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If $f(x)=2^{100} x+1, g(x)=3^{100} x+1$, then the set of real numbers $x$ such that $f[g(x)\}=x$ is
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The correct answer is:
a singleton
Given, $\quad f(x)=2^{100} \cdot x+1$
$g(x)=3^{300} \cdot x+1$
Now, $f\{g(x)\}=x$
$\Rightarrow \quad f\left(3^{100}. x+1\right)=x$
$\Rightarrow \quad 2^{100}\left\{3^{100} \cdot x+1\right\}+1=x$
$\Rightarrow \quad 6^{100} \cdot x+2^{100}+1=x$
$\Rightarrow \quad x\left(1-6^{100}\right)=\left(1+2^{100}\right)$
$\Rightarrow \quad x=\frac{1+2^{100}}{1-6^{100}}$
Hence, fog $(x)=x$ represent a singleton set.
$g(x)=3^{300} \cdot x+1$
Now, $f\{g(x)\}=x$
$\Rightarrow \quad f\left(3^{100}. x+1\right)=x$
$\Rightarrow \quad 2^{100}\left\{3^{100} \cdot x+1\right\}+1=x$
$\Rightarrow \quad 6^{100} \cdot x+2^{100}+1=x$
$\Rightarrow \quad x\left(1-6^{100}\right)=\left(1+2^{100}\right)$
$\Rightarrow \quad x=\frac{1+2^{100}}{1-6^{100}}$
Hence, fog $(x)=x$ represent a singleton set.
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