Question

Let $f\left( x \right)={{x}^{2}},x\in \mathbb{R}$. For any $A\subset R$, define $g\left( A \right)=\left\{ x\in \mathbb{R}:g\left( x \right)\in S \right\}$. If $S=\left[ 0,4 \right]$ then which one of the following statements is not true?
[a] $f\left( g\left( S \right) \right)\ne f\left( S \right)$
[b] $f\left( g\left( S \right) \right)=S$
[c] $g\left( f\left( S \right) \right)=f\left( S \right)$
[d] $g\left( f\left( S \right) \right)\ne S$

Answer 1

Hint: Use the fact that if $A=B,$ then $A\subseteq B$ and $B\subseteq A$. Use the fact that if $A\subseteq B$, then $\forall x\in A\Rightarrow x\in B$. Use the fact that $f\left( A \right)=\left\{ f\left( x \right):x\in A \right\}$ . Hence verify which of the options is correct and which of the options is incorrect.

Complete step-by-step answer:
We have $S=\left[ 0,4 \right]$
Hence, we have
$f\left( S \right)=\left\{ f\left( x \right):x\in \left[ 0,4 \right] \right\}$
Claim: $f\left( S \right)=\left[ 0,16 \right]$
Proof:
Let $x\in \left[ 0,16 \right]\Rightarrow 0\le x\le 16$
Hence, we have
$\exists y\in \left[ 0,4 \right],y=\sqrt{x}$ such that ${{y}^{2}}=x$
Therefore, we have
$\exists y\in S$ such that $f\left( y \right)=x$
Hence, we conclude that
$x\in f\left( S \right)$
Since x was arbitrary, we have
$\forall x\in \left[ 0,16 \right]\Rightarrow x\in f\left( S \right)$
Hence, we have
$\left[ 0,16 \right]\subseteq f\left( S \right)$
Now let $x\in f\left( S \right)\Rightarrow \exists y\in S$ such that $f\left( y \right)=x$
Hence, we have
$y\in S,{{y}^{2}}=x$
Since $y\in S,0\le y\le 4\Rightarrow 0\le {{y}^{2}}\le 16\Rightarrow 0\le x\le 16$
Hence, we have
$x\in \left[ 0,16 \right]$
Since x was arbitrary, we have
$\forall x\in f\left( S \right)\Rightarrow x\in \left[ 0,16 \right]$
Hence, we have
$f\left( S \right)\subseteq \left[ 0,16 \right]$
Combining the results $f\left( S \right)\subseteq \left[ 0,16 \right]$ and $\left[ 0,16 \right]\subseteq f\left( S \right)$, we have
$f\left( S \right)=\left[ 0,16 \right]$
We have $g\left( A \right)=\left\{ x\in \mathbb{R}:f\left( x \right)\in A \right\}$
We claim that $g\left( S \right)=\left[ -2,2 \right]$
Proof:
Let $x\in \left[ -2,2 \right]\Rightarrow -2\le x\le 2\Rightarrow 0\le {{x}^{2}}\le 4\Rightarrow 0\le f\left( x \right)\le 4$
Hence, we have
$f\left( x \right)\in S\Rightarrow x\in g\left( S \right)$
Since x was arbitrary, we have
$\forall x\in \left[ -2,2 \right]\Rightarrow x\in g\left( S \right)$
Hence, we have
$\left[ -2,2 \right]\subseteq g\left( S \right)$
Now let $x\in g\left( S \right)\Rightarrow f\left( x \right)\in S\Rightarrow 0\le {{x}^{2}}\le 4$
Hence, we have
$-2\le x\le 2\Rightarrow x\in \left[ -2,2 \right]$
Since x was arbitrary, we have
$\forall x\in g\left( S \right)\Rightarrow x\in \left[ -2,2 \right]$
Hence, we have
$g\left( S \right)\subseteq \left[ -2,2 \right]$
Hence, we have
$g\left( S \right)=\left[ -2,2 \right]$
Claim: $f\left( g\left( S \right) \right)=\left[ 0,4 \right]=S$
Proof:
Let $x\in \left[ 0,4 \right]\Rightarrow \exists y\in \left[ -2,2 \right],y=\sqrt{x}$ such that ${{y}^{2}}=x$
$\Rightarrow f\left( y \right)=x,y\in g\left( S \right)$
Hence, we have$x\in f\left( g\left( S \right) \right)$
Hence, we have
$S\subseteq f\left( g\left( S \right) \right)$
Now let $x\in f\left( g\left( S \right) \right)\Rightarrow \exists y\in \left[ -2,2 \right]$ such that $f\left( y \right)=x$
$\Rightarrow -2\le y\le 2,x={{y}^{2}}\Rightarrow 0\le x\le 4$
Hence, we have
$x\in \left[ 0,4 \right]$
Hence, we have $f\left( g\left( S \right) \right)\subseteq S$
Hence, we have
$f\left( g\left( S \right) \right)=S$
Claim: $g\left( f\left( S \right) \right)=\left[ -4,4 \right]$
Proof:
Let $x\in \left[ -4,4 \right]\Rightarrow -4\le x\le 4\Rightarrow 0\le {{x}^{2}}\le 16\Rightarrow 0\le f\left( x \right)\le 16$
Hence, we have
$f\left( x \right)\in f\left( S \right)\Rightarrow x\in g\left( f\left( S \right) \right)$
Hence, we have
$\left[ -4,4 \right]\subseteq g\left( f\left( S \right) \right)$
Now let $x\in g\left( f\left( S \right) \right)\Rightarrow f\left( x \right)\in f\left( S \right)\Rightarrow 0\le {{x}^{2}}\le 16$
Hence, we have
$-4\le x\le 4\Rightarrow x\in \left[ -4,4 \right]$
Hence, we have
$g\left( f\left( S \right) \right)\subseteq \left[ -4,4 \right]$
Hence, we have
$g\left( f\left( S \right) \right)=\left[ -4,4 \right]$
Hence the only option which is incorrect is option [c].

So, the correct answer is “Option c”.

Note: [1] Many students make a mistake by claiming that $x\in A\Rightarrow x\in B$ means $A=B$ which is incorrect. This leads to incorrect solutions.
[2] The set $g\left( S \right)$ is denoted by ${{f}^{-1}}\left( S \right)$. If $f\left( g\left( S \right) \right)=g\left( f\left( S \right) \right)\forall S\subseteq \mathbb{R}$ then the function f(x) is a one-one function