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(D) Given the function $f(x) = \frac{e^{|x|} - e^{-x}}{e^x + e^{-x}}$.Case $I$: If $x \geq 0$,then $|x| = x$. Thus,$f(x) = \frac{e^x - e^{-x}}{e^x + e

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$f$ is neither one-one nor onto

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(D) Given the function $f(x) = \frac{e^{|x|} - e^{-x}}{e^x + e^{-x}}$.Case $I$: If $x \geq 0$,then $|x| = x$. Thus,$f(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} = 	anh(x)$. As $x$ increases from $0$ to $\infty$,$f(x)$ increases from $0$ to $1$.Case $II$: If $x Since $f(x) = 0$ for all $x Since the range of $f$ is $[0, 1)$,which is a proper subset of the codomain $R$,the function is not onto.Therefore,$f$ is neither one-one nor onto.

Let $f: R \rightarrow R$ be a function defined by $f(x) = \frac{e^{|x|} - e^{-x}}{e^x + e^{-x}}$,then

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