Statement-1 : `f(x) = x^(4) - 3x^(2) + 4x-1` is many one into in `R rarr R`.
and
Statement-2 : If `f : R rarr R` is a polynomial of even degree it will neither be injective nor surjective.

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.
C. Statement -1 is False, Statement -2 is False
D. Statement -1 is False, Statement -2 is True

Answer

To determine the validity of the statements regarding the function \( f(x) = x^4 - 3x^2 + 4x - 1 \), we will analyze both statements step by step.

### Step 1: Analyze Statement 1
**Statement 1:** \( f(x) = x^4 - 3x^2 + 4x - 1 \) is many-one into \( \mathbb{R} \to \mathbb{R} \).

1. **Find the derivative of \( f(x) \)**:
\[ f'(x) = \frac{d}{dx}(x^4 - 3x^2 + 4x - 1) = 4x^3 - 6x + 4 \]

2. **Set the derivative to zero to find critical points**:
\[ 4x^3 - 6x + 4 = 0 \]
This is a cubic equation. To analyze the behavior of \( f'(x) \), we can find the roots of this equation.

3. **Use the Rational Root Theorem or numerical methods** to find the roots. For simplicity, let's check for possible rational roots:
- Testing \( x = -1 \):
\[ f'(-1) = 4(-1)^3 - 6(-1) + 4 = -4 + 6 + 4 = 6 \quad (\text{not a root}) \]
- Testing \( x = 0 \):
\[ f'(0) = 4(0)^3 - 6(0) + 4 = 4 \quad (\text{not a root}) \]
- Testing \( x = 1 \):
\[ f'(1) = 4(1)^3 - 6(1) + 4 = 4 - 6 + 4 = 2 \quad (\text{not a root}) \]
- Testing \( x = -2 \):
\[ f'(-2) = 4(-2)^3 - 6(-2) + 4 = -32 + 12 + 4 = -16 \quad (\text{not a root}) \]

Since we do not find simple rational roots, we can analyze the sign of \( f'(x) \) to determine the behavior of \( f(x) \).

4. **Analyze the sign of \( f'(x) \)**:
- As \( x \to -\infty \), \( f'(x) \to -\infty \) (since the leading term \( 4x^3 \) dominates).
- As \( x \to +\infty \), \( f'(x) \to +\infty \).
- Since \( f'(x) \) is a cubic polynomial, it can change signs, indicating that \( f(x) \) has at least one local maximum and one local minimum.

5. **Conclusion for Statement 1**:
Since \( f(x) \) has at least one local maximum and one local minimum, it is not one-one (injective), hence it is many-one.

### Step 2: Analyze Statement 2
**Statement 2:** If \( f: \mathbb{R} \to \mathbb{R} \) is a polynomial of even degree, it will neither be injective nor surjective.

1. **Understanding even degree polynomials**:
- An even degree polynomial \( f(x) \) can be expressed as \( a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \) where \( n \) is even.
- Such polynomials tend to have a "U" shape or an inverted "U" shape.

2. **Injectivity**:
- Since even degree polynomials can have local maxima and minima, they will not be injective (one-one). They can take the same value for different inputs.

3. **Surjectivity**:
- The range of an even degree polynomial is limited; it cannot cover all real numbers. For example, a polynomial that opens upwards will have a minimum value and will not reach negative infinity.

4. **Conclusion for Statement 2**:
Therefore, statement 2 is correct: even degree polynomials are neither injective nor surjective.

### Final Conclusion
Both statements are correct:
- Statement 1 is true: \( f(x) \) is many-one into \( \mathbb{R} \to \mathbb{R} \).
- Statement 2 is true: Even degree polynomials are neither injective nor surjective.