The range of the function `y=(x^2)/(1+x^4)` is :

Options

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

Answer

To find the range of the function \( y = \frac{x^2}{1 + x^4} \), we can follow these steps:

### Step 1: Rewrite the function
We start with the function:
\[ y = \frac{x^2}{1 + x^4} \]
We can cross-multiply to rearrange the equation:
\[ y(1 + x^4) = x^2 \]
This simplifies to:
\[ yx^4 + y = x^2 \]

### Step 2: Rearrange into a standard form
Rearranging gives us a quadratic equation in terms of \( x^2 \):
\[ yx^4 - x^2 + y = 0 \]
Let \( z = x^2 \). Then the equation becomes:
\[ yz^2 - z + y = 0 \]

### Step 3: Identify coefficients for the quadratic
This is a standard quadratic equation of the form \( Az^2 + Bz + C = 0 \), where:
- \( A = y \)
- \( B = -1 \)
- \( C = y \)

### Step 4: Use the discriminant
For the quadratic to have real solutions, the discriminant must be non-negative:
\[ D = B^2 - 4AC \]
Substituting in our values:
\[ D = (-1)^2 - 4(y)(y) = 1 - 4y^2 \]
We require:
\[ 1 - 4y^2 \geq 0 \]

### Step 5: Solve the inequality
Rearranging the inequality gives:
\[ 1 \geq 4y^2 \]
Dividing both sides by 4:
\[ \frac{1}{4} \geq y^2 \]
Taking the square root:
\[ -\frac{1}{2} \leq y \leq \frac{1}{2} \]

### Step 6: Consider the function's behavior
Since \( y = \frac{x^2}{1 + x^4} \) is always non-negative (as both \( x^2 \) and \( 1 + x^4 \) are non-negative), we restrict our range to:
\[ 0 \leq y \leq \frac{1}{2} \]

### Final Step: Conclusion
Thus, the range of the function \( y = \frac{x^2}{1 + x^4} \) is:
\[ \text{Range} = [0, \frac{1}{2}] \]