Let `f(x)` be an non - zero polynomial of degree 4. Extremum points of `f(x)` are `0,-1,1`. If `f(k)=f(0)` then,

Options

A. k has one rational and two irrational roots
B. k has four rational roots
C. k has four irrational roots
D. k has three irrational roots.

Answer

To solve the problem, we need to find the values of \( k \) such that \( f(k) = f(0) \) for a polynomial \( f(x) \) of degree 4 with extremum points at \( 0, -1, \) and \( 1 \).

### Step-by-Step Solution:

1. **Understanding the Extremum Points**:
Since the extremum points of \( f(x) \) are \( 0, -1, \) and \( 1 \), we know that the derivative \( f'(x) \) will have roots at these points. Therefore, we can express \( f'(x) \) as:
\[ f'(x) = \lambda (x)(x + 1)(x - 1) \]
where \( \lambda \) is a non-zero constant.

2. **Finding \( f(x) \)**:
To find \( f(x) \), we need to integrate \( f'(x) \):
\[ f(x) = \int f'(x) \, dx = \int \lambda (x)(x + 1)(x - 1) \, dx \]
Expanding \( f'(x) \):
\[ f'(x) = \lambda (x)(x^2 - 1) = \lambda (x^3 - x) \]
Now, integrating:
\[ f(x) = \lambda \left( \frac{x^4}{4} - \frac{x^2}{2} \right) + C \]
where \( C \) is the constant of integration.

3. **Setting \( f(k) = f(0) \)**:
We need to find \( k \) such that \( f(k) = f(0) \):
\[ f(0) = \lambda \left( \frac{0^4}{4} - \frac{0^2}{2} \right) + C = C \]
Now, substituting \( k \) into \( f(x) \):
\[ f(k) = \lambda \left( \frac{k^4}{4} - \frac{k^2}{2} \right) + C \]
Setting \( f(k) = f(0) \):
\[ \lambda \left( \frac{k^4}{4} - \frac{k^2}{2} \right) + C = C \]
This simplifies to:
\[ \lambda \left( \frac{k^4}{4} - \frac{k^2}{2} \right) = 0 \]

4. **Solving for \( k \)**:
Since \( \lambda \neq 0 \), we can divide both sides by \( \lambda \):
\[ \frac{k^4}{4} - \frac{k^2}{2} = 0 \]
Factoring out \( k^2 \):
\[ k^2 \left( \frac{k^2}{4} - \frac{1}{2} \right) = 0 \]
This gives us two cases:
- \( k^2 = 0 \) which implies \( k = 0 \).
- \( \frac{k^2}{4} - \frac{1}{2} = 0 \) which simplifies to \( k^2 = 2 \) giving \( k = \pm \sqrt{2} \).

5. **Final Values of \( k \)**:
The values of \( k \) are:
\[ k = 0, \sqrt{2}, -\sqrt{2} \]
Thus, we have:
- Two rational roots: \( 0 \)
- Two irrational roots: \( \sqrt{2}, -\sqrt{2} \)

### Conclusion:
The final answer indicates that \( k \) has 1 rational root and 2 irrational roots. Therefore, the correct option is:
- **Option 1**: \( k \) has 1 rational and 2 irrational roots.