The value of `k + f(0) ` so that `f (x)={{:((sin (4k-1)x)/(3x)"," , x lt 0),((tan (4k +1)x )/(5x)"," , 0 lt x lt (pi)/(2)), ( 1"," , x =0):}` can bemade continous at `x=0` is:

Options

A. 1
B. 2
C. `5/4`
D. `0`

Answer

To determine the value of \( k + f(0) \) such that the function

\[ f(x) = \begin{cases} \frac{\sin((4k-1)x)}{3x} & \text{if } x < 0 \\ \frac{\tan((4k+1)x)}{5x} & \text{if } 0 < x < \frac{\pi}{2} \\ 1 & \text{if } x = 0 \end{cases} \]

is continuous at \( x = 0 \), we need to ensure that the left-hand limit, the value of the function at \( x = 0 \), and the right-hand limit are all equal.

### Step 1: Find the left-hand limit as \( x \to 0^- \)

The left-hand limit is given by:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sin((4k-1)x)}{3x} \]

Using the limit property \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \), we can rewrite this limit as:

\[ \lim_{x \to 0^-} \frac{\sin((4k-1)x)}{(4k-1)x} \cdot \frac{(4k-1)}{3} = \frac{4k-1}{3} \]

### Step 2: Find the right-hand limit as \( x \to 0^+ \)

The right-hand limit is given by:

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\tan((4k+1)x)}{5x} \]

Using the limit property \( \lim_{x \to 0} \frac{\tan x}{x} = 1 \), we can rewrite this limit as:

\[ \lim_{x \to 0^+} \frac{\tan((4k+1)x)}{(4k+1)x} \cdot \frac{(4k+1)}{5} = \frac{4k+1}{5} \]

### Step 3: Set the limits equal to \( f(0) \)

Since \( f(0) = 1 \), we set up the following equations:

1. Left-hand limit: \( \frac{4k-1}{3} = 1 \)
2. Right-hand limit: \( \frac{4k+1}{5} = 1 \)

### Step 4: Solve the equations

**From the left-hand limit:**

\[ \frac{4k-1}{3} = 1 \implies 4k - 1 = 3 \implies 4k = 4 \implies k = 1 \]

**From the right-hand limit:**

\[ \frac{4k+1}{5} = 1 \implies 4k + 1 = 5 \implies 4k = 4 \implies k = 1 \]

Both limits give us \( k = 1 \).

### Step 5: Calculate \( k + f(0) \)

Now we find \( k + f(0) \):

\[ k + f(0) = 1 + 1 = 2 \]

### Final Answer

Thus, the value of \( k + f(0) \) that makes the function continuous at \( x = 0 \) is:

\[ \boxed{2} \]