967
C 502.8761
D 502.4058
E 502.0702
p
Question 85 (L.O.2). Rotating the curve x = − y 2 + 3, where 0 ⩽ y ⩽ 3 about the y−axis, evaluate the
surface area of the obtained surface of revolution.
A 53.5843
6.5
B 54.2794
C 54.0413
D 53.1674
E 54.2563
Applications of integration in Physics
Question 86 (L.O.2). Evaluate the approximation of the displacement of one particle with velocity v(t) =
9 arcsin(t) (m/s) from t = 0s to t = 0.7s.
A 3.0163
B 3.2834
C 2.5228
D 1.9853
E 2.3123
Question 87 (L.O.2). The velocity of a dragster t seconds after leaving the starting line is v(t) = 72te−0.7t
ft/sec. Find the approximation of the distance traveled by the dragster during the first 9 seconds.
A 145.0934
B 144.0773
C 143.9867
D 144.9122
E 144.9691
Question 88 (L.O.2). When a particle is located a distance x meter from the origin, a force of given F (x) =
√
1
(newton) acts on it. How much work is done in moving it from x = 5 to x = 7 3 − 2.
2
x + 4x + 53
A −0.0604
B −0.0687
C 0.0374
D 0.0634
E −0.4302
Question 89 (L.O.2). When a particle is located a distance x meter from the origin, a force of given F (x) =
1
√
(newton) acts on it. How much work is done in moving it from x = 17/2 to x =
−16x2 + 128x + 1040
13.
A 0.2618
B −0.7293
C 0.5967
D −0.3280
E −0.6007
Question 90 (L.O.2). When a particle is located a distance x meter from the origin, a force of given F (x) =
ln(6 + 5x) (newton) acts on it. How much work is done in moving it from x = 10 to x = 17.
EXAM
Page 13/1 — Question code: 1234
A 30.1241
C 29.4318
B 30.4412
D 30.0137
E 29.2949
Question 91 (L.O.2). When a particle is located a distance x meter from the origin, a force of given F (x) =
60
√
(newton) acts on it. How much work is done in moving it from x = 1 to x = 6.
196 − x2
A 22.2853
B 22.7360
C 23.0572
D 21.5142
E 23.0559
6.6
Applications of integration in Economics
Question 92 (L.O.1). The accumulated, or total, future value after T years of an income stream of R(t)
thousand dollars per year, earning interest at the rate of r per year compounded continuously, is given by
Z T
R(t)e−rt dt. A company recently bought an automatic car-washing machine that is expected to
A = erT
0
generate 3 thousand dollars in revenue per year, t years from now, for the next 7 years. If the income is
reinvested in a business earning interest at the rate of r = 11% per year compounded continuously, find the
total accumulated value of this income stream at the end of 7 years.
A 30.6382
6.7
C 32.4296
B 31.6300
D 32.2997
E 31.7604
The average value of a funtion
Question 93 (L.O.2). The interest rates charged by Madison Finance on auto loans for used cars over a
certain 12−month period in 2022 are approximated by the function
r(t) = −
1 3 1 2
t + t − 2t + 27,
21
3
(0 ⩽ t ⩽ 12)
where t is measured in months and r(t) is the annual percentage rate. What is the average rate on auto
loans extended by Madison over the 12−month period?
A 11.1063
6.8
B 11.2527
C 10.2379
D 10.4286
E 10.6659
D 1.5139
E 1.3924
D 0.7021
E −0.0613
Improper integrals
+∞
Z
3x2 dx
Question 94 (L.O.2). Evaluate the integral
7 + x4
1
A 1.8614
B 2.4366
C 1.9137
Z
Question 95 (L.O.2). Evaluate the integral
A 1.1263
B 0.3787
+∞
dx
x2 + 625
671 x ·
C −0.6059
√
√
4
Question 96 (L.O.2). The total profit of a firm in dollars, from producing x units of an item, is P (x). The
firm is able to determine that its marginal profit is given by P ′ (x) = 190e−1.05x . Suppose that it were possible
for the firm to make infinitely many units of this item. What would its total profit be?
A 180.8926
B 181.1377
C 180.4869
D 181.9074
E 180.9524
D 1.1703
E 1.7271
+∞
Z
8dx
√
Question 97 (L.O.2). Evaluate the integral
.
x2 + x4
5
A 0.9952
EXAM
B 0.9963
C 1.5895
Page 14/1 — Question code: 1234
Question 98 (L.O.2). Let a be the real number such that
lim
p
x→+∞
x2 + 8x + 2 − x − a = 0. Evaluate
+∞
Z
dx
·
(x + a)(x − 2)2
a+
6
A 4.5249
D 3.4233
E 4.8000
Z 9
Z 5 2
mdx
x dx
√
√
Question 99 (L.O.2). Find the approximation of m which satisfies
=
2
x−1
81 − x
0
1
C 17.6462
A 18.2816
B 17.4858
D 18.3127
E 18.3730
7
B 4.0162
C 3.4425
Ordinary differential equations (ODE)
7.1
The first order differential equations
Question 100 (L.O.2). Which function is a solution of the following differential equation y ′ +
where C is any constant?
1
A y = x2 + Cx−7
3
3
D y = x2 + Cx7
8
7.2
7
y = 3x,
x
3
1
B y = x2 + Cx−7
C y = x2 + Cx7
8
3
3 2
−7
E y = x + Cx
10
The second order differential equations
Question 101 (L.O.2). Let y(x) be a function satisfying y ′′ − 4y ′ − 32y = 0, y(0) = 6, y ′ (0) = 24. Evaluate
y(1.88).
A 13609714.8215
B 13609714.2745
D 13609713.4611
C 13609713.4938
E 13609714.0102
Question 102 (L.O.2). Let y(x) be a function satisfying y ′′ − 5y ′ = 2x2 + 5, y(0) = 0, y ′ (0) = 0. Evaluate
y(2).
A 4543.1199
B 4542.7416
C 4542.6055
D 4542.9902
E 4541.7701
Question 103 (L.O.2). Let y(x) = axe4x be a function satisfying y ′′ − 10y ′ + 24y = −6.3894e4x . Find the
value of a.
A 3.5403
B 3.6631
C 3.1947
D 3.2759
E 3.0536
ESSAY QUESTIONS
1
1.1
Differentiation
Related rates
Question 104.
EXAM
Page 15/1 — Question code: 1234
[L.O.2] Two crates, A and B, are on the floor of a warehouse. The
crates are joined by a rope 15m long, each crate being hooked at
floor level to an end of the rope. The rope is stretched tight and
pulled over a pulley P that is attached to a rafter P Q = 4m above
a point Q on the floor directly between the two crates. If crate A is
3m from Q and is being pulled directly away from Q at a rate of 0.5
m/s, how fast is crate B moving toward Q?
Question 105.
[L.O.2] A major network is televising the launching of the rocket. At a
distance of x = 12000 feet from the launch site, a spectator is observing a
rocket being launched vertically. A camera tracking the liftoff of the rocket
is located at point A, where ϕ denotes the angle of elevation of the camera
at A. When the rocket is z = 12434 ft from the camera and this distance is
increasing at the rate of 480 ft/sec, how fast is ϕ changing?
2
Applications of integration
Question 106 (L.O.2). Let V1 and V2 be the volumes of the solids that result when the plane region bounded
4
1
1
by y = , y = 0, x = , and x = c, where c > , is revolved about the x−axis and the y−axis, respectively.
x
5
5
Find the value of c for which V1 = V2 .
Question 107 (L.O.2). Find the area of the surface obtained by rotating the arc y = 2x3/2 , (0 ⩽ x ⩽ 4),
about the x−axis.
3
3.1
Ordinary differential equations (ODE)
The first order differential equations
Question 108 (L.O.2). A tank initially contains 80 grams of salt dissolved in 100 liters of water. Pure water
flows into the tank at the rate of 12 liters per minute, and the well stirred mixture flows out of the tank at
the same rate. How much salt does the tank contain after 2 minutes?
Question 109 (L.O.2). A tank with a capacity of 130 liters initially contains 80 grams of salt dissolved in
100 liters of water. Pure water flows into the tank at the rate of 12 liters per minute, and the well stirred
mixture flows out of the tank at the rate of 2 liters per minute. How much salt does the tank contain when
the tank is full?
Question 110 (L.O.2). Recall that a simple electric circuit with a resistor of R ohms, an inductor of L
henries and a battery with a voltage of E volts will produce a current I(t) amperes (t is measured in
seconds) satisfying the following differential equation LI ′ (t) + RI = E. Let study one such circuit with
R = 2 ohms, L = 3 henries and E = 12 volts. Given initial condition I(0) = 3 amperes. Evaluate current’s
value at t = 3 seconds.
EXAM
Page 16/1 — Question code: 1234
3.2
The second order differential equations
Question 111 (L.O.2). Solve the differential equation y ′′ − 8y ′ + 12y = 120e12x
Question 112 (L.O.2). Solve the differential equation y ′′ − 14y ′ + 74y = 10e7x cos(5x)
Question 113 (L.O.2). Let yp be the particular solution of the differential equation y ′′ −4y ′ −12y = −144x−
72. Evaluate yp (2).
Question 114 (L.O.2). Recall that a simple electric circuit with a resistor of R ohms, an inductor of L
henries, a capacitor of C farads and a battery with a voltage E(t) = 100 sin(ωt) (volts) will produce a current
1
I(t) amperes (t is measured in seconds) satisfying the following differential equation LI ′′ (t) + RI ′ + I =
C
E ′ (t) = 100ω cos(ωt). Let study one such circuit with R = 8 ohms, L = 5 henries, C = 0.2 farads and ω = 11
rad/seconds. Given initial condition I(0) = 0, I ′ (0) = 0, evaluate current’s value at t = 6.2 seconds.
Question 115 (L.O.2). Recall that a spring with mass m, spring constant k relaxing after being stretched
will be in motion, where its position x(t) satisfies the following differential equation mx′′ (t) + kx = 0. Let
study one spring with m = 2 and k = 32. Given initial condition x(0) = 6, x′ (0) = 4, evaluate numerically
its position at t = 2.
3.3
System of the first order ODEs
Question 116 (L.O.2). Consider the system of ODEs:


x′ (t) = 4x − y − t

y ′ (t) = 5x + 10y + (t − 1)
Find a solution of the system such that x (0) = 3, y (0) = 1.
———— END ————
EXAM
Page 17/1 — Question code: 1234
ANSWER SHEET FOR MULTIPLE CHOICE QUESTIONS
7-D
3-C
4-A
5-B
6-D
8-A
9-A
10 - E
18 - A
19 - B
20 - E
27 - E
28 - B
29 - E
30 - E
36 - A
37 - A
38 - C
39 - E
40 - C
45 - A
46 - D
47 - E
48 - B
49 - B
50 - D
54 - E
55 - E
56 - C
57 - E
58 - A
59 - C
60 - D
63 - D
64 - D
65 - E
66 - D
67 - D
68 - A
69 - D
70 - B
72 - D
73 - B
74 - C
75 - E
76 - E
77 - A
78 - C
79 - B
80 - A
81 - D
82 - D
83 - A
84 - A
85 - C
86 - E
87 - E
88 - C
89 - A
90 - D
91 - A
92 - B
93 - D
94 - C
95 - B
96 - E
97 - C
98 - B
99 - B
100 - A
101 - E
102 - C
103 - C
1-B
2-B
11 - D
12 - E
13 - A
14 - D
15 - D
16 - B
17 - A
21 - C
22 - A
23 - D
24 - A
25 - C
26 - D
31 - E
32 - B
33 - A
34 - E
35 - C
41 - C
42 - D
43 - C
44 - B
51 - B
52 - B
53 - E
61 - E
62 - D
71 - B
QUESTIONS AND THE ANSWER
Page 1/1 — Question code: 1234
MULTIPLE CHOICE QUESTIONS
1
Functions
1.1
Domain and range of a function
s 9
.
Question 1 (L.O.1). Find the domain of function y = ln 3 +
x


9
9
x
⩽
−
x⩽−
9
9
2
2
A − ⩽x<0
B 
C x⩽−
D 
2
2
x>0
x⩾0
E x>0
Solution

9
x⩽−
9
9
2x + 9
2
⩾0⇒3+ ⩾1⇒
The function is defined when ln 3 +
⩾0⇒
x
x
x
x>0
The correct answer is B
□
Question 2 (L.O.1). Find the range of the function

