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 will analyze the left-hand limit and the right-hand limit of \( f(x) \) as \( x \) approaches \(-1\).

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

### Step 2: Calculate the right-hand limit as \( x \to -1^+ \)
For \( x > -1 \), we have:
\[ \lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} b([x]^2 + [x] + 1) \]
As \( x \) approaches \(-1\) from the right, \([x] = -1\). Thus:
\[ \lim_{x \to -1^+} f(x) = b((-1)^2 + (-1) + 1) = b(1 - 1 + 1) = b(1) = b \]

### Step 3: Calculate the left-hand limit as \( x \to -1^- \)
For \( x < -1 \), we have:
\[ \lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} \sin(\pi(x + a)) \]
Substituting \( x = -1 - h \) where \( h \to 0^+ \):
\[ \lim_{h \to 0} \sin(\pi(-1 - h + a)) = \sin(\pi(-1 + a) - \pi h) \]
As \( h \to 0 \), this simplifies to:
\[ \sin(\pi(a - 1)) \]

### Step 4: 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:
\[ b = \sin(\pi(a - 1)) \]

### Step 5: Analyze the sine function
The sine function can take values from \(-1\) to \(1\). Therefore, we can express \( b \) in terms of \( a \):
\[ b = \sin(\pi(a - 1)) \]
To find the values of \( a \) for which \( \sin(\pi(a - 1)) = -1 \):
\[ \sin(\pi(a - 1)) = -1 \implies \pi(a - 1) = \frac{3\pi}{2} + 2n\pi \quad (n \in \mathbb{Z}) \]
This simplifies to:
\[ a - 1 = \frac{3}{2} + 2n \implies a = 2n + \frac{5}{2} \]

### Step 6: Conclusion
The values of \( a \) can be expressed as:
\[ a = 2n + \frac{5}{2} \quad (n \in \mathbb{Z}) \]
And \( b \) can take any real number since it is defined by the sine function.

### Final Answer
The function \( f(x) \) is continuous at \( x = -1 \) for values of \( a \) given by \( a = 2n + \frac{5}{2} \) for any integer \( n \), and \( b \) can be any real number.