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

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(D) Given that,$f: R \rightarrow R$ and $f(x)=x^{4}$.For $f$ to be one-one,$f(x_{1}) = f(x_{2})$ must imply $x_{1} = x_{2}$.Here,$x_{1}^{4} = x_{2}^{4

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$f$ is neither one-one nor onto

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(D) Given that,$f: R ightarrow R$ and $f(x)=x^{4}$.For $f$ to be one-one,$f(x_{1}) = f(x_{2})$ must imply $x_{1} = x_{2}$.Here,$x_{1}^{4} = x_{2}^{4} \Rightarrow x_{1} = \pm x_{2}$.For example,$f(1) = 1^{4} = 1$ and $f(-1) = (-1)^{4} = 1$.Since $f(1) = f(-1)$ but $1 
eq -1$,$f$ is not one-one.For $f$ to be onto,the range must equal the codomain $R$.Since $x^{4} \geq 0$ for all $x \in R$,the range is $[0, \infty)$.Since the range $[0, \infty) 
eq R$,$f$ is not onto.Therefore,$f$ is neither one-one nor onto.

Let $f: R \rightarrow R$ be defined by $f(x)=x^{4}$,then

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