Let `f(x)={{:("sin 4x"/"log (1+3x)",","x ne 0),("A+1",","x=0):}` The value of A for f to be continuous at x=0 is

Options

A. `2/3`
B. `1/3`
C. `1`
D. `3/4`

Answer

To determine the value of \( A \) for the function

\[ f(x) = \begin{cases} \frac{\sin 4x}{\log(1 + 3x)} & \text{if } x \neq 0 \\ A + 1 & \text{if } x = 0 \end{cases} \]

to be continuous at \( x = 0 \), we need to ensure that

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

### Step 1: Calculate \( f(0) \)
Since \( f(0) = A + 1 \), we have:

\[ f(0) = A + 1. \]

### Step 2: Calculate \( \lim_{x \to 0} f(x) \)
For \( x \neq 0 \):

\[ f(x) = \frac{\sin 4x}{\log(1 + 3x)}. \]

We need to find:

\[ \lim_{x \to 0} \frac{\sin 4x}{\log(1 + 3x)}. \]

### Step 3: Apply L'Hôpital's Rule
As \( x \to 0 \), both the numerator and denominator approach 0. Thus, we can apply L'Hôpital's Rule:

\[ \lim_{x \to 0} \frac{\sin 4x}{\log(1 + 3x)} = \lim_{x \to 0} \frac{4 \cos 4x}{\frac{3}{1 + 3x}}. \]

### Step 4: Simplify the limit
Now, substituting \( x = 0 \):

\[ = \lim_{x \to 0} \frac{4 \cos 4x (1 + 3x)}{3} = \frac{4 \cdot 1 \cdot 1}{3} = \frac{4}{3}. \]

### Step 5: Set the limit equal to \( f(0) \)
For continuity at \( x = 0 \):

\[ \lim_{x \to 0} f(x) = f(0) \implies \frac{4}{3} = A + 1. \]

### Step 6: Solve for \( A \)
Rearranging gives:

\[ A = \frac{4}{3} - 1 = \frac{4}{3} - \frac{3}{3} = \frac{1}{3}. \]

### Conclusion
Thus, the value of \( A \) for \( f \) to be continuous at \( x = 0 \) is

\[ \boxed{\frac{1}{3}}. \]