BdMO Online Forum

Shad wrote: ↑

Wed Dec 23, 2020 12:39 pm

Sorry the total run is 358

Well that doesn't matter much. The answer is still same.

Putting y= 0 we get, $f(x^4)=x^3f(x) $ How? putting $y = 0$ gives, $\ f(x^4)=x^3f(x) + f(f(0))$ and you didn't prove that $f(f(0)) = 0$. Or, $f(x^4)=f(t)=x^3f(x)\cdots (2)$[Let $x^4=t$] Now from the question we get, $f(x^4+y)=x^3f(x)+f(f(y))$ Or, $f(t+y)=f(t)+f(y)$[Using (1) and (2)] Which is the c...

Already posted: viewtopic.php?f=17&t=3262&p=15746#p15746

I am not sure this solution is correct. Bhaya plz check it if right then I will post a new problem if not then I will probably try more to solve it. Ifx=0 in the given equation, $f(y)=f(f(y))$ Let, $f(t)=q$ for some real t,q Putting y=t in (1) we get, $f(t)=f(f(t))$ Or, $q=f(q)$ Thus, $f(x)=x$ is t...

The answer is $33$. If all of the $11$ batsmen scored 32 runs, then the total score would be $11 \times 32 = 352$. Since the total score is $358$, at least one batsman needs to score more than $32$ runs. So, the run scored by the highest scorer should be at least $33$. Now, if $5$ batsmen score $32$...

One of the differences will be $0$. So the product will also become $0$.

If you wanna insert image of a geometric figure, you can draw it using https://www.geogebra.org/?lang=en and then attach the file with your post. You can attach any kind of files like below:

Anindya Biswas wrote: ↑

Mon Dec 14, 2020 11:09 pm

Assume there are $3$ peoples. $a$, $b$ and $c$. If $a$ knows $b$, $b$ knows $c$ and $c$ knows $a$. Then the first condition holds but the second condition doesn't!

Assume that knowing is mutual.

একটি কমিটিতে $2n+1$ সদস্য আছে। তাদের যে কোনো $n$ জনকে বাছাই করলে, বাকি $n+1$ জনের মধ্যে অন্তত একজন থাকবে যে এই $n$ জনের সবাইকে চিনে। প্রমাণ করো, কমিটিতে একজন আছে যিনি বাকি সবাইকে চিনে। Don't post problems or solutions in Bangla in 'Olympiad Level' sub-forum. I am shifting your post to 'Higher Secon...

[N.B I am not sure the solution is correct. If somebody would kindly check it, I will be grateful.] Given that, $f(xf(x)+f(y))=f(x)^2+y \cdots (1)$ Let, $x = 0$ then, $f(0f(0)+f(y))=f(0)^2+y$ Or, $f(f(y))=f(0)^2+y$ Or, $f(f(y))=k^2+y [\text{Let, f(0)=k}] $ Now proving $f^2(y)$ is onto function, pro...