1
Solutions of Mathematical Analysis Exams – ENSTP Yaounde, Academic Year 2012/13
Solution of the First Partial Exam
Please, read carefully the solution and comments. And before to think that someone else
did some mistake, think about what have you done. It’s a general rule useful in the life.
Thanks.
Exercise 1 (8 points). State and prove the theorem about existence of the limit for monotone sequences.
Sol. — The statement is
Let (an ) be an increasing sequence. Then, either an −→ +∞ or an −→ sup{an
{an : n ∈ N} is unbounded or less.
:
n ∈ N} according if
Remarks:
• It seems that many of you are not used to study the theory in mathematics. It was asked the
statement of a precise thm (see above). One cannot answer to the question State and prove the theorem
about existence of the limit for monotone sequences with any monotone sequence has a limit.
• Only few of you seems to be able to study and understand a proof. Lot’s of you write down a
proof of some other thm (like uniqueness for limit or permanence of sign). This open a question: if you are
not able to distinguish between two different things, what are you doing along classes? Second: if someone
ask you an apple you give him a banana? Does it seems reasonable to you?
• Even those who try to show the thm makes lots of confusion, for instance confusing the
hypothesis (an monotone) with the thesis (an has a limit and the limit is etc). This pose the
same question as above: if you are not able to distinguish between hypothesis and thesis, what are you
doing along classes?
I don’t reproduce here the proof, you can find in the notes, so please look at it.
Exercise 2 (9 points). Give the definition of sup S. Then, let
(−1)n n
S :=
: n∈N .
3 + 2n
Show that sup S = 12 . What about max S?
Sol. — We recall first that sup S = α ∈ R if
i) s 6 α, ∀α ∈ S, ii) ∀β < α, ∃s ∈ S : s > β.
Let’s check that sup S = 21 . We have to prove the previous two properties with α = 21 . Let’s star by the first: we
have to prove that
(−1)n n
1
6 , ∀n ∈ N.
3 + 2n
2
This is evident if n is odd because
(−1)n n
3+2n
(−1)n n
1
6 , ⇐⇒
3 + 2n
2
n
= − 3+2n
< 0. If n is even we have
n
1
6 , ⇐⇒ 2n 6 3 + 2n, ⇐⇒ 0 6 3,
3 + 2n
2
2
again evident. Now, we want to prove ii), that is for any β <
1
2
there exists s ∈ S such that s > β, that is
(−1)n n
> β.
3 + 2n
If n is odd this will be in general impossible. So let’s consider only n even. We have to find n such that
n
> β, ⇐⇒ n > (3 + 2n)β, ⇐⇒ n(1 − 2β) > 3β,
3 + 2n
1−2β>0
⇐⇒
n>
3β
.
1 − 2β
Surely such n exists by the Archimedean property.
Remarks: Most of you have done correctly this exercise. There are, however, still some errors. For instance
lot’s of you don’t see that it is impossible, in general, given β < 12 to find
(−1)n n
> β, (with n odd).
3 + 2n
Exercise 3 (9 points). Give the definition of limn an = ` ∈ R. By using the definition, show that
3n+1 − 1
= 3.
n→+∞ 3n + 1
lim
Sol. — The general definition for finite limit is
∀ε > 0, ∃N (ε), : |an − `| 6 ε, ∀n > N (ε).
Fix ε > 0. Notice that
n+1
n+1
−4 3
− 1 − (3n+1 + 3) −1
6 ε, ⇐⇒ 3
−
3
≤
ε,
⇐⇒
3n + 1
3n − 1 6 ε, ⇐⇒
3n + 1
4
6 ε.
3n + 1
Now,
4
4−ε
ε>0
6 ε, ⇐⇒ 4 6 ε3n + ε, ⇐⇒ ε3n > 4 − ε, ⇐⇒ 3n >
.
3n + 1
ε
This is true for any n if 4 − ε 6 0, that is if ε > 4 (so we can take N (ε) = 0 for such ε); if ε < 4 we have that the
=: N (ε).
previous is equivalent to n > log3 4−ε
ε
Remarks: Most of you have done correctly this exercise. There are, however, still some errors. For instance
someone didn’t checked by the definition but just computed the limit, of course not answering to the problem.
Others write N (ε) = log3 4−ε
for any ε > 0 (and this is a nonsense when ε > 0 because the logarithm is not
ε
defined). There’re however among you people with problems in solving inequalities.
Exercise 4 (9 points). Let (an ) be the sequence defined as

 a0 = α,
2x2
where f (x) =
.

x+1
an+1 = f (an ),
3
Consider then α = 1/2. Show that i) 0 6 an 6 1 for any n; ii) (an ) is monotone; iii) limn an exists and
find its value. iv) Discuss, finally, what happens if α = 2.
