Let f(x)= `{{:(|x| cos (1//x)+9x^2, xne 0),(k, x=0):}`
then f is continuous if the value k is

Options

A. 9
B. 6
C. 0
D. none of these

Answer

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

\[ f(x) = \begin{cases} |x| \cos\left(\frac{1}{x}\right) + 9x^2 & \text{if } x \neq 0 \\ k & \text{if } x = 0 \end{cases} \]

is continuous at \( x = 0 \), we need to ensure that the limit of \( f(x) \) as \( x \) approaches 0 from both sides equals \( f(0) \).

### Step 1: Find the left-hand limit as \( x \) approaches 0
For \( x < 0 \), we have:

\[ f(x) = -x \cos\left(\frac{1}{x}\right) + 9x^2 \]

We need to calculate:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \left(-x \cos\left(\frac{1}{x}\right) + 9x^2\right) \]

As \( x \to 0^- \), \( -x \) approaches 0 and \( \cos\left(\frac{1}{x}\right) \) oscillates between -1 and 1. Therefore, the term \( -x \cos\left(\frac{1}{x}\right) \) will approach 0.

The term \( 9x^2 \) also approaches 0. Thus:

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

### Step 2: Find the right-hand limit as \( x \) approaches 0
For \( x > 0 \), we have:

\[ f(x) = x \cos\left(\frac{1}{x}\right) + 9x^2 \]

We need to calculate:

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \left(x \cos\left(\frac{1}{x}\right) + 9x^2\right) \]

As \( x \to 0^+ \), \( x \) approaches 0 and \( \cos\left(\frac{1}{x}\right) \) oscillates between -1 and 1. Therefore, the term \( x \cos\left(\frac{1}{x}\right) \) will approach 0.

The term \( 9x^2 \) also approaches 0. Thus:

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

### Step 3: Set the limits equal to \( f(0) \)
For \( f(x) \) to be continuous at \( x = 0 \), we need:

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

From our calculations, we have:

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

Thus, we need:

\[ f(0) = k = 0 \]

### Conclusion
The value of \( k \) for which \( f(x) \) is continuous at \( x = 0 \) is:

\[ \boxed{0} \]