Question

The set of all a∈ R for which the equation x|x-1|+|x-2|+a=0 has exactly one real root is

• A. (-∞,-3)
• B. (-6,∞)
• C. (-∞,∞)
• D. (-6,-3)

Answer 1

The Correct Option is C

Define \[ f(x) = x|x - 1| + |x + 2| + a. \] We want to find all \(a \in \mathbb{R}\) for which \(f(x) = 0\) has exactly one real solution. Observe that \[ x|x - 1| \quad \text{and} \quad |x + 2| \] are both piecewise linear but each is non-decreasing in certain intervals and non-decreasing overall when appropriately pieced together. A more detailed analysis (or plotting) shows that \(f(x)\) is strictly increasing as a function of \(x\). Since \(f(x)\) is strictly increasing in \(x\), the equation \(f(x) = 0\) intersects the \(x\)-axis exactly once, regardless of the value of \(a\). Consequently, for any real \(a\), there is exactly one solution to \(f(x) = 0\). Hence, the set of all such \(a\) is \[ (-\infty, \infty). \]