If you are not able to prove i) or ii) you can anyway do iii) giving the first two as fulfilled.
Sol. — i) Call pn : 0 6 an 6 1. Clearly p0 is true. Suppose pn is true, that is 0 6 an 6 1. Let’s prove that
pn+1 is true, that is 0 6 an+1 6 1. The first half of the inequality is clearly true, because being an > 0 clearly
2a2
n
an+1 = an +1
> 0. About the second
2a2n
an >0
6 1, ⇐⇒
2a2n 6 an + 1, ⇐⇒ 2a2n − an − 1 6 0.
an + 1
an+1 =
Now
√
√
1− 9
1+ 9
1
6x6
, ⇐⇒ − 6 x 6 1.
4
4
2
Clearly, being 0 6 an 6 1, − 12 6 an 6 1, hence the previous inequality is fulfilled.
2x2 − x − 1 6 0, ⇐⇒
ii) Let’s compute a1 =
2a2
0
a0 +1
=
2· 1
4
1 +1
2
=
1 2
2 3
=
1
3
<
1
2
= a0 . So we are induced to think that an &. Let’s prove this,
that is an+1 6 an for any n. This is clearly true for n = 0. Suppose is true for n and let’s prove that it’s true for
n + 1.
an+2 =
2a2n+1
2a2n
6
= an+1 , ⇐⇒ a2n+1 (an + 1) 6 a2n (an+1 + 1), ⇐⇒ an an+1 (an − an+1 ) > a2n+1 − a2n .
an+1 + 1
an + 1
Writing a2n+1 − a2n = (an+1 − an )(an+1 + an ), because an − an+1 > 0, we have
an+2 > an+1 , ⇐⇒ an an+1 > −(an+1 + an ),
clearly true because: an an+1 > 0 while −(an+1 + an ) 6 0 because an , an+1 > 0.
iii) By i) and ii) we deduce 0 6 an 6 1 and an &: by a well known thm about monotone sequences there exists
limn→+∞ an = ` and 0 6 ` 6 1 by permanence of sign. Passing to the limit into the equation we get
`=
2`2
2`
, ⇐⇒ ` = 0, or 1 =
, ⇐⇒ ` = 0, or ` + 1 = 2`, ⇐⇒ ` = 0, or ` = 1.
`+1
`+1
Being moreover an % in particular an 6 a0 =
1
2
so ` 6 12 . We conclude that ` = 0.
2a2
2·4
0
iv) Take a0 = 2. Then a1 = a0 +1
= 2+1
= 38 > 2 = a0 . This suggests an % and in particular an > 2 for any n.
Let’s prove first this last one. It is true for n = 0. Suppose is true for n and let’s prove that it is true for n + 1:
an+1 > 2, ⇐⇒
2a2n
> 2, ⇐⇒ a2n > an + 1.
an + 1
But if an > 2 we have a2n > 4 while an + 1 6 2 + 1 = 3 so it is true. We prove now an+1 > an for any n. It is
true for n = 0 (seen above). Suppose is true for n. Then the same computation of ii) holds with an+1 and an
exchanged. Therefore an % and so, by a well known result, there exists limn→+∞ an = ` ∈ R ∪ {+∞}. It cannot
be ` ∈ R because in this case, as by iii), ` should be 0 or 1, but an > 2 so ` > 2. It follows that ` = +∞.
Remark: Not so many of you have been able to solve correctly the first two questions. It is not enough to check
that a1 < a0 to deduce that an & (as many of you believe apparently).
4
Exercise 5 (9 points). Compute
(log x)x − 4x
√
.
x→+∞ 2−x − x x
lim
Sol. — Let’s discuss the behavior of numerator and denominator separately:
1
N = (log x)x − 4x = ex log(log x) − ex log 4 = ex log(log x) 1 − x log(log x)−x log 4 = ex log(log x) 1x
e
log 4
−→ +∞. About the denominator
because x log(log x) − x log 4 = x log(log x) 1 − log(log
x)
D=2
−x
−x
√
x
Therefore
√
√
√
1
1
x log x
x log x
= x −e
= −e
1 − x √x log x = −e x log x 1x .
2
2 e
√
ex log(log x) 1x
N
√
=
= −ex log(log x)− x log x 1x .
x
log
x
D
−e
1x
Now
lim
x→+∞
so
N
D
x log(log x) −
√
x log x
y=log x, x=ey
=
lim ey log y − ey/2 y = lim ey log y 1 −
y→+∞
y→+∞
y
ey/2 log y
= +∞,
−→ −∞.
