Question:

<div class="solution"><p><strong>Key Idea:</strong> Two simplifications: $\sin([x]\pi)=0$ and $\sqrt{x(x-1)+\frac{1}{4}} = |x-\frac{1}{2}|$.</p><p><strong>Step 1:</strong> First fraction vanishes since $[x]\in\mathbb{Z}$.</p><p><strong>Step 2:</strong> <span class="math-block">$$f(x) = 2x-1+\left|x-\tfrac{1}{2}\right| = \begin{cases}x-\frac{1}{2}, & x<\frac{1}{2}\\ 3x-\frac{3}{2}, & x\ge\frac{1}{2}\end{cases}$$</p><p><strong>Step 3:</strong> Both branches strictly increasing, range is all $\mathbb{R}$. Hence bijective.</p><p><strong>Answer: One-one and onto</strong></p><div class="trap-box"><strong>Trap:</strong> Do not leave the radical as a square root -- it is exactly an absolute value.<div class="key-concept"><strong>Key Concept:</strong> $\sin(n\pi)=0$ and perfect-square recognition under radical