Let `f:NrarrN ` in such that `f(n+1)gtf(f(n))` for all `n in N` then

Options

A. (a) `f(n)=n^(2)-n+1`
B. (b) `f(n)=n-1`
C. (c) `f(n)=n^(2)+1`
D. (d) none of the above

Answer

To solve the problem, we need to determine a function \( f: \mathbb{N} \to \mathbb{N} \) such that \( f(n+1) > f(f(n)) \) for all \( n \in \mathbb{N} \). We will check the options provided to find the correct function.

### Step 1: Analyze the options
We will check each option one by one to see if they satisfy the condition \( f(n+1) > f(f(n)) \).

### Step 2: Check Option A: \( f(n) = n^2 - n + 1 \)

1. **Calculate \( f(n+1) \)**:
\[ f(n+1) = (n+1)^2 - (n+1) + 1 = n^2 + 2n + 1 - n - 1 + 1 = n^2 + n + 1 \]

2. **Calculate \( f(f(n)) \)**:
\[ f(n) = n^2 - n + 1 \]
Now, substituting \( f(n) \) into itself:
\[ f(f(n)) = f(n^2 - n + 1) = (n^2 - n + 1)^2 - (n^2 - n + 1) + 1 \]
Expanding this:
\[ = n^4 - 2n^3 + 3n^2 - 2n + 1 \]

3. **Compare \( f(n+1) \) and \( f(f(n)) \)**:
We need to check if \( n^2 + n + 1 > n^4 - 2n^3 + 3n^2 - 2n + 1 \).
This inequality does not hold for large \( n \), so Option A is incorrect.

### Step 3: Check Option B: \( f(n) = n - 1 \)

1. **Calculate \( f(n+1) \)**:
\[ f(n+1) = (n+1) - 1 = n \]

2. **Calculate \( f(f(n)) \)**:
\[ f(n) = n - 1 \implies f(f(n)) = f(n - 1) = (n - 1) - 1 = n - 2 \]

3. **Compare \( f(n+1) \) and \( f(f(n)) \)**:
We need to check if \( n > n - 2 \), which is always true for \( n \in \mathbb{N} \). Thus, Option B is correct.

### Step 4: Check Option C: \( f(n) = n^2 + 1 \)

1. **Calculate \( f(n+1) \)**:
\[ f(n+1) = (n+1)^2 + 1 = n^2 + 2n + 1 + 1 = n^2 + 2n + 2 \]

2. **Calculate \( f(f(n)) \)**:
\[ f(n) = n^2 + 1 \implies f(f(n)) = f(n^2 + 1) = (n^2 + 1)^2 + 1 = n^4 + 2n^2 + 1 + 1 = n^4 + 2n^2 + 2 \]

3. **Compare \( f(n+1) \) and \( f(f(n)) \)**:
We need to check if \( n^2 + 2n + 2 > n^4 + 2n^2 + 2 \).
This inequality does not hold for large \( n \), so Option C is incorrect.

### Conclusion
The only option that satisfies the condition \( f(n+1) > f(f(n)) \) for all \( n \in \mathbb{N} \) is:
**Option B: \( f(n) = n - 1 \)**.