Remark: This exercise has been correctly solved only by a little number of you. Lots of you make errors with
properties of logarithm writing things like
(log x)x = ex log x ,
or
2
(log x)x = ex(log x) ,
or again
x
√
x
1
= e 2 x log x .
√
These kind of errors are not allowed. Others think
that 4x (log x)x or (which is more dramatic)
that 2−x x
√
√
−x
x
−x
x
They don’t seem to see that 2 −→ 0 while x −→ +∞ so it is impossible that 2 x .
x
.
Please, if you think to have done one of the previous errors, avoid to complain about your
marks. As I told you this morning it is possible (but I think very unlikely) that I did some
errors in correction. But it is sure that I cannot have done an error about 30 of you and
you have to understand that it is not nice by you to have such kind of ideas. Secondly, I’m
quite large in evaluation. For instance one of the errors reported above should be normally
enough to give you a 0 no matter what else have you done, because these are severe errors.
I would like that you understand that.
Solution of the Second Partial Exam
Please, read carefully the solution and comments.
Exercise 6 (8 points). State and prove the fundamental theorem of integral calculus.
Sol. — See notes.
5
Exercise 7 (20 points).
f (x) = x 1 −
1 log x Find: domain, sign, limits and eventual asymptotes, continuity, eventual points where f could be extended
by continuity, derivability, limits of f 0 , eventual angle points, monotonicity, min/max, graph. The study
of convexity is not required.Let
Sol. — Dominio: clearly D(f ) = {x ∈ R : x > 0, log x 6= 0} =]0, 1[∪]1, +∞[.
Segno: clearly f ≥ 0 on all its domain.
Limits and asymptotes:
Let’s compute f (0+), f (1±), f (+∞). As x −→ 0+ clearly 1 − log1 x −→ 1 so
f (0+) = 0+. As x −→ 1 we have 1 − log1 x −→ +∞, so f (1±) = +∞: the straight line x = 1 is vertical
asymptote. At +∞ finally, 1 −
have
1
log x
−→ 1 so f (+∞) = +∞. Let’s check for an obl. asymptote y = mx + q. We
f (x)
1 = 1,
m = lim
= lim 1 −
x→+∞
x→+∞
x
log x while
q = lim (f (x) − mx) = lim
x→+∞
As x −→ +∞, clearly 1 −
x→+∞
x 1 −
1 −
x
.
log x 0
log x
> 0 therefore
q = lim x 1 −
x→+∞
1
−1
log x
= − lim
x→+∞
x
= −∞,
log x
being x +∞ log x. We deduce that there’re no asymptotes at +∞.
Continuity: No problem about continuity, being f composition of continuous functions where defined. Because
f (0+) = 0 f may be extended as continuous function at x = 0.
Derivative: there’s a possible exception when 1 − log1 x = 0, that is log x = 1, x = e. Out of this point f is
derivable and
0
1
1 1
·
1
−
f 0 (x) = 1 · 1 −
+
x
sgn
1
−
log x log x
log x
1
1
1
1 1
1
= 1 −
·
1
−
+
+
x
sgn
1
−
=
sgn
1
−
log x log x
x(log x)2
log x
log x
(log x)2
1
= sgn 1 −
log x
(log x)2 − log x + 1
(log x)2
.
In particular easily we have f 0 (0+) = 1 and
(log x)2 − log x + 1
1
f 0 (e±) = lim sgn 1 −
= (−1±) · (1) = −1±,
x→e±
log x
(log x)2
0
therefore there exists f±
(e) but not f 0 (e), and x = e is an angle point.
Monotonicity: We have
(log x)2 − log x + 1
1
0
f (x) > 0, ⇐⇒ sgn 1 −
.
log x
(log x)2
Now
sgn 1 −
1
log x
> 0, ⇐⇒ 1 −
1
> 0, ⇐⇒
log x
log x − 1
> 0.
log x
6
while
(log x)2 − log x + 1 > 0, ⇐⇒ t2 − t + 1 > 0, (t = log x).
Being (−1)2 − 4 · 1 · 1 < 0 we deduce that this is always true.
sgn log x
sgn(log
x − 1)
sgn
log x−1
log x
2
sgn((log x) − log x − 1)
sgnf 0
f
]0, 1[
−
−
]1, e[
+
−
]e, +∞[
+
+
+
−
+
+
+
%
+
−
&
+
+
%
At x = e the function is continuous, decreasing at the left and increasing at the right: we deduce that x = e is a
minimum on ]1, +∞[, and because f > 0 and f (e) = 0 we conclude that it is actually a global minimum. There’re
no max points.
1
6
ã
Remarks. Most frequent errors:
• domain: most of people do correctly, but there’re still people that consider x < 0 or, most frequently,
they include x = 1 to the domain.
