Key Concepts and Formulas

• Continuity of a function: A function $f(x)$ is continuous at a point $c$ if the following three conditions are met:
  1. $f(c)$ is defined.
  2. $\lim_{x \to c} f(x)$ exists. This means the left-hand limit and the right-hand limit are equal: $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)$.
  3. $\lim_{x \to c} f(x) = f(c)$.
• Piecewise functions: For a function defined in pieces, continuity at the points where the definition changes requires the limits from both sides to be equal to the function's value at that point.

Step-by-Step Solution

Step 1: Understand the condition for continuity. The problem states that $f(x)$ is continuous on $\mathbb{R}$. This means $f(x)$ must be continuous at $x = -1$ and $x = 1$, the points where the definition of the function changes. For continuity at a point $c$, the left-hand limit, the right-hand limit, and the function value at $c$ must all be equal: $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$.

Step 2: Apply the continuity condition at $x = -1$. We need to evaluate the left-hand limit, the right-hand limit, and the function value at $x = -1$.

• Left-hand limit at $x = -1$: For $x < -1$, $f(x) = 2\sin\left(-\frac{\pi x}{2}\right)$. $\lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} 2\sin\left(-\frac{\pi x}{2}\right)$ Substitute $x = -1$: $2\sin\left(-\frac{\pi (-1)}{2}\right) = 2\sin\left(\frac{\pi}{2}\right) = 2(1) = 2$

• Right-hand limit at $x = -1$: For $-1 \le x \le 1$, $f(x) = |ax^2 + x + b|$. $\lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} |ax^2 + x + b|$ Substitute $x = -1$: $|a(-1)^2 + (-1) + b| = |a - 1 + b| = |a + b - 1|$

• Function value at $x = -1$: For $-1 \le x \le 1$, $f(x) = |ax^2 + x + b|$. $f(-1) = |a(-1)^2 + (-1) + b| = |a - 1 + b| = |a + b - 1|$

For continuity at $x = -1$, we must have $\lim_{x \to -1^-} f(x) = \lim_{x \to -1^+} f(x) = f(-1)$. Therefore, we get the equation: $2 = |a + b - 1| \quad (*)$

Step 3: Apply the continuity condition at $x = 1$. We need to evaluate the left-hand limit, the right-hand limit, and the function value at $x = 1$.

• Left-hand limit at $x = 1$: For $-1 \le x \le 1$, $f(x) = |ax^2 + x + b|$. $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} |ax^2 + x + b|$ Substitute $x = 1$: $|a(1)^2 + 1 + b| = |a + 1 + b| = |a + b + 1|$

• Right-hand limit at $x = 1$: For $x > 1$, $f(x) = \sin(\pi x)$. $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \sin(\pi x)$ Substitute $x = 1$: $\sin(\pi \cdot 1) = \sin(\pi) = 0$

• Function value at $x = 1$: For $-1 \le x \le 1$, $f(x) = |ax^2 + x + b|$. $f(1) = |a(1)^2 + 1 + b| = |a + 1 + b| = |a + b + 1|$

For continuity at $x = 1$, we must have $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$. Therefore, we get the equation: $|a + b + 1| = 0$ This implies: $a + b + 1 = 0$ $a + b = -1 \quad (**)$

Step 4: Solve the system of equations. We have two equations:

1. $|a + b - 1| = 2$ (from Step 2)
2. $a + b = -1$ (from Step 3)

Substitute the value of $a + b$ from equation () into equation (*): $|(-1) - 1| = 2$ $|-2| = 2$ $2 = 2$ This equation is consistent. We are asked to find the value of $a+b$. From equation (), we have already found that $a+b = -1$.

Step 5: Verify the solution. We found $a+b = -1$. Let's check if this satisfies the condition $|a+b-1|=2$. Substituting $a+b = -1$ into $|a+b-1|$, we get $|-1-1| = |-2| = 2$, which is true. Also, $a+b+1 = -1+1 = 0$, so $|a+b+1|=0$, which is also true.

Thus, the condition $a+b = -1$ is consistent with the continuity requirements.

Common Mistakes & Tips

• Absolute Value: Remember that $|x| = k$ implies $x = k$ or $x = -k$. In this problem, the equation $|a+b-1|=2$ could have led to two cases: $a+b-1=2$ or $a+b-1=-2$. However, the second continuity condition at $x=1$ directly gives $a+b=-1$, which simplifies the problem significantly.
• Careful with Signs: Pay close attention to the signs when substituting values into the function definitions, especially when dealing with negative numbers and absolute values.
• Check Both Points: Ensure you apply the continuity conditions at all transition points ($x=-1$ and $x=1$ in this case).

Summary

To ensure the function $f(x)$ is continuous on $\mathbb{R}$, we applied the condition for continuity at the points where the function definition changes, namely $x = -1$ and $x = 1$. At $x = -1$, the equality of the left-hand limit and the right-hand limit yielded $|a+b-1|=2$. At $x = 1$, the equality of the left-hand limit and the right-hand limit gave $|a+b+1|=0$, which simplified to $a+b=-1$. Substituting $a+b=-1$ into the first equation $|a+b-1|=2$ resulted in $|-1-1|=|-2|=2$, confirming consistency. The problem asks for the value of $a+b$, which we found directly from the continuity condition at $x=1$ to be $-1$.

The final answer is $\boxed{-1}$.