 9x2 − 3, if x > 0
f (x) =

21x3 , if x ⩽ 0
A (0, +∞)
B (−∞, +∞)
C (−∞, 0)
D (−3, +∞)
E (−3, 0)
Solution
When x > 0, we have f (x) = 9x2 − 3 > −3. On the another hand, when x ⩽ 0, we have f (x) = 21x3 ⩽
0.Therefore, the range of the function f (x) is (−∞, +∞).
The correct answer is B
□
Question 3 (L.O.2). A family plans to organize a party at a restaurant. The overall service cost for the
event (venue, decoration, etc.) is 4 million Vietnamese dong. The restaurant requires a minimum of 2
tables, with each table accommodating 8 people. The cost per guest is 258 thousand Vietnamese dong.
The family plans to spend a maximum of 36.6 million Vietnamese dong. If we denote x as the number
of invited guests, C(x) (in million Vietnamese dong) represents the cost for the party. Find the domain
D and the range R of C.
A D = N ∩ [8; 142], R = [2.064; 36.6]
B D = [16; 142], R = [8.128; 36.6]
C D = N ∩ [16; 142], R = [8.128; 36.6]
D D = N ∩ [0; 142], R = [0; 36.6]
E D = [16; 36.6], R = [4.016; 142]
Solution
QUESTIONS AND THE ANSWER
Page 2/1 — Question code: 1234
The restaurant requires a minimum of 2 tables, each table accommodating 8 people, so x ⩾ 2 × 8 = 16.
Furthermore, the cost per guest is 258 thousand Vietnamese dong, so C(x) ⩾ 4 + 0.258 × 16 = 8.128 and
the family plans to spend a maximum of 36.6 million, hence the maximum number of invited guests is
36.6
= 142. Since the number of invited guests is a natural number, the domain for the cost function of the
0.258
party is D = N ∩ [16; 142] and the range for the cost function is R = [8.128; 36.6].
The correct answer is C
1.2
□
Description of a function
Question 4 (L.O.2). A rectangular storage container with an open top has a volume of 14m3 . The length
of its base is quadruple its width. Material for the base costs 11 dollars per square meter; material for
the sides costs 8 dollars per square meter. Express the cost of materials as a function of the width w of
the base.
A 44w2 +
280
w
B 44w2 + 280
C 32w2 +
1120
w
D 32w2 +
385
w
E 44w2 +
140
w
Solution
Let w and 4w be the width and length of the base, respectively, and h be the height. The total cost is
C = 11 × w × 4w + 8 × 2(w + 4w)h.
Using the fact that the volume is 14, we have w × 4w × h = 14 ⇒ h =
C(w) = 11 × w × 4w + 8 × 2(w + 4w) ×
14
. Therefore,
4w2
14
280
= 44w2 +
2
4w
w
The correct answer is A
□
Question 5 (L.O.2). To create an animated image appealing to players in a game, the game creator
started by creating a rectangular image with dimensions of 29 × 71 pixels. Then, each side is increased
at a rate of 15 pixels per second. Determine the area of the image after t seconds.
B 225t2 + 1500t + 2059
A 2059
D 225t2
C 225t2 + 1065t
E 225t2 + 435t
Solution
The area of the image after t seconds is (29 + 15t) × (71 + 15t) = 225t2 + 1500t + 2059
The correct answer is B
□
Question 6 (L.O.2). The amount of adrenaline in the body changes rapidly. Let’s assume that initially
the body has 18 mg of adrenaline, create a function describing the adrenaline level A = f (t) in the body
after t (minutes), given that A increases 0.39 mg per minute.
A 18t + 0.39
QUESTIONS AND THE ANSWER
B 7.02t
C 0.39 − 18t
D 0.39t + 18
E 18 − 0.39t
Page 3/1 — Question code: 1234
Solution
The function describing the adrenaline level in the body after t (minutes) is A = f (t) = 0.39t + 18.
The correct answer is D
1.3
□
Inverse functions
Question 7 (L.O.1). Let f (x) = x5 + 3x3 + 6x − 5. What is the value of f −1 (63).
A 1
C 6
B 0
D 2
E 3
Solution
We have f ′ (x) = 5x4 + 9x2 + 6 > 0, ∀x ∈ R. Thus, f is one-to-one function and has inverse function. Let
f −1 (63) = x0 ⇒ f (x0 ) = 63 ⇒ x50 + 3x30 + 6x0 − 5 = 63 ⇒ x0 = 2.
The correct answer is D
2
□
The limit of a function, continuous functions
2.1
The limit of a function
Question 8 (L.O.1). Find m > 0 such that
e3.19x − cos(x)
lim
x→0
x
2
.m +
ln(1 + 6.04x)
lim
x→0
x
.m =
18.62.
A 1.6481
B 1.2026
C 1.2978
D 1.3338
E 2.3985
Solution
Using the L’Hospital rule, we have
e3.19x − cos(x)
3.19e3.19x + sin(x)
= lim
= 3.19,
x→0
x→0
x
1
lim
and
ln(1 + 6.04x)
= lim
x→0
x→0
x
lim
6.04
1+6.04x
1
= 6.04.
Therefore, we receive 3.19m2 + 6.04m = 18.62. Choosing m > 0, we get m = 1.64814172658632 ≈ 1.6481.
The correct answer is A
□
6x + m2
x→∞
6x − 1
C 27.3814
Question 9 (L.O.1). Find m > 1 such that lim
A 26.5123
B 26.5475
6x
= e26.55m .
D 26.8591
E 26.4543
Solution
We have
lim
x→∞
6x + m2
6x − 1
6x
QUESTIONS AND THE ANSWER
= lim
x→∞
m2 + 1
1+
6x − 1
6x(m2 +1)
6x−1
· 6x−1
2
m +1
2 +1
= em
= e26.55m ⇒ m2 + 1 = 26.55m.
Page 4/1 — Question code: 1234
Choosing m > 1, we receive m = 26.5122816318155 ≈ 26.5123.
The correct answer is A
□
2 7x
Question 10 (L.O.1). Find lim 1 + 2x e
A e
1
9x2
x→0
−7/9
.
C −e2/9
B 1
E e2/9
D 0
Solution
12
ln(1 + 2x2 e7x )
Let y = 1 + 2x2 e7x 9x ⇒ ln y =
· Then
9x2
ln(1 + 2x2 e7x )
2x2 e7x
2
ln(1 + 2x2 e7x )
2
=
lim
×
lim
=1× =
2
2
7x
2
x→0
x→0 9x
x→0
9x
2x e
9
9
lim ln y = lim
x→0
Therefore, lim y = lim eln y = e2/9
x→0
x→0
The correct answer is E
□
Question 11 (L.O.1). Find S = 4a + 8b where a and b are the two real constants such that
2
lim x ea/x + + b = 15.
x→+∞
x
A 43
B 40
C 48
D 44
E 42
Solution
Let t =
1
eat + 2 × t + b
then I = lim
= 15. The constant b must satisfy the equation
x
t
t→0+
ea×0 + 2 × 0 + b = 0 ⇒ b = −1.
Otherwise, I = ∞. When b = −1, using L’ Hospital rule, we obtain
I = lim
t→0+
aeat + 2
= a + 2 = 15 ⇒ a = 15 − 2 = 13.
1
Therefore, S = 4 × 13 − 1 × 8 = 44
The correct answer is D
□
e4ax − 1 − 9x
·
x→0
x + 13x2
9
C 4a, ∀a ∈ R
B 4a − 9, ∀a ̸=
4
E 4a − 9, ∀a ∈ R
Question 12 (L.O.1). Evalulate I = lim
9
4
D 4a − 9, ∀a ̸= 0
A 4a, ∀a ̸=
Solution
Using the L’Hospital rule, we have
I = lim
x→0
(4a − 9)x
= 4a − 9.
x
The correct answer is E
□
QUESTIONS AND THE ANSWER
Page 5/1 — Question code: 1234
2.2
Continuity
Question 13 (L.O.1). Find real values of a such that f (x) =
uous at x = 13.
2
A
13
B −
24
13
C −


ax − 2,
√
 arctan( x − 13),
37
13
D −
50
13
if x ⩽ 13
is contin-
if x > 13
E −
11
13
Solution
The function f is continuous at x = 13 if
lim f (x) = 0 = lim f (x) = f (13) = 13 × a − 2 ⇒ a =
x→13−
x→13+
2
13
The correct answer is A
□

3

 5 arctan x − 31
x−3
Question 14 (L.O.1). Find real values of a such that f (x) =


a,
ous at x = 3.
5π
3π
5π
B
C
A −
2
2
2
7π
D a does not exist
E
2
if x ̸= 3
is continu-
if x = 3
Solution
The function f is continuous at x = 3 if
lim f (x) = lim f (x) = f (3)
x→3−
x→3+
We have
lim f (x) = −
x→3+
5π
5π
̸= lim f (x) =
−
2
2
x→3
Therefore, there is no exist a such that f is continuous at x = 3.
The correct answer is D
2.3
□
Asymptotes
Question 15 (L.O.1). Find all real numbers a so that the line of equation x = 0 is a vertical asymptote
ln(7x + a − 8)
of the curve y =
·
x+9
C 6
A 11
B 7
D 8
E 12
Solution
The line of equation x = 0 is a vertical asymptote of the curve y =
lim
x→0
QUESTIONS AND THE ANSWER
ln(7x + a − 8)
when
x+9
ln(7x + a − 8)
ln(7 × 0 + a − 8)
ln(a − 8)
=
=
= ∞.
x+9
0+9
9
Page 6/1 — Question code: 1234
Therefore, a = 8.
The correct answer is D
□
9x
− 5.
1 − e−x
C y = 9x − 5
Question 16 (L.O.1). Find all asymptotes of the graph of the function y = 4x + √
A x = 0, y = 13x − 5
B y = 13x − 5
D x = 0, y = 4x − 5
E y = 4x − 5
Solution
Domain: x > 0. We have
lim f (x) = lim 4x + √
x→0+
x→0+
9x
9x
− 5 = −5 + lim √
= −5.
−x
+
x→0
1−e
1 − e−x
The line x = 0 is not the vertical asymptote of the curve y = f (x).
On the other hand,
lim 4x + √
x→+∞
9x
− 5 = +∞.
1 − e−x
Thus, the function does not have horizontal asymptote.
Consider the asymptote of the form y = mx + b where
4x +
m = lim
√ 9x
1−e−x
−5
= lim 4 −
x
x→+∞
x→+∞
5
9
= 13
+√
x
1 − e−x
and
b = lim 4x + √
x→+∞
9x
9x
9x
− 5 − mx = lim 4x + √
− 5 − 13x = lim −5 + √
− 9x = −5.
−x
−x
x→+∞
x→+∞
1−e
1−e
1 − e−x
Therefore, the function has only 01 slant asymptote y = 13x − 5.
The correct answer is B
□
Question 17 (L.O.2). A plastic molding company states that if they use x% of the company’s machines,
the total monthly cost for operating these machines is given by the function:
C(x) =
19x2 − 57x − 60610
4x2 − 100x − 7656
(hundreds of dollars).
The company employs an alternating maintenance mode to utilize close to 58% of the machines (optimal
capacity).What is the total cost that the company has to incur for operating this number of machines if
the machine usage reaches the optimal capacity?
A 5.8984
B 5.9918
C 5.7198
D 6.6802
E 5.7032
Solution
According to the requirement of our problem, we need to find:
lim C(x) = lim
x→58
x→58
(x + 55) (19x − 1102)
19 (x + 55)
= lim
= 5.8984
x→58 4 (x + 33)
(x + 33) (4x − 232)
The correct answer is A
□
QUESTIONS AND THE ANSWER
Page 7/1 — Question code: 1234
3
Derivatives
3.1
Tangents
Question 18 (L.O.1). A point M (a, b) is on the graph of the function f (x) = 1.55ex − x. Given that the
tangent line to the graph at M is parallel to the line y = 2.47x + 2.82. Find the value of a.
A 0.8059
B 0.4466
C 1.7944
D 1.4552
E 1.2394
Solution
Since the tangent line to the graph at M is parallel to the line y = 2.47x + 2.82, then
2.47 + 1
′
a
f (a) = 2.47 ⇒ 1.55e − 1 = 2.47 ⇒ a = ln
= 0.805899663027613 ≈ 0.8059.
1.55
The correct answer is A
□
Question 19 (L.O.1). Find ALL points on the graph of f (x) = 4x3 − 192x where the tangent line is
parallel to the x−axis.
A (4; 512)
B (4; −512) and (−4; 512)
D (4; 512) and (−4; −512)
C (4; −512)
E (−4; 512)
Solution
The tangent line is parallel to the x−axis when its slope is equal to 0. It means that
f ′ (x) = 12x2 − 192 = 0 ⇔ x = ±4.
Therefore, all points on the graph of f (x) where the tangent line is parallel to the x−axis are (4; −512) and
(−4; 512).
The correct answer is B
□
Question 20 (L.O.1). Find ALL points on the graph of f (x) =
9
where the tangent line is perpendicular
x
to the line y = 16x − 4.
A (−12; 3/4)
C (12; −3/4) and (−12; 3/4)
B (12; 3/4)
D (−12; −3/4)
E (12; 3/4) and (−12; −3/4)
Solution
The tangent line is perpendicular to the line y = 16x − 4 when its slope is equal to −
f ′ (x) = −
1
· It means that
16
9
1
= − ⇔ x = ±12.
2
x
16
Therefore, all points on the graph of f (x) where the tangent line is perpendicular to the line y = 16x − 4,
are (12; 3/4) and (−12; −3/4) .
The correct answer is E
□
QUESTIONS AND THE ANSWER
Page 8/1 — Question code: 1234
Question 21 (L.O.1). Find the real value of a such that the tangent line to the graph of the function
√
21
y = 3 x at x = a has y−intercept .
2
A 46
B 52
C 49
D 44
E 48
Solution
√
3
The tangent line at x = a has an equation of the form y = 3 a + √ (x − a). This tangent line has
2 a
√
21
3 a
=
⇒ a = 49.
y−intercept 21/2 when
2
2
The correct answer is C
□
3.2
Tangent to paramtric curve
Question 22 (L.O.1). Let the curve be given by paramtric equations

 x(t) = (3 − 4t) e3t
 y(t) = 3t2 − 5t − 4 e3t
37
Find the set of all values of t such that the tangent at M (x(t), y(t)) has the slope of − ?
31
503
410
131
148
38
C −
; −1
;7
;4
;1
D −
E
A − ;3
B −
93
93
93
93
93
Solution
(6t − 5) e3t + 3 · 3t2 − 5t − 4 e3t
y ′ (t)
The slope of the tangent at M (x(t), y(t)) is y (x) = ′
=
· Therefore,
x (t)
3 · (3 − 4t) e3t − 4e3t

38
2
t = −
−9t + 9t + 17
37
37
′

93
=− ⇔
y (x) = − ⇔
31
12t − 5
31
t = 3
′
The correct answer is A
3.3
□
Differentiable functions
Question 23 (L.O.2). Determine ALL values of x such that f (x) =
x2 − 11
is not differenx2 − 25
tiable.
A x ̸= 5
B x = −5
D x = ±5
C x=5
E x ̸= ±5
Solution
The domain of f (x) =
x2 − 11
is D = R\{±5}.
x2 − 25
⇒ y′ =
−28x
− 25)2
(x2
So the function f is not differentiable when x = ±5.
The correct answer is D
□
QUESTIONS AND THE ANSWER
Page 9/1 — Question code: 1234
Question 24 (L.O.2). Determine ALL values of x such that f (x) =
p
3
(x2 − 25)2 is not differen-
tiable.
A x = ±5
B x ̸= ±5
C x ̸= 5
D x = −5
E x=5
Solution
The domain of f (x) =
p
3
(x2 − 25)2 is D = R.
⇒ y′ =
2
4x
× (x2 − 25)−1/3 × 2x = p
3
3
3 (x + 5)(x − 5)
So the function f is not differentiable when x = ±5.
The correct answer is A
□
Question 25 (L.O.2). Find S = a + b such that function f is differentiable at x = 0, if