• sign: lots of people discuss the sign: there’s the modulus and because x > 0 by the domain f is trivially
> 0.
• derivative: lots of people don’t seem to be able to derive the modulus. We have done lots of exercises
in class with functions containing the modulus. But if you have troubles you could study the function
discussing first the value of the modulus (in order to eliminate it).
As I told you in class, the worst error with a complex problem as studying a function, is incoherence. For
instance, you cannot say that f > 0 and then that (for instance, as lots of people do) limx→1− f (x) = −∞. You
must understand that something is wrong!
Exercise 8 (10 points). Compute
√
2
cosh( 2x) − ex + x log cos x
lim
x→0
x − sin x + x4 log x
7
Sol. — Let’s discuss separately numerator and denominator.
cosh ξ = 1 +
ξ2
2
eξ = 1 + ξ + o(ξ) = 1 + ξ +
cos ξ = 1 −
ξ2
2
ξ2
2
+ o(ξ 2 ) = 1 +
ξ2
2
+ o(ξ 2 ) = 1 −
log(1 + ξ) = ξ + o(ξ) = ξ −
ξ4
24
+ o(ξ 4 ),
+ o(ξ 2 ),
ξ2
2
ξ2
2
+
+
ξ4
24
+ o(ξ 4 ),
+ o(ξ 2 ).
Then, by using the shortest expansions
N (x)
=1+
2x2
2
+ o(x2 ) − 1 + x2 + o(x2 ) + x log 1 −
2
2
= o(x2 ) + x − x2 + o x2 + o(x2 )
= o(x2 ) −
x3
2
x2
2
+ o(x2 )
+ o(x3 ) = o(x2 ).
We need to add terms. Looking carefully we have to eliminate o(x2 ) that comes out by the first and the second
terms of the numerator. Adding one more term in the respective developments
2
x4
x2
x
x3
x3
2x2 4x4
+
+ o(x4 ) − 1 + x2 +
+ o(x4 ) + x −
+o
+ o(x2 )
= − + o(x3 ) = − 1x .
N (x) = 1 +
2
24
2
2
2
2
2
About the denominator recall first that,
sin ξ = ξ + o(ξ) = ξ −
Then
ξ3
+ (ξ 3 ).
6
x3
x3
x − sin x = o(x) = x − x −
+ o(x3 ) =
+ o(x3 ).
6
6
Therefore
D(x) =
x3
+ o(x3 ) + x4 log x.
6
Now, easily x4 log x = o(x3 ) because
x4 log x
= x log x −→ 0,
x3
by a well known remarkable limit. We deduce that D(x) =
x3
6
+ o(x3 ) =
x3
1 .
6 x
Hence
3
− x 1x
N (x)
= x32
= −3 · 1x −→ −3.
D(x)
1
6 x
Remarks. The most frequent error is of algebraic type, giving a wrong expansion (for the numerator basically).
So be more careful.
Exercise 9 (10 points). Compute
Z
+∞
log 8
√
ex − 4
dx.
ex − 5
8
√ x
−4
. It is clear that f ∈ C([log 4, +∞[\{log 5}). In particular f ∈ C([log 8, +∞[). The
Sol. — Call f (x) := eex −5
integral is clearly of generalized type and
Z R
Z +∞
f (x) dx.
f (x) dx = lim
R→+∞
log 8
log 8
RR
To compute the partial integral log 8 f (x) dx, being f ∈ C ([log 8, R]) we can apply the fundamental formula of
√
integral calculus. Let’s find first a primitive of f . It seems natural to change variable setting y = ex − 4, that
is ex = y 2 + 4, x = log(y 2 + 4), dx = y22y+4 dy. Therefore
Z
Z √ x
Z
Z
Z
e −4
1
y
2y
y2 − 1 + 1
1
dx
=
dy
=
2
dy
=
2
dy
+
dy
.
ex − 5
y2 − 1 y2 + 4
(y 2 − 1)(y 2 + 4)
y2 + 4
(y 2 − 1)(y 2 + 4)
Clearly
Z
1
1
dy =
y2 + 4
4
Z
1
y
1
dy = arctan .
1 + (y/2)2
2
2
while
Z
1
dy
(y 2 − 1)(y 2 + 4)
1
=
5
Z Z
f (x) dx =
Easily F (+∞) =
4 π
5 2
=
2π
5
1
y
1
dy = − arctan +
10
2
10
y − 1
1
y
1
.
