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If $$f \colon \mathbb{R} \rightarrow \mathbb{R}$$ is a function defined by $$f(x) = \frac{x^2}{x^2 + 1}$$ then the number of integral values taken by $$f$$ is
Since $$0 \le x^2 < x^2 + 1$$, $$0 \le f(x) < 1$$ for every real $$x$$. The only integer in $$[0, 1)$$ is 0, taken at $$x = 0$$. So $$f$$ takes exactly 1 integral value.
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