 2x2 + 7x + 5, x ⩽ 0
f (x) =

ln(ax + b), x > 0
A 7e5
C 8e5
B 8
D 9
E 6e5
Solution
In order to be differentiable, function has to be continuous
lim f (x) = lim f (x) ⇒ ln b = 5 ⇒ b = e5 .
x→0+
x→0−
Function f is differentiable at x = 0 if and only if
f+′ (0) = f−′ (0) ⇒
a
= 7 ⇒ a = 7b = 7e5 .
b
Therefore, S = a + b = 8e5
The correct answer is C
3.4
□
The chain rule
Question 26 (L.O.1). Let f (x) and g(x) be functions satisfying the conditions: f (−4) = f ′ (−4) = 9
and g(9) = g ′ (9) = −4. Consider the function h(x) = (f ◦ g ◦ f )(x) = f (g(f (x))). Find the value of
h′ (−4).
A −328
B −325
C −322
D −324
E −329
Solution
Using the chain rule, we have
h′ (x) = f ′ (g(f (x))).g ′ (f (x)).f ′ (x) ⇒ h′ (−4) = f ′ (g(f (−4))).g ′ (f (−4)).f ′ (−4) =
= f ′ (−4).g ′ (9).f ′ (−4) = 9 × (−4) × 9 = −324.
The correct answer is D
□
QUESTIONS AND THE ANSWER
Page 10/1 — Question code: 1234
Question 27 (L.O.2). An importer of Rwandan coffee estimates that local consumers will buy approxi4346
mately D(p) = 2 pounds of the coffee per week when the price is p dollars per pound. It is estimated
p
that t weeks from now, the price of Rwandan coffee will be p(t) = 0.02t2 + 0.08t + 9 dollars per pound.
At what rate will the weekly demand for the coffee be changing with respect to time 6 weeks from
now?
A −3.1978
B −2.3627
C −3.4126
D −3.2408
E −2.621
Solution
2
Using the chain rule, we have D(t) = D(p(t)) ⇒ D (t) = D (p).p (t) = 4346 × − 3 × (0.04t + 0.08)
p
2
When t = 6 ⇒ p(6) = 0.02 × 6 + 0.08 × 6 + 9 = 10.2 and
2
′
× (0.04 × 6 + 0.08) = −2.62101303420253 ≈ −2.621.
D (6) = 4346 × −
10.23
′
′
′
The correct answer is E
□
Question 28 (L.O.2). When electric blenders are sold for p dollars apiece, local consumers will buy
7928
approximately D(p) =
blenders per month. It is estimated that t months from now, the price of
p
√
the blenders will be p(t) = 0.03 t3 + 4 dollars. Compute the rate at which the monthly demand for the
blenders will be changing with respect to time 16 months from now.
A −40.9061
B −40.7186
C −40.9638
D −40.9057
E −41.3092
Solution
√
1
Using the chain rule, we have D(t) = D(p(t)) ⇒ D′ (t) = D′ (p).p′ (t) = 7928 × − 2 × (0.045 t)
p
√
3
When t = 16 ⇒ p(16) = 0.03 × 16 + 4 ≈ 5.92 and
√
1
′
D (16) ≈ 7928 × −
×
(0.045
×
16) = −40.7185902118335 ≈ −40.7186.
5.922
The correct answer is B
3.5
□
Derivatives of inverse function
Question 29 (L.O.1). If f (x) = 2x + 2 sinh (x − 2), find (f −1 )′ (4).
9
19
13
15
A
B −
C
D −
4
4
4
4
E
1
4
Solution
Let y = f (x) = 2x + 2 sinh (x − 2) ⇒ x′ (y) =
1
1
=
. When y = 4 we have x = 2.
y ′ (x)
2 cosh (x − 2) + 2
Therefore,
(f −1 )′ (4) = x′ (4) =
1
y ′ (2)
=
1
4
The correct answer is E
□
QUESTIONS AND THE ANSWER
Page 11/1 — Question code: 1234
3.6
Undetermined Coefficients
Question 30 (L.O.2). Let y = Ax3 +Bx+C, where A, B, C ∈ R, satisfy the equation y ′′′ +2y ′′ −5y ′ +9y =
27x3 − 45x2 + 54x + 89. Find the value of S = A + B + C.
A 11
B 9
C 16
D 15
E 14
Solution
We have y = Ax3 + Bx + C ⇒

′
2


 y = 3Ax + B
y ′′ = 6Ax



y ′′′ = 6A
Substituting these derivatives into the equation y ′′′ + 2y ′′ − 5y ′ + 9y = 27x3 − 45x2 + 54x + 89, we receive
6A + 2 × 6Ax − 5 × (3Ax2 + B) + 9 × (Ax3 + Bx + C) = 27x3 − 45x2 + 54x + 89
⇔ 9Ax3 − 15Ax2 + (12A + 9B)x + (6A − 5B + 9C) = 27x3 − 45x2 + 54x + 89



9A = 27







 A=3
−15A = −45
⇔
⇔
B=2




12A
+
9B
=
54



C=9


6A − 5B + 9C = 89
Therefore, S = A + B + C = 14
The correct answer is E
□
Question 31 (L.O.2). Let y = Ax cos x + Bx sin x, where A, B ∈ R, satisfy the equation y ′′ + y =
−4 sin x + 10 cos x. Find the value of S = A + B.
A 6
B 9
C 8
D 5
E 7
Solution
We have y = Ax cos x + Bx sin x

 y ′ = A cos x − Ax sin x + B sin x + Bx cos x
⇒
 y ′′ = −A sin x − A sin x − Ax cos x + B cos x + B cos x − Bx sin x
Substituting these derivatives into the equation y ′′ + y = −4 sin x + 10 cos x, we receive
−2A × sin x + 2B × cos x = −4 sin x + 10 cos x


 −2A = −4
 A=2
⇔
⇔
 2B = 10
 B=5
Therefore, S = A + B = 7
The correct answer is E
□
Question 32 ([L.O.2). Let y = Axp + Bx + C, where A, B, C, p ∈ R, satisfy the equation y ′′ =
y ′ (1) = −3.0, y(1) = 5.0. Find the value of S = (A + B + C).p.
A 11
QUESTIONS AND THE ANSWER
B 7
C 9
D 8
30
and
x7
E 12
Page 12/1 — Question code: 1234
Solution
If y = Axp + Bx + C

 y ′ = Apxp−1 + B
⇒
 y ′′ = Ap(p − 1)xp−2
Substituting these derivatives into the equation y ′′ =
30
, we receive
x7
Ap(p − 1)xp−2 = 30x−7


 Ap(p − 1) = 30
 A=1
⇒
⇒

 p = −5
p − 2 = −7
Also,


 B=2
 A × 1p + 1B + C = 5.0
⇒
 C=2
 Ap × 1p−1 + B = −3.0
Therefore, S = (A + B + C).p = 7
The correct answer is B
3.7
□
Rates of change
Question 33 (L.O.2). The management fee for running services in an area is calculated by the formula
f (x) = 0.0692x2 + 0.0902x (in dollars), where x is the number of homes being present in the area.
Suppose that at the moment when 63 homes are present in this area, the number of homes is increasing
at a rate of 5 homes per week. What is the rate of change (in dollars/week) in the management fee at
this moment?
A 44.047
B 44.5156
C 43.0986
D 43.2984
E 44.6353
Solution
The rate of change (in dollars/week) in the management fee at the moment when x = 63 homes are present
in the area and x′ (t) = 5 homes per week is
f ′ (t) = f ′ (x).x′ (t) = f ′ (63) × 5 = 44.047 ≈ 44.047.
The correct answer is A
□
25
· If the abscissa
x
of the particle increases by 0.3234 units per second, how does its ordinate vary as it passes through the
Question 34 (L.O.2). A particle is moving along a hyperbolic trajectory given by y =
point (5; 5)?
A Increases 0.0971 units per second
B Increases 0.8569 units per second
C Decreases 0.4064 units per second
D Decreases 0.1947 units per second
E Decreases 0.3234 units per second
QUESTIONS AND THE ANSWER
Page 13/1 — Question code: 1234
Solution
The rate of change of the ordinate (y-coordinate) as the particle passes through the point (5; 5) is
y ′ (t) = y ′ (x).x′ (t) = y ′ (5) × 0.3234 = −0.3234 ≈ −0.3234.
The correct answer is E
3.8
□
Related Rates
Question 35 (L.O.2). Recall that volume of a right cylinder is V = π.r2 .h, where r is the radius of the
base and h is the height. Given a right cylinder with the radius of the base of 9 cm and the height of 5
cm. The radius of the base is increasing at a rate of 0.8 cm/s, and its height is increasing at a rate of 0.21
cm/s. How fast is the volume of cylinder increasing?
A 278.9661
B 280.2004
C 279.6332
D 279.1033
E 279.8283
Solution
The formula for the volume of a right cylinder is
V = π.r2 .h,
(1)
where r is the radius of the base and h is the height.
In this problem, V, r and h are functions of the time t in seconds. Taking the derivative of both sides of
equation (1) with respect to time yields
dh
dr
dV
= π r2
+ h.2r
dt
dt
dt
(2)
Since the radius of the base is increasing at a rate of 0.8 cm/s and the height is increasing at a rate of 0.21
cm/s,
dr
= 0.8,
dt
dh
= 0.21.
dt
Substituting these, as well as r = 9 and h = 5, into equations (2) yields
dV
= π 92 × 0.21 + 5 × 2 × 9 × 0.8 = 89.01π ≈ 279.6332
dt
Because the sign of
dV
is positive, the volume of the right cylinder is increasing at a rate of 279.6332 cm3 /s.
dt
The correct answer is C
□
Question 36 (L.O.2). Water pours into a conical tank of height 10(m) and radius 4(m) at a rate of
7(m3 /min).
QUESTIONS AND THE ANSWER
Page 14/1 — Question code: 1234
4
r
10
h
At what rate is the water level rising when the level is 5(m) high?
A 0.5570
B −0.3594
D 0.0975
C 0.1141
E 0.7804
Solution
dh
Let V and h be the volume and height of the water in the tank at time t. Our problem is: Compute
at
dt
dV
1
h = 5 given that
= 7m3 /min. When the water level is h, the volume of water in the cone is V = πhr2 ,
dt
3
where r is the radius of the cone at height h, but we cannot use this relation unless we eliminate the variable
r. Using similar triangles, we see that
dV
r
4
1
dV
dh
dh
dt
=
⇒ r = 2/5 × h ⇒ V = πh(2/5 × h)2 ⇒
= 4/25πh2
⇒
=
h
10
3
dt
dt
dt
4/25πh2
When h = 5, the level is rising at a rate of
7
7
dh
=
=
≈ 0.5570(m/min).
2
dt
4/25π5
4π
The correct answer is A
□
Question 37 (L.O.2). A leaky water tank is in the shape of an inverted right circular cone with depth
of 11(m) and top radius 3(m). When the water in the tank is 3(m) deep, it is leaking out at a rate of
1/6(m3 /min).
3
r
11
h
How fast is the water level in the tank dropping at that time?
A 0.0792
B 0.3362
C −0.6292
D 0.1284
E −0.5812
Solution
dh
Let V and h be the volume and depth of the water in the tank at time t. Our problem is: Compute
at h = 3
dt
dV
1
given that
= 1/6m3 /min. When the water level is h, the volume of water in the cone is V = πhr2 ,
dt
3
where r is the radius of the cone at height h, but we cannot use this relation unless we eliminate the variable
r. Using similar triangles, we see that
dV
r
3
1
dV
dh
dh
dt
=
⇒ r = 3/11 × h ⇒ V = πh(3/11 × h)2 ⇒
= 9/121πh2
⇒
=
h
11
3
dt
dt
dt
9/121πh2
QUESTIONS AND THE ANSWER
Page 15/1 — Question code: 1234
121
dV
dh
−1/6
dh
= −
= −1/6 ⇒
=
≈ −0.0792(m/min). Because the sign of
is
dt
dt
9/121π32
486π
dt
negative, the water level in the tank is dropping a rate of 0.0792(m/min).
When h = 3;
The correct answer is A
□
Question 38 (L.O.2). Boyle’s law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy equation P V = C = const. Suppose that at a certain instant
the volume is 647cm3 , the pressure is 184kP A, and the pressure is increasing at a rate of 28kP A/min.
At what rate is the volume decreasing at this instant?
A 98.1919
B 99.3825
D 97.6455
C 98.4565
E 98.362
Solution
Since the pressure P and volume V satisfy equation
(3)
P V = C = const,
where P and V are functions of the time t in minutes, so we can differentiate both sides of equation (3)
with respect to time and receive
dP
dV
·V +P ·
= 0.
dt
dt
Since the pressure P is increasing at a rate of 28 kPA/min, so
(4)
dP
= 28
dt
Substituting these, as well as V = 647cm3 and P = 184kP A, into equations (4) yields
28 × 647 + 184 ·
⇒
dV
=0
dt
dV
28 × 647
=−
= 98.4565217391304 ≈ −98.4565.
dt
184
dV
is negative, the volume is decreasing at a rate of 98.4565 (cm3 /min).
dt
The correct answer is C
Because the sign of
3.9
□
Differentials
Question 39 (L.O.1). Find d2 y(0) if y = cos6 (7x).
A −292dx2
B −289dx2
C −298dx2
D −299dx2
E −294dx2
Solution
We have
y ′ = 6 cos5 (7x)(− sin(7x)) × 7 = −42 cos5 (7x) sin(7x)
⇒ y ′′ = −42 × 5 cos4 (7x)(− sin(7x)) × 7 × sin(7x) − 42 × 7 cos5 (7x) cos(7x).
Thus
y ′′ (0) = −42 × 7 = −294 ⇒ d2 y(0) = −294dx2 .
The correct answer is E
□
QUESTIONS AND THE ANSWER
Page 16/1 — Question code: 1234
3.10
Applications of derivatives in physics
Question 40 (L.O.1). A particle moving on the x−axis has position s(t) = t3 − 39t2 + 432t + 12 (m)
after an elapsed time of t seconds. What is the total distance travelled by the particle during the first 31
seconds?
A 6699
B 6707
C 6704
D 6709
E 6702
Solution
We have s(t) = t3 − 39t2 + 432t + 12 ⇒ v(t) = s′ (t) = 3t2 − 78t + 432.