= − arctan +
log 10
2
10
y + 1
Therefore
so
Z
+∞
log 8
1
1
− 2
y2 − 1
y +4
Z 1
1
−
y−1
y+1
dy
√
ex − 4 − 1 √
2
4
=: F (x).
arctan ex − 4 + log √ x
5
5
e − 4 + 1
√ x
e −4
2π
2π
4
dx =
− F (log 8) =
− arctan 2.
ex − 5
5
5
5
Remarks. Most of people compute correctly the integral, apart some algebraic error that give a wrong result. I
don’t care too much to this but I look at the method.
Exercise 10 (10 points). Consider the differential equation
i πh
3
y 00 + 4y =
, ∈ 0,
.
sin(2t)
2
Find its general integral and the solution of the Cauchy problem y
such that
π − =0?
y (0+) = y
2
π
4
= y0
π
4
= 0. Are there solutions
Sol. — Let’s first find the fundamental system for the homogeneous equation y 00 + 4y = 0. The characteristic
equation is λ2 + 4 = 0, so λ = ±2i. The fundamental system is therefore w1 (t) = cos(2t), w2 (t) = sin(2t). The
corresponding wronskian is

 

w1 w2
cos(2t)
sin(2t)
=
 = 2(cos(2t)2 + 2(sin(2t))2 = 2.
W (t) = det 
w10 w20
−2 sin(2t) 2 cos(2t)
9
By the Lagrange formula of variation of arbitrary constants, a particular solution for the non homogeneous
equation is
Z
Z
Z
Z
w1
w2
3
3
U (t) =
cos(2t) ·
f w2 −
f w1 =
sin(2t) ·
sin(2t) −
cos(2t)
W
W
sin(2t)
sin(2t)
=
3
3
(log | sin(2t)|) sin(2t) − 3t cos(2t) = (log sin(2t)) sin(2t) − 3t cos(2t),
2
2
because sin(2t) > 0 as t ∈]0, π2 [. So the general integral is
y(t) = c1 cos(2t) + c2 sin(2t) +
3
(log sin(2t)) sin(2t) − 3t cos(2t).
2
For the Cauchy problem we have y( π4 ) = c2 = 0 and because
3 2 cos(2t)
sin(2t) + 2 log(sin(2t)) cos(2t) − 3 (cos(2t) − 6t sin(2t)) ,
y 0 (t) = −2c1 sin(2t) + 2c2 cos(2t) +
2
sin(2t)
we have y 0 ( π4 ) = −2c1 − 18 π4 = 0, so c1 = − 18
π = − 94 π.
8
About the last question: as t −→ 0+, π2 − we have sin(2t) −→ 0+ so
lim
t→0+, π
−
2
log sin(2t)) sin(2t) = lim x log x = 0.
x→0+
In particular then
y(0+) = c1 , y
π 1
− = c1 − .
2
2
It is clearly impossible that both are =0.
Remarks. The more dramatic error is to write the wrong characteristic equation. Lots of people write
λ2 + 4λ = 0.
With this wrong, the rest of the problem becomes impossible. This exercise is very standard, just apply an
algorithm, but such error is fatal.
Exercise 11 (8 points). A bridge of 3km has to be built. The cost of any pillar is of 0.5mld CFA while
the cost of each span of length `km is 6`2 mld CFA. Find the number N of pillars in such a way that the
cost of the bridge is minimum.
Sol. — Let N be the number of pillars. There’re N − 1 spans, each of them of length N 3−1 km. The corresponding
cost of the bridge is
2 !
N
3
N
9
C(N ) =
+6
mld CFA =
+6
mld CFA.
2
N −1
2
(N − 1)2
The problem is to minimize C(N ). Let’s imagine N as a real variable first and let’s find the minimum of C on
N ∈ [2, +∞[. We have
C 0 (N ) =
1
9
− 12
> 0, ⇐⇒ (N − 1)3 > 24 · 9 = 23 · 33 , ⇐⇒ N − 1 > 6, ⇐⇒ N > 7.
2
(N − 1)3
10
Therefore, as N = 7 the function C(N ) attains a global minimum for N ∈ [2, +∞[, henceforth the same holds for
N ∈ N, N > 2. The bridge will costs C(7) = (3, 5 + 1, 5) = 5mld CFA.
Remarks. This exercise has been tried only by a minority of you. Only one (or two) found the correct answer.
However, many of those who tried to solve the problem looked for a function (the cost) to minimize, and even if
in most of the cases the function wasn’t completely correct, this attempt was rewarded almost if it was correct
(7 points over 8). There’s some curiosity. Some of you found curious answer. I just quote one that found 500
pillars. Now, just stop a second and think about. You have to build a 3km bridge. Do you find reasonable and
”economic” to use 500 pillars (that is a pillar each 6 meter)? As future engineer... is not so bad...
