Let `f(x) = {{:((2^(x)+2^(3-x) - 6)/(sqrt(2^(-x))-2^(1-x))",","if",x gt 2),((x^(2) - 4)/(x - sqrt(3x - 2))",","if",x lt 2):}`,then

Options

A. (a)`f(2) = 8 rArr f` is continuous at x = 2
B. (b)`f(2) = 16 rArr f` is continuous at x = 2
C. (c)`f(2^(-)) ne f(2^(+)) rArr f` is discontinuous
D. (d)f has a removable discontinuity at x = 2

Answer

To solve the given problem, we need to analyze the function \( f(x) \) defined piecewise and determine its continuity at \( x = 2 \). The function is defined as follows:

\[ f(x) = \begin{cases} \frac{2^x + 2^{3-x} - 6}{\sqrt{2^{-x}} - 2^{1-x}} & \text{if } x > 2 \\ \frac{x^2 - 4}{x - \sqrt{3x - 2}} & \text{if } x < 2 \end{cases} \]

### Step 1: Calculate the Left-Hand Limit as \( x \) approaches 2
We need to find the left-hand limit \( \lim_{x \to 2^-} f(x) \).

For \( x < 2 \):
\[ f(x) = \frac{x^2 - 4}{x - \sqrt{3x - 2}} \]

Substituting \( x = 2 \):
\[ f(2) = \frac{2^2 - 4}{2 - \sqrt{3(2) - 2}} = \frac{0}{0} \]
This is an indeterminate form, so we will simplify it.

### Step 2: Simplify the Left-Hand Limit
We can factor the numerator:
\[ x^2 - 4 = (x - 2)(x + 2) \]

Now we rewrite the limit:
\[ \lim_{x \to 2^-} \frac{(x - 2)(x + 2)}{x - \sqrt{3x - 2}} \]

Next, we rationalize the denominator:
\[ x - \sqrt{3x - 2} = \frac{(x - \sqrt{3x - 2})(x + \sqrt{3x - 2})}{x + \sqrt{3x - 2}} = \frac{x^2 - (3x - 2)}{x + \sqrt{3x - 2}} = \frac{-2x + 2}{x + \sqrt{3x - 2}} = \frac{2(1 - x)}{x + \sqrt{3x - 2}} \]

Substituting this back into the limit:
\[ \lim_{x \to 2^-} \frac{(x - 2)(x + 2)}{\frac{2(1 - x)}{x + \sqrt{3x - 2}}} \]

This simplifies to:
\[ \lim_{x \to 2^-} \frac{(x - 2)(x + 2)(x + \sqrt{3x - 2})}{2(1 - x)} \]

### Step 3: Evaluate the Limit
As \( x \to 2 \):
- The term \( (x - 2) \) approaches \( 0 \).
- The term \( (1 - x) \) also approaches \( 0 \).

Thus, we can cancel \( (x - 2) \) and \( (1 - x) \) (which is \( -(x - 2) \)):
\[ \lim_{x \to 2^-} \frac{-(x + 2)(x + \sqrt{3x - 2})}{2} \]

Substituting \( x = 2 \):
\[ = \frac{-(2 + 2)(2 + \sqrt{4 - 2})}{2} = \frac{-4 \cdot 3}{2} = -6 \]

### Step 4: Calculate the Right-Hand Limit as \( x \) approaches 2
Now, we find the right-hand limit \( \lim_{x \to 2^+} f(x) \).

For \( x > 2 \):
\[ f(x) = \frac{2^x + 2^{3-x} - 6}{\sqrt{2^{-x}} - 2^{1-x}} \]

Substituting \( x = 2 \):
\[ f(2) = \frac{2^2 + 2^{3-2} - 6}{\sqrt{2^{-2}} - 2^{1-2}} = \frac{4 + 2 - 6}{\sqrt{\frac{1}{4}} - \frac{1}{2}} = \frac{0}{0} \]
This is again an indeterminate form, so we apply L'Hôpital's Rule.

### Step 5: Apply L'Hôpital's Rule
Differentiate the numerator and denominator:
- Numerator: \( \frac{d}{dx}(2^x + 2^{3-x} - 6) = 2^x \ln(2) - 2^{3-x} \ln(2) \)
- Denominator: \( \frac{d}{dx}(\sqrt{2^{-x}} - 2^{1-x}) = -\frac{1}{2} \cdot 2^{-x} \ln(2) + 2^{1-x} \ln(2) \)

Now we compute the limit:
\[ \lim_{x \to 2^+} \frac{2^x \ln(2) - 2^{3-x} \ln(2)}{-\frac{1}{2} \cdot 2^{-x} \ln(2) + 2^{1-x} \ln(2)} \]

Evaluating this limit gives us a value of \( 8 \).

### Step 6: Conclusion
Since the left-hand limit \( \lim_{x \to 2^-} f(x) = -6 \) and the right-hand limit \( \lim_{x \to 2^+} f(x) = 8 \) are not equal, the function \( f(x) \) is discontinuous at \( x = 2 \).

### Final Answer
The function \( f(x) \) is discontinuous at \( x = 2 \).