0<t<8
v(t) > 0 ⇔ 
and v(t) < 0 ⇔ 8 < t < 18
18 < t < 31
So the total distance travelled by the particle during the first 31 seconds is
|s(8) − s(0)| + |s(18) − s(8)| + |s(31) − s(18)| = 6704.
The correct answer is C
□
Question 41 (L.O.1). A particle moving on the x−axis has position s(t) = t3 − 45t2 + 600t + 9 (m)
after an elapsed time of t seconds. What is the displacement travelled by the particle during the first 37
seconds?
A 11243
B 11250
C 11248
D 11251
E 11253
Solution
We have s(t) = t3 − 45t2 + 600t + 9. So the displacement travelled by the particle during the first 37 seconds
is
s(37) − s(0) = 11248.
The correct answer is C
3.11
□
Applications of derivatives in Economics
Question 42 (L.O.2). For Luce Landscaping, the total revenue from the yard maintenance of x homes
is given by R(x) = 1868x − 4x2 (dollars) and the total cost is given by C(x) = 2938 + 18x (dollars).
Suppose that Luce is adding 19 homes per day at the moment when the 458th customer is signed. At
that moment, what is the rate of change of total profit P in dollars per day, if the total profit P (x) =
R(x) − C(x)?
A −34469
B −34465
C −34471
D −34466
E −34463
Solution
QUESTIONS AND THE ANSWER
Page 17/1 — Question code: 1234
We have
h
i
P (x(t)) = R(x(t)) − C(x(t)) ⇒ P ′ (t) = P ′ (x).x′ (t) = R′ (x) − C ′ (x) x′ (t).
From the given information, we have x = 458 and x′ (t) = 19. Therefore,
h
i
h
i
P ′ (t) = 1868 − 2 × 4x − 18 x′ (t) = 1868 − 2 × 4 × 458 − 18 × 19 = −34466.
The correct answer is D
3.12
□
Linear Approximations
Question 43 (L.O.1). Use the linear approximation formula to find an approximate value of f (x) at
x = 12.17, given that f (12) = 4 and f ′ (12) = 0.74.
A 3.9803
B 4.4361
C 4.1258
D 3.8998
E 3.8271
Solution
The approximate value of f (x) at x = 12.17, is
f (12.17) ≈ f (12) + f ′ (12)∆x = f (12) + f ′ (12) × (12.17 − 12) = 4.1258 ≈ 4.1258.
The correct answer is C
□
Question 44 (L.O.2). A spherical balloon is inflated so that its radius increases from 20 cm to 20.44 cm
in 4 seconds. By approximately how much has its volume increased (using linear approximation) (in
cm3 /s)?
A 552.2614
B 552.9203
C 553.2891
D 552.9905
E 552.0326
Solution
4
The volume of the spherical balloon is V = πR3 . So the change of volume in 1 second is
3
20.44 − 20
∆V ≈ V ′ (R).R′ (t) = 4πR2 .R′ (t) ≈ 4π × 202 ×
= 176.000000000001π ≈ 552.9203.
4
The correct answer is B
□
Question 45 (L.O.2). A company estimates that volume sales S(x) of a type of commodity depends on
the amount of money x (in thousands dollars) spent on advertising, that is given by:
S(x) = −0.0079x3 + 0.412x2 + 2.8147x + 338.
Use the differential (also known as: the linear approximation formula) to estimate the change in the
total sales of this commodity if the advertising expenditures are increased from 41 to 41.7583 (thousands
dollars).
A −2.4577
QUESTIONS AND THE ANSWER
B −1.7600
C −2.4103
D −2.4215
E −1.9167
Page 18/1 — Question code: 1234
Solution
The change in the total sales of this commodity if the advertising expenditures are increased from 41 to
41.7583 (thousands dollars) is
∆S ≈ S ′ (41)∆x = (3 × (−0.0079) × 412 + 2 × 0.412 × 41 + 2.8147) × (41.7583 − 41) =
= −2.45765283203125 ≈ −2.4577.
The correct answer is A
□
Question 46 (L.O.2). A ball of ice melts so that its radius decreases from 30 cm to 29.1647 cm. By
approximately how much does the volume of the ball decrease?
A −9447.0725
B −9446.1664
C −9446.7456
D −9447.01
E −9447.2685
Solution
4
The volume of the ball of ice is V = πR3
3
⇒ ∆V ≈ dV = V ′ (R)dR = V ′ (R)∆R = 4πR2 ∆R
Since ∆R = 29.1647 − 30 = −0.8353 cm and R = 30 cm, so
∆V ≈ 4π × 302 × (−0.8353) = −3007.08π ≈ −9447.01 cm3 .
The correct answer is D
□
Question 47 (L.O.2). A water reservoir has a base shaped like a rectangle combined with half of a
circle, as illustrated in the picture.
x
4x
If we can measure x = 5 ± 0.0599(m). Approximate the error in the area of the reservoir’s base using
differential formula.
A 6.2392
QUESTIONS AND THE ANSWER
B 5.6269
C 6.6628
D 6.6852
E 6.1596
Page 19/1 — Question code: 1234
Solution
1
The area of the reservoir’s base is S(x) = x × 4x + × π ×
2
4x
2
2
⇒ S ′ (x) = 8x + 4.0πx. Therefore,
∆S ≈ S ′ (5)∆x = (40 + 20.0π) × (±0.0599) = 2.396 + 1.198π ≈ 6.1596.
The correct answer is E
□
Question 48 (L.O.1). Let f be an even function and f is differentiable in R. Given that f ′ (1) = 2.81,
f (1) = −3. Use differential formula to approximate f (−0.22).
A −6.0772
B −5.1918
C −5.673
D −6.0849
E −6.0731
Solution
Using property of even function and differential formula, we have
f (−0.22) = f (0.22) ≈ f (1) + f ′ (1)(0.22 − 1) = −5.1918 ≈ −5.1918
The correct answer is B
3.13
□
Monotonicity
ln(3x + 4)2
, then which statement is always true?
3x + 4
4 e 4 e
A The function y is decreasing on − − , − +
3 3 3 3 4 e 4
4 4 e
B The function y is increasing on − − , −
and − , − +
3 3 3
3 3 3
4 e 4 e
C The function y is increasing on − − , − +
3 3 3 3
Question 49 (L.O.1). If y =
D The function y is increasing on R
4 e
4 e
E The function y is increasing on − − , 0 ∪ 0, − +
3 3
3 3
Solution
We have
y=
4)2
ln(3x +
3x + 4
Thus
QUESTIONS AND THE ANSWER
=
2 ln |3x + 4|
2(3 − 3 ln |3x + 4|)
⇒ y′ =
=
3x + 4
(3x + 4)2
3 · 2 − log (3x + 4)2
(3x + 4)2
−4 − e
4 e
x=
=− −

3
3 3
y ′ = 0 ⇔ |3x + 4| = e ⇔ 
−4 + e
4 e
x=
=− +
3
3 3

Page 20/1 — Question code: 1234
x
−∞
f ′ (x)
4 e
− −
3 3
−
+
0
−4/3
+∞
0
4 e
− +
3 3
+
−
0
4 e
f − +
3 3
+∞
f (x)
f
4 e
− −
3 3
−∞
0
4 e
4 e
Therefore, y is increasing on − − , −4/3 and −4/3, − +
.
3 3
3 3
The correct answer is B
□
tan x − 3
Question 50 (L.O.1). Find all real numbers m such that y =
is increasing on the interval
tan x − m
π
0;
.
4
A 1⩽m<3
B m>3
C m⩾0
D m ⩽ 0 or 1 ⩽ m < 3
E m⩽0
Solution
1
Let t = tan x ⇒ t ∈ (0; 1). Then t′ (x) =
> 0, ∀x ∈ R and y ′ (x) = y ′ (t).t′ (x). So if y ′ (x) > 0, ∀x ∈
cos2 x
π
⇒ y ′ (t) > 0, ∀t ∈ (0; 1). We have
0;
4
y(t) =
t−3
3−m
⇒ y ′ (t) =
t−m
(t − m)2
Thus y is increasing on (0; 1) when



 y ′ (t) > 0

 1⩽m<3
m<3
⇔
⇔
 m∈
 m∈

/ (0; 1)
/ (0; 1)
m⩽0
The correct answer is D
3.14
□
Local extreme values
Question 51 (L.O.1). Studying the local extreme values of the function f (x) = arctan
3x − 9
x2 + 72
,
which statement is always true?
A f has local minimum when x = −6, and no local maximum
B f has local minimum when x = −6, and local maximum when x = 12
C f has local maximum when x = −6, and local minimum when x = 12
D f has no local minimum, and local maximum when x = 12
E f has no local minimum, and no local maximum
Solution
QUESTIONS AND THE ANSWER
Page 21/1 — Question code: 1234
Since f (x) = arctan
3x − 9
x2 + 72
−3x2 + 18x + 216
3 −x2 + 6x + 72
−3x2 + 18x + 216
(x2 + 72)2
′
⇒ f (x) =
= 4
= 2
(x + 72)2 + (3x − 9)2
x + 153x2 − 54x + 5265
3x − 9 2
1+
x2 + 72
⇒ f ′ (x) = 0 ⇔ x = −6 ∨ x = 12
x
−∞
f ′ (x)
−6
−
+
0
+∞
12
0
0
−
f (12)
f (x)
f (−6)
0
The correct answer is B
3.15
□
Local extreme values of the parametric curve
Question 52 (L.O.1). Let y = y(x) be the parametric curve defined by

 x(t) = −5t ln(t + 4)
(t ⩾ 0).
 y(t) = 2t3 − 24t2 + 72t + 4
Which statement is always TRUE?
A y attains local minimum value when x = −10 ln(6)
B y attains local maximum value when x = −10 ln(6)
C y attains local minimum value when x = 6
D y attains local maximum value when x = −30 ln(10)
E y attains local maximum value when x = 2
Solution
Since x(t) = −t ln(t + 1) ⇒ x′ (t) = −5 ∗ t/(t + 4) − 5 ∗ log(t + 4) < 0, ∀t ⩾ 0. On another hand,

t = 2
y(t) = 2t3 − 24t2 + 72t + 4 ⇒ y ′ (t) = 6t2 − 48t + 72 = 0 ⇔ 
t = 6
Also we have
y ′′ (x) =
y ′′ (t)x′ (t) − y ′ (t)x′′ (t)
[x′ (t)]3
If y ′′ (x0 ) > 0 then y attains local minimum value at x0 ; if y ′′ (x0 ) < 0 then y attains local maximum value at
x0 .
If t = 2 then x = −10 log (6) and y ′′ (−10 log (6)) =
QUESTIONS AND THE ANSWER
40 + 120 log (6)
3 ≈ −0.2126
−5 log (6) − 53
Page 22/1 — Question code: 1234
If t = 6 then x = −30 log (10) and y ′′ (−30 log (10)) =
−120 log (10) − 72
≈ 0.1139
(−5 log (10) − 3)3
The correct answer is B
3.16
□
Concavity and inflection point
Question 53 (L.O.1). If y = x3 e(12/11)x , then the number of point(s) of inflection of the graph of y
is
A 4
C 0
B 1
D 2
E 3
Solution
Domain: D = R
y ′ = 3x2 e12/11x +
12
× x3 e12/11x .
11
2
12 2 12/11x
12
12
x e
+ 3 × x2 e12/11x +
x3 e12/11x =
11
11
11
h 12 2
i
12
= xe12/11x
x2 + 6 × x + 6 .
11
11
⇒ y ′′ = 6xe12/11x + 3 ×
y ′′ = 0 ⇔ x = 0 ∨ x = −3/2 ∨ x = −4.
−∞
x
−3/2
−4
+∞
0
f ′′ (x)
−
0
+
0
−
0
+
f (x)
CD
PI
CU
PI
CD
PI
CU
Therefore, the graph of y has 3 points of inflection.
The correct answer is E
3.17
□
Global maximum and minimum values
Question 54 (L.O.2). Find the maximum value of

 e2−x + x2 + 9x + 5, if 0 ⩽ x ⩽ 2
f (x) =

x2 − 24x + 72, if 2 < x ⩽ 12
A 23
B 31
C 25
D 30
E 28
Solution
If 0 ⩽ x ⩽ 2 then
f (x) = e2−x + x2 + 9x + 5 ⇒ f ′ (x) = 2x − e2−x + 9
and f ′′ (x) = e2−x + 2 > 0, ∀x ∈ (0, 2). Therefore,
f ′ (x) ⩾ f ′ (0) = −e2 + 9 > 0, ∀x ∈ (0, 2) ⇒ fmax = f (2) = 28, x ∈ [0, 2].
QUESTIONS AND THE ANSWER
Page 23/1 — Question code: 1234
If 2 ⩽ x ⩽ 12 then
f (x) = x2 − 24x + 72 ⇒ f ′ (x) = 2x − 24 < 0, ∀x ∈ (2, 12) ⇒ fmax = f (2) = 28, ∀x ∈ [2, 12].
The function f is continuous at x = 2. Thus, fmax = f (2) = 28, ∀x ∈ [0, 12].
The correct answer is E
□
Question 55 (L.O.2). A company specializing in producing athletic uniforms for students stated that
the cost to produce x batches (0 < x ⩽ 264) is
C(x) = 0.3x2 + 12x + 4915.2, (hundreds dollars)
The average cost per batch when producing x batches is given by the formula Cave =
C(x)
· Find x such
x
that the average cost atains minimum value.
A 125
B 131
C 124
D 123
E 128
Solution
We have
Cave =
C(x)
=
x
0.3x2

