If `x in R ,` the numbers `2^(1+x)+2^(1-x), b//2,36^x+36^(-x)` form an A.P., then `b` may lie in the interval `[12 ,oo)` b. `[6,oo)` c. `(-oo,6]` d. `[6, 12]`

Options

A. `[12, oo)`
B. `[6, oo)`
C. `(-oo, 6]`
D. `[6, 12]`

Explanation

To solve the problem, we need to determine the value of \( b \) such that the numbers \( 2^{1+x} + 2^{1-x} \), \( \frac{b}{2} \), and \( 36^x + 36^{-x} \) form an arithmetic progression (A.P.).

### Step-by-step Solution:

1. **Identify the terms of the A.P.**:
Let:
- \( a = 2^{1+x} + 2^{1-x} \)
- \( b = \frac{b}{2} \)
- \( c = 36^x + 36^{-x} \)

2. **Use the property of A.P.**:
For three numbers to be in A.P., the condition is:
\[ 2b = a + c \]
Substituting the values of \( a \) and \( c \):
\[ 2 \cdot \frac{b}{2} = (2^{1+x} + 2^{1-x}) + (36^x + 36^{-x}) \]
Simplifying this gives:
\[ b = 2^{1+x} + 2^{1-x} + 36^x + 36^{-x} \]

3. **Simplify \( a \)**:
We can rewrite \( a \):
\[ a = 2^{1+x} + 2^{1-x} = 2 \cdot (2^x + 2^{-x}) = 2 \cdot 2 \cosh(x \ln 2) = 4 \cosh(x \ln 2) \]

4. **Simplify \( c \)**:
Similarly, for \( c \):
\[ c = 36^x + 36^{-x} = 2 \cdot (6^x + 6^{-x}) = 2 \cdot 2 \cosh(x \ln 6) = 4 \cosh(x \ln 6) \]

5. **Combine the results**:
Now substituting back, we have:
\[ b = 4 \cosh(x \ln 2) + 4 \cosh(x \ln 6) \]
\[ b = 4 (\cosh(x \ln 2) + \cosh(x \ln 6)) \]

6. **Determine the minimum value of \( b \)**:
The minimum value of \( \cosh(t) \) is 1, which occurs when \( t = 0 \). Therefore:
\[ \cosh(x \ln 2) \geq 1 \quad \text{and} \quad \cosh(x \ln 6) \geq 1 \]
Thus:
\[ b \geq 4(1 + 1) = 8 \]

7. **Conclusion**:
Therefore, \( b \) must be at least 8. Hence, the possible intervals for \( b \) are:
\[ b \in [8, \infty) \]

### Final Answer:
The correct option is that \( b \) may lie in the interval \([6, \infty)\) because \( [8, \infty) \) is a subset of \([6, \infty)\).