This problem

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

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

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 Note by U Z
5 years, 9 months ago

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

- 5 years, 9 months ago

What is your reason behind the claim f(e)=1993e

- 5 years, 9 months ago

Sorry I misunderstood it

- 5 years, 9 months ago

Wanna know the Wolfram|Alpha approach for this?

- 5 years, 9 months ago

nope

- 5 years, 9 months ago

@U Z Hmm, I wish more people had read A=B and were exposed to EM

- 5 years, 9 months ago

Should I delete this note because in a way I am discussing a problem

- 5 years, 9 months ago

@U Z Just delete that solution comment

- 5 years, 9 months ago

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

- 5 years, 9 months ago

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.

- 5 years, 8 months ago

@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

- 5 years, 8 months ago

@ mention currently only works in comments. It currently does not work in notes or problems

Staff - 5 years, 8 months ago

Sometimes it also causes trouble in comments. Lemme try here. @Pranjal Jain

- 5 years, 8 months ago

I solved it see a comment is deleted after krishna's comment there i wrote the solution , thanks for helping

- 5 years, 8 months ago