If the domain of $$f(x)$$ is the largest possible subset of natural numbers such that $$f(x) = \sqrt{676x - x^2 - 2019}$$ is defined then the number of elements in the domain of $$f(x)$$ is

Solution

We need $$676x - x^2 - 2019 \ge 0$$, i.e. $$x^2 - 676x + 2019 \le 0$$. The roots are $$x = \frac{676 \pm \sqrt{676^2 - 4 \times 2019}}{2} = \frac{676 \pm \sqrt{448900}}{2} = \frac{676 \pm 670}{2}$$, i.e. $$3$$ and $$673$$. So $$3 \le x \le 673$$, which contains $$673 - 3 + 1 = 671$$ natural numbers.

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