Let f, g: R rightarrow R be defined as: f(x)= | vedclass

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(A) We are given $f(x) = |x-1|$ and $g(x) = \begin{cases} e^x, & x \geq 0 \ x+1, & x \leq 0 \end{cases}$.To find $f(g(x))$,we substitute $g(x)$ into

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neither one-one nor onto.

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(A) We are given $f(x) = |x-1|$ and $g(x) = \begin{cases} e^x, & x \geq 0 \ x+1, & x \leq 0 \end{cases}$.To find $f(g(x))$,we substitute $g(x)$ into $f(x)$:$f(g(x)) = |g(x) - 1| = \begin{cases} |e^x - 1|, & x \geq 0 \ |(x+1) - 1|, & x \leq 0 \end{cases} = \begin{cases} e^x - 1, & x \geq 0 \ |x|, & x \leq 0 \end{cases} = \begin{cases} e^x - 1, & x \geq 0 \ -x, & x \leq 0 \end{cases}$.Now,let $h(x) = f(g(x))$.For $x \geq 0$,$h(x) = e^x - 1$. As $x$ increases from $0$ to $\infty$,$h(x)$ increases from $0$ to $\infty$.For $x \leq 0$,$h(x) = -x$. As $x$ decreases from $0$ to $-\infty$,$h(x)$ increases from $0$ to $\infty$.Since $h(x)$ takes the same positive values for both positive and negative $x$ (e.g.,$h(1) = e-1$ and $h(-(e-1)) = e-1$),the function is not one-one.Since the range of $h(x)$ is $[0, \infty)$,which is a proper subset of the codomain $R$,the function is not onto.Therefore,the function is neither one-one nor onto.

Let $f, g: R \rightarrow R$ be defined as: $f(x)=|x-1|$ and $g(x)=\begin{cases} e^x, & x \geq 0 \\ x+1, & x \leq 0 \end{cases}$. Then the function $f(g(x))$ is

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