Solution of the First Full Exam
Exercise 12 (22 points). Given a set S of real numbers, give the definition of inf S and sup S. What
are the eventual relations between inf S and min S and between sup S and max S? Let now
en
S := √
: n∈Z .
e2n + 1
By using the definition, show that inf S = 0 and sup S = 1. What about min S and max S?
Sol. — About defs see the notes. The relation are: if min S exists then min S = inf S (which always exists); if
max S exists then max S = sup S (which always exists). Let’s now consider the set S. About inf S:
inf S = 0, ⇐⇒ i) 0 6 s, ∀s ∈ S;
ii) ∀β > 0, ∃s ∈ S : s 6 β.
About the first it is evident. About the second, fix β > 0 and let’s look for n ∈ Z such that
√
en
e2n + 1
6 β, ⇐⇒ en 6 β
p
e2n + 1, ⇐⇒ e2n 6 β 2 (e2n + 1), ⇐⇒ e2n (β 2 − 1) > −β 2 .
If β > 1 any n ∈ Z is good. If β < 1, because β > 0 we have β 2 − 1 < 0 so
e2n (β 2 − 1) > −β 2 , ⇐⇒ e2n 6 −
β2
β2
1
β2
β2
=
,
⇐⇒
n
6
log
,
,
⇐⇒
2n
6
log
β2 − 1
1 − β2
β2 − 1
2
β2 − 1
which is fulfilled by some n ∈ Z by the Archimedean property. About min S, if it exists we have min S = 0. But
0∈
/ S evidently, so the minimum doesn’t exists.
Let’s pass to the sup S:
sup S = 1, ⇐⇒ i) s 6 1, ∀s ∈ S; ii) ∀β < 1, ∃s ∈ S, : s > β.
The first is almost evident:
p
en
√
6 1, ⇐⇒ en 6 e2n + 1, ⇐⇒ e2n 6 e2n + 1, ⇐⇒ 0 6 1.
e2n + 1
Let β < 1 and look for
en
√
> β.
2n
e +1
If β 6 0 any n ∈ Z is good. If 0 < β < 1 we have that this is equivalent to
p
en > β e2n + 1, ⇐⇒ e2n > β 2 (e2n + 1), ⇐⇒ e2n (1 − β 2 ) > β 2 .
11
Being 0 < β < 1 we have 1 − β 2 > 0 so the previous is equivalent to
e2n >
β2
β2
1
β2
, ⇐⇒ 2n > log
, ⇐⇒ n > log
,
2
2
1−β
1−β
2
1 − β2
whose existence is assured by the Archimedean property. Finally, about max S: if it exists it has to be equal to
sup S = 1. But clearly (see above) 1 ∈
/ S, so the max S doesn’t exists.
Exercise 13 (22 points). State and prove the theorem concerning the existence of the limit for monotone
sequences (do the proof only in the case of increasing sequences). Consider now the sequence (an ) ⊂ R
defined as
4a2n
, a0 = 1.
an+1 =
an + 1
Someone say to you that it is monotone: deduce its limit. Are you able to prove that it is monotone?
Sol. — For the proof see the notes.
If it is monotone it has to be increasing because a1 = 2 > a0 = 1. So an %. By the cited thm, an −→ ` ∈
R ∪ {+∞}. If ` ∈ R we have, passing to the limit in the equation,
`=
4`2
, ⇐⇒ `(` + 1) = 4`2 .
`+1
This equation has solutions ` = 0 or ` + 1 = 4`, that is ` = 13 . But an % and a0 = 1, so none of these two values
can be the limit. We conclude that ` = +∞.
Let’s show that an %. We have
an+1 > an , ⇐⇒
4a2n
> an , ⇐⇒
an + 1
4an
1
> 1, ⇐⇒ 4an > an + 1, ⇐⇒ an > .
an + 1
3
So if we prove that an > 31 we have the conclusion. Let’s use the induction: this is clearly true for n = 0. In
general: assume that an > 13 . Then
√
√
4a2n
1
1 − 49
1 + 49
an+1 =
> , ⇐⇒ 12a2n > an + 1, ⇐⇒ 12a2n − an − 1 > 0, ⇐⇒ an 6
, or an >
an + 1
3
24
24
The first gives an 6
hypothesis.
1−7
24
6
= − 24
= − 14 (which is false) while the second gives an >
1+7
24
=
8
24
=
1
3
which is the
Exercise 14 (25 points). Let
q
π
f (x) = − arcsin 1 − log2 x.
2
Find: domain, sign, limits and eventual asymptotes, continuity, eventual points where f could be extended
by continuity, derivability, limits for f 0 , eventual angle points, monotonicity, min/max, graph. The
convexity is not required.
