Let f: R rightarrow R be defined as f(x)=x^{4 | vedclass

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(C) Given $f: R \rightarrow R$ is defined as $f(x) = x^{4}$.To check for one-one:Let $x, y \in R$ such that $f(x) = f(y)$.$\Rightarrow x^{4} = y^{4}$$

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$f$ is neither one-one nor onto

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(C) Given $f: R ightarrow R$ is defined as $f(x) = x^{4}$.To check for one-one:Let $x, y \in R$ such that $f(x) = f(y)$.$\Rightarrow x^{4} = y^{4}$$\Rightarrow x = \pm y$.Since $f(1) = 1^{4} = 1$ and $f(-1) = (-1)^{4} = 1$,we have $f(1) = f(-1)$ but $1 
eq -1$.Therefore,$f$ is not one-one.To check for onto:$A$ function is onto if its range equals its co-domain. Here,the co-domain is $R$.Since $x^{4} \geq 0$ for all $x \in R$,the range of $f$ is $[0, \infty)$.Since the range $[0, \infty) 
eq R$,the function is not onto.For example,there is no $x \in R$ such that $f(x) = -2$.Thus,$f$ is neither one-one nor onto.The correct answer is $C$.

Let $f: R \rightarrow R$ be defined as $f(x)=x^{4}$. Choose the correct answer.

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