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@Apu
Note for OP
+solved @user to close the thread when your doubt is solved. Mention the user who helped you solve the doubt. This will be added to their stats.hello! can you share an attempt please?
Approach hi nhi click ho rhi mujhe...
Got it. Any inputs? @JEE
Vo x belongs to complex likha hai? 👀
Yep
try adding and subtracting the two equations
the prefactoral terms of A(x) and B(x) would simplify a bit
Using the two equations you can prove that A(x) and B(x) are in forms of [x^2+1]f(x)
would you like me to attach a solution? or do you want to solve it yourself
Yeah send the solution plz
let me know if you need any clarification anywhere
(the first image is adding and subtracting the two equations)
(the second image is a bit of a trick i caught that would help with one of the equations formed)
rip slowmode
A(x) + 2B(x) = [x^2 + 1]*[C(x) - D(x)]/2
which means that the function A(x) + 2B(x) is divisible by x^2 + 1
Wtf is that
Ohh yess...lemme try
the point of this was to bring C(x) + D(x) in terms of x(g(x))
because in equation 3 we had xA(x) + xB(x)
overall, C(x) + D(x) is also a polynomial, but if C(0) + D(0) = 0, that would mean that constant of C(x) + D(x) = 0. So, i was able to take x common.
this was a random thought that came to mind that helped solve the question, im not sure if there's a more orthodox way @Vish maybe enlighten me 🥺
Alr...tqsm
Really appreciate tnx!
no problem 🫂
yo thanks everyone for helping! And apologies for the jee ping 😄 Will make a new role where folks who want to help out with solving doubts - only they will be pinged haha.
@psinfinity@MindfulMaven theres a better way to tell C(x) + D(x) is in terms of xg(x)
What's it..?
share 🙏
from here we get
A(x) + B(x) = -(x^2+1)[ C(x) + D(x) ] / 2x
and since A(x) and B(x) are polynomials, C(x) + D(x) must be in terms of x* g(x)
As a polynomial has non-negative integer exponents
Make sense to me..
true
Umm...then can I say both are divisible directly?
No from here we conclude that their sum is divisible by x2+1
For eg A(x) = x+1 and B(x) = x2-x
neither of them is divisible
@iTeachChem mai 11th mai hu
May vala batch join karunga
Apart from allen modules chem ke liye aur kya padhna must hai till 11th?
Ahh right...then can I conclude option D is correct?
Nah
Yo can you make another post about this? Let us focus on the math q here 😄 Post on https://discord.com/channels/1226379612238385242/1226385032038453249 please
Also, check here https://discord.com/channels/1226379612238385242/1226389969237446741, that should help with your q.
thats why we found 2 equations to check
overall it should be A
Exactly so what was the significance of like finding both are divisible by x^2+1?
what do you mean? thats what the question is asking for
God...yep my bad.
Apologise for oversight...I just thought they're separately...
Tnxx @psinfinity @Linarp
M confused as their sum is divisible by x^2+1...well it doesn't mean that they both are...but why m Makin this step..as it doesn't give any clear idea right?
Perhaps you'll understand what m saying...
you can take an example as well
here ( Ax + Bx ) is divisible by x2+1
but individually Ax and Bx are not divisible by x2+1
similarly If Ax= 2x2+2 and Bx= -x2-1
their sum is divisible by x2+1 and
Ax and Bx are also individually divisible by x2+1
from (Ax + Bx) = (x2+1) x*gx we conclude that the sum of Ax and Bx is divisible by x2+1 this does not give any information about the function Ax and Bx
⬆️ thats why 2 equations
(subtract the two equations in the second screenshot)
all jee role ping?
sensational
is the question solved or - still looking for answer
+solved @MindfulMaven
Answered
yawr cool banana tha
Solved
+solved
You need to mention a member to mark the thread as solved.
+solved @user. This will be added to their stats.+Solved @psinfinity
Command not found! Use
+help to view all available commandschota s i think
Hahaha..okayyyy
+solved @psinfinity
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