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

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(D) The function is defined as $f(x) = 5^{-|x|} + \operatorname{sgn}(5^{-x})$.Since $5^{-x} > 0$ for all $x \in R$, the signum function $\operatorname

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neither one-one nor onto

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(D) The function is defined as $f(x) = 5^{-|x|} + \operatorname{sgn}(5^{-x})$.Since $5^{-x} > 0$ for all $x \in R$, the signum function $\operatorname{sgn}(5^{-x}) = 1$ for all $x \in R$.Thus, the function simplifies to $f(x) = 5^{-|x|} + 1$.For $x \geq 0$, $f(x) = 5^{-x} + 1$, which is a strictly decreasing function with range $(1, 2]$.For $x Since $f(x) = f(-x)$ for all $x$, the function is many-one.Also, the range of the function is $(1, 2]$, which is a proper subset of the codomain $R$, so the function is not onto.Therefore, $f$ is neither one-one nor onto.

Let $f: R \rightarrow R$ be defined by $f(x) = 5^{-|x|} + \operatorname{sgn}(5^{-x})$, where $\operatorname{sgn}(x)$ denotes the signum function of $x$. Then $f$ is

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