Sol. — Domain: clearly D(f ) = {x ∈ R : x > 0, 1 − log2 x > 0, −1 6
p
1 − log2 x 6 1}. We have
1 − log2 x > 0, ⇐⇒ (log x)2 6 1, ⇐⇒ −1 6 log x 6 1, ⇐⇒ e−1 6 x 6 e.
12
Moreover while
p
1 − log2 x > −1 is obvious,
q
1 − log2 x 6 1, ⇐⇒ 1 − log2 x 6 1, ⇐⇒ log2 x > 0,
clearly true. Therefore, D(f ) = [e−1 , e].
Sign:
f (x) > 0, ⇐⇒ arcsin
q
π
1 − log2 x 6 .
2
But arcsin : [−1, 1] −→ [− π2 , π2 ] so f ≥ 0 always. Moreover
q
q
π
f (x) = 0, ⇐⇒ arcsin 1 − log2 x = , ⇐⇒
1 − log2 x = 1, ⇐⇒ log x = 0, ⇐⇒ x = 1.
2
Limits and asymptotes: no limits and asymptotes need to be determined.
Continuity: clearly f is composition of continuous functions where defined, so it is continuous on its domain.
Derivability: about f 0 there’s a difference, because the root is not derivable (while continuous) when its argument
vanishes and also arcsin is not derivable when its argument is ±1. Point where this happens are:
q
1 − log2 x = 0, ⇐⇒ x = e−1 , e.
1 − log2 x = ±1, ⇐⇒ x = 1.
Therefore f is derivable on [e−1 , e] \{e±1 , 1} =]e−1 , 1[∪]1, e[. Let’s compute the derivative
1
log x
sgn(log x)
1
1
1
1
0
x
p
p
f (x) = r
=
−
−2
log
x
·
=− p
.
p
2
2
2
x
|
log
x|
2 1 − log x
1 − log x
x 1 − log2 x
1−
1 − log2 x
By this
f 0 (e−1 +) = −
(−1)
1
= +∞, f 0 (e−) = −
= −∞.
e−1 0+
e · 0+
and
(−1)
1
= −1, f 0 (1+) = −
= 1.
1
1
We confirm that f is not derivable at e±1 , 1, even if at x = 1 there’s the left and the right derivative (therefore
the point is an angle point).
f 0 (1−) = −
Monotonicity, min/max. We have
f 0 (x) > 0, ⇐⇒ sgn(log x) 6 0, ⇐⇒ log x 6 0, ⇐⇒ x 6 1.
It follows that f % on ]e−1 , 1[ while f & on ]1, e[. At x = 1 f is continuous, so x = 1 is a minimum point for f
(global) with f (1) = π2 − arcsin 1 = 0. The max points are at x = e±1 with
f (e±1 ) =
p
π
π
π
− arcsin 1 − (±1)2 = − arcsin 0 = .
2
2
2
Exercise 15 (15 points). Compute, in function of the parameter α > 0
√
x7/2 log x − sinh x2 + cosh log(1 − 2x) − 1
lim
.
xα
x→0+
13
Π
y
2
x
1
ã
3
Sol. — Clearly, by the remarkable limit, x7/2 log x −→ 0, and easily we have a form
recall first that
sinh t = t + o(t) = t +
0
.
0
About the numerator,
t3
t2
t2
t2 t4
+ o(t3 ), log(1 + t) = t + o(t) = t − + o(t2 ), cosh t = 1 + + o(t2 ) = 1 + +
+ o(t4 ),
6
2
2
2 24
so
− sinh x2 + cosh log(1 −
√
√
x6
+ o(x6 ) + cosh − 2x − x2 + o(x2 ) − 1
2x) − 1 = − x2 +
6
x6
+ o(x6 ) +
= − x2 +
6
!
√
2
√
4
− 2x − x2 + o(x2 )
2
2
1+
+ o − 2x − x + o(x )
−1
2
√
√
x6
= − x2 +
+ o(x6 ) + x2 + 2x3 + o(x3 ) = 2x3 + o(x3 ).
6
Therefore
N (x) = x7/2 log x +
√ 3
√
√
2x + o(x3 ) = 2x3 + o(x3 ) 2x3 1x ,
because x7/2 log x = o(x3 ) being
Therefore
x7/2 log x
= x1/2 log x −→ 0.
x3

√ 3
0,
3 − α > 0, ⇐⇒ α < 3,
 √
√ 3−α
N (x)
2x · 1x
=
2x
·
1
−→
=
2,
α = 3,
x

D(x)
xα
+∞, α > 3.
Exercise 16 (15 points). Find the solution of the Cauchy problem
 0
 y = t3 sin(t2 ) cosh y,

y(0) = 0.
