Let `f(x)={{:((sin(a+1)x+sinx)/(x)",",xlt0),(c",",x=0),((sqrt(x+bx^(2)))/(bx^((3)/(2)))",",xgt0):}`
If f(x) is continous at x = 0, then

Options

A. `a+c=0,b=1`
B. `a+c=1, b in R`
C. `a+c=-1, b in R`
D. `a+c=-1, b=-1`

Answer

To determine the values of \( a \), \( b \), and \( c \) such that the function \( f(x) \) is continuous at \( x = 0 \), we need to ensure that the left-hand limit and right-hand limit at \( x = 0 \) are equal to \( f(0) \).

The function \( f(x) \) is defined as follows:
\[ f(x) = \begin{cases} \frac{\sin((a+1)x) + \sin x}{x} & \text{if } x < 0 \\ c & \text{if } x = 0 \\ \frac{\sqrt{x + bx^2}}{bx^{3/2}} & \text{if } x > 0 \end{cases} \]

### Step 1: Calculate the Left-Hand Limit as \( x \to 0^- \)

For \( x < 0 \):
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sin((a+1)x) + \sin x}{x} \]

Using the limit property \( \lim_{x \to 0} \frac{\sin(kx)}{x} = k \) for any constant \( k \):
\[ = \lim_{x \to 0^-} \left( (a+1) + \frac{\sin x}{x} \right) = (a+1) + 1 = a + 2 \]

### Step 2: Calculate the Right-Hand Limit as \( x \to 0^+ \)

For \( x > 0 \):
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sqrt{x + bx^2}}{bx^{3/2}} \]

Factoring out \( x \) from the square root:
\[ = \lim_{x \to 0^+} \frac{\sqrt{x(1 + b x)}}{bx^{3/2}} = \lim_{x \to 0^+} \frac{\sqrt{x} \sqrt{1 + b x}}{b x^{3/2}} = \lim_{x \to 0^+} \frac{\sqrt{1 + b x}}{b x} = \frac{1}{b} \lim_{x \to 0^+} \frac{\sqrt{1 + b x}}{\sqrt{x}} \]

As \( x \to 0 \):
\[ = \frac{1}{b} \cdot \frac{1}{\sqrt{0}} = \text{undefined} \]

Instead, we can rewrite:
\[ = \lim_{x \to 0^+} \frac{\sqrt{1 + b x} - 1}{b x} \cdot \frac{\sqrt{1 + b x} + 1}{\sqrt{1 + b x} + 1} = \lim_{x \to 0^+} \frac{b x}{b x (\sqrt{1 + b x} + 1)} = \lim_{x \to 0^+} \frac{1}{\sqrt{1 + b x} + 1} = \frac{1}{2} \]

### Step 3: Set the Limits Equal to Each Other

For continuity at \( x = 0 \):
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \]
\[ a + 2 = c = \frac{1}{2} \]

### Step 4: Solve for \( a \) and \( c \)

From \( c = \frac{1}{2} \):
\[ a + 2 = \frac{1}{2} \implies a = \frac{1}{2} - 2 = -\frac{3}{2} \]

### Step 5: Conclusion

Thus, we have:
\[ a = -\frac{3}{2}, \quad b \text{ can be any non-zero value}, \quad c = \frac{1}{2} \]