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If $\mathrm{f}(\mathrm{x})$ and $\mathrm{g}(\mathrm{x})$ are two real valued functions such that $\mathrm{f}(\mathrm{g}(\mathrm{x}+\mathrm{y}))=\mathrm{f}(\mathrm{g}(\mathrm{x}))+\mathrm{f}(\mathrm{g}(\mathrm{y})), \mathrm{g}(1)=2$ and $\mathrm{f}(2)=1$, then the function $g(f(x))$ is discontinuous on the set
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$\mathrm{f}(\mathrm{g}(\mathrm{x}+\mathrm{y}))=\mathrm{f}(\mathrm{g}(\mathrm{x}))+\mathrm{f}(\mathrm{g}(\mathrm{y}))$ ...(i)
and $f(2)=1, g(1)=2$
Now, from equation (i) :
$\mathrm{f}(\mathrm{g}(2))=\mathrm{f}(\mathrm{g}(1+1))=\mathrm{f}(\mathrm{g}(1))+\mathrm{f}(\mathrm{g}(1))$
$=1+1=2$
$\begin{aligned} & \mathrm{f}(\mathrm{g}(3))=\mathrm{f}(\mathrm{g}(2+1))=\mathrm{f}(\mathrm{g}(2))+\mathrm{f}(\mathrm{g}(1))=2+1=3 \\ & \mathrm{f}(\mathrm{g}(1))=\mathrm{f}(\mathrm{g}(2-1))=\mathrm{f}(\mathrm{g}(2))+\mathrm{f}(\mathrm{g}(-1))\end{aligned}$
$\begin{aligned} & \Rightarrow 1=2+\mathrm{f}(\mathrm{g}(-1)) \Rightarrow \mathrm{f}(\mathrm{g}(-1))=-1 \\ & \therefore \mathrm{f}(\mathrm{g}(\mathrm{n}))=\mathrm{x}\end{aligned}$
So, $f$ and $g$ are inverse of each other.
Then $f(g(x))=g(f(x))=x$.
$\therefore \mathrm{g}(\mathrm{f}(\mathrm{x}))$ is continuous on IR.
$\Rightarrow \mathrm{g}(\mathrm{f}(\mathrm{x})$ is discontinuous on $\phi$
and $f(2)=1, g(1)=2$
Now, from equation (i) :
$\mathrm{f}(\mathrm{g}(2))=\mathrm{f}(\mathrm{g}(1+1))=\mathrm{f}(\mathrm{g}(1))+\mathrm{f}(\mathrm{g}(1))$
$=1+1=2$
$\begin{aligned} & \mathrm{f}(\mathrm{g}(3))=\mathrm{f}(\mathrm{g}(2+1))=\mathrm{f}(\mathrm{g}(2))+\mathrm{f}(\mathrm{g}(1))=2+1=3 \\ & \mathrm{f}(\mathrm{g}(1))=\mathrm{f}(\mathrm{g}(2-1))=\mathrm{f}(\mathrm{g}(2))+\mathrm{f}(\mathrm{g}(-1))\end{aligned}$
$\begin{aligned} & \Rightarrow 1=2+\mathrm{f}(\mathrm{g}(-1)) \Rightarrow \mathrm{f}(\mathrm{g}(-1))=-1 \\ & \therefore \mathrm{f}(\mathrm{g}(\mathrm{n}))=\mathrm{x}\end{aligned}$
So, $f$ and $g$ are inverse of each other.
Then $f(g(x))=g(f(x))=x$.
$\therefore \mathrm{g}(\mathrm{f}(\mathrm{x}))$ is continuous on IR.
$\Rightarrow \mathrm{g}(\mathrm{f}(\mathrm{x})$ is discontinuous on $\phi$
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