If f(x) is a non-zero polynomial of degree four, having loca extreme points at `x=-1,0,1` then the set `S={x in R:f(x)=f(0)}` contains exactly

Options

A. four rational numbers
B. two irrational and two rational numbers
C. two irrational and one rational number
D. four irrational numbers

Explanation

To solve the problem, we need to analyze the given polynomial function \( f(x) \) of degree four, which has local extreme points at \( x = -1, 0, 1 \).

### Step-by-step Solution:

1. **Understanding the Polynomial and Its Derivative**:
Since \( f(x) \) is a polynomial of degree four and has local extrema at \( x = -1, 0, 1 \), we know that the derivative \( f'(x) \) must have roots at these points. Therefore, we can express the derivative as:
\[ f'(x) = k(x + 1)(x)(x - 1) = k(x^3 - x) \]
where \( k \) is a non-zero constant.

2. **Integrating to Find \( f(x) \)**:
To find \( f(x) \), we integrate \( f'(x) \):
\[ f(x) = \int f'(x) \, dx = \int k(x^3 - x) \, dx = k\left(\frac{x^4}{4} - \frac{x^2}{2}\right) + C \]
where \( C \) is a constant of integration.

3. **Setting Up the Condition**:
We need to find the set \( S = \{ x \in \mathbb{R} : f(x) = f(0) \} \). First, we calculate \( f(0) \):
\[ f(0) = k\left(\frac{0^4}{4} - \frac{0^2}{2}\right) + C = C \]
Thus, we need to solve the equation:
\[ f(x) = C \]
This leads to:
\[ k\left(\frac{x^4}{4} - \frac{x^2}{2}\right) + C = C \]
Simplifying this gives:
\[ k\left(\frac{x^4}{4} - \frac{x^2}{2}\right) = 0 \]

4. **Factoring the Equation**:
Since \( k \neq 0 \), we can divide both sides by \( k \):
\[ \frac{x^4}{4} - \frac{x^2}{2} = 0 \]
Multiplying through by 4 to eliminate the fraction:
\[ x^4 - 2x^2 = 0 \]
Factoring out \( x^2 \):
\[ x^2(x^2 - 2) = 0 \]

5. **Finding the Roots**:
Setting each factor to zero gives:
\[ x^2 = 0 \quad \text{or} \quad x^2 - 2 = 0 \]
From \( x^2 = 0 \), we have:
\[ x = 0 \]
From \( x^2 - 2 = 0 \), we have:
\[ x = \pm \sqrt{2} \]

6. **Identifying the Set \( S \)**:
The solutions we found are:
\[ x = 0, \quad x = \sqrt{2}, \quad x = -\sqrt{2} \]
Therefore, the set \( S \) contains the elements \( \{ 0, \sqrt{2}, -\sqrt{2} \} \).

7. **Counting the Elements**:
The set \( S \) contains:
- 1 rational number: \( 0 \)
- 2 irrational numbers: \( \sqrt{2} \) and \( -\sqrt{2} \)

Thus, the set \( S \) contains exactly **two irrational numbers and one rational number**.

### Final Answer:
The answer is that the set \( S \) contains exactly **two irrational and one rational number**.