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
3 years, 3 months ago

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

- 3 years, 3 months ago

Comment deleted Jan 01, 2015

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

$$f(11) + f(-7)$$

- 3 years, 3 months ago

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

- 3 years, 3 months ago

by writing f(x) = .... , I mean 1,2,3,c are the roots , now for f(0) = d (which we get by initial equation , and by second equation we get 6c = d

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

- 3 years, 3 months ago

@U Z I know that you mean c is a root of f(x). But we still don't have to determine its value. I don't really understand why d must be zero if f(c)=0. I agree with Krishna Sharma

- 3 years, 3 months ago

@U Z You only have three equations , $$f(0) = d$$ , $$f(c) = 0$$ and $$6c = d$$. How do you get $$d = 0$$?

- 3 years, 3 months ago

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

- 3 years, 3 months ago

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

- 3 years, 3 months ago

Sorry I misunderstood it

- 3 years, 3 months ago

Wanna know the Wolfram|Alpha approach for this?

- 3 years, 3 months ago

nope

- 3 years, 3 months ago

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

- 3 years, 3 months ago

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

- 3 years, 3 months ago

@U Z Just delete that solution comment

- 3 years, 3 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.

- 3 years, 3 months ago

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

- 3 years, 3 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

- 3 years, 3 months ago

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

Staff - 3 years, 3 months ago

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

- 3 years, 3 months ago