JEE MAIN - Mathematics (2024 - 1st February Morning Shift - No. 11)
Let $f: \mathbf{R} \rightarrow \mathbf{R}$ and $g: \mathbf{R} \rightarrow \mathbf{R}$ be defined as
$f(x)=\left\{\begin{array}{ll}\log _{\mathrm{e}} x, & x>0 \\ \mathrm{e}^{-x}, & x \leq 0\end{array}\right.$ and
$g(x)=\left\{\begin{array}{ll}x, & x \geqslant 0 \\ \mathrm{e}^x, & x<0\end{array}\right.$. Then, gof : $\mathbf{R} \rightarrow \mathbf{R}$ is :
$f(x)=\left\{\begin{array}{ll}\log _{\mathrm{e}} x, & x>0 \\ \mathrm{e}^{-x}, & x \leq 0\end{array}\right.$ and
$g(x)=\left\{\begin{array}{ll}x, & x \geqslant 0 \\ \mathrm{e}^x, & x<0\end{array}\right.$. Then, gof : $\mathbf{R} \rightarrow \mathbf{R}$ is :
one-one but not onto
neither one-one nor onto
onto but not one-one
both one-one and onto
Explanation
Given, $f(x)=\left\{\begin{array}{ll}\log _{\mathrm{e}} x, & x>0 \\ \mathrm{e}^{-x}, & x \leq 0\end{array}\right.$ and
$g(x)=\left\{\begin{array}{ll}x, & x \geqslant 0 \\ \mathrm{e}^x, & x<0\end{array}\right.$
then $g \circ f(x)$ $=g(f(x))$
$\begin{aligned} & \mathrm{g}(\mathrm{f}(\mathrm{x}))=\left\{\begin{array}{l}f(x), f(x) \geq 0 \\ e^{f(x)}, f(x)<0\end{array}\right. \\\\ & \mathrm{g}(\mathrm{f}(\mathrm{x}))=\left\{\begin{array}{l}e^{-x},(-\infty, 0] \\ e^{\ln x},(0,1) \\ \ln x,[1, \infty)\end{array}\right.\end{aligned}$
$g(f(x))=\left\{\begin{array}{c}e^{-x},(-\infty, 0] \\ x,(0,1) \\ \ln x,[1, \infty)\end{array}\right.$
_1st_February_Morning_Shift_en_11_1.png)
From the graph of $g(f(x))$, we can say
$\mathrm{g}(\mathrm{f}(\mathrm{x})) \Rightarrow$ Many one into
$g(x)=\left\{\begin{array}{ll}x, & x \geqslant 0 \\ \mathrm{e}^x, & x<0\end{array}\right.$
then $g \circ f(x)$ $=g(f(x))$
$\begin{aligned} & \mathrm{g}(\mathrm{f}(\mathrm{x}))=\left\{\begin{array}{l}f(x), f(x) \geq 0 \\ e^{f(x)}, f(x)<0\end{array}\right. \\\\ & \mathrm{g}(\mathrm{f}(\mathrm{x}))=\left\{\begin{array}{l}e^{-x},(-\infty, 0] \\ e^{\ln x},(0,1) \\ \ln x,[1, \infty)\end{array}\right.\end{aligned}$
$g(f(x))=\left\{\begin{array}{c}e^{-x},(-\infty, 0] \\ x,(0,1) \\ \ln x,[1, \infty)\end{array}\right.$
From the graph of $g(f(x))$, we can say
$\mathrm{g}(\mathrm{f}(\mathrm{x})) \Rightarrow$ Many one into
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