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-to-one or onto, we will analyze the function step by step.

### Step 1: Check if \( f \) is one-to-one (injective)

A function is one-to-one if different inputs produce different outputs. In other words, 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:
\[ a^2 + a^2 b^4 = b^2 + b^2 a^4 \]
Rearranging yields:
\[ a^2 - b^2 = b^2 a^4 - a^2 b^4 \]
Factoring gives:
\[ (a - b)(a + b) = (b^2 a^4 - a^2 b^4) \]
This shows that if \( a \neq b \), then \( f(a) \) can equal \( f(b) \). For example, if we take \( a = 2 \) and \( b = \frac{1}{2} \):
\[ f(2) = \frac{2^2}{1 + 2^4} = \frac{4}{17} \]
\[ f\left(\frac{1}{2}\right) = \frac{\left(\frac{1}{2}\right)^2}{1 + \left(\frac{1}{2}\right)^4} = \frac{\frac{1}{4}}{1 + \frac{1}{16}} = \frac{\frac{1}{4}}{\frac{17}{16}} = \frac{4}{17} \]
Here, \( f(2) = f\left(\frac{1}{2}\right) \) but \( 2 \neq \frac{1}{2} \). Thus, \( f \) is not one-to-one.

### Step 2: Check if \( f \) 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 \( x \geq 0 \) such that:
\[ y = \frac{x^2}{1 + x^4} \]
Rearranging gives:
\[ y(1 + x^4) = x^2 \]
This leads to:
\[ y + yx^4 = x^2 \implies yx^4 - x^2 + y = 0 \]
This is a quadratic equation in \( x^2 \):
\[ yx^4 - x^2 + y = 0 \]
Applying the quadratic formula:
\[ x^2 = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(y)(y)}}{2(y)} = \frac{1 \pm \sqrt{1 - 4y^2}}{2y} \]
For \( x^2 \) to be non-negative, the discriminant must be non-negative:
\[ 1 - 4y^2 \geq 0 \implies 1 \geq 4y^2 \implies y^2 \leq \frac{1}{4} \implies |y| \leq \frac{1}{2} \]
Since \( y \) must be in \( [0, \infty) \), this means \( y \) can only take values in the interval \( [0, \frac{1}{2}] \). Therefore, there are values in the codomain \( [0, \infty) \) (for example, \( y = 4 \)) that do not have a corresponding \( x \) in the domain.

### Conclusion

Since \( f \) is neither one-to-one nor onto, we conclude that the function \( f(x) = \frac{x^2}{1 + x^4} \) is neither injective nor surjective.