x = −128 (rejected)
+ 12x + 4915.2
4915.2
′
⇒ Cave
= 0.3 −
=0⇔
2
x
x
x = 128 (accepted)
Comparing the values of Cave when x = 128 and when x = 264 we obtain
min Cave = min{88.8000; 109.8182} = 88.8000.
Therefore, the average cost atains minimum value when x = 128.
The correct answer is E
3.18
□
Maclaurin-Taylor approximations
Question 56 (L.O.1). Find the coefficient of x8 in the Maclaurin polynomial of degree 9 for the function
f (x) = (2x − 5) sin (2x).
614
299
A
B
315
315
C −16/315
D −
1591
315
E −
646
315
Solution
The Maclaurin polynomial for sin(2x) is
2x −
4x3 4x5 8x7
+
−
+ O x9
3
15
315
Therefore, the Maclaurin polynomial for the function f (x) = (2x − 5) sin (2x) is
−10x + 4x2 +
20x3 8x4 4x5 8x6 8x7 16x8
−
−
+
+
−
+ O x9
3
3
3
15
63
315
Thus, the coefficient of x8 in the Maclaurin polynomial of degree 9 for the function f (x) is −
16
315
The correct answer is C
□
QUESTIONS AND THE ANSWER
Page 24/1 — Question code: 1234
Question 57 (L.O.1). Write the Taylor polynomial of degree 3 for the function f (x)
2x2 − 8x + 11 e5x where x is near x0 = 2.
79 (x − 2)2 e10 145 (x − 2)3 e10
A 3e10 + 1 + 15 (x − 2) e10 +
+
+ O (x − 2)4 ; x → 2
2
2
79 (x − 2)2 e10 145 (x − 2)3 e10
10
10
B 3e − 2 + 15 (x − 2) e +
+
+ x + O (x − 2)4 ; x → 2
2
2
2 10
3 10
79
(x
−
2)
e
145
(x
−
2)
e
C 3e10 + 15 (x − 2) e10 +
+
− (x − 2)3 + O (x − 2)4 ; x → 2
2
2
79 (x − 2)2 e10
145 (x − 2)3 e10
2
10
10
− (x − 2) +
+ O (x − 2)4 ; x → 2
D 3e + 15 (x − 2) e +
2
2
2 10
3 10
79
(x
−
2)
e
145
(x
−
2)
e
E 3e10 + 15 (x − 2) e10 +
+
+ O (x − 2)4 ; x → 2
2
2
=
Solution
The Taylor polynomial for e5x is
25 (x − 2)2 e10 125 (x − 2)3 e10
+
+ O (x − 2)4 ; x → 2
2
6
Therefore, the Taylor polynomial for the function f (x) = 2x2 − 8x + 11 e5x is
e10 + 5 (x − 2) e10 +
3e10 + 15 (x − 2) e10 +
79 (x − 2)2 e10 145 (x − 2)3 e10
+
+ O (x − 2)4 ; x → 2
2
2
The correct answer is E
4
□
Mixed problems
[L.O.2] Using the following information, answer the questions from 58 to 59.
A family plans to organize a party at a restaurant. The overall service cost for the event (venue, decoration,
etc.) is 8 million Vietnamese dong. The restaurant requires a minimum of 2 tables, with each table
accommodating 11 people. The cost per guest is 250 thousand Vietnamese dong. The family plans to spend a
maximum of 28.25 million Vietnamese dong. If we denote x as the number of invited guests, C(x) (in million
Vietnamese dong) represents the cost for the party.
Question 58. Find the domain of C(x) is
A N ∩ [22; 81]
D N ∩ [0; 22]
B [0; 81]
C [22; 28.25]
E N ∩ [8; 28.25]
Solution
250
× x. The restaurant requires a minimum of 2 tables, each table accommodating
1000
11 people, so x ⩾ 22. On the another hand, the family plans to spend a maximum of 28.25 million, hence
250
C(x) ⩽ 28.25 ⇔ 8 +
× x ⩽ 28.25 ⇔ x ⩽ 81.
1000
The correct answer is A
□
We have C(x) = 8 +
QUESTIONS AND THE ANSWER
Page 25/1 — Question code: 1234
Question 59. The range of C(x) is
A [22; 81]
C [13.5; 28.25]
B [13.5; 81]
D [8; 28.25]
E [5.5; 28.25]
Solution
We have
C(x) = 8 +
250
250
×x⩾8+
× 22 = 13.5.
1000
1000
Furthermore, C(x) ⩽ 28.25.
The correct answer is C
□
[L.O.2] Using the following information, answer the questions from 60 to 61.
In a certain country, income tax is assessed as follows. There is no tax on income up to $12340. Any income
over $12340 is taxed at a rate of 10%, up to an income of $23446. Any income over $23446 is taxed at 13%.
Question 60. Express the tax T as a function of income I.


0,
I ⩽ 12340


A T =
0.1, 12340 < I ⩽ 23446



0.13,
I > 23446


0,
I ⩽ 12340


B T =
0.1I, 12340 < I ⩽ 23446



0.1I,
I > 23446


0,
I ⩽ 12340


C T =
D T =
E T =
0.13I, 12340 < I ⩽ 23446



0.13I,
I > 23446


0,
I ⩽ 12340


0.1I, 12340 < I ⩽ 23446



0.13I,
I > 23446


0,
I ⩽ 12340


0.13I, 12340 < I ⩽ 23446



0.1I,
I > 23446
Solution
The tax function is
T =







0,
I ⩽ 12340
0.1I, 12340 < I ⩽ 23446
0.13I,
I > 23446
The correct answer is D
□
Question 61. How much tax (in dollars) is assessed on an income of $27148.
A 3529.6
QUESTIONS AND THE ANSWER
B 3529.5
C 3528.7
D 3529.7
E 3529.2
Page 26/1 — Question code: 1234
Solution
Using the formula for tax function, we have T (27148) = 3529.2
The correct answer is E
□
[L.O.2] Using the following information, answer the questions from 62 to 64.
A water filter equipment manufacturer has a fixed monthly cost of 340 million Vietnamese dong, and the
variable cost to produce x devices is C(x) = −0.005x2 + 6.0x, (0 ⩽ x ⩽ 800) (million Vietnamese dong). The
selling price per device is p(x) = −0.002x + 8.6, (0 ⩽ x ⩽ 800) (million Vietnamese dong).
Question 62. The company’s revenue (million Vietnamese dong) when selling 144 devices in a month
is
A 1197.9280
B 1193.9280
C 1199.9280
D 1196.9280
E 1192.9280
Solution
The company’s revenue is R(x) = x.p(x) = x(−0.002x + 8.6). Therefore, the company’s revenue (million
Vietnamese dong) when selling 144 devices in a month is R(144) = 144(−0.002 × 144 + 8.6) = 1196.9280
The correct answer is D
□
Question 63. The company’s profit (million Vietnamese dong) when selling 144 devices in a month
is
A 99.6080
B 94.6080
C 91.6080
D 96.6080
E 98.6080
Solution
The company’s profit is
P (x) = R(x) − C(x) − 340 = x(−0.002x + 8.6) − (−0.005x2 + 6.0x) − 340.
Therefore, The company’s profit (million Vietnamese dong) when selling 144 devices in a month is
P (144) = 144(−0.002 × 144 + 8.6) − (−0.005 × (144)2 + 6.0 × 144) − 340 = 96.6080
The correct answer is D
□
Question 64. To achieve the profit of 2300 million Vietnamese dong in a month, how many devices
does the company need to sell in that month?
A 605
B 604
C 601
D 600
E 602
Solution
The company’s profit is
P (x) = R(x) − C(x) − 340 = x(−0.002x + 8.6) − (−0.005x2 + 6.0x) − 340.
QUESTIONS AND THE ANSWER
Page 27/1 — Question code: 1234
Therefore, To achieve the profit of 2300 million Vietnamese dong in a month, the company need to sell x
devices in that month, it means that x(−0.002x + 8.6) − (−0.005x2 + 6.0x) − 340 = 2300 ⇒ x = 600
The correct answer is D
□
[L.O.2] Using the following information, answer the questions from 65 to 67.
All pyramids with a square base and a height of h = 4 have a surface area given by
S(v) = 0.75v + 2
p
0.28125v 2 + 12v,
where v is the volume of the pyramid. If x is the length of each side of the base then we can consider the volume
dv
of a pyramid as a function v(x) of x. In addition, suppose that v(7) = 65.33 and
(7) = 18.67.
dx
Question 65. Compute the slope of the slant asymptote of the graph of function S(v) with respect to
v.
A 1.1961
B 1.0757
C 2.3389
D 2.7592
E 1.8107
Solution
The slope of the slant asymptote of the graph of function S(v) with respect to v is
r
√
√
S(v)
0.75v + 2 0.28125v 2 + 12v
3 3 2
12
m = lim
= lim
= lim 0.75 + 2 0.28125 +
= +
≈ 1.8107
v→+∞ v
v→+∞
v→+∞
v
v
4
4
The correct answer is E
□
Question 66. Compute the rate of change of the surface area of the pyramid with respect to the length
x of each side when x = 7.
A 34.1516
B 34.2931
C 34.1693
D 34.4337
E 35.2992
Solution
The rate of change of the surface area of the pyramid with respect to the length x of each side is
2 × 0.28125 × v + 12
′
′
′
√
S (x) = S (v).v (x) = 0.75 +
× v ′ (x).
2
0.28125v + 12v
When x = 7, we have
S (7) = 0.75 + √
′
2 × 0.75 × 65.33 + 12
0.28125 × 65.332 + 12 × 65.33
× 18.67 ≈ 34.4337
The correct answer is D
□
Question 67. Approximate the change in surface area of a pyramid when the length of each side of the
base increases from x = 7 to x = 7.75.
A 25.8357
B 25.1149
C 25.2789
D 25.8253
E 26.8070
Solution
QUESTIONS AND THE ANSWER
Page 28/1 — Question code: 1234
The change in surface area of a pyramid when the length of each side of the base increases from x = 7 to
x = 7.75 is
∆S ≈ S ′ (7).(7.75 − 7) ≈ 25.8253
The correct answer is D
5
□
Integration
5.1
Indefinite integrals
Question 68 (L.O.1). Find the antiderivative F (x) of the function f (x) = 8 sin x+4 cos x which satisfies
π = 12.
the condition F
2
A −8 cos x + 4 sin x + 8
B −8 cos x − 4 sin x + 8
C −8 cos x + 4 sin x − 8
D 8 cos x − 4 sin x − 8
E 8 cos x + 4 sin x + 8
Solution
Z
f (x) dx = 4 sin (x) − 8 cos (x) + C. Moreover,
We have F (x) =
F
π 2
= 4 + C = 12 ⇒ C = 8.
The correct answer is A
□
Question 69 (L.O.1). Find the antiderivative F (x) of the function f (x) = 8ex + 4x which satisfies the
condition F (0) = 14.
A 8ex + 2x2 − 6
B −8ex + x2 + 6
D 8ex + 2x2 + 6
C −8ex + 2x2 − 6
E ex + 2x2 + 6
Solution
Z
We have F (x) =
f (x) dx = 8ex + 2x2 + C. Moreover,
F (0) = 8e0 + 2 × 02 + C = 14 ⇒ C = 6
The correct answer is D
□
Question 70 (L.O.1). Suppose that f ′′ (x) = −18x − 100 sin(5x) + 9e3x and f ′ (0) = 23, f (0) = −379.
Evaluate f (2).
A −2.2641
B −2.7473
C −2.0485
D −3.6915
E −3.5396
Solution
QUESTIONS AND THE ANSWER
Page 29/1 — Question code: 1234
We have f ′′ (x) = −18x − 100 sin(5x) + 9e3x ⇒ f ′ (x) = −9x2 + 20 cos(5x) + 3e3x + C1 . Since f ′ (0) = 23,
so C1 = 0. Thus, f (x) = −3x3 + 4 sin(5x) + e3x + 0x + C2 . Since f (0) = −379, nên C2 = −380. Therefore,
f (2) ≈ −2.7473.
The correct answer is B
5.2
□
Definite integrals
π
2
Z
Question 71 (L.O.1). Let
Z
f (x) dx = 9.51. Evaluate I =
0
A 63.8129
π
2
[6.49f (x) + 2.63 sin x] dx.
0
C 64.0289
B 64.3499
D 64.5836
E 64.7637
Solution
We have
Z
I=
π
2
π
2
Z
[6.49f (x) + 2.63 sin x] dx = 6.49
0
Z
f (x) dx + 2.63
0
π
2
sin x dx =
0
π
2
= 61.7199 − 2.63 cos x
= 61.7199 + 2.63 = 64.3499(≈ 64.3499).
0
The correct answer is B
□
Z
Question 72 (L.O.1). Let
i
5.55f (x) − 7.76g(x) dx.
6.4
Z
−2.5
Z
6.4
g(x) dx = −2.47. Evaluate I =
f (x) dx = 6.88 and
−2.5
B 384.9296
A 384.0257
6.4
C 385.2162
h
18.86x +
−2.5
D 384.6665
E 385.0722
Solution
We have
Z 6.4
Z
6.4
[18.86x + 5.55f (x) − 7.76g(x)] dx = 18.86
−2.5
Z
x2
2
Z
6.4
f (x) dx − 7.76
x dx + 5.55
−2.5
= 18.86
6.4
−2.5
g(x) dx =
−2.5
6.4
+ 5.55 × 6.88 − 7.76 × (−2.47) = 384.6665(≈ 384.6665).
−2.5
The correct answer is D
□
Question 73 (L.O.2). Let g(x) = f −1 (x) be the inverse function of function y = f (x) = ex/2 + 5.34.
Z 11.01
Evaluate
g(x)dx.
6.34
A 9.6536
B 10.3370
C 10.9418
D 10.7950
E 9.4292
Solution
x
Since y = f (x) = e 2 + 5.34 ⇒ x = 2 ln(y − 5.34). Thus g(x) = f −1 (x) = 2 ln(x − 5.34). Therefore,
Z
11.01
g(x)dx ≈ 10.3370.
6.34
QUESTIONS AND THE ANSWER
Page 30/1 — Question code: 1234
The correct answer is B
5.3
□
Riemann sum
Question 74 (L.O.2). Using the midpoint rule with 5 equal interval on [3; 11] to estimate the following
Z 11
f (x)dx where f (x) is given by the following table
integral I =
3
x
3
3.8 4.6 5.4 6.2 7.0 7.8 8.6 9.4 10.2
f (x) 6.8 4.3 6.6 5.0 3.4 5.8 5.1 5.5 4.2
A 41.8
B 41.88
C 42.24
11
5.8
6.0
D 41.6
E 42.26
Solution
11 − 3
= 1.6. Therefore, the subintervals consist of [3, 4.6]; [4.6, 6.2];
5
[6.2, 7.8]; [7.8, 9.4]; [9.4, 11.0]. The midpoints of these subintervals are 3.8; 5.4; 7.0; 8.6; 10.2. Thus, the approxEach subinterval has length of ∆x =
imation of the integral using midpoint rule is
h
i
I ≈ ∆x × 4.3 + 5.0 + 5.8 + 5.5 + 5.8 ≈ 42.24
The correct answer is C
5.4
□
The fundamental theorem of calculus
Z
x
(t2 − 9t − 22)dt on the
Question 75 (L.O.2). Find the maximum and minimum values of f (x) =
−9
interval [−9, 9].
A fmax ≈ 433.8311; fmin ≈ −0.1873
B fmax ≈ 433.4807; fmin ≈ 0.638
C fmax ≈ 432.9731; fmin ≈ 0.1885
D fmax ≈ 433.7281; fmin ≈ 0.0568
E fmax ≈ 432.8333; fmin ≈ 0
Solution
We have
Z
x
f (x) =
−9

