Let f: R rightarrow R be defined by f(x)=5x^4 | vedclass

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(D) Given the function $f: R \rightarrow R$ defined by $f(x) = 5x^4 + 2$. To check for one-one: Consider $f(x_1) = f(x_2)$. $5x_1^4 + 2 = 5x_2^4 + 2$

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$f$ is neither one-one nor onto

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(D) Given the function $f: R ightarrow R$ defined by $f(x) = 5x^4 + 2$. To check for one-one: Consider $f(x_1) = f(x_2)$. $5x_1^4 + 2 = 5x_2^4 + 2$ $x_1^4 = x_2^4$ $x_1 = \pm x_2$. Since $f(1) = 5(1)^4 + 2 = 7$ and $f(-1) = 5(-1)^4 + 2 = 7$,we have $f(1) = f(-1)$ but $1 
eq -1$. Therefore,$f$ is not one-one. To check for onto: Since $x^4 \geq 0$ for all $x \in R$,it follows that $5x^4 \geq 0$. Thus,$f(x) = 5x^4 + 2 \geq 2$. The range of $f$ is $[2, \infty)$,which is not equal to the codomain $R$. Therefore,$f$ is not onto. Hence,$f$ is neither one-one nor onto.

Let $f: R \rightarrow R$ be defined by $f(x)=5x^4+2$. Then

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