If `f(x)={{:((5|x|+4tanx)/(x),xne0),(k,x=0):}`, then `f(x)` is continuous at x = 0 for

Options

A. k = 9
B. `k=-1`
C. no value of k
D. `k=2`

Answer

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

\[ f(x) = \begin{cases} \frac{5|x| + 4\tan x}{x} & \text{if } x \neq 0 \\ k & \text{if } x = 0 \end{cases} \]

is continuous at \( x = 0 \), we need to check the left-hand limit (LHL) and right-hand limit (RHL) as \( x \) approaches 0, and ensure they are equal to \( f(0) \).

### Step 1: Calculate the Left-Hand Limit (LHL)

The left-hand limit as \( x \) approaches 0 is given by:

\[ \text{LHL} = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{5|x| + 4\tan x}{x} \]

Since \( x \) is approaching 0 from the left, \( |x| = -x \). Thus, we have:

\[ \text{LHL} = \lim_{x \to 0^-} \frac{5(-x) + 4\tan x}{x} = \lim_{x \to 0^-} \frac{-5x + 4\tan x}{x} \]

This simplifies to:

\[ \text{LHL} = \lim_{x \to 0^-} \left(-5 + \frac{4\tan x}{x}\right) \]

Using the fact that \( \lim_{x \to 0} \frac{\tan x}{x} = 1 \):

\[ \text{LHL} = -5 + 4 \cdot 1 = -5 \]

### Step 2: Calculate the Right-Hand Limit (RHL)

The right-hand limit as \( x \) approaches 0 is given by:

\[ \text{RHL} = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{5|x| + 4\tan x}{x} \]

Since \( x \) is approaching 0 from the right, \( |x| = x \). Thus, we have:

\[ \text{RHL} = \lim_{x \to 0^+} \frac{5x + 4\tan x}{x} = \lim_{x \to 0^+} \left(5 + \frac{4\tan x}{x}\right) \]

Again, using the fact that \( \lim_{x \to 0} \frac{\tan x}{x} = 1 \):

\[ \text{RHL} = 5 + 4 \cdot 1 = 9 \]

### Step 3: Set the Limits Equal to Each Other

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

\[ \text{LHL} = \text{RHL} = f(0) \]

Thus, we require:

\[ -5 = 9 \quad \text{(not possible)} \]

This indicates that the left-hand limit and right-hand limit are not equal, meaning \( f(x) \) cannot be continuous at \( x = 0 \) for any value of \( k \).

### Conclusion

The function \( f(x) \) is not continuous at \( x = 0 \) for any value of \( k \).