# which one of the following statements is not true?

Question:

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?

1. (1) $g(f(S)) \neq S$

2. (2) $\mathrm{f}(\mathrm{g}(\mathrm{S}))=\mathrm{S}$

3. (3) $\mathrm{g}(\mathrm{f}(\mathrm{S}))=\mathrm{g}(\mathrm{S})$

4. (d) $\mathrm{f}(\mathrm{g}(\mathrm{S})) \neq \mathrm{f}(\mathrm{S})$

Correct Option: , 3

Solution:

$f(x)=x^{2} ; x \in \mathrm{R}$

$g(\mathrm{~A})=\{x \in \mathrm{R}: f(x) \in \mathrm{A}\} \mathrm{S}=[0,4]$

$g(\mathrm{~S})=\{x \in \mathrm{R}: f(x) \in \mathrm{S}\}$

$=\left\{x \in \mathrm{R}: 0 \leq x^{2} \leq 4\right\}=\{x \in \mathrm{R}:-2 \leq x \leq 2\}$

$\therefore g(\mathrm{~S}) \neq \mathrm{S} \therefore f(g(\mathrm{~S})) \neq f(\mathrm{~S})$

$g(f(\mathrm{~S}))=\{x \in \mathrm{R}: f(x) \in f(\mathrm{~S})\}$

$=\left\{x \in \mathrm{R}: x^{2} \in \mathrm{S}^{2}\right\}=\left\{x \in \mathrm{R}: 0 \leq x^{2} \leq 16\right\}$

$=\{x \in \mathrm{R}:-4 \leq x \leq 4\}$

$\therefore g(f(\mathrm{~S})) \neq g(\mathrm{~S})$

$\therefore g(f(\mathrm{~S}))=g(\mathrm{~S})$ is incorrect.