If `f(x)` is continuous at `x=0`, where `f(x)={:{((sin 4x)/(5x)+a", for " x gt 0),(x+4-b", for " x lt 0),(1", for " x=0):}`, then

Options

A. `a=1/5, b=3`
B. `a=(-1)/(5), b=-3`
C. `a=1/5, b=-3`
D. `a=( -1)/(5), b=3`

Answer

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

Given the piecewise function:
\[ f(x) = \begin{cases} \frac{\sin(4x)}{5x} + a & \text{for } x > 0 \\ x + 4 - b & \text{for } x < 0 \\ 1 & \text{for } x = 0 \end{cases} \]

### Step 1: Calculate the Left-Hand Limit (LHL) as \( x \) approaches 0
For \( x < 0 \):
\[ \text{LHL} = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x + 4 - b) = 0 + 4 - b = 4 - b \]

### Step 2: Calculate the Right-Hand Limit (RHL) as \( x \) approaches 0
For \( x > 0 \):
\[ \text{RHL} = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \left(\frac{\sin(4x)}{5x} + a\right) \]
Using the limit property \( \lim_{x \to 0} \frac{\sin(kx)}{kx} = 1 \):
\[ = \frac{1}{5} \cdot 4 + a = \frac{4}{5} + a \]

### Step 3: Set the Left-Hand Limit equal to the Right-Hand Limit
For continuity at \( x = 0 \), we need:
\[ 4 - b = \frac{4}{5} + a \tag{1} \]

### Step 4: Set the Left-Hand Limit equal to the value of the function at \( x = 0 \)
We also know:
\[ f(0) = 1 \]
Thus, we set:
\[ 4 - b = 1 \tag{2} \]

### Step 5: Solve the equations
From equation (2):
\[ 4 - b = 1 \implies b = 4 - 1 = 3 \]

Substituting \( b = 3 \) into equation (1):
\[ 4 - 3 = \frac{4}{5} + a \implies 1 = \frac{4}{5} + a \]
Rearranging gives:
\[ a = 1 - \frac{4}{5} = \frac{1}{5} \]

### Final Values
Thus, the values of \( a \) and \( b \) are:
\[ a = \frac{1}{5}, \quad b = 3 \]

### Summary
The final answer is:
\[ \boxed{a = \frac{1}{5}, \quad b = 3} \]