AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
and 
be defined as
,
its trivial for 
is where it gets confusing. is invertible if and only if it is bijective. This is because the condition
implies that is injective, and the condition
implies that is surjective. isn't invertible because isn't surjective since never reaches values 
in 
the domain it won't produce values for for 
, but I think that argument is wrong but I can't point it out. The only argument I could provide was
, then 
and 
, regardless of whether the codomain is or 
.
or 
precisely to define inverses — so maybe it's fair to come up with an inverse here too.
which verify relation![\[
\left( f'(x) \right)^2 + f''(x) \leq 0, \forall x \in \mathbb{R}.
\]](http://latex.artofproblemsolving.com/3/a/2/3a235e8be4c61c32f376999e61d64973db21dc75.png)
having 
and satisfying the equation
Show that this sequence is convergent and find its limit.
is 
, maybe even with rather elementary means? As usual, 
denotes the 
-th prime number. The problem of that limit came up in my partial solution of this problem: https://artofproblemsolving.com/community/c7h3495516. be a function taking in positive integers and outputting nonnegative integers, defined as follows:
is the number of positive integers with 
such that the equation 
has an integer solution 
.
such that
and 
Adit Aggarwal, India.
such that the following inequality holds for all 
:![\[
ab + bc + ca \leq \frac{(a + b + c)^2}{3} + k \cdot \max \{ (a - b)^2, (b - c)^2, (c - a)^2 \} \leq a^2 + b^2 + c^2
\]](http://latex.artofproblemsolving.com/1/c/e/1ce66163068b499383adb4a4ecec86c5f4edaec3.png)
are smaller than the square root of , i.e., 
for all proper positive divisors 
of different from .
and 
for 
. Let 
. Prove that there exist infinitely many natural numbers 
for which 
is divisible by 
. Does there exist for which 
is divisible by ?
satisfy,
Where {
} is a permutation of . Prove that 
be a fixed square and 
a variable point on segment 
. The square 
is constructed such that 
is on segment 
and 
is on segment 
. Let 
be the intersection point of lines 
and 
. Find the locus of as varies along segment .
holds 
. does equality occur?
do there exist infinitely many positive integers such that 
divides 
?
be a function such that
, 
, and
, 
, are continuous is also continuous.
, with the following property: there exists an interval
, such that, for any 
in 
, the function 
, 
, is continuous.[/list]
be a function such that, , and, , are continuous is also continuous., with the following property: there exists an interval, such that, for any in , the function , , is continuous.
has just one point of discontinuity. Moreover, the interval must not be degenerated (to a point).
and 
for 
, with 
, for which 
is identically null for any 
.
in terms of 
and 
and since is continuous so 
is also continuous. So
and hence is continuous.![\[
f(x) = \begin{cases}
1, x\geq 0\\
-1, x<0
\end{cases}
\]](http://latex.artofproblemsolving.com/4/a/9/4a9a01ca493a67ba83416180119104ebb834c126.png)
.
therefore : 
,
are continuous we get that