Let `f:R->R` be a function defined by `f(x)=x^2-(x^2)/(1+x^2)`. Then:

Options

A. (a) `f` is one-one but not onto
B. (b) `f` is onto but not one-one
C. (c) `f` is both one-one and onto
D. (d) `f` is neither one-one nor onto

Answer

To solve the problem, we need to analyze the function \( f(x) = x^2 - \frac{x^2}{1+x^2} \) and determine whether it is one-one (injective) and/or onto (surjective).

### Step 1: Simplify the Function
First, we simplify the function \( f(x) \):

\[ f(x) = x^2 - \frac{x^2}{1+x^2} \]

To combine the terms, we can rewrite it with a common denominator:

\[ f(x) = \frac{x^2(1+x^2) - x^2}{1+x^2} = \frac{x^4 + x^2 - x^2}{1+x^2} = \frac{x^4}{1+x^2} \]

### Step 2: Check if the Function is One-One
A function is one-one if different inputs lead to different outputs. We need to check if \( f(a) = f(b) \) implies \( a = b \).

Assume \( f(a) = f(b) \):

\[ \frac{a^4}{1+a^2} = \frac{b^4}{1+b^2} \]

Cross-multiplying gives:

\[ a^4(1+b^2) = b^4(1+a^2) \]

This simplifies to:

\[ a^4 + a^4b^2 = b^4 + b^4a^2 \]

Rearranging terms leads to:

\[ a^4 - b^4 + a^4b^2 - b^4a^2 = 0 \]

Factoring gives:

\[ (a^2 - b^2)(a^2 + b^2) + b^2(a^4 - b^4) = 0 \]

This indicates that if \( a \neq b \), then \( f(a) \) can equal \( f(b) \). For example, \( f(1) = f(-1) = \frac{1}{2} \). Thus, \( f \) is not one-one.

### Step 3: Check if the Function is Onto
A function is onto if every element in the codomain has a pre-image in the domain. The codomain here is \( \mathbb{R} \).

To check if \( f(x) \) can take any real value, we analyze the expression:

\[ f(x) = \frac{x^4}{1+x^2} \]

Since \( x^4 \geq 0 \) and \( 1+x^2 > 0 \), it follows that \( f(x) \geq 0 \) for all \( x \). Therefore, \( f(x) \) cannot take negative values, which means it cannot be onto \( \mathbb{R} \).

### Conclusion
Since \( f \) is not one-one and not onto, the correct answer is:

**Option d: f is neither one-one nor onto.**

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