If `f(x)={{:((sin2x)/(cx)+(x)/((sqrt(x+a^(2))-a)),,,xne 0"," (alt0)),(b,,,x=0","(bne0)):}`
and f(x) is continuous at x = 0, then the value of bc is equal to

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous at \( x = 0 \). The function is defined as follows:

\[ f(x) = \begin{cases} \frac{\sin(2x)}{cx} + \frac{x}{\sqrt{x + a^2} - a} & \text{if } x \neq 0 \\ b & \text{if } x = 0 \end{cases} \]

Given that \( a < 0 \) and \( b \neq 0 \), we need to find the value of \( bc \).

### Step 1: Calculate the limit of \( f(x) \) as \( x \) approaches 0

To check for continuity at \( x = 0 \), we need to find:

\[ \lim_{x \to 0} f(x) \]

This limit must equal \( f(0) = b \).

### Step 2: Analyze the first term \( \frac{\sin(2x)}{cx} \)

Using the limit property \( \lim_{x \to 0} \frac{\sin(kx)}{kx} = 1 \):

\[ \lim_{x \to 0} \frac{\sin(2x)}{cx} = \lim_{x \to 0} \frac{2 \sin(2x)}{2cx} = \frac{2}{c} \cdot 1 = \frac{2}{c} \]

### Step 3: Analyze the second term \( \frac{x}{\sqrt{x + a^2} - a} \)

We can simplify this limit by multiplying the numerator and denominator by the conjugate:

\[ \frac{x(\sqrt{x + a^2} + a)}{(\sqrt{x + a^2} - a)(\sqrt{x + a^2} + a)} = \frac{x(\sqrt{x + a^2} + a)}{x + a^2 - a^2} = \frac{x(\sqrt{x + a^2} + a)}{x} \]

This simplifies to:

\[ \sqrt{x + a^2} + a \]

Now, taking the limit as \( x \to 0 \):

\[ \lim_{x \to 0} (\sqrt{x + a^2} + a) = \sqrt{0 + a^2} + a = |a| + a \]

Since \( a < 0 \), \( |a| = -a \). Thus:

\[ |a| + a = -a + a = 0 \]

### Step 4: Combine the limits

Now we can combine the limits:

\[ \lim_{x \to 0} f(x) = \frac{2}{c} + 0 = \frac{2}{c} \]

### Step 5: Set the limit equal to \( b \)

For continuity at \( x = 0 \):

\[ \frac{2}{c} = b \]

### Step 6: Solve for \( bc \)

Multiplying both sides by \( c \):

\[ bc = 2 \]

### Final Answer

Thus, the value of \( bc \) is:

\[ \boxed{2} \]