Let $f(x)=x^{2}, x \in \mathrm{R} .$ For any $\mathrm{A} \subseteq \mathrm{R}$, define $\mathrm{g}(\mathrm{A})=$ $\{x \in \mathrm{R}: f(x) \in \mathrm{A}\}$. If $\mathrm{S}=[0,4]$, then which one of the
following statements is not true?
Correct Option: , 3
$f(x)=x^{2} ; x \in \mathrm{R}$
$g(\mathrm{~A})=\{x \in \mathrm{R}: f(x) \in \mathrm{A}\} \mathrm{S}=[0,4]$
$g(\mathrm{~S})=\{x \in \mathrm{R}: f(x) \in \mathrm{S}\}$
$=\left\{x \in \mathrm{R}: 0 \leq x^{2} \leq 4\right\}=\{x \in \mathrm{R}:-2 \leq x \leq 2\}$
$\therefore g(\mathrm{~S}) \neq \mathrm{S} \therefore f(g(\mathrm{~S})) \neq f(\mathrm{~S})$
$g(f(\mathrm{~S}))=\{x \in \mathrm{R}: f(x) \in f(\mathrm{~S})\}$
$=\left\{x \in \mathrm{R}: x^{2} \in \mathrm{S}^{2}\right\}=\left\{x \in \mathrm{R}: 0 \leq x^{2} \leq 16\right\}$
$=\{x \in \mathrm{R}:-4 \leq x \leq 4\}$
$\therefore g(f(\mathrm{~S})) \neq g(\mathrm{~S})$
$\therefore g(f(\mathrm{~S}))=g(\mathrm{~S})$ is incorrect.
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