Let f: R rightarrow R be defined by f(x) = x^ | vedclass

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(D) We have,$f(x) = x^{2} - \frac{x^{2}}{1+x^{2}}$.First,check for one-one property:$f(-x) = (-x)^{2} - \frac{(-x)^{2}}{1+(-x)^{2}} = x^{2} - \frac{x^

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$f$ is neither one-one nor onto

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(D) We have,$f(x) = x^{2} - \frac{x^{2}}{1+x^{2}}$.First,check for one-one property:$f(-x) = (-x)^{2} - \frac{(-x)^{2}}{1+(-x)^{2}} = x^{2} - \frac{x^{2}}{1+x^{2}} = f(x)$.Since $f(-x) = f(x)$ for all $x \in R$,the function is not one-one (it is many-one).Next,simplify the expression for range:$f(x) = \frac{x^{2}(1+x^{2}) - x^{2}}{1+x^{2}} = \frac{x^{2} + x^{4} - x^{2}}{1+x^{2}} = \frac{x^{4}}{1+x^{2}}$.Since $x^{4} \ge 0$ and $1+x^{2} > 0$ for all $x \in R$,$f(x) \ge 0$.The range of $f(x)$ is $[0, \infty)$.Since the co-domain is $R$ and the range $[0, \infty) 
eq R$,the function is not onto.Therefore,$f$ is neither one-one nor onto.

Let $f: R \rightarrow R$ be defined by $f(x) = x^{2} - \frac{x^{2}}{1+x^{2}}$ for all $x \in R$. Then,

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