BdMO National 2008: Secondary

Discussion on Bangladesh Mathematical Olympiad (BdMO) National
User avatar
Kazi_Zareer
Posts:86
Joined:Thu Aug 20, 2015 7:11 pm
Location:Malibagh,Dhaka-1217
BdMO National 2008: Secondary

Unread post by Kazi_Zareer » Mon Dec 19, 2016 9:01 pm

The function $f (x) $ is a complicated nonlinear function. It satisfies, $ f(x) + f(1-x) = 1 $ . Evaluate $ \int_{0}^{1} f(x)dx $.
We cannot solve our problems with the same thinking we used when we create them.

User avatar
ahmedittihad
Posts:181
Joined:Mon Mar 28, 2016 6:21 pm

Re: BdMO National 2008: Secondary

Unread post by ahmedittihad » Fri Jan 13, 2017 6:37 pm

Is that a deriviative sign?
Frankly, my dear, I don't give a damn.

User avatar
Thanic Nur Samin
Posts:176
Joined:Sun Dec 01, 2013 11:02 am

Re: BdMO National 2008: Secondary

Unread post by Thanic Nur Samin » Fri Jan 13, 2017 10:44 pm

ahmedittihad wrote:Is that a deriviative sign?
That is a integreation sign.

$$\displaystyle \int_0^1f(x)dx=\dfrac{1}{2}\left(\int_0^1f(x)dx+\int_0^1f(1-x)dx\right)$$

$$\displaystyle =\dfrac{1}{2}\left(\int_0^1(f(x)+f(1-x))dx\right)$$

$$\displaystyle =\dfrac{1}{2}\int_0^1 dx=\dfrac{1}{2}$$
Hammer with tact.

Because destroying everything mindlessly isn't cool enough.

User avatar
ahmedittihad
Posts:181
Joined:Mon Mar 28, 2016 6:21 pm

Re: BdMO National 2008: Secondary

Unread post by ahmedittihad » Sat Jan 14, 2017 12:46 pm

Do we need calculus in Math Olympiads?!
Frankly, my dear, I don't give a damn.

User avatar
Thanic Nur Samin
Posts:176
Joined:Sun Dec 01, 2013 11:02 am

Re: BdMO National 2008: Secondary

Unread post by Thanic Nur Samin » Sat Jan 14, 2017 1:42 pm

I think not. This problem just used the very basics. I can't say anything about the nationals, but in IMO, APMO and other contests, you won't need it.
Hammer with tact.

Because destroying everything mindlessly isn't cool enough.

samiul_samin
Posts:1007
Joined:Sat Dec 09, 2017 1:32 pm

Re: BdMO National 2008: Secondary

Unread post by samiul_samin » Wed Feb 21, 2018 6:59 pm

A particular formula kills the problem

$\int^a_0 f(x)dx= \int^a_0 f(a-x)dx$

Now put,$a=1$

Post Reply