Key Concepts and Formulas

• Absolute Value Definition: $|x| = x$ if $x \geq 0$ and $|x| = -x$ if $x < 0$.
• Piecewise Functions: A function defined by multiple sub-functions, each applying to a certain interval of the main function's domain.
• Graphical Interpretation of Roots: The real roots of an equation $f(x) = k$ correspond to the x-coordinates of the intersection points of the graph $y = f(x)$ and the horizontal line $y = k$.

Step-by-Step Solution

Step 1: Rewrite the Equation

• What & Why: We want to isolate the parameter 'a' on one side of the equation. This allows us to analyze the function on the other side and relate its range to the possible values of 'a' that result in a specific number of roots.
• Math: Given $x|x-1|+|x+2|+a=0$, we rewrite it as $x|x-1|+|x+2| = -a$.
• Reasoning: Let $f(x) = x|x-1|+|x+2|$. The problem now becomes finding the values of $a$ for which $f(x) = -a$ has exactly one real root. This is equivalent to finding the values of $-a$ for which the horizontal line $y = -a$ intersects the graph of $y = f(x)$ exactly once.

Step 2: Identify Critical Points

• What & Why: The critical points of the absolute value expressions determine the intervals where we need to define the piecewise function.
• Math: The absolute value expressions are $|x-1|$ and $|x+2|$. Their critical points are found by setting the expressions inside the absolute values to zero: $x-1 = 0 \Rightarrow x = 1$ $x+2 = 0 \Rightarrow x = -2$
• Reasoning: These points divide the real number line into three intervals: $x < -2$, $-2 \leq x < 1$, and $x \geq 1$.

Step 3: Define the Piecewise Function

• What & Why: We now define $f(x)$ explicitly in each of the intervals determined by the critical points by removing the absolute value signs using the absolute value definition.

Case I: $x < -2$

• What & Why: In this interval, both $(x-1)$ and $(x+2)$ are negative.
• Math: $|x-1| = -(x-1) = 1-x$ and $|x+2| = -(x+2)$. Therefore, $f(x) = x(1-x) - (x+2) = x - x^2 - x - 2 = -x^2 - 2$
• Reasoning: We've simplified the expression for $f(x)$ in this interval.

Case II: $-2 \leq x < 1$

• What & Why: In this interval, $(x-1)$ is negative and $(x+2)$ is non-negative.
• Math: $|x-1| = -(x-1) = 1-x$ and $|x+2| = x+2$. Therefore, $f(x) = x(1-x) + (x+2) = x - x^2 + x + 2 = -x^2 + 2x + 2$
• Reasoning: We've simplified the expression for $f(x)$ in this interval.

Case III: $x \geq 1$

• What & Why: In this interval, both $(x-1)$ and $(x+2)$ are non-negative.
• Math: $|x-1| = x-1$ and $|x+2| = x+2$. Therefore, $f(x) = x(x-1) + (x+2) = x^2 - x + x + 2 = x^2 + 2$
• Reasoning: We've simplified the expression for $f(x)$ in this interval.

Step 4: Analyze the Function's Range and Continuity

• What & Why: We need to know the range of $f(x)$ to determine what values $-a$ can take. Checking continuity ensures a smooth transition between the piecewise definitions.

• Math:

  • For $x < -2$, $f(x) = -x^2 - 2$. As $x \to -2^-$, $f(x) \to -(-2)^2 - 2 = -6$. As $x \to -\infty$, $f(x) \to -\infty$. Thus, the range is $(-\infty, -6)$.
  • For $-2 \leq x < 1$, $f(x) = -x^2 + 2x + 2$. At $x = -2$, $f(-2) = -(-2)^2 + 2(-2) + 2 = -6$. As $x \to 1^-$, $f(x) \to -(1)^2 + 2(1) + 2 = 3$. Thus, the range is $[-6, 3)$.
  • For $x \geq 1$, $f(x) = x^2 + 2$. At $x = 1$, $f(1) = (1)^2 + 2 = 3$. As $x \to \infty$, $f(x) \to \infty$. Thus, the range is $[3, \infty)$.

• Reasoning: We observe that $f(x)$ is continuous at $x = -2$ and $x = 1$ since the piecewise definitions agree at these points. The overall range of $f(x)$ is $(-\infty, -6) \cup [-6, 3) \cup [3, \infty) = (-\infty, \infty)$. This means $f(x)$ can take any real value.

Step 5: Determine the Number of Roots

• What & Why: Since $f(x)$ is continuous and its range is $(-\infty, \infty)$, any horizontal line $y = -a$, where $a \in \mathbb{R}$, will intersect the graph of $f(x)$ at least once. We need to argue that the intersection is exactly once.
• Math: Consider the derivatives of the piecewise function:
  • For $x < -2$, $f'(x) = -2x > 0$ (since $x < -2$).
  • For $-2 < x < 1$, $f'(x) = -2x + 2 = -2(x-1) > 0$ (since $x < 1$).
  • For $x > 1$, $f'(x) = 2x > 0$ (since $x > 1$).
• Reasoning: The derivative is strictly positive in each interval. This, along with the continuity of $f(x)$, implies that $f(x)$ is strictly increasing over its entire domain. A strictly increasing function can intersect any horizontal line at most once. Since the range of $f(x)$ is $(-\infty, \infty)$, it must intersect any horizontal line exactly once. Therefore, for any real number $a$, the equation $f(x) = -a$ has exactly one real root.

Step 6: Conclusion

• What & Why: State the final answer based on the analysis.
• Reasoning: The set of all $a \in \mathbb{R}$ for which the equation has exactly one real root is $(-\infty, \infty)$.

Common Mistakes & Tips

• Incorrect Absolute Value Handling: Pay close attention to the signs when removing absolute values in different intervals. A sign error can lead to a completely different function.
• Forgetting to Check Continuity: Always verify that the piecewise function is continuous at the critical points. Discontinuities can drastically alter the number of roots.
• Assuming Monotonicity without Proof: Do not assume the function is monotonic; prove it using derivatives or by analyzing the behavior of the individual components.

Summary

We analyzed the equation $x|x-1|+|x+2|+a=0$ by rewriting it as $f(x) = -a$, where $f(x) = x|x-1|+|x+2|$. By considering the intervals defined by the critical points of the absolute value expressions, we defined $f(x)$ as a piecewise function. We then showed that $f(x)$ is continuous and strictly increasing with a range of $(-\infty, \infty)$. This guarantees that for any real value of $a$, the equation $f(x) = -a$ has exactly one real root.

Final Answer

The final answer is $\boxed{(-\infty, \infty)}$, which corresponds to option (A).