I recently learnt a way to tackle these types of problems - Here I Learnt it, explained to me by @Pranjal Jain

\( f(x) = x^4 + ax^3 + bx^2 + cx + d\)

We can observe that -

\( f(e) = e\times1993\)

thus - \(f(11) - 11\times1993 = 0\) , \(f(-7) + 7\times1993 = 0\)

\( f(11) + f(-7) = 1993(11 - 7) = 1993\times4\)

\(\dfrac{f(11) + f(-7)}{4} = 1993\)

Where I am wrong , please don't provide me the answer , just comment where I am wrong.

Did I started wrongly i.e - \( f(e) = e\times1993\) ?

Thank you

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## Comments

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TopNewest\(f(x) \neq 1993x, x \in \mathbb R\)

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Comment deleted Jan 01, 2015

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No need to find the 'c' (root of\( f(x) - 1993x\)) it will cancel out in

\( f(11) + f(-7) \)

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Why \(c=d=0\)? In my opinion, we can't directly determine the value of \(c\). In fact, we don't have to.

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now since c is the root of f(x) thus f(c) = 0 , then d must be zero

Correct me if I am wrong , Thanks for asking the question

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Thank you! because of your note I also got a chance to learn .

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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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Hey! You misunderstood it!! See. \(f(x)-1993x\) has roots \(1, 2\) and \(3\). So we can say that \[f(x)-1993x=(x-1)(x-2)(x-3)(x-\alpha)\] (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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I solved it see a comment is deleted after krishna's comment there i wrote the solution , thanks for helping

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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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