<s>[INST] If $f(x) $ is a continuous function with a given identity, how do I find $f(\sqrt{3})$? 
If $f(x) $ Is a continuous function $   \forall\ x \in R$  and satisfies
  $x^2+x \{f(x)\} - 3 = \sqrt{3} \ f(x)   \  \forall\ x \in R$
Find $f(\sqrt{3})$.

$\{ t\}$ is the fractional part of $t$.
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My attempt: I substituted $\sqrt{3}$ in that and could only conclude that $0 \leq f(\sqrt{3}) < 1$
 [/INST] By substituting $\sqrt{3}$ in there, you get that $f(x) - \{f(x)\} = 0$, which implies that $0 \leq a < 1$ as you concluded.
Now suppose $f(\sqrt{3}) = a, 0 < a < 1$. Then, by continuity, $f$ satisfies the equation $x^2 + xf(x) - 3 = \sqrt{3} f(x)$ in a neighborhood of $x = \sqrt{3}$, so that $$f(x) = \frac{x^2 - 3}{\sqrt{3} - x} = -(\sqrt{3} + x),$$
which implies that $f(\sqrt{3}) = - 2\sqrt{3} < 0$ (a contradiction because $a > 0$). Therefore, we must have $f(\sqrt{3}) =0$.
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