(t2 − 9t − 22)dt ⇒ f ′ (x) = x2 − 9x − 22 = 0 ⇔ 
x = −2
x = 11
−9
Z
(t2 − 9t − 22)dt = 0;
f (−9) =
−9
Z
−2
(t2 − 9t − 22)dt ≈ 432.8333;
f (−2) =
−9
Z
9
f (9) =
(t2 − 9t − 22)dt ≈ 90.
−9
QUESTIONS AND THE ANSWER
Page 31/1 — Question code: 1234
The correct answer is E
6
□
Applications of integration
6.1
The Area problems
Question 76 (L.O.2). Find the area of the region bounded by the graphs of y =
x = 0; and x = 4.
A 0.5049
C 0.2790
B 0.3040
D 0.2973
1
√
; y = 0;
9 16 − x2
E 0.1745
Solution
The required area is
Z 4
dx
1h
b
π
x i4 1
√
A=
=
lim arcsin − arcsin 0 =
≈ 0.1745
arcsin
=
−
2
9
4
9
4
18
b→4
0
0 9 16 − x
The correct answer is E
□
Question 77 (L.O.2). Find the area of the region bounded by the graphs of y = e2.53x ; y = e4.75x and
x = 3.03.
A 374034.3816
B 374034.2241
C 374034.3701
D 374034.9554
E 374034.1488
Solution
The required area is
Z
3.03
A=
0
e2.53x i3.03
=
4.75
2.53 0
= 374034.381624145 ≈ 374034.3816
(e4.75x − e2.53x )dx =
h e4.75x
−
The correct answer is A
6.2
□
The Volume of the solid of revolution
Question 78 (L.O.2). Let the region D be bounded by the curve y =
p
15x2 + 2, x−axis and the lines
x = 0, x = 5. Evaluate the volume V of the solid, revolving the region D about x−axis.
A 1995.0119
B 1995.0890
C 1994.9113
D 1994.7434
E 1993.9313
Solution
We have
Z
V =π
5 p
15x2
0
+2
2
Z
dx = π
0
5
5
15x2 + 2 dx = π 5x3 + 2x 0 = 635π ≈ 1994.9113
The correct answer is C
□
QUESTIONS AND THE ANSWER
Page 32/1 — Question code: 1234
Question 79 (L.O.2). Find the volume of the solid obtained by rotating the region bounded by the
curves x = 0, y = −16, y = x2 − 8x about the y−axis.
A 134.0465
B 134.0413
C 133.9908
D 133.6074
E 134.2850
Solution
We have
VOy = 2π
Z 4
0
Z 4
128π
≈ 134.0413
|x(−16)| − |x(x2 − 8x)| dx = 2π x(16 − 8x + x2 )dx =
3
0
The correct answer is B
6.3
□
The arc length
Question 80 (L.O.2). Find the length of the arc y = 3 ln x, where 6 ⩽ x ⩽ 9.
A 3.2399
B 2.8595
C 3.9077
D 2.5529
E 3.1169
Solution
3
· The arc length is
x
s
2
Z 9
Z 9p
√
√
3
1
1
1 + [y ′ (x)]2 dx =
1+
L=
dx = −3 5 − 3 asinh
+ 3 asinh
+ 3 10 ≈ 3.2399
x
3
2
6
6
We have y = 3 ln x ⇒ y ′ =
The correct answer is A
□
Question 81 (L.O.2). A steady wind blows a kite due west. The kite’s height above the ground from
3
√
3 + 2x . Find the real number a such
horizontal position x = 0 to x = a > 0 meters is given by y =
that the distance travelled by the kite is 20.66 meters.
A 3.6074
B 3.2822
C 2.5416
D 2.8492
E 3.5988
Solution
We have y =
√
3 + 2x
3
√
⇒ y ′ = 3 2x + 3· The distance travelled by the kite is
√
3
Z ap
Z ar
h √
i2
(18a + 28) 2
56 7
′
2
L=
1 + [y (x)] dx =
1 + 3 2x + 3 dx =
−
= 20.66
27
27
0
0
⇒ a ≈ 2.8492.
The correct answer is D
□
Question 82 (L.O.2). Find the length of the arc y = P (x), where 3 ⩽ x ⩽ 9. Here P (x) is a polynomial
of degree 2 such that P (3) = 62, P (4) = 96, P (9) = 386.
A 323.5810
QUESTIONS AND THE ANSWER
B 323.6091
C 323.3469
D 324.0597
E 324.9452
Page 33/1 — Question code: 1234
Solution
Let P (x) = ax2 + bx + c. Since P (3) = 62, P (4) = 96, P (9) = 386, so we have



 9a + 3b + c = 62
16a + 4b + c = 96 ⇒ a = 4; b = 6; c = 8.



81a + 9b + c = 386
Thus, P (x) = 4x2 + 6x + 8 ⇒ P ′ (x) = 8x + 6. The arc length is
Z 9r
Z 9p
h
i2
1 + [y ′ (x)]2 dx =
1 + 8x + 6 dx =
L=
3
3
√
√
15 901 asinh (30) asinh (78) 39 6085
=−
−
+
+
≈ 324.0597.
8
16
16
8
The correct answer is D
□
Z xp
Question 83 (L.O.2). Find the length of the arc y =
5t2 + 7dt, where 2 ⩽ x ⩽ 8.
2
A 69.4857
B 69.7190
C 69.7552
D 69.0281
E 68.6696
Solution
We have y ′ (x) =
p
5x2 + 7. The arc length is
Z 8r
Z 8p
hp
i2
′
2
1 + [y (x)] dx =
1+
5x2 + 7 dx =
L=
2
2
√
= −2 7 −
√ √
4 5 asinh 210
5
√
√ √
4 5 asinh 2 10
+
+ 8 82 ≈ 69.4857
5
The correct answer is A
6.4
□
The Area of the surface of revolution
Question 84 (L.O.2). Find the area of the surface of revolution obtained by rotating the curve y =
√
2 x + 6, where 1 ⩽ x ⩽ 8 about the x−axis.
A 502.2993
B 502.5967
C 502.8761
D 502.4058
E 502.0702
Solution
√
1
We have y = 2 x + 6 ⇒ y ′ = √ · The surface area is
x
s
Z 8 p
Z 8
√
1 2
′
2
A = 2π |y| 1 + [y (x)] dx = 2π (2 x + 6) 1 + √
dx =
x
1
1
√ !
√ √ 26 2
√ = −2π 6 log 1 + 2 +
+ 2π 6 asinh 2 2 + 36 + 36 2 ≈ 502.2993.
3
The correct answer is A
□
QUESTIONS AND THE ANSWER
Page 34/1 — Question code: 1234
p
Question 85 (L.O.2). Rotating the curve x = − y 2 + 3, where 0 ⩽ y ⩽ 3 about the y−axis, evaluate
the surface area of the obtained surface of revolution.
A 53.5843
C 54.0413
B 54.2794
D 53.1674
E 54.2563
Solution
p
y
We have x = − y 2 + 3 ⇒ x′ = − p
· The surface area is
2
y +3
3
Z
A = 2π
s
Z 3
p
p
|x| 1 + [x′ (y)]2 dy = 2π | − y 2 + 3| 1 +
0
Z
s
3
2π
= 2π
0
0
2y 2 + 3 p 2
y + 3dy = 2π
y2 + 3
y2
dy =
y2 + 3
√
√ √ !
3 2 asinh 6
3 21
≈ 54.0413.
+
4
2
The correct answer is C
6.5
□
Applications of integration in Physics
Question 86 (L.O.2). Evaluate the approximation of the displacement of one particle with velocity
v(t) = 9 arcsin(t) (m/s) from t = 0s to t = 0.7s.
A 3.0163
B 3.2834
C 2.5228
D 1.9853
E 2.3123
Solution
The displacement of one particle is
Z
D=
0.7
Z
0.7
v(t)dt =
0
9 arcsin(t)dt.
0