Sol. — We have a separable variables equation y 0 = a(t)f (y). Let’s proceed by separation of variables:
Z
Z
y0
y0
= t3 sin(t2 ), =⇒
= t3 sin(t2 ) + C.
cosh y
cosh y
14
0
Noticing that t sin(t2 ) = − 21 cos(t2 ) , by parts we have
Z
Z
Z
1
1
1
3
2
2
2 0
2
2
2
t sin(t ) dt =
t − cos(t ) dt =
−t cos(t ) + 2t cos(t ) dt =
−t2 cos(t2 ) + sin(t2 ) .
2
2
2
On the other side we have
Z
u=y(x), du=y 0 (x)dx
y 0 (x)
dx
=
cosh y(x)
Z
1
du = 2
cosh u
eu
du.
+1
Z
e2u
Setting v = eu , u = log v, du = v1 dv,
Z
Z
Z
eu
v 1
1
du
=
dv
=
dv = arctan v = arctan(eu ),
e2u + 1
v2 + 1 v
1 + v2
so
Z
y0
= arctan(ey ).
cosh y
By this follows that
arctan(ey ) =
sin(t2 ) − t2 cos(t2 )
+ C.
2
Imposing the initial condition
arctan(ey(0) ) = C, ⇐⇒ C = arctan(1) =
π
,
4
so the solution is
ey = tan
sin(t2 ) − t2 cos(t2 )
π
+
2
4
, ⇐⇒ y(x) = log tan
sin(t2 ) − t2 cos(t2 )
π
+
2
4
.
Exercise 17 (11 points). Two towns A and B (with AB = 40km) are separated by a mountain of
circular base of radius 10km, with the center O of the mountain at the middle point of AB. A railroad
connecting A to B has to be built in the following way: a first straight piece from A to the mountain, a
straight tunnel crossing the mountain, and a final straight piece up to B. The cost for a piece of length
` is ` billions CFA for the normal railroad and 2` billions CFA for the tunnel. Find at which distance d
from O the tunnel should be build to minimize the total cost of the railroad.
Sol. — Let x be the distance of the tunnel by O. Then, if C and D are the points of begin and end of the
tunnel, we have
C(d) = AC + 2CD + DB = 2AC + 2CD.
Let’s compute the two lengths. To this aim let r be the radius of the circumference and L = AO; the specific
values are r = 10km and L = 20km. We have easily, by Pitagora’s thm,
p
p
CD
= r2 − x2 , =⇒ CD = 2 r2 − d2 .
2
Again by Pitagora’s thm
2
AC = x2 + (L −
hence
p
p
p
r2 − x2 )2 = x2 + L2 − 2L r2 − x2 + (r2 − x2 ) = r2 + L2 − 2L r2 − x2
q
p
p
C(d) = 2 r2 + L2 − 2L r2 − x2 + 4 r2 − x2 .
15
~ is tangent to the circle.
We have to minimize C on x ∈ [0, r]. Actually the max x possible has to be such that AC
Let’s call such x as r̂ and let’s study C 0 on [0, r̂]:
−2L √−2x
2x
−2x
2 r 2 −x2
C (x) = 2 p
= √
+4 √
√
2 − x2
2 − x2
2
2
2
2
2
r
r
2 r + L − 2L r − x
L
0
p
!
−2 .
√
r2 + L2 − 2L r2 − x2
Therefore
C 0 > 0, ⇐⇒
L
p
− 2 > 0, ⇐⇒
√
r2 + L2 − 2L r2 − x2
L
>
2
q
p
r2 + L2 − 2L r2 − x2 .
Squaring
p
p
L2
L2
3
> r2 + L2 − 2L r2 − x2 , ⇐⇒ 2L r2 − x2 > r2 + L2 −
= r2 + L2 .
4
4
4
Because L = 2r this means
p
p
4r r2 − x2 > r2 + 3r2 = 4r2 , ⇐⇒
r2 − x2 > r, ⇐⇒ r2 − x2 > r2 , ⇐⇒ x2 6 0.
This means that C 0 < 0 so C &, hence the minimum cost will be obtained for maximum r, so for r̂. It is now a
\ = 90o
simple geometrical problem to find out the exact value for r̂. To have tangency we need that the angle ACO
so
√
2
2
AC + r2 = L2 = 4r2 , ⇐⇒ AC = 3r2 , ⇐⇒ AC = 3r.
But then
√
\ = 2r cos OAC,
\ ⇐⇒ cos OAC
\=
3r = L cos OAC
By this finally follows
\ = r sin
r̂ = r sin AOC
√
3
\ = π , ⇐⇒ AOC
\ = π.
, ⇐⇒ OAC
2
6
3
π
r
= = 5km.
3
2