If `a_0 = x,a_(n+1)= f(a_n)`, where `n = 0, 1, 2, ...,` then answer thefollowing questions. If `f (x) = msqrt(a-x^m),x lt0,m leq 2,m in N`,then

Options

A. `a_(n)=x, n=2k+1,` where k is an integer
B. `a_(n)=f(x) " if " n=2k,` where k is an integer
C. The inverse of `a_(n)` exists for any value of n and m
D. None of these

Explanation

To solve the problem step by step, let's break down the given information and analyze it systematically.

### Given:
- \( a_0 = x \)
- \( a_{n+1} = f(a_n) \)
- \( f(x) = \sqrt{a - x^m} \) where \( x < 0 \), \( m \leq 2 \), and \( m \in \mathbb{N} \)

### Step 1: Calculate \( a_1 \)
Using the definition of the sequence:
\[ a_1 = f(a_0) = f(x) = \sqrt{a - x^m} \]

### Step 2: Calculate \( a_2 \)
Now we calculate \( a_2 \):
\[ a_2 = f(a_1) = f(\sqrt{a - x^m}) = \sqrt{a - (\sqrt{a - x^m})^m} \]
Since \( (\sqrt{a - x^m})^m = (a - x^m)^{m/2} \):
\[ a_2 = \sqrt{a - (a - x^m)^{m/2}} \]

### Step 3: Calculate \( a_3 \)
Next, we find \( a_3 \):
\[ a_3 = f(a_2) = f\left(\sqrt{a - (a - x^m)^{m/2}}\right) \]
This will be more complex, but we can denote it as:
\[ a_3 = \sqrt{a - \left(\sqrt{a - (a - x^m)^{m/2}}\right)^m} \]

### Step 4: Identify the pattern
From the calculations above, we can see that each term \( a_n \) is derived from the previous term using the function \( f \). The values will depend on \( x \) and \( m \).

### Step 5: Analyze the conditions
1. **For odd \( n \)**: We can observe that \( a_1, a_3, a_5, \ldots \) will be derived from \( f(x) \) and will likely depend on the initial value \( x \).
2. **For even \( n \)**: The terms \( a_2, a_4, a_6, \ldots \) will also depend on the previous terms but will follow a different pattern.

### Conclusion
From the calculations and the observations:
- \( a_n = x \) when \( n \) is of the form \( 2k + 2 \) (even).
- The inverse of \( a_n \) exists only under certain conditions, which depend on the nature of the function \( f \).

### Final Answer
The correct conclusion is that none of the options provided are true, leading us to select option D: "None of these".