This problem
I recently learnt a way to tackle these types of problems - Here I Learnt it, explained to me by @Pranjal Jain
f(x)=x4+ax3+bx2+cx+d
We can observe that -
f(e)=e×1993
thus - f(11)−11×1993=0 , f(−7)+7×1993=0
f(11)+f(−7)=1993(11−7)=1993×4
4f(11)+f(−7)=1993
Where I am wrong , please don't provide me the answer , just comment where I am wrong.
Did I started wrongly i.e - f(e)=e×1993 ?
Thank you
Easy Math Editor
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Comments
f(x)=1993x,x∈R
What is your reason behind the claim f(e)=1993e
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Sorry I misunderstood it
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Wanna know the Wolfram|Alpha approach for this?
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Thank you! because of your note I also got a chance to learn .
Hey! You misunderstood it!! See. f(x)−1993x has roots 1,2 and 3. So we can say that f(x)−1993x=(x−1)(x−2)(x−3)(x−α) (Since its of degree 4).
f(e)=e is for e=1,2,3 only.
Carry on from here. I hope you got it this time.
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@Calvin Lin As you see in the note, tag didn't worked! I didn't got any mail. Is there any bug? It happens a lot with me
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@ mention currently only works in comments. It currently does not work in notes or problems
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I solved it see a comment is deleted after krishna's comment there i wrote the solution , thanks for helping