If `f(x) = {{:(b([x]^(2)+[x])+1",","for",x gt -1),(sin(pi(x + a))",","for",x lt -1):}`, where [x] denotes the integral part of x, then for what values of a, b, the function is continuous at x = - 1?

Options

A. `a = 2n + (3)/(2), b in R, n in I`
B. `a = 4n + 2, b in R, n in I`
C. `a = 4n + (3)/(2), b in R^(+), n in I`
D. `a = 4n + 1, b in R^(+), n in I`

Answer

To determine the values of \( a \) and \( b \) for which the function \( f(x) \) is continuous at \( x = -1 \), we need to ensure that the left-hand limit and the right-hand limit at \( x = -1 \) are equal to the value of the function at that point.

The function is defined as follows:
\[ f(x) = \begin{cases} b([x]^2 + [x] + 1) & \text{for } x > -1 \\ \sin(\pi(x + a)) & \text{for } x < -1 \end{cases} \]
where \([x]\) denotes the integral part of \( x \).

### Step 1: Calculate the Right-Hand Limit as \( x \) Approaches -1
For \( x > -1 \), the function is given by:
\[ f(x) = b([x]^2 + [x] + 1) \]
At \( x = -1 \), \([x] = -1\). Thus:
\[ f(-1) = b((-1)^2 + (-1) + 1) = b(1 - 1 + 1) = b(1) = b \]

### Step 2: Calculate the Left-Hand Limit as \( x \) Approaches -1
For \( x < -1 \), the function is given by:
\[ f(x) = \sin(\pi(x + a)) \]
Taking the limit as \( x \) approaches -1 from the left:
\[ \lim_{x \to -1^-} f(x) = \sin(\pi(-1 + a)) = \sin(\pi(a - 1)) \]

### Step 3: Set the Limits Equal for Continuity
For \( f(x) \) to be continuous at \( x = -1 \), we need:
\[ \lim_{x \to -1^-} f(x) = \lim_{x \to -1^+} f(x) \]
Thus, we set the two limits equal:
\[ \sin(\pi(a - 1)) = b \]

### Step 4: Analyze the Condition for Continuity
We have:
1. \( b = \sin(\pi(a - 1)) \)

### Step 5: Determine Values of \( a \) and \( b \)
The sine function can take values from -1 to 1. Therefore, \( b \) must satisfy:
\[ -1 \leq b \leq 1 \]
This implies:
\[ -1 \leq \sin(\pi(a - 1)) \leq 1 \]

### Step 6: Solve for \( a \)
To find the possible values of \( a \), we need to analyze the equation:
\[ \sin(\pi(a - 1)) = b \]
The general solution for \( \sin(x) = k \) is:
\[ x = n\pi + (-1)^n \arcsin(k) \quad \text{for } n \in \mathbb{Z} \]
Thus:
\[ \pi(a - 1) = n\pi + (-1)^n \arcsin(b) \]
This leads to:
\[ a = 1 + n + \frac{(-1)^n \arcsin(b)}{\pi} \]

### Conclusion
The values of \( a \) can be expressed in terms of \( n \) and \( b \) as:
\[ a = 1 + n + \frac{(-1)^n \arcsin(b)}{\pi} \]
where \( b \) is constrained to the interval \([-1, 1]\).