AoPS Academy
![\[
I = 2 \int_{0}^{\pi/4} \frac{1}{\cos 2\phi} \cdot 2 \ln(\cot \phi) \cdot 2 \, d\phi
\]](http://latex.artofproblemsolving.com/7/a/8/7a82b62346bf5636f262c4b27f3a4cf462d8deba.png)
which satisfy the simultaneous equations
and
where 
and 
are integers such that 
and 
.
, s.t. ecuation: 
have 2 real solutions .
, 
. Are they isolated?
be a ring with unity such that for every 
there exist 
such that 
. Prove that
then 
such that 
then the result from a) may no longer hold.
and 
be fixed positive integers. Show that the set of primes that divide at least one of the terms of the sequence 
is infinite.
such that 
.
then using Mean Value Theorem, there exist an such that 
case, suppose 
and assume 
on the whole interval 
. Then by the mean value theorem, there is some such that 
so 
. That by itself isn't a contradiction, but we can do something similar to get a contradiction. The intuitive idea is that since we're right on the "edge" of a contradiction, we have no room to move the graph of 
. If 
and , that forces 
on the interval ![$[0,1/2]$](http://latex.artofproblemsolving.com/8/2/4/8240ac9beaadf3f2654c3ba09a3987a7305539c5.png)
. Moving the graph up or down at any point makes the slope larger than 
somewhere. Similarly, 
on the interval ![$[1/2,1]$](http://latex.artofproblemsolving.com/5/5/8/5581510e950d8ceb9bcbcfd0dc988d9c4538c684.png)
. for every ![$x\in [0,1/2]$](http://latex.artofproblemsolving.com/0/f/d/0fd16eeab1cb0c53e329c87cea809db55ea5f207.png)
. On any interval ![$[0,a]$](http://latex.artofproblemsolving.com/6/8/b/68bacdc4b96052586744d16169ccd8c3233ed09f.png)
with 
, we have 
for some 
, depending on . Since we're assuming 
is always bounded by , this means that 
so 
for all ![$a\in[0,1/2]$](http://latex.artofproblemsolving.com/e/0/c/e0c5302f45677b07a169c44aa1085cf1dace8463.png)
.
, then we could use the mean value theorem on the interval ![$[a,1/2]$](http://latex.artofproblemsolving.com/f/7/c/f7c796537cc023b8e3d0dd550cf1c3bdfc0278d4.png)
to get 
But the left side is larger than , and the right side is at most , a contradiction. on the interval , so 
However, this function is not differentiable at 
, so we are done.
:
Assume towards a contradiction that 
.
and 
:![\[\usepackage{mathtools}
\tfrac {1-f(x)}{\frac 12-x}\le 2\xRightarrow{\hspace{40pt}}\left(\forall 0<x<\tfrac 12\right):f(x)\ \ge\ 2x\tag{1}\]](http://latex.artofproblemsolving.com/4/6/d/46d07da0b23e15de91ca1912093a406842150f21.png)
LMVT on and 
:![\[\usepackage{mathtools}
\tfrac {f(x)-1}{x-\frac 12}\ge -2\xRightarrow{\hspace{40pt}}\left(\forall \tfrac 12<x< 1\right):f(x)\ \ge\ 2-2x\tag{2}\]](http://latex.artofproblemsolving.com/9/2/c/92c82c921fad2971d842ec2dba87a5e4725c3e97.png)
Differentiability at 
implies 
is not possible.
and 
, reflecting about the line if necessary, we may assume 
.
and 
:![\[\usepackage{mathtools}
\tfrac {2\alpha+\eta-f(x)}{\alpha-x}\le 2\xRightarrow{\hspace{40pt}}\left(\forall 0<x<\alpha\right):f(x)\ \ge\ 2x+\eta\tag{3}\]](http://latex.artofproblemsolving.com/8/3/5/83584e1961403c7cf5c6891281a4cbb97346d193.png)
continuity at 
:![\[(\forall \epsilon>0)(\exists\delta>0):0<x<\delta\xRightarrow{\hspace{40pt}}\big|f(x)\big|<\epsilon\tag{4} \]](http://latex.artofproblemsolving.com/9/d/a/9da78481512b68b5c3e9ed771a83552d9878e137.png)
contradicts 
for 
.
is differentiable on . For 
and 
you can hopefully do!! put the value of 
in the two equations,So clearly visible that not differentiable on ....
and 
is differentiable on . For and you can hopefully do!! put the value of in the two equations,So clearly visible that not differentiable on .... and
is differentiable on . For and you can hopefully do!! put the value of in the two equations,So clearly visible that not differentiable on .... and 
![$$\int_0^{\pi/2} \Bigl[
\log\bigl(\sqrt{5}-\sin\theta+1\bigr)
+\log\bigl(\sqrt{5}+\sin\theta-1\bigr)
-\log\bigl(\sqrt{5}-\sin\theta-1\bigr)
-\log\bigl(\sqrt{5}+\sin\theta+1\bigr)
\Bigr]\,d\theta.
$$](http://latex.artofproblemsolving.com/a/f/f/affd7cb3d94d4fbcab6f9912e71ebcffeb86d960.png)
be a function which is continuous on ![\( [0,1] \)](http://latex.artofproblemsolving.com/0/1/0/01026be294289be8f090c7b2586dde64a0d5b479.png)
and differentiable on 
,
.
such that 
.
such that 
