If f(x)=2^{100} x+1 and g(x)=3^{100} x+1, the | vedclass

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(B) Given, $f(x)=2^{100} x+1$ and $g(x)=3^{100} x+1$. We need to find $x$ such that $f(g(x))=x$. Substituting $g(x)$ into $f(x)$, we get: $f(3^{100} x

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a singleton

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(B) Given, $f(x)=2^{100} x+1$ and $g(x)=3^{100} x+1$. We need to find $x$ such that $f(g(x))=x$. Substituting $g(x)$ into $f(x)$, we get: $f(3^{100} x+1) = x$ $2^{100}(3^{100} x+1) + 1 = x$ $2^{100} \cdot 3^{100} x + 2^{100} + 1 = x$ $6^{100} x + 2^{100} + 1 = x$ $x(6^{100} - 1) = -(2^{100} + 1)$ $x = -\frac{2^{100} + 1}{6^{100} - 1}$ Since there is exactly one value of $x$ that satisfies the equation, the set of real numbers $x$ is a singleton set.

If $f(x)=2^{100} x+1$ and $g(x)=3^{100} x+1$, then the set of real numbers $x$ such that $f(g(x))=x$ is

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