If `f(x) = {{:((36^(x) - 9^(x) -4^(x)+1)/(sqrt(2)- sqrt(1+ cos x)), x ne 0),(k, x =0):}`, is continuous at x = 0, then k equals

Options

A. 16log2 log 3
B. `16sqrt(2)` log 6
C. `16sqrt(2)` log 2 log 3
D. none of these

Answer

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

\[ f(x) = \frac{36^x - 9^x - 4^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}}, \quad x \neq 0 \]

is continuous at \( x = 0 \), we need to ensure that

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

### Step 1: Calculate \( f(0) \)

Since \( f(0) = k \), we have:

\[ f(0) = k. \]

### Step 2: Calculate the limit \( \lim_{x \to 0} f(x) \)

We need to evaluate the limit:

\[ \lim_{x \to 0} \frac{36^x - 9^x - 4^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}}. \]

### Step 3: Simplify the numerator

Using the exponential property \( a^x = e^{x \ln a} \), we can rewrite the terms in the numerator:

\[ 36^x = e^{x \ln 36}, \quad 9^x = e^{x \ln 9}, \quad 4^x = e^{x \ln 4}. \]

As \( x \to 0 \), we can use the Taylor expansion:

\[ e^{x \ln a} \approx 1 + x \ln a \quad \text{for small } x. \]

Thus, we have:

\[ 36^x - 9^x - 4^x + 1 \approx (1 + x \ln 36) - (1 + x \ln 9) - (1 + x \ln 4) + 1. \]

This simplifies to:

\[ x \ln 36 - x \ln 9 - x \ln 4 = x (\ln 36 - \ln 9 - \ln 4). \]

### Step 4: Calculate \( \ln 36 - \ln 9 - \ln 4 \)

Using properties of logarithms:

\[ \ln 36 = \ln(9 \cdot 4) = \ln 9 + \ln 4. \]

Thus,

\[ \ln 36 - \ln 9 - \ln 4 = 0. \]

This means the numerator approaches \( 0 \) as \( x \to 0 \).

### Step 5: Simplify the denominator

For the denominator, we can use the identity \( 1 + \cos x = 2 \cos^2\left(\frac{x}{2}\right) \):

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

Thus,

\[ \sqrt{2} - \sqrt{1 + \cos x} = \sqrt{2} - \sqrt{2} \cos\left(\frac{x}{2}\right) = \sqrt{2}(1 - \cos\left(\frac{x}{2}\right)). \]

Using the limit \( 1 - \cos\left(\frac{x}{2}\right) \approx \frac{x^2}{8} \) as \( x \to 0 \):

\[ \sqrt{2}(1 - \cos\left(\frac{x}{2}\right)) \approx \sqrt{2} \cdot \frac{x^2}{8} = \frac{\sqrt{2}}{8} x^2. \]

### Step 6: Combine results

Now we have:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{x (\ln 36 - \ln 9 - \ln 4)}{\frac{\sqrt{2}}{8} x^2} = \lim_{x \to 0} \frac{0}{\frac{\sqrt{2}}{8} x^2} = 0. \]

### Step 7: Set the limit equal to \( k \)

Since \( \lim_{x \to 0} f(x) = 0 \), we have:

\[ k = 0. \]

### Conclusion

Thus, the value of \( k \) for which \( f(x) \) is continuous at \( x = 0 \) is:

\[ \boxed{0}. \]