# 1. Find the real numbers x for which the power series ∑ (4x − 12) n

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```MÄLARDALEN UNIVERSITY
School of Education, Culture and Communication
Department of Applied Mathematics
EXAMINATION IN MATHEMATICS
MAA151 Single Variable Calculus, TEN2
Date: 2017-03-24
Write time: 3 hours
Aid: Writing materials, ruler
This examination is intended for the examination part TEN2. The examination consists of five randomly ordered problems
each of which is worth at maximum 4 points. The pass-marks 3, 4 and 5 require a minimum of 9, 13 and 17 points respectively.
The minimum points for the ECTS-marks E, D, C, B and A are 9, 10, 13, 16 and 20 respectively. If the obtained sum of points
is denoted S2 , and that obtained at examination TEN1 S1 , the mark for a completed course is according to the following:
S1 ≥ 11, S2 ≥ 9
and
S1 + 2S2 ≤ 41
→
3
S1 ≥ 11, S2 ≥ 9
and
42 ≤ S1 + 2S2 ≤ 53
→
4
54 ≤ S1 + 2S2
→
5
S1 ≥ 11, S2 ≥ 9
and
S1 + 2S2 ≤ 32
→
E
S1 ≥ 11, S2 ≥ 9
and
33 ≤ S1 + 2S2 ≤ 41
→
D
S1 ≥ 11, S2 ≥ 9
and
42 ≤ S1 + 2S2 ≤ 51
→
C
52 ≤ S1 + 2S2 ≤ 60
→
B
61 ≤ S1 + 2S2
→
A
Solutions are supposed to include rigorous justifications and clear answers. All sheets of solutions must be sorted in the order
the problems are given in.
1.
Find the real numbers x for which the power series
∞
X
(4x − 12)n
n=1
n
.
is convergent. Are there any of those x for which the series is not absolutely
convergent, i.e. is (only) conditionally convergent?
(
y 0 = (xy)3 ,
2.
Solve the initial-value problem
3.
A closed (with bottom and lock) cylindrical tin is to hold π liters. Find the
radius and height of the can so that the consumption of material (the number
of square decimeter thin sheet) becomes a minimum.
(
y(1) = − 21 .
x = 12 cos2 (θ) ,
4.
Find the length of the curve
5.
Evaluate the generalized integral
Z ∞
y = 21 sin2 (θ) ,
9/2
0 ≤ θ ≤ π.
dx
,
4x(x − 6) + 45
and write the result in as simple form as possible.
MÄLARDALENS HÖGSKOLA
Akademin för utbildning, kultur och kommunikation
TENTAMEN I MATEMATIK
MAA151 Envariabelkalkyl, TEN2
Datum: 2017-03-24
Skrivtid: 3 timmar
Hjälpmedel: Skrivdon, linjal
Denna tentamen är avsedd för examinationsmomentet TEN2. Provet består av fem stycken om varannat slumpmässigt ordnade
uppgifter som vardera kan ge maximalt 4 poäng. För godkänd-betygen 3, 4 och 5 krävs erhållna poängsummor om minst 9,
13 respektive 17 poäng. Om den erhållna poängen benämns S2 , och den vid tentamen TEN1 erhållna S1 , bestäms graden av
sammanfattningsbetyg på en slutförd kurs enligt följande:
S1 ≥ 11, S2 ≥ 9
och
S1 + 2S2 ≤ 41
→
3
S1 ≥ 11, S2 ≥ 9
och
42 ≤ S1 + 2S2 ≤ 53
→
4
54 ≤ S1 + 2S2
→
5
Lösningar förutsätts innefatta ordentliga motiveringar och tydliga svar. Samtliga lösningsblad skall vid inlämning vara sorterade
i den ordning som uppgifterna är givna i.
1.
Bestäm de reella tal x för vilka potensserien
∞
X
(4x − 12)n
n
n=1
.
är konvergent. Är det några av dessa x för vilka serien inte är absolutkonvergent,
dvs. är (endast) betingat konvergent?
(
y 0 = (xy)3 ,
2.
Lös begynnelsevärdesproblemet
3.
En sluten (med botten och lock) cylinderformad konservburk ska rymma π liter.
Bestäm burkens radie och höjd så att åtgången av material (antalet kvadratdecimeter tunn plåt) blir minimal.
(
y(1) = − 21 .
x = 21 cos2 (θ) ,
4.
Bestäm längden av kurvan
5.
Z ∞
y = 21 sin2 (θ) ,
9/2
0 ≤ θ ≤ π.
dx
,
4x(x − 6) + 45
och skriv resultatet på en så enkel form som möjligt.
If you prefer the problems formulated in English, please turn the page.
MÄLARDALEN UNIVERSITY
School of Education, Culture and Communication
Department of Applied Mathematics
Examination TEN2 – 2017-03-24
EXAMINATION IN MATHEMATICS
MAA151 Single Variable Calculus
EVALUATION PRINCIPLES with POINT RANGES
Maximum points for subparts of the problems in the final examination
1.
The series is convergent for 114 ≤ x < 134 . 1p: Correctly, by e.g. the ratio test, found that the series is
Of those x , it is only conditionally
absolutely convergent for | 4( x − 3) | < 1 , i.e. for
11
13
11
convergent for x = 4 .
, and hopefully correctly mentioned that the
4 < x< 4
series definitely is divergent for | x − 3 | > 14
1p: Correctly found that the series is convergent for x = 114
1p: Correctly found that the series is divergent for x = 134
1p: Correctly stressed that the series is only conditionally
convergent for x = 114
2.
y=−
3.
4.
2
9 − x4
1
dm
2
2
height = 3 dm = 3 4 dm
2
1p: Correctly for the optimization problem formulated a
function of one variable
2p: Correctly found and concluded about the local extreme
points of the function
1p: Correctly stated the radius and height which minimizes
the consumption of material
------------------------------------- One scenario ------------------------------------------------l.u. if the curve has been interpreted
as the path back and forth between the 1p: Correctly formulated an integral for the length of the
curve (an integral with explicit expressions for the
points P : ( 12 ,0) and Q : (0, 12 )
derivatives dx dθ and dy dθ )
1p: Correctly rewrited the integrand into 2 cos(θ ) sin(θ )
1 2 l.u. if the curve has been inter1p: Correctly treated the absolute values when integrating
preted as the path between the points
1
1
1p: Correctly found an antiderivative of the integrand, and
P : ( 2 ,0) and Q : (0, 2 )
correctly found the length of the curve
2
accepted provided they have been justified.
5.
1p: Correctly identified the differential equation as nonlinear
and separable, and correctly found general antiderivatives
of both sides of the separated differential equation
1p: Correctly adapted the integrated equation to the IV
1p: Correctly solved for y 2
1p: Correctly solved for y (the IV had to be used once more)
π
24
---------------------------------- Another scenario ----------------------------------------------
2p: Correctly noticed that the curve is the straight line
x + y = 12 back and forth between two points, P and Q
1p: Correctly found the length of the straight line between
the points P and Q
1p: Correctly found the length of the path back and forth
between the points P and Q
1p: Correctly rewrited the denominator of the integrand into
(2 x − 6) 2 + 9 , and correctly by the substitution
2( x − 3) = 3u translated the integrand into 1 (6(1 + u 2 ))
1p: Correctly translated the limits of the integral in
connection with the substitution 2( x − 3) = 3u
1p: Correctly found an antiderivative of the integrand
1p: Correctly evaluated the antiderivative at the limits, and
by that also correctly found the value of the integral
1 (1)
```