IIT JEE , AIEEE Forum - Mathematics , Algebra

man111

(1) if f(x²+x+3)+2f(x²-3x+5) = 6x2-10x+17,

then find the value of f(85).

hsbhatt

Not so hard actually.

You can easily prove by comparison of coeffs on both sides that deg f(x) = 1.

So f(x) = ax+b

$\therefore a(x^2+x+3)+2a(x^2-3x+5) + 3b \equiv 6x^2-10x+17$

This gives a = 2 and b = -3

Hence f(x) = 2x-3 which gives $f(85) = \boxed{167}$

norton

quite easy. i guess no use solving it whole as bhatt sir has done it already. just compare coefficieents of x² and the constant terms and you'll get the answer. f(85)= 167

rohitkuruvila

how can you say that f(x) is of degree 1

kaymant

is it known that f(x) isactually a polynomial?

hsbhatt

actually no, but almost any function has a Taylor expansion

kaymant

but taylor' expansion only approximates a smooth function by a polynomial in the close proximity of a point.. in this particular case we have no idea at all

hsbhatt

Yeah, got to agree with you. But, the students sometimes neglect to mention everything relevant to the problem. Sometimes we clarify whether our assumption was what was meant. Otherwise, if we are convinced that our assumption is right, we just go ahead.

One clue here is that this is not a functional equation. He appears to have only a single valued function in mind. So that was the motivation for my assuming a polynomial.

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