Let `f(x)=({:((3|x|+4tanx)/(x),",",x ne0),(k,",",x=0):}` then f(x) is continuous at x = 0 for :

Options

A. k = 7
B. k = 1
C. No values of k
D. k = 2

Explanation

To determine the value of \( k \) for which the function

\[ f(x) = \begin{cases} \frac{3|x| + 4\tan x}{x} & \text{if } x \neq 0 \\ k & \text{if } x = 0 \end{cases} \]

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

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

For \( x \to 0^- \) (approaching 0 from the left), \( |x| = -x \). Therefore, we have:

\[ f(x) = \frac{3(-x) + 4\tan x}{x} = \frac{-3x + 4\tan x}{x} \]

This simplifies to:

\[ f(x) = -3 + \frac{4\tan x}{x} \]

Now, we need to find the limit:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \left(-3 + \frac{4\tan x}{x}\right) \]

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

\[ \lim_{x \to 0^-} f(x) = -3 + 4 \cdot 1 = 1 \]

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

For \( x \to 0^+ \) (approaching 0 from the right), \( |x| = x \). Thus, we have:

\[ f(x) = \frac{3x + 4\tan x}{x} = \frac{3x + 4\tan x}{x} \]

This simplifies to:

\[ f(x) = 3 + \frac{4\tan x}{x} \]

Now we find the limit:

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \left(3 + \frac{4\tan x}{x}\right) \]

Using the same limit property, we find:

\[ \lim_{x \to 0^+} f(x) = 3 + 4 \cdot 1 = 7 \]

### Step 3: Set the Limits Equal to the Function Value at \( x = 0 \)

For \( f(x) \) to be continuous at \( x = 0 \), we need:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \]

This gives us:

\[ 1 = 7 = k \]

Since \( 1 \neq 7 \), there is no value of \( k \) that can make the function continuous at \( x = 0 \).

### Conclusion

The function \( f(x) \) is discontinuous at \( x = 0 \) for all values of \( k \).

---