AoPS Academy
is clearly not true. Even a 1st grader can see that. And what does "a number not a product" mean? A product is a number? Literally makes 0 sense.
.
as 
? What's wrong with them? Even a 1st grader can see that's so easy to simplify. Not to mention them acting all high and mighty since they memorized the cosine addition formula... so annoying.
if you do go around, acting like some kind of god again, "anticodon." Everyone will worship you finally for being able to copy from Google correctly!...
+3 w
is differentiable and 
for all 
. Show that for some 
, 
.
on the real line such that for all 
, ![\[ \left( f(x) \right)^2 = \displaystyle\int_0^x \left[ \left( f(t) \right)^2 + \left( f'(t) \right)^2 \right] \, \mathrm{d}t + 1990. \]](http://latex.artofproblemsolving.com/e/7/9/e79b1d8b17010c15a0f6397c8775e4e268a66a2e.png)
and the distribution function is 
then
and find the expression of 
be a function whose derivative is continuous on ![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
. Show that![$$\lim_{n \to \infty} \sum^n_{k = 1}\left[f\left(\frac{k}{n}\right) - f\left(\frac{2k - 1}{2n}\right)\right] = \frac{f(1) - f(0)}{2}.$$](http://latex.artofproblemsolving.com/3/1/4/314a1597acae030381f980e6e26c12432f31e069.png)
, where both and 
are integers between 
and 
inclusive, such that 
.
can be factored as 
. Thus, we consider three cases: 
, 
, or both.
pairs. The second case yields 
pairs. The third case yields 
pair. This yields 
pairs.
is a solution then divide by 
, sub 
to get 
, so 
. Casework gives 
.
at the end
and 
can equal 
can equal 
so we have to subtract for overcount.
,assume some constant 
allows us to factor by grouping, thus the expression becomes 
or 
. In order to factor we want 
,we find 
and we can factor and solve as above
We get 
or 
so 
,assume some constant allows us to factor by grouping, thus the expression becomes or . In order to factor we want ,we find and we can factor and solve as above
, where both and are integers between and inclusive, such that .) this equation is equivalent to:
We now have two mutually exclusive cases:
. Since 
, we can constraint 
which rearranges to 
. Likewise, since 
, 
. All such 
with will produce a valid and unique 
satisfying the original equation, so it just remains to count the number of solutions for which is 
.
and 
and we proceed to bound . Since 
, we have 
, and similarly 
. All such 
with will produce a valid and unique solution for except for 
, which violates the constraint. Thus our answer for this case is (subtracting at the end so as not to count ) 
.
solutions.
, and solve over , we have![\[
x=\frac{-y\pm17y}{24}
\]](http://latex.artofproblemsolving.com/a/6/c/a6c330741d1e09ecbd859c435f9b3dbae664ebbc.png)
so either 
, or 
.
if , then we have 
, hence there are 
solutions. Similarly for , there are 
solutions.
as a solution, so the answer is 
.
of
2. 
3. 
Bracket 2
2. 
3. 
Bracket 3![$$\int_0^{1}\frac{dx}{1-\sqrt[3]{1-\sqrt{x}}}$$](http://latex.artofproblemsolving.com/d/5/e/d5e8db00985944b2eca341e37a94e34661cb5c86.png)
2. 
3. 
Bracket 4
2. 
3. 
Bracket 5
2. 
3. 
Bracket 6
2. 
3. 
Bracket 7
2. 
3. 
Bracket 8
2. 
3. 
![$$\int_0^{2\pi}\sqrt[4]{\cos^4(x)-\cos(2x)}dx$$](http://latex.artofproblemsolving.com/7/e/e/7eebb4a488fec702e0130a253ef87f0f6dfedb2d.png)
2. ![$$\int_0^1 \frac{x^5}{\sqrt[3]{1-x^3}}dx$$](http://latex.artofproblemsolving.com/6/0/5/605bb9d81d0ccff54a276c9f7bf1c1726c136398.png)
3. 
Bracket 2
2. 
3. 
Bracket 3
2. ![$$\int_0^1 \frac{dx}{(\sqrt{x}+\sqrt[3]{x})^2}$$](http://latex.artofproblemsolving.com/6/9/8/69869c0ed754c0dfacf412ac6411742ffa164a6d.png)
3. 
Bracket 4
2. 
3. 
2. 
3. 
Bracket 2
2. 
3. 
2. 
3. 
2. 
3. 
4. 
5. 
,
or 
.
,
,
possible ordered pairs. If it is divisible by 
but not 
,
non-zero values of 
ordered pairs.
non-zero values of 
ordered pairs.
non-zero values of 
ordered pairs.
.
.
to 
.
.
,
The pairs of integers that satisfy this equation are 
and 
for 
For the first type of solution, we have 
so the number of solutions of that type are 
Then, for the second type, we get 
so the number of solutions of the second type is 
However, note that the solution 
So when we get that our first case is 3x+2y and 
that then we get 
but we over count (0,0) so we have 
then just find x and y and make sure not to double counted the zero 
