If a and b are chosen randomly from the set consisting of number 1, 2, 3, 4, 5, 6 with replacement. Then the probability that `lim_(x to 0)[(a^(x)+b^(x))//2]^(2//x)=6` is

Options

A. `1//3`
B. `1//4`
C. `1//9`
D. `2//9`

Explanation

To solve the problem, we need to find the probability that the limit \( \lim_{x \to 0} \left( \frac{a^x + b^x}{2} \right)^{\frac{2}{x}} = 6 \) when \( a \) and \( b \) are chosen randomly from the set \( \{1, 2, 3, 4, 5, 6\} \) with replacement.

### Step-by-Step Solution:

1. **Understanding the Limit**:
We start with the expression:
\[ y = \left( \frac{a^x + b^x}{2} \right)^{\frac{2}{x}} \]
As \( x \to 0 \), both \( a^x \) and \( b^x \) approach 1, leading to an indeterminate form \( 1^{\infty} \).

2. **Rewriting the Limit**:
To resolve the indeterminate form, we can use the exponential limit:
\[ y = e^{\lim_{x \to 0} \frac{2}{x} \left( \frac{a^x + b^x}{2} - 1 \right)} \]

3. **Simplifying the Expression**:
We can simplify the limit:
\[ \lim_{x \to 0} \frac{a^x + b^x - 2}{x} \]
This is a \( \frac{0}{0} \) form, so we can apply L'Hôpital's Rule.

4. **Applying L'Hôpital's Rule**:
Differentiate the numerator and denominator:
\[ \text{Numerator: } \frac{d}{dx}(a^x + b^x - 2) = a^x \ln a + b^x \ln b \]
\[ \text{Denominator: } \frac{d}{dx}(x) = 1 \]
Thus, we have:
\[ \lim_{x \to 0} (a^x \ln a + b^x \ln b) = \ln a + \ln b \]

5. **Final Limit Calculation**:
Therefore, substituting back, we find:
\[ y = e^{2(\ln a + \ln b)} = e^{\ln(a^2 b^2)} = (ab)^2 \]
We need this to equal 6:
\[ (ab)^2 = 6 \implies ab = \sqrt{6} \]

6. **Finding the Probability**:
Now, we need to find the probability that \( ab = 6 \) when \( a \) and \( b \) are chosen from \( \{1, 2, 3, 4, 5, 6\} \).

The pairs \( (a, b) \) that satisfy \( ab = 6 \) are:
- \( (1, 6) \)
- \( (2, 3) \)
- \( (3, 2) \)
- \( (6, 1) \)

Thus, there are 4 favorable outcomes.

7. **Total Outcomes**:
Since \( a \) and \( b \) can each take any of the 6 values, the total number of outcomes is:
\[ 6 \times 6 = 36 \]

8. **Calculating the Probability**:
The probability \( P \) that \( ab = 6 \) is given by:
\[ P = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{4}{36} = \frac{1}{9} \]

### Final Answer:
The probability that \( \lim_{x \to 0} \left( \frac{a^x + b^x}{2} \right)^{\frac{2}{x}} = 6 \) is \( \frac{1}{9} \).