If `f : [0, oo) rarr [0, oo)` and `f(x) = (x^(2))/(1+x^(4))`, then f is

Options

A. One - one and onto
B. One - one but not onto
C. Onto but not one-one
D. Neither one-one nor onto

Answer

To determine whether the function \( f(x) = \frac{x^2}{1 + x^4} \) is one-one, onto, both, or neither, we will analyze the function step by step.

### Step 1: Check if the function is one-one (injective)

A function is one-one if different inputs produce different outputs. To check this, we need to see if \( f(a) = f(b) \) implies \( a = b \).

1. Assume \( f(a) = f(b) \):
\[ \frac{a^2}{1 + a^4} = \frac{b^2}{1 + b^4} \]
Cross-multiplying gives:
\[ a^2(1 + b^4) = b^2(1 + a^4) \]
Expanding both sides:
\[ a^2 + a^2b^4 = b^2 + b^2a^4 \]
Rearranging:
\[ a^2 - b^2 = b^2a^4 - a^2b^4 \]
Factoring the right side:
\[ a^2 - b^2 = (b^2 - a^2)(a^2 + b^2) \]
This can be simplified to:
\[ (a^2 - b^2)(1 + a^2 + b^2) = 0 \]
Thus, either \( a^2 - b^2 = 0 \) or \( 1 + a^2 + b^2 = 0 \). The second equation cannot hold for non-negative \( a \) and \( b \). Therefore, \( a^2 - b^2 = 0 \) implies \( a = b \).

Since we found that \( f(a) = f(b) \) implies \( a = b \), the function \( f \) is one-one.

### Step 2: Check if the function is onto (surjective)

A function is onto if every element in the codomain has a pre-image in the domain.

1. Let \( y \) be an arbitrary element in the codomain \( [0, \infty) \). We need to find \( x \) such that:
\[ y = \frac{x^2}{1 + x^4} \]
Rearranging gives:
\[ y(1 + x^4) = x^2 \]
This simplifies to:
\[ y + yx^4 = x^2 \]
Rearranging:
\[ yx^4 - x^2 + y = 0 \]
This is a quadratic equation in terms of \( x^2 \):
\[ yx^4 - x^2 + y = 0 \]
Using the quadratic formula:
\[ x^2 = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(y)(y)}}{2y} = \frac{1 \pm \sqrt{1 - 4y^2}}{2y} \]
For \( x^2 \) to be non-negative, the discriminant must be non-negative:
\[ 1 - 4y^2 \geq 0 \implies 4y^2 \leq 1 \implies y^2 \leq \frac{1}{4} \implies |y| \leq \frac{1}{2} \]
Since \( y \) can take values from \( 0 \) to \( \infty \), this means that not every \( y \) in the codomain has a corresponding \( x \) in the domain. Thus, \( f \) is not onto.

### Conclusion

Since \( f \) is one-one but not onto, the final answer is that \( f \) is **one-one but not onto**.