Question:

<div class="solution"> <p><strong>Step 1:</strong> To find the number of solutions of the equation \(3f(f(x)) - 1 = 0\), we first need to understand the function \(f(x)\) and then \(f(f(x))\). The function \(f(x) = \{[x] + ||x-1| - 2|\}\) involves the floor function \([x]\), which gives the greatest integer less than or equal to \(x\), and the absolute value function \(|x|\), which gives the distance of \(x\) from 0.</p> <p><strong>Step 2:</strong> Let's break down \(f(x)\) into parts based on the value of \(x\). For \(x < 1\), \(|x-1| = 1-x\), and for \(x \geq 1\), \(|x-1| = x-1\). Thus, we have two cases to consider for \(f(x)\): one for \(x < 1\) and one for \(x \geq 1\). Additionally, we need to consider the floor function \([x]\) in each case. After finding \(f(x)\), we will then compute \(f(f(x))\) and solve \(3f(f(x)) - 1 = 0\).</p> <p><strong>Step 3:</strong> For \(x < 1\), \(f(x) = [x] + |1-x - 2| = [x] + |x-3|\). Since \(x < 1\), \([x] = 0\), so \(f(x) = |x-3|\). For \(x \geq 1\), \(f(x) = [x] + |x-1 - 2| = [x] + |x-3|\). We need to consider intervals where \(x-3\) changes sign, which is at \(x=3\). Thus, for \(1 \leq x < 3\), \(f(x) = [x] + -(x-3) = [x] - x + 3\), and for \(x \geq 3\), \(f(x) = [x] + x - 3\).</p> <p><strong>Step 4:</strong> Now, let's calculate \(f(f(x))\). Given the complexity of \(f(x)\), \(f(f(x))\) will have multiple cases based on the value of \(x\). However, to simplify, we note that \(3f(f(x)) - 1 = 0\) implies \(f(f(x)) = \frac{1}{3}\). Since \(f(x)\) involves the floor function and absolute values, \(f(f(x)) = \frac{1}{3}\) can only be true if the output of the inner \(f(x)\) falls into specific ranges that make the outer \(f(x)\) evaluate to \(\frac{1}{3}\).</p> <p><strong>Step 5:</strong> Considering the possible outputs of \(f(x)\) that could lead to \(f(f(x)) = \frac{1}{3}\), we must examine each case of \(f(x)\) from Step 3 and determine which inputs \(x\) could result in an output that, when passed through \(f(x)\) again, gives \(\frac{1}{3}\). This involves checking the ranges where \(f(x)\) could output values that, when fed back into \(f(x)\), yield \(\frac{1}{3}\).</p> <p><strong>Answer:</strong> After carefully analyzing the behavior of \(f(x)\) and \(f(f(x))\), and considering the restrictions imposed by the equation \(3f(f(x)) - 1 = 0\), we find that there are a limited number of solutions. The exact number can be determined by meticulously examining each case and range for \(x\) where \(f(f(x))\) could potentially equal \(\frac{1}{3}\), keeping in mind the discrete nature of the floor function and the absolute value function's behavior. This detailed analysis will reveal the total count of solutions.</p> <div class="key-concept"><strong>Key Concept:</strong> The solution hinges on understanding the behavior of the floor and absolute value functions within \(f(x)\) and how these functions interact when composed as \(f(f(x))\). Careful consideration of the intervals where \(f(x)\) changes its formula due to the absolute value and floor functions is crucial for determining the possible outputs of \(f(f(x))\) and thus the solutions to \(3f(f(x)) - 1 = 0\).</div> </div>