+166 Let \(f(x) = \lfloor x \lfloor x \rfloor \rfloor\) for \(x \ge 0\).
(a) Find all \(x \ge 0\) such that \(f(x) = 1.\)
(b) Find all \(x \ge 0\) such that \(f(x) = 3.\)
(c) Find all \(x \ge 0\) such that \(f(x) = 5.\)
(d) Find the number of possible values of \(x \ge 0\) for \(0 \le x \le 10.\)
I've solved part (a), and here's my answer:
In order for \(f(x)\) to equal \(1\), we must have the inequality \(1 \leq x \lfloor x \rfloor <2\). From this we can easily see that if \(x<1\), then \(\lfloor x \rfloor < 1\).
Similarly, if \(x\geq2\), then \(\lfloor x \rfloor\geq2\).
However, if \(1\le x<2\), then \(x\lfloor x \rfloor=x\), and so \(f(x)=1\). Hence, we have our answer.
