If `x in R` and `alpha = (x^(2))/((1+x^(4)))`, thens

Options

A. `0 le alpha le 2`
B. `0 le alpha le 1`
C. `0 le alpha le (1)/(4)`
D. `0 le alpha le (1)/(2)`

Answer

To find the range of the expression \(\alpha = \frac{x^2}{1 + x^4}\) where \(x\) is a real number, we will use the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-Step Solution:

1. **Identify the expression**: We have \(\alpha = \frac{x^2}{1 + x^4}\).

2. **Apply the AM-GM Inequality**: According to the AM-GM inequality, for any non-negative real numbers \(a\) and \(b\):
\[ \frac{a + b}{2} \geq \sqrt{ab} \]
We will let \(a = 1\) and \(b = x^4\).

3. **Set up the inequality**:
\[ \frac{1 + x^4}{2} \geq \sqrt{1 \cdot x^4} \]
This simplifies to:
\[ \frac{1 + x^4}{2} \geq x^2 \]

4. **Multiply both sides by 2** (since 2 is positive, the inequality remains the same):
\[ 1 + x^4 \geq 2x^2 \]

5. **Rearrange the inequality**:
\[ x^4 - 2x^2 + 1 \geq 0 \]

6. **Factor the left-hand side**:
\[ (x^2 - 1)^2 \geq 0 \]
This inequality holds for all \(x\) since a square is always non-negative.

7. **Find the maximum value of \(\alpha\)**: Since \((x^2 - 1)^2 \geq 0\), we can conclude that:
\[ 1 + x^4 \geq 2x^2 \implies \frac{x^2}{1 + x^4} \leq \frac{1}{2} \]
Thus, \(\alpha \leq \frac{1}{2}\).

8. **Determine the minimum value of \(\alpha\)**: Since \(x^2\) is always non-negative and \(1 + x^4\) is always positive, \(\alpha\) is always non-negative:
\[ \alpha \geq 0 \]

9. **Combine the results**: Therefore, the range of \(\alpha\) is:
\[ 0 \leq \alpha \leq \frac{1}{2} \]

### Final Answer:
The range of \(\alpha\) is \([0, \frac{1}{2}]\).