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Question: Let \(f\left( x \right)={{x}^{2}},x\in \mathbb{R}\). For any \(A\subset R\), define \(g\left( A \rig...

Let f(x)=x2,x∈Rf\left( x \right)={{x}^{2}},x\in \mathbb{R}. For any A⊂RA\subset R, define g\left( A \right)=\left\\{ x\in \mathbb{R}:g\left( x \right)\in S \right\\}. If S=[0,4]S=\left[ 0,4 \right] then which one of the following statements is not true?
[a] f(g(S))≠f(S)f\left( g\left( S \right) \right)\ne f\left( S \right)
[b] f(g(S))=Sf\left( g\left( S \right) \right)=S
[c] g(f(S))=f(S)g\left( f\left( S \right) \right)=f\left( S \right)
[d] g(f(S))≠Sg\left( f\left( S \right) \right)\ne S

Explanation

Solution

Use the fact that if A=B,A=B, then A⊆BA\subseteq B and B⊆AB\subseteq A. Use the fact that if A⊆BA\subseteq B, then ∀x∈A⇒x∈B\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=[0,4]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(S)=[0,16]f\left( S \right)=\left[ 0,16 \right]
Proof:
Let x∈[0,16]⇒0≤x≤16x\in \left[ 0,16 \right]\Rightarrow 0\le x\le 16
Hence, we have
∃y∈[0,4],y=x\exists y\in \left[ 0,4 \right],y=\sqrt{x} such that y2=x{{y}^{2}}=x
Therefore, we have
∃y∈S\exists y\in S such that f(y)=xf\left( y \right)=x
Hence, we conclude that
x∈f(S)x\in f\left( S \right)
Since x was arbitrary, we have
∀x∈[0,16]⇒x∈f(S)\forall x\in \left[ 0,16 \right]\Rightarrow x\in f\left( S \right)
Hence, we have
[0,16]⊆f(S)\left[ 0,16 \right]\subseteq f\left( S \right)
Now let x∈f(S)⇒∃y∈Sx\in f\left( S \right)\Rightarrow \exists y\in S such that f(y)=xf\left( y \right)=x
Hence, we have
y∈S,y2=xy\in S,{{y}^{2}}=x
Since y∈S,0≤y≤4⇒0≤y2≤16⇒0≤x≤16y\in S,0\le y\le 4\Rightarrow 0\le {{y}^{2}}\le 16\Rightarrow 0\le x\le 16
Hence, we have
x∈[0,16]x\in \left[ 0,16 \right]
Since x was arbitrary, we have
∀x∈f(S)⇒x∈[0,16]\forall x\in f\left( S \right)\Rightarrow x\in \left[ 0,16 \right]
Hence, we have
f(S)⊆[0,16]f\left( S \right)\subseteq \left[ 0,16 \right]
Combining the results f(S)⊆[0,16]f\left( S \right)\subseteq \left[ 0,16 \right] and [0,16]⊆f(S)\left[ 0,16 \right]\subseteq f\left( S \right), we have
f(S)=[0,16]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(S)=[−2,2]g\left( S \right)=\left[ -2,2 \right]
Proof:
Let x∈[−2,2]⇒−2≤x≤2⇒0≤x2≤4⇒0≤f(x)≤4x\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(x)∈S⇒x∈g(S)f\left( x \right)\in S\Rightarrow x\in g\left( S \right)
Since x was arbitrary, we have
∀x∈[−2,2]⇒x∈g(S)\forall x\in \left[ -2,2 \right]\Rightarrow x\in g\left( S \right)
Hence, we have
[−2,2]⊆g(S)\left[ -2,2 \right]\subseteq g\left( S \right)
Now let x∈g(S)⇒f(x)∈S⇒0≤x2≤4x\in g\left( S \right)\Rightarrow f\left( x \right)\in S\Rightarrow 0\le {{x}^{2}}\le 4
Hence, we have
−2≤x≤2⇒x∈[−2,2]-2\le x\le 2\Rightarrow x\in \left[ -2,2 \right]
Since x was arbitrary, we have
∀x∈g(S)⇒x∈[−2,2]\forall x\in g\left( S \right)\Rightarrow x\in \left[ -2,2 \right]
Hence, we have
g(S)⊆[−2,2]g\left( S \right)\subseteq \left[ -2,2 \right]
Hence, we have
g(S)=[−2,2]g\left( S \right)=\left[ -2,2 \right]
Claim: f(g(S))=[0,4]=Sf\left( g\left( S \right) \right)=\left[ 0,4 \right]=S
Proof:
Let x∈[0,4]⇒∃y∈[−2,2],y=xx\in \left[ 0,4 \right]\Rightarrow \exists y\in \left[ -2,2 \right],y=\sqrt{x} such that y2=x{{y}^{2}}=x
⇒f(y)=x,y∈g(S)\Rightarrow f\left( y \right)=x,y\in g\left( S \right)
Hence, we havex∈f(g(S))x\in f\left( g\left( S \right) \right)
Hence, we have
S⊆f(g(S))S\subseteq f\left( g\left( S \right) \right)
Now let x∈f(g(S))⇒∃y∈[−2,2]x\in f\left( g\left( S \right) \right)\Rightarrow \exists y\in \left[ -2,2 \right] such that f(y)=xf\left( y \right)=x
⇒−2≤y≤2,x=y2⇒0≤x≤4\Rightarrow -2\le y\le 2,x={{y}^{2}}\Rightarrow 0\le x\le 4
Hence, we have
x∈[0,4]x\in \left[ 0,4 \right]
Hence, we have f(g(S))⊆Sf\left( g\left( S \right) \right)\subseteq S
Hence, we have
f(g(S))=Sf\left( g\left( S \right) \right)=S
Claim: g(f(S))=[−4,4]g\left( f\left( S \right) \right)=\left[ -4,4 \right]
Proof:
Let x∈[−4,4]⇒−4≤x≤4⇒0≤x2≤16⇒0≤f(x)≤16x\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(x)∈f(S)⇒x∈g(f(S))f\left( x \right)\in f\left( S \right)\Rightarrow x\in g\left( f\left( S \right) \right)
Hence, we have
[−4,4]⊆g(f(S))\left[ -4,4 \right]\subseteq g\left( f\left( S \right) \right)
Now let x∈g(f(S))⇒f(x)∈f(S)⇒0≤x2≤16x\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≤x≤4⇒x∈[−4,4]-4\le x\le 4\Rightarrow x\in \left[ -4,4 \right]
Hence, we have
g(f(S))⊆[−4,4]g\left( f\left( S \right) \right)\subseteq \left[ -4,4 \right]
Hence, we have
g(f(S))=[−4,4]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∈A⇒x∈Bx\in A\Rightarrow x\in B means A=BA=B which is incorrect. This leads to incorrect solutions.
[2] The set g(S)g\left( S \right) is denoted by f−1(S){{f}^{-1}}\left( S \right). If f(g(S))=g(f(S))∀S⊆Rf\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