If a real polynomial of degree n satisfies the relation `f(x)=f(x)f''(x)"for all "x in R" Then "f"R to R`

Options

A. an onto function
B. an into function
C. always a one function
D. always a many one function.

Explanation

To solve the problem, we need to analyze the given relation \( f(x) = f'(x) f''(x) \) for a polynomial \( f(x) \) of degree \( n \).

### Step-by-Step Solution:

1. **Understanding the Degrees of the Polynomials**:
- Let the degree of the polynomial \( f(x) \) be \( n \).
- The derivative \( f'(x) \) will have a degree of \( n-1 \).
- The second derivative \( f''(x) \) will have a degree of \( n-2 \).

2. **Setting Up the Degree Equation**:
- The left-hand side \( f(x) \) has a degree of \( n \).
- The right-hand side \( f'(x) f''(x) \) will have a degree of \( (n-1) + (n-2) = 2n - 3 \).

3. **Equating the Degrees**:
- Since the equation \( f(x) = f'(x) f''(x) \) must hold for all \( x \), we can equate the degrees:
\[ n = 2n - 3 \]

4. **Solving for \( n \)**:
- Rearranging the equation gives:
\[ 3 = n \]
- Thus, the degree \( n \) of the polynomial \( f(x) \) is 3.

5. **Determining the Nature of the Polynomial**:
- A polynomial of degree 3 is a cubic polynomial. Cubic polynomials can take all real values (i.e., they are onto functions).
- Therefore, \( f: \mathbb{R} \to \mathbb{R} \) is onto.

### Conclusion:
The polynomial \( f(x) \) is a cubic polynomial, and it is an onto function.

### Final Answer:
The function \( f: \mathbb{R} \to \mathbb{R} \) is onto.

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