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- #1

- Thread starter jacks
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- Thread starter
- #1

- Jan 26, 2012

- 890

Continuous and rational implies that \(f(x)\) is a constant.if $f(x)$ be a continous and assumes only rational values so that $ f(2010) =1. $ then roots of

the equation $f(1)x^2 + 2f(2)x + 3f(3) =0$ are

CB

- Thread starter
- #3

- Jan 26, 2012

- 890

If \(f(x)\) is continuous it satisfies the intermediate value principle, that is \(f(x)\) takes on all values between \(f(a)\) and \(f(b)\) for any distinct reals \(a\) and \(b\).means $x^2+2x+3=0\Leftrightarrow (x+1)^2+2>0\forall x\in \mathbb{R}$

Means no real Roots.

but I did not understand the line if $f(x)$ is Conti. and assume only rational values .then it must be Constant

Thanks

We are told that \(f(a)\) and \(f(b)\) are rational, and if they are not equal there is an irrational \(\rho\) between them and a \(c \in (a,b)\) such that \(f(c)=\rho\) which contradicts \(f(x)\) only taking rational values, so for any two real \(a, b\) \(f(a)=f(b)\) hence \(f(x)\) is a constant.

CB