If `x in R " and "alpha=(x^(2))/(1+x^(4))`, then `alpha` lies in the interval

Options

A. [0,2]
B. [0,1]
C. `[0,1//2]`
D. `[1//2,1]`

Explanation

To solve the problem, we need to determine the range of the variable α, defined as:

\[ \alpha = \frac{x^2}{1 + x^4} \]

where \( x \in \mathbb{R} \).

### Step 1: Analyze the expression for α

We start with the expression for α:

\[ \alpha = \frac{x^2}{1 + x^4} \]

### Step 2: Apply the AM-GM Inequality

To find the maximum value of α, we can apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality. We consider the numbers 1 and \( x^4 \):

\[ \frac{1 + x^4}{2} \geq \sqrt{1 \cdot x^4} \]

This simplifies to:

\[ \frac{1 + x^4}{2} \geq x^2 \]

### Step 3: Rearranging the inequality

Multiplying both sides by 2 gives:

\[ 1 + x^4 \geq 2x^2 \]

Rearranging this, we have:

\[ x^4 - 2x^2 + 1 \geq 0 \]

### Step 4: Factor the quadratic

Let \( y = x^2 \). Then the inequality becomes:

\[ y^2 - 2y + 1 \geq 0 \]

This can be factored as:

\[ (y - 1)^2 \geq 0 \]

This inequality holds for all real numbers \( y \), and it is equal to zero when \( y = 1 \) (i.e., \( x^2 = 1 \) or \( x = \pm 1 \)).

### Step 5: Finding the maximum value of α

Since \( (y - 1)^2 \geq 0 \), we can conclude that:

\[ 1 + x^4 \geq 2x^2 \]

Thus, taking the reciprocal (and noting that both sides are positive), we have:

\[ \frac{1}{2} \geq \frac{x^2}{1 + x^4} \]

This means:

\[ \alpha \leq \frac{1}{2} \]

### Step 6: Establishing the lower bound of α

Next, we note that since \( x^2 \geq 0 \) and \( 1 + x^4 > 0 \) for all \( x \in \mathbb{R} \), we have:

\[ \alpha \geq 0 \]

### Conclusion

Combining both results, we find that:

\[ 0 \leq \alpha \leq \frac{1}{2} \]

Thus, α lies in the interval:

\[ \alpha \in [0, \frac{1}{2}] \]

### Final Answer

The final answer is:

\[ \alpha \text{ lies in the interval } [0, \frac{1}{2}] \]