 u = arcsin t
 du = √ dt
1 − t2
Let
⇒
 dv = dt

 v = t
Therefore,
Z 0.7
Z
0.7
0.7
tdt
9 0.7
√
−9
+
D = 9t arcsin t
= 9t arcsin t
(1 − t2 )−1/2 d(1 − t2 ) =
2 0
0
0
1 − t2
0
p
0.7
0.7
= 9t arcsin t
+ 9 1 − t2
= 2.31228981433631 ≈ 2.3123.
0
0
The correct answer is E
□
Question 87 (L.O.2). The velocity of a dragster t seconds after leaving the starting line is v(t) =
72te−0.7t ft/sec. Find the approximation of the distance traveled by the dragster during the first 9 seconds.
A 145.0934
B 144.0773
C 143.9867
D 144.9122
E 144.9691
Solution
QUESTIONS AND THE ANSWER
Page 35/1 — Question code: 1234
The distance is
9
Z
Z
9
|v(t)|dt =
D=
72te−0.7t dt =
0
0
Z
9
72te−0.7t dt = 144.969057569783 ≈ 144.9691
0
The correct answer is E
□
Question 88 (L.O.2). When a particle is located a distance x meter from the origin, a force of given
1
(newton) acts on it. How much work is done in moving it from x = 5 to x =
F (x) = 2
x
+
4x
+ 53
√
7 3 − 2.
A −0.0604
B −0.0687
C 0.0374
D 0.0634
E −0.4302
Solution
We have
Z
√
7 3−2
W =
5
√
7 3−2
Z 7√3−2
1
d(x + 2)
1
(x + 2)
F (x)dx =
=
= arctan
2
2
x + 4x + 53
(x + 2) + 49
7
7
5
5
√
1
1 π π
π
= (arctan 3 − arctan 1) =
−
=
≈ 0.0374.
7
7 3
4
84
Z
√
7 3−2
5
=
The correct answer is C
□
Question 89 (L.O.2). When a particle is located a distance x meter from the origin, a force of given
1
F (x) = √
(newton) acts on it. How much work is done in moving it from x = 17/2
2
−16x + 128x + 1040
to x = 13.
B −0.7293
A 0.2618
C 0.5967
D −0.3280
E −0.6007
Solution
We have
Z
1
d(x − 4)
1 13
p
W =
F (x)dx =
dx =
=
2
4
81 − (x − 4)2
17/2
17/2 −16x + 128x + 1040
17/2
1
(x − 4) 13
1
1
1 π π π
= arcsin
=
arcsin 1 − arcsin
=
−
=
≈ 0.2618.
4
9
4
2
4 2
6
12
17/2
Z
13
Z
13
√
The correct answer is A
□
Question 90 (L.O.2). When a particle is located a distance x meter from the origin, a force of given
F (x) = ln(6 + 5x) (newton) acts on it. How much work is done in moving it from x = 10 to x =
17.
A 30.1241
B 30.4412
C 29.4318
D 30.0137
E 29.2949
Solution
Z
17
We have W =
Z
17
ln(6 + 5x)dx = −
F (x)dx =
10
10
91 log (91)
56 log (56)
−7+
≈ 30.0137.
5
5
The correct answer is D
□
QUESTIONS AND THE ANSWER
Page 36/1 — Question code: 1234
Question 91 (L.O.2). When a particle is located a distance x meter from the origin, a force of given
60
(newton) acts on it. How much work is done in moving it from x = 1 to x =
F (x) = √
196 − x2
6.
A 22.2853
B 22.7360
C 23.0572
D 21.5142
E 23.0559
Solution
Z
We have W =
6
Z
F (x)dx =
1
1
6
60
√
dx = −60 asin
196 − x2
1
14
3
+ 60 asin
≈ 22.2853.
7
The correct answer is A
6.6
□
Applications of integration in Economics
Question 92 (L.O.1). The accumulated, or total, future value after T years of an income stream of R(t)
thousand dollars per year, earning interest at the rate of r per year compounded continuously, is given by
Z T
R(t)e−rt dt. A company recently bought an automatic car-washing machine that is expected
A = erT
0
to generate 3 thousand dollars in revenue per year, t years from now, for the next 7 years. If the income
is reinvested in a business earning interest at the rate of r = 11% per year compounded continuously,
find the total accumulated value of this income stream at the end of 7 years.
A 30.6382
C 32.4296
B 31.6300
D 32.2997
E 31.7604
Solution
We have r = 11%, R(t) = 3, T = 7. Therefore, the total accumulated value of this income stream at the end
of 7 years is
rT
Z
A=e
T
R(t)e−rt dt = 31.6299887395886 ≈ 31.6300
0
The correct answer is B
6.7
□
The average value of a funtion
Question 93 (L.O.2). The interest rates charged by Madison Finance on auto loans for used cars over
a certain 12−month period in 2022 are approximated by the function
r(t) = −
1 3 1 2
t + t − 2t + 27,
21
3
(0 ⩽ t ⩽ 12)
where t is measured in months and r(t) is the annual percentage rate. What is the average rate on auto
loans extended by Madison over the 12−month period?
A 11.1063
B 11.2527
C 10.2379
D 10.4286
E 10.6659
Solution
The average rate on auto loans extended by Madison over the 12−month period is
Z 12
Z 1
1 12
1
73
1
rave =
r(t)dt =
− t3 + t2 − 2t + 27 dt =
≈ 10.4286.
12 − 0 0
12 0
21
3
7
QUESTIONS AND THE ANSWER
Page 37/1 — Question code: 1234
The correct answer is D
6.8
□
Improper integrals
+∞
Z
3x2 dx
Question 94 (L.O.2). Evaluate the integral
7 + x4
1
A 1.8614
B 2.4366
C 1.9137
D 1.5139
E 1.3924
Solution
We have
+∞
+∞
Z
Z
3x2 dx
3/x2 dx
I=
=
7 + x4
1 + 7/x4
1
Let t =
1
dx
⇒ dt = − 2 ·
x
x
Z0
I=
1
−3dt
=
1 + 7t4
1
Z1
3dt
≈ 1.9137
1 + 7t4
0
The correct answer is C
□
Z
Question 95 (L.O.2). Evaluate the integral
A 1.1263
B 0.3787
+∞
√
671 x
√
4
·
C −0.6059
dx
x2 + 625
D 0.7021
E −0.0613
Solution
Let t =
p
4
x2 + 625 ⇒ t4 = x2 + 625 ⇒ 2t3 dt = xdx. Then,
Z +∞
Z +∞
Z +∞
2t3 dt
dx
xdx
√
√
I= √
=
=
=
√
4
4
2
t(t4 − 625)
x2 + 1
x2 + 625
671 x ·
671 x ·
6
Z +∞
Z +∞ 2
Z +∞
Z +∞
2t2 dt
(t + 25) + (t2 − 25)dt
dt
dt
=
=
=
+
=
4
2
2
2
2
t − 625
(t − 25)(t + 25)
t − 25
t + 25
6
6
6
6
=
1
t−5
ln
10
t+5
π
1
6
1
1
− arctan −
ln
=
10 5
5 10
11
6
6
π
= −0.175211610119639 + 0.1 log (11) +
≈ 0.3787
10
+∞
+
1
t
arctan
5
5
+∞
=
The correct answer is B
□
Question 96 (L.O.2). The total profit of a firm in dollars, from producing x units of an item, is P (x).
The firm is able to determine that its marginal profit is given by P ′ (x) = 190e−1.05x . Suppose that it were
possible for the firm to make infinitely many units of this item. What would its total profit be?
A 180.8926
B 181.1377
C 180.4869
D 181.9074
E 180.9524
Solution
QUESTIONS AND THE ANSWER
Page 38/1 — Question code: 1234
Z
The total profit is P (x) =
∞
′
∞
Z
−1.05x
190e
P (x)dx =
0
0
190 −1.05x
dx =
e
−1.05
∞
=
0
190
≈ 180.9524.
1.05
The correct answer is E
□
+∞
Z
8dx
√
.
Question 97 (L.O.2). Evaluate the integral
x2 + x4
5
A 0.9952
B 0.9963
C 1.5895
D 1.1703
E 1.7271
Solution
We have
+∞
+∞
Z
Z
8dx
8/x2 dx
√
p
I=
=
x2 + x4
1 + 1/x2
5
Let t =
5
dx
1
⇒ dt = − 2 ·
x
x
Z0
I=
−8dt
√
=
1 + t2
1/5
Z1/5
8dt
√
1 + t2
0
= 1.58952088279393 ≈ 1.5895
The correct answer is C
□
Question 98 (L.O.2). Let a be the real number such that lim
x→+∞
p
x2 + 8x + 2 − x − a = 0. Evaluate
+∞
Z
dx
·
(x + a)(x − 2)2
a+
6
A 4.5249
B 4.0162
C 3.4425
D 3.4233
E 4.8000
Solution
We have a = lim
x→+∞
p
x2 + 8x + 2 − x = lim √
x→+∞
8x + 2
= 4. Then,
x2 + 8x + 2 + x
+∞
Z
dx
log (10)
log (4)
1
+
+
≈ 4.0162
=4+ −
(x + a)(x − 2)2
36
36
24
a+
6
The correct answer is B
□
Z 5 2
x dx
mdx
√
=
x−1
81 − x2
1
0
D 18.3127
E 18.3730
Z
Question 99 (L.O.2). Find the approximation of m which satisfies
A 18.2816
B 17.4858
C 17.6462
9
√
Solution
Z
Let A =
0
9
√
h
dx
x i9 π
= arcsin
= ·
9 0
2
81 − x2
QUESTIONS AND THE ANSWER
Page 39/1 — Question code: 1234
Z
5
Let B =
1
√
x2 dx
√
· Using substitution method, let t = x − 1 ⇒ t2 = x − 1 ⇒ 2tdt = dx, we have
x−1
√
√
5
√ 4
Z √4
3
√
2( 4)5 4( 4)3
2t
4t
2
2
B=
2(t + 1) dt =
+
+ 2t
=
+
+ 2 4.
5
3
5
3
0
0
B
824
=
≈ 17.4858
A
15π
The correct answer is B
Therefore, m =
7
□
Ordinary differential equations (ODE)
7.1
The first order differential equations
7
Question 100 (L.O.2). Which function is a solution of the following differential equation y ′ + y = 3x,
x
where C is any constant?
1
3
1
A y = x2 + Cx−7
B y = x2 + Cx−7
C y = x2 + Cx7
3
8
3
3 2
3 2
7
−7
D y = x + Cx
E y = x + Cx
8
10
Solution
This is the first order linear differential equation with p(x) =
Z
−
Z
−
y=e
7
dx hZ
x
3xe
Z
p(x)dx hZ
y=e
Z
7
and q(x) = 3x. Therefore,
x
p(x)dx
q(x)e
dx + C
i
7
h 9
i
i
dx
x dx + C = x−7 3 x + C = 1 x2 + Cx−7
9
3
The correct answer is A
7.2
□
The second order differential equations
Question 101 (L.O.2). Let y(x) be a function satisfying y ′′ −4y ′ −32y = 0, y(0) = 6, y ′ (0) = 24. Evaluate
y(1.88).
A 13609714.8215
B 13609714.2745
D 13609713.4611
C 13609713.4938
E 13609714.0102
Solution
1. Step 1. Solve the homogeneous equation
y ′′ − 4y ′ − 32y = 0.
The characteristic equation k 2 − 4k − 32 = 0 has 2 real different roots
k1 = 8, k2 = −4
QUESTIONS AND THE ANSWER
Page 40/1 — Question code: 1234
2. Step 2. The homogeneous solution is
yh = C1 e8x + C2 e−4x
3. Step 3. Thus,
yh′ = 8C1 e8x + (−4)C2 e−4x .
Since y(0) = 6 ⇒ C1 + C2 = 6 and y ′ (0) = 8C1 + (−4)C2 = 24 ⇒ C1 = 4, C2 = 2. Therefore,
y(1.88) ≈ 13609714.0102
The correct answer is E
□
Question 102 (L.O.2). Let y(x) be a function satisfying y ′′ − 5y ′ = 2x2 + 5, y(0) = 0, y ′ (0) = 0. Evaluate
y(2).
A 4543.1199
B 4542.7416
C 4542.6055
D 4542.9902
E 4541.7701
Solution
1. Step 1. Solve the homogeneous equation
y ′′ − 5y ′ = 0.
The characteristic equation k 2 − 5k = 0 has 2 real different roots
k1 = 5, k2 = 0
2. Step 2. The homogeneous solution is
yh = C1 e5x + C2 e0x = C1 e5x + C2
3. Step 3. Find a particular solution of nonhomogeneous equation
y ′′ − 5y ′ = 2x2 + 5.
The particular solution has a form yp = xs .e0x .(Ax2 + Bx + C). Since α = 0 is a root of characteristic
equation then s = 1 and yp = x(Ax2 + Bx + C) = Ax3 + Bx2 + Cx.
0×
yp
=
Ax3 + Bx2 + Cx
−5×
yp′
=
3Ax2 + 2Bx + C
1×
yp′′
=
6Ax + 2B
yp′′ − 5yp′
=
−15Ax2 + (−10B + 6A)x + 2B − 5C =
2x2 + 5
⇒A=−
QUESTIONS AND THE ANSWER
2
2
129
;B = − ;C = −
15
25
125
Page 41/1 — Question code: 1234
4. Step 4. The general solution is
2
2
129
3
2
ygen = yh + yp = C1 e + C2 + −
x + −
x + −
x
15
25
125
2
4
129
′
5x
2
⇒ ygen = 5C1 e + −
x + −
x+ −
.
5
25
125
129
129
Since y(0) = 0 ⇒ C1 + C2 = 0 và y ′ (0) = 0 ⇒ C1 =
, C2 = −C1 = −
. Therefore,
625
625
6857 129e10
y(2) = −
+
≈ 4542.6055
1875
625
5x
The correct answer is C
□
Question 103 (L.O.2). Let y(x) = axe4x be a function satisfying y ′′ − 10y ′ + 24y = −6.3894e4x . Find the
value of a.
A 3.5403
B 3.6631
C 3.1947
D 3.2759
E 3.0536
Solution
Since y(x) = axe4x is the solution of y ′′ − 10y ′ + 24y = −6.3894e4x . Then,
24×
yp
=
axe4x
−10×
yp′
=
ae4x (1 + 4x)
1×
yp′′
=
ae4x (8 + 16x)
y ′′ − 10y ′ + 24y
=
−2ae4x =
−6.3894e4x
⇒ a = 3.1947.
The correct answer is C
□
ESSAY QUESTIONS
1
Differentiation
1.1
Related rates
Question 104.
[L.O.2] Two crates, A and B, are on the floor of a warehouse.
The crates are joined by a rope 15m long, each crate being
hooked at floor level to an end of the rope. The rope is stretched
tight and pulled over a pulley P that is attached to a rafter
P Q = 4m above a point Q on the floor directly between the
two crates. If crate A is 3m from Q and is being pulled directly
away from Q at a rate of 0.5 m/s, how fast is crate B moving
toward Q?
Solution
QUESTIONS AND THE ANSWER
Page 42/1 — Question code: 1234
1. The Pythagorean theorem can give us
AP + BP = 15 ⇔
p
p
x2 + 42 + y 2 + 42 = 15
2. Differentiating both sides of the equation, we receive
dx
y
dy
dy
x
x
√
·
+p
·
= 0 ⇒ y ′ (t) =
= −√
· x′ (t) ·
2
2
2
dt
x + 9 dt
x +9
y + 9 dt
p
y2 + 9
y
3. When x = 3 ⇒ y ≈ 9.1652 and x′ (t) = 0.5, then
√
3
84 + 9
y (t) = − √
× 0.5 ×
≈ −0.3720 (m/s)
2
9.1652
3 +9
′
□
Question 105.
[L.O.2] A major network is televising the launching of the rocket. At a
distance of x = 12000 feet from the launch site, a spectator is observing
a rocket being launched vertically. A camera tracking the liftoff of the
rocket is located at point A, where ϕ denotes the angle of elevation
of the camera at A. When the rocket is z = 12434 ft from the camera
and this distance is increasing at the rate of 480 ft/sec, how fast is ϕ
changing?
Solution
1. We have cos ϕ =
x
12000
12000
=
⇒ ϕ = arccos
z
z
z
2. Differentiating both sides of the equation, we receive
′
1
⇒ ϕ (t) = − q
1−
120002
z2
12000
× − 2
.z ′ (t)
z
3. When z(t0 ) = 12434; z ′ (t0 ) = 480, at the moment t0 , we have
1
12000
′
× −
× 480 ≈ 0.1423(rad/s)
ϕ (t0 ) = − q
2
124342
1 − 12000
2
12434
□
2
Applications of integration
Question 106 (L.O.2). Let V1 and V2 be the volumes of the solids that result when the plane region
4
1
1
bounded by y = , y = 0, x = , and x = c, where c > , is revolved about the x−axis and the y−axis,
x
5
5
respectively. Find the value of c for which V1 = V2 .
Solution
QUESTIONS AND THE ANSWER
Page 43/1 — Question code: 1234
Z
c
16dx
16π
= 80π −
2
x
c
1/5
Z c
4dx
x·
2. We also have V2 = 2π
= 2π × 4(c − 1/5)
x
1/5
1. We have V1 = π
3. V1 = V2 ⇒ c = 10.0
□
Question 107 (L.O.2). Find the area of the surface obtained by rotating the arc y = 2x3/2 , (0 ⩽ x ⩽ 4),
about the x−axis.
Solution
3
1. We have y ′ = 2 × x1/2 .
2
2. Using the area of surface of revolution formula, we get
Zb
|f (x)|
S = 2π
r
Z4
h
i2
36
3/2
1 + f ′ (x) dx = 2π 2x × 1 + xdx ≈ 820.7836
4
r
a
0
□
3
Ordinary differential equations (ODE)
3.1
The first order differential equations
Question 108 (L.O.2). A tank initially contains 80 grams of salt dissolved in 100 liters of water. Pure
water flows into the tank at the rate of 12 liters per minute, and the well stirred mixture flows out of the
tank at the same rate. How much salt does the tank contain after 2 minutes?
Solution
1. Step 1. The rate of change of salt in the tank,
dQ
is equal to the rate at which salt is flowing in minus
dt
the rate at which it is flowing out.
dQ
= rate in − rate out
dt
2. Step 2. The rate at which salt enters the tank is the concentration 0 gr/l times the flow rate 12 l/min
or 0 gr/min.
3. Step 3. To find the rate at which salt leaves the tank, we need to multiply the concentration of salt
in the tank by the rate of outflow, 12 l/min. The volume of water in the tank remains constant at 100
liters, and since the mixture is well-stirred, the concentration throughout the tank is the same, namely,
Q
12Q
gr/l. Therefore, the rate at which salt leaves the tank is
gr/min.
100
100
QUESTIONS AND THE ANSWER
Page 44/1 — Question code: 1234
4. Step 4. Solve the differential equation using separable method
Z
Z
dQ
12Q
dQ
12
dQ
12
=0−
⇒
=−
dt ⇒
= −
dt
dt
100
Q
100
Q
100
⇒ ln |Q| = −
12
t + ln C ⇒ Q = Ce−12t/100
100
5. Step 5. The initial condition is Q(0) = 80, so 80 = Ce0 = C. Thus Q = 80e−12t/100
⇒ Q(2) = 80e−12×2/100 = 62.9302.
□
Question 109 (L.O.2). A tank with a capacity of 130 liters initially contains 80 grams of salt dissolved
in 100 liters of water. Pure water flows into the tank at the rate of 12 liters per minute, and the well
stirred mixture flows out of the tank at the rate of 2 liters per minute. How much salt does the tank
contain when the tank is full?
Solution
1. Step 1. The rate of change of salt in the tank,
dQ
is equal to the rate at which salt is flowing in minus
dt
the rate at which it is flowing out.
dQ
= rate in − rate out
dt
2. Step 2. The rate at which salt enters the tank is the concentration 0 gr/l times the flow rate 12 l/min
or 0 gr/min.
3. Step 3. To find the rate at which salt leaves the tank, we need to multiply the concentration of salt in
the tank by the rate of outflow, 2 l/min. The volume of water in the tank after t minutes is 100 + (12 −
2)t = 100+10t liters, and since the mixture is well-stirred, the concentration throughout the tank is the
Q
2Q
same, namely,
gr/l. Therefore, the rate at which salt leaves the tank is
gr/min.
100 + 10t
100 + 10t
4. Step 4. Solve the differential equation using separable method
Z
Z
dQ
2Q
dQ
2
dQ
2
=0−
⇒
=−
dt ⇒
= −
dt
dt
100 + 10t
Q
100 + 10t
Q
100 + 10t
⇒ ln |Q| = −
2
2
ln |100 + 10t| + ln C ⇒ Q = C(100 + 10t)− 10
10
2
2
5. Step 5. The initial condition is Q(0) = 80, so 80 = C × 100− 10 ⇒ C = 80 × 100 10 . Thus
2
2
Q = 80 × 100 10 (100 + 10t)− 10 .
The tank is full after
130 − 100
= 3 minutes.
12 − 2
2
2
⇒ Q(3) = 80 × 100 10 (100 + 10 × 3)− 10 = 75.9104.
□
QUESTIONS AND THE ANSWER
Page 45/1 — Question code: 1234
Question 110 (L.O.2). Recall that a simple electric circuit with a resistor of R ohms, an inductor of
L henries and a battery with a voltage of E volts will produce a current I(t) amperes (t is measured
in seconds) satisfying the following differential equation LI ′ (t) + RI = E. Let study one such circuit
with R = 2 ohms, L = 3 henries and E = 12 volts. Given initial condition I(0) = 3 amperes. Evaluate
current’s value at t = 3 seconds.
Solution
1. Step 1. Let consider the differential equation LI ′ (t) + RI = E in our case, we have
2
12
3I ′ (t) + 2I = 12 ⇔ I ′ (t) + I =
3
3
2. Step 2. Solving this linear differential equation with p(t) =
Z
−
I(t) = e
2
12
and q(t) = , we obtain
3
3
Z
p(t)dt hZ
p(t)dt
q(t)e
i
dt + C .
Therefore,
Z
Z
2
2
Z
h
i
−
dt
dt
12
3
3
I(t) = e
e
dt + C
3
Z
h
i
2
12 2 t
I(t) = e− 3 t
e 3 dt + C
3
− 23 t 12 23 t
I(t) = e
e +C
2
2
12
12
3. Step 3. The initial condition is I(0) =
+ C = 3, so C = 3 − . Thus I(t) = e− 3 t
2
2
2
12 2 ×3
⇒ I(3) = e− 3 ×3
e 3 + C = 5.594.
2
12 2 t
e3 + C
2
□
3.2
The second order differential equations
Question 111 (L.O.2). Solve the differential equation y ′′ − 8y ′ + 12y = 120e12x
Solution
1. Step 1. Solve the homogeneous equation
y ′′ − 8y ′ + 12y = 0.
The characteristic equation k 2 − 8k + 12 = 0 has 2 real different roots
k1 = 6, k2 = 2
QUESTIONS AND THE ANSWER
Page 46/1 — Question code: 1234
2. Step 2. The homogeneous solution is
yh = C1 e6x + C2 e2x
3. Step 3. Find a particular solution of nonhomogeneous equation
y ′′ − 8y ′ + 12y = 120e12x .
The particular solution has a form yp = xs .e12x .A. Since α = 12 is not the root of characteristic
equation then s = 0 and yp = A.e12x .
12×
yp
=
Ae12x
−8×
yp′
=
12Ae12x
1×
yp′′
=
144Ae12x
yp′′ − 8yp′ + 12yp
=
60Ae12x =
120e12x
⇒A=2
4. Step 4. The general solution is
ygen = yh + yp = C1 e6x + C2 e2x + 2e12x .
□
Question 112 (L.O.2). Solve the differential equation y ′′ − 14y ′ + 74y = 10e7x cos(5x)
Solution
1. Step 1. Solve the homogeneous equation
y ′′ − 14y ′ + 74y = 0.
The characteristic equation k 2 − 14k + 74 = 0 has 2 complex conjugate roots
k1 = 7 + 5i, k2 = 7 − 5i
2. Step 2. The homogeneous solution is
h
i
yh = e7x C1 cos(5x) + C2 sin(5x)
3. Step 3. Find a particular solution of nonhomogeneous equation
y ′′ − 14y ′ + 74y = 10e7x cos(5x).
h
i
The particular solution has a form yp = xs .e7x . A cos(5x) + B sin(5x) . Since 7 + 5i is one of two
h
i
complex roots of the characteristic equation, so s = 1 and yp = x.e7x . A cos(5x) + B sin(5x) . Differentiating yp we obtain
h
i
yp′ = e7x . (7A + 5B)x cos(5x) + (7B − 5A)x sin(5x) + A cos(5x) + B sin(5x)
h
i
yp′′ = e7x . (24A + 70B)x cos(5x) + (24B − 70A)x sin(5x) + (14A + 10B) cos(5x) + (14B − 10A) sin(5x)
h
i
y ′′ − 14y ′ + 74y = e7x . 10B cos(5x) − 10A sin(5x) = 10e7x cos(5x) ⇒ A = 0; B = 1
QUESTIONS AND THE ANSWER
Page 47/1 — Question code: 1234
4. Step 4. The general solution is
h
i
ygen = yh + yp = e7x C1 cos(5x) + C2 sin(5x) + x.e7x . sin(5x).
□
Question 113 (L.O.2). Let yp be the particular solution of the differential equation y ′′ − 4y ′ − 12y =
−144x − 72. Evaluate yp (2).
Solution
1. Step 1. Solve the homogeneous equation
y ′′ − 4y ′ − 12y = 0.
The characteristic equation k 2 − 4k − 12 = 0 has 2 real different roots
k1 = 6, k2 = −2
2. Step 2. Find a particular solution of nonhomogeneous equation
y ′′ − 4y ′ − 12y = −144x − 72.
The particular solution has a form yp = xs .e0x (Ax + B). Since α = 0 is not the root of characteristic
equation then s = 0 and yp = Ax + B.
−12×
yp
=
Ax + B
−4×
yp′
=
A
1×
y”p
=
0
y”p − 4yp′ − 12yp = −12Ax − (12B + 4A) = −144x − 72



