If f: R rightarrow R, f(x) is a differentiable bijective function , then which of the following may be true?

Options

A. `(f(x) - x) f''(x) lt 0 forall x in R`
B. `(f(x) - x) f''(x) gt 0 forall x in R`
C. `(f(x) - x) f''(x) gt 0 , then f(x) = f^(-1) (x) `has no solution
D. `(f(x) - x) f''(x) gt 0 , then f(x) = f^(-1) (x)` has at least a real solution

Answer

To solve the problem, we need to analyze the given conditions about the function \( f: \mathbb{R} \to \mathbb{R} \) which is differentiable and bijective. We will evaluate the options provided in the question step by step.

### Step 1: Understanding the Properties of \( f \)

Since \( f \) is a bijective function, it means that it is both injective (one-to-one) and surjective (onto). This implies that for every \( y \in \mathbb{R} \), there exists a unique \( x \in \mathbb{R} \) such that \( f(x) = y \).

### Step 2: Analyzing the Options

We will analyze each option provided in the question.

#### Option 1: \( f(x) - x \cdot f''(x) < 0 \) for all \( x \in \mathbb{R} \)

1. **Assumption**: Let \( f(x) - x > 0 \) and \( f''(x) < 0 \).
- If \( f(x) - x > 0 \), then \( f(x) > x \).
- If \( f''(x) < 0 \), it indicates that \( f'(x) \) is decreasing.

2. **Contradiction**: If \( f(x) > x \) and \( f'(x) < 1 \) (since \( f''(x) < 0 \)), it leads to a contradiction because \( f(x) \) cannot remain greater than \( x \) for all \( x \).

3. **Conclusion**: This option cannot hold true for all \( x \) in \( \mathbb{R} \).

#### Option 2: \( (f(x) - x) \cdot f''(x) > 0 \) for all \( x \in \mathbb{R} \)

1. **Case 1**: Assume \( f(x) - x > 0 \) and \( f''(x) > 0 \).
- Both terms are positive, hence their product is positive.

2. **Case 2**: Assume \( f(x) - x < 0 \) and \( f''(x) < 0 \).
- Both terms are negative, hence their product is also positive.

3. **Conclusion**: This option can hold true under the given conditions.

#### Option 3: If \( (f(x) - x) \cdot f'(x) > 0 \), then \( f(x) = f^{-1}(x) \) has no solution.

1. **Interpretation**: If \( f(x) - x > 0 \) and \( f'(x) > 0 \), it suggests that \( f(x) \) is increasing and above the line \( y = x \).

2. **Conclusion**: Since \( f(x) \) is increasing and does not intersect \( y = x \), this option is valid.

#### Option 4: If \( (f(x) - x) \cdot f''(x) > 0 \), then \( f(x) = f^{-1}(x) \) has at least one real solution.

1. **Contradiction**: If \( f(x) - x > 0 \) and \( f''(x) > 0 \), it implies that \( f(x) \) is strictly increasing and does not intersect \( y = x \).

2. **Conclusion**: This option cannot be true as it contradicts the properties of a bijective function.

### Final Conclusion

After analyzing all options, we conclude that:

- **Option 2** is true.
- **Option 3** is true.
- **Option 1** and **Option 4** are false.