Let f: R rightarrow R and g: R rightarrow R b | vedclass

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(B) We are given $f(x) = \begin{cases} \ln x & x > 0 \ e^{-x} & x \leq 0 \end{cases}$ and $g(x) = \begin{cases} x & x \geq 0 \ e^x & x < 0 \end{case

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neither one-one nor onto

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(B) We are given $f(x) = \begin{cases} \ln x & x > 0 \ e^{-x} & x \leq 0 \end{cases}$ and $g(x) = \begin{cases} x & x \geq 0 \ e^x & x To find $(gof)(x) = g(f(x))$,we analyze the cases for $f(x)$:Case $1$: $x \leq 0$. Then $f(x) = e^{-x}$. Since $e^{-x} > 0$ for all $x$,$g(f(x)) = f(x) = e^{-x}$.Case $2$: $x > 0$. Then $f(x) = \ln x$.Subcase 2a: $0 Subcase 2b: $x \geq 1$. Then $f(x) = \ln x \geq 0$. Thus $g(f(x)) = f(x) = \ln x$.Combining these,$(gof)(x) = \begin{cases} e^{-x} & x \leq 0 \ x & 0 Checking one-one: For $x \leq 0$,$g(f(x)) = e^{-x} \in [1, \infty)$. For $0 Since the range of $g(f(x))$ is $[0, \infty)$,it is not onto (as the codomain is $R$).Also,for $x \leq 0$,$g(f(x)) \geq 1$,and for $x \geq 1$,$g(f(x)) \geq 0$. There exist values in the range that are mapped to by multiple $x$,and the function is not injective. Thus,it is neither one-one nor onto.

Let $f: R \rightarrow R$ and $g: R \rightarrow R$ be defined as $f(x)=\begin{cases} \log _e x & , x>0 \\ e^{-x} & , x \leq 0 \end{cases}$ and $g(x)=\begin{cases} x & , x \geq 0 \\ e^{x} & , x < 0 \end{cases}$. Then $gof: R \to R$ is . . . .

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