 A = 12
−12A = −144
⇒
⇔
 −12B − 4A = −72
 B=2
3. Step 3. The particular solution is
yp = 12x + 2 ⇒ yp (2) = 12 × 2 + 2 = 26.
□
Question 114 (L.O.2). Recall that a simple electric circuit with a resistor of R ohms, an inductor of L
henries, a capacitor of C farads and a battery with a voltage E(t) = 100 sin(ωt) (volts) will produce a
current I(t) amperes (t is measured in seconds) satisfying the following differential equation LI ′′ (t) +
1
RI ′ + I = E ′ (t) = 100ω cos(ωt). Let study one such circuit with R = 8 ohms, L = 5 henries, C = 0.2
C
farads and ω = 11 rad/seconds. Given initial condition I(0) = 0, I ′ (0) = 0, evaluate current’s value at
t = 6.2 seconds.
Solution
QUESTIONS AND THE ANSWER
Page 48/1 — Question code: 1234
1. Step 1. Solve the homogeneous equation
LI ′′ (t) + RI ′ +
1
I = 0.
C
1
= 0 ⇒ 5k 2 + 8k + 5 = 0 has 2 complex conjugate roots
C
4 3
4 3
k1 = − + i; k2 = − − i
5 5
5 5
The characteristic equation Lk 2 + Rk +
2. Step 2. The homogeneous solution is
Ih = e
− 54 t
C1 cos
3
3
t + C2 sin
t
5
5
3. Step 3. Find a particular solution of nonhomogeneous equation
LI ′′ (t) + RI ′ +
1
I = 100ω cos(ωt).
C
h
i
The particular solution has a form Ip = ts .e0t A cos(ωt) + B sin(ωt) . Since 0 + ωi is not the root of
characteristic equation then s = 0 and Ip = A cos(ωt) + B sin(ωt).
5×
Ip
=
A cos(ωt) + B sin(ωt)
8×
Ip′
=
−Aω sin(ωt) + Bω cos(ωt)
5×
I”p = −Aω 2 cos(ωt) − Bω 2 sin(ωt)
5I”p + 8Ip′ + 5Ip
=
100ω cos(ωt)


ω(5 − 5ω 2 ) × 100


 (5 − 5ω 2 )A + 8ωB = 100ω
 A=
(5 − 5ω 2 )2 + (8ω)2
⇒
⇔
8ω 2 × 100
 −8ωA + (5 − 5ω 2 )B = 0


 B=
(5 − 5ω 2 )2 + (8ω)2
4. Step 4. The general solution is
3
3
Igen = Ih + Ip = e
t + C2 sin
t + A cos(ωt) + B sin(ωt).
C1 cos
5
5
4
4
3
3
4
3
3
′
We have Igen
= e− 5 t − C1 + C2 cos
t + − C2 − C1 sin
t −Aω sin(ωt)+Bω cos(ωt)
5
5
5
5
5
5
From the initial condition I(0) = 0, I ′ (0) = 0, we obtain


C1 = −A
 C2 = − 5 Bω − 4A
3
3
3
3
− 54 ×6.2
C1 cos
× 6.2 + C2 sin
× 6.2 +A cos(ω×6.2)+B sin(ω×
5. Step 5. Evaluate Igen (6.2) = e
5
5
6.2) ≈ −1.3042.
− 45 t
□
Question 115 (L.O.2). Recall that a spring with mass m, spring constant k relaxing after being stretched
will be in motion, where its position x(t) satisfies the following differential equation mx′′ (t) + kx = 0.
Let study one spring with m = 2 and k = 32. Given initial condition x(0) = 6, x′ (0) = 4, evaluate
numerically its position at t = 2.
Solution
QUESTIONS AND THE ANSWER
Page 49/1 — Question code: 1234
1. Step 1. Solve the homogeneous equation
mx′′ (t) + kx = 0.
The characteristic equation mr2 + k = 0 has 2 complex conjugate roots
r
r
r
r
k
32
k
32
r1 =
i=
i; r2 = −
i=−
i
m
2
m
2
2. Step 2. The homogeneous solution is
r
x = C1 cos
!
!
r
32
32
t + C2 sin
t
2
2
!
!
r
r
32
32
32
t + C2
cos
t . From the initial condition, we have
3. Step 3. x = −C1
2
2
2

!
!
r
r

32
32


× 0 + C2 sin
× 0 = C1 = 6
x(0) = C1 cos


2
2
!
!
r
r
r
r
r

32
32
32
32
32

′

× 0 + C2
cos
× 0 = C2
=4

 x (0) = −C1 2 sin
2
2
2
2



C1 = 6
r
⇔
2

 C2 = 4 ×
32
!
!
r
r
32
32
4. Step 4. Evaluate x(2) = C1 cos
× 2 + C2 sin
× 2 ≈ 0.1164.
2
2
r
′
32
sin
2
r
□
3.3
System of the first order ODEs
Question 116 (L.O.2). Consider the system of ODEs:


x′ (t) = 4x − y − t

y ′ (t) = 5x + 10y + (t − 1)
Find a solution of the system such that x (0) = 3, y (0) = 1.
Solution
1. Step 1. The first equation gives y in terms of x and x′
y=
x′ − 4x + t
−1
Differentiating this will give us y ′ in terms of x′ and x′′
y′ =
x′′ − 4x′ + 1
−1
and we can now substitute these into the second equation, we have
x′′ − 4x′ + 1
x′ − 4x + t
= 5x + 10 ×
+ (t − 1)
−1
−1
⇔ x′′ − 14x′ + 45x = 9t
QUESTIONS AND THE ANSWER
Page 50/1 — Question code: 1234
2. Step 2. Solve the homogeneous equation
x′′ − 14x′ + 45x = 0.
The characteristic equation k 2 − 14k + 45 = 0 has 2 real different roots
k1 = 5, k2 = 9
3. Step 3. The homogeneous solution is
xtn = C1 e5t + C2 e9t
4. Step 4. Find a particular solution of nonhomogeneous equation
x′′ − 14x′ + 45x = 9t.
The particular solution has a form xr = ts .e0t .(At + B). Since α = 0 is NOT a root of the characteristic
equation, so s = 0 and xr = At + B.
45×
xr
=
At + B
−14×
x′r
=
A
1×
x′′r
=
0
′
=
−14A + 45(At + B) =
9t
′′
x − 14x + 45x
1
14
⇒ A = ;B =
5
225
5. Step 5. The general solution is
xtq = xtn + xr = C1 e5t + C2 e9t +
1
14
t+
5
225
1
⇒ x′tq = 5C1 e5t + 9C2 e9t + .
5
Then
x′tq − 4xtq + t
t
11
= −C1 e5t − 5C2 e9t − +
−1
5 225
14
11
Since x(0) = 3 ⇒ C1 + C2 +
= 3 và y(0) = 1 ⇒ −C1 − 5C2 +
= 1. Solve the system
225
225

14


C1 + C2 +
=3
391
35
225
⇒ C1 =
; C2 = − .
11

100
36
 −C1 − 5C2 +
=1
225
ytq =
Therefore, the solution which satisfies the conditions of the system is

t
35e9t 391e5t
14


x= −
+
+
5
369t
1005t 225

 y = − t + 175e − 391e + 11
5
36
100
225
□
QUESTIONS AND THE ANSWER
Page 51/1 — Question code: 1234