question_answer

Let X be a non-empty set and P(X) be its power set. Let '*' be an operation defined on elements of P(X) by

\[A*B=A\,\cap B,\] \[\forall A,\] \[B\in P(X).\] Then,

(i) Prove that '*' is a binary operation in P(X).

(ii) Prove that '*' is commutative.

(iii) Prove that '*' is associative.

(iv) If 'o' is another binary operation defriended on P(X) as \[AoB=A\cup B,\] then verify that 'o' distributes over '*'.

OR

Find \[{{f}^{-1}},\] If \[f:R\to (-\,1,\,1)\] is defined by

\[f(x)=\frac{{{\sqrt{7}}^{x}}-{{\sqrt{7}}^{-x}}}{{{\sqrt{7}}^{x}}+{{\sqrt{7}}^{\,-\,x}}}.\]

Answer:

Given, for any non-empty set X, P(X) is the power set, i.e. set of all possible subsets of set X.

Also, \[A*B=A\,\cap B,\] \[\forall A,\] \[B\in P(X)\]

(i) For any two elements of P(X), the intersection is again the element P(X).

\[\therefore \] '*' is a binary operation in P(X).

(ii) For any two sets A and \[B\in P(X),\]

\[A\cap B=B\cap A\]

\[\Rightarrow \]   \[A*B=B*A,\] \[\forall \,A,\] \[B\in P(X),\]

\[\therefore \]      ?*? is commutative.

(iii) For any three sets A, B and C,

\[(A\cap B)\cap C=A\cap (B\cap C)\]

\[\Rightarrow \]   \[(A*B)*C=A*(B*C),\] \[\forall \,A,\]

\[B,\,C\in P(X)\]

\[\therefore \]      '*' is associative.

(iv) We have, \[AoB=A\cup B\]

and we want to verify that,

\[Ao(B*C)=(AoB)*(AoC),\] \[\forall \,A,\,B,\,C\in P(X)\]

For any three sets A, B and C,

\[A\cup (B\cap C)=(A\cup B)\cap (A\cup C)\]

\[\therefore \]      \[Ao(B*C)=(AoB)*(AoC)\]

Hence, 'o' distributes over'*'.

Hence proved.

OR

Given a function \[f:R\to (-1,\,\,1),\] defined as

\[f(x)=\frac{{{\sqrt{7}}^{x}}-{{\sqrt{7}}^{-x}}}{{{\sqrt{7}}^{x}}+{{\sqrt{7}}^{-x}}}\]

For \[f\]  to be one-one

Let \[{{x}_{1}},\] \[{{x}_{2}}\in R\] such that \[f({{x}_{1}})=f({{x}_{2}})\]

\[\Rightarrow \]   \[\frac{{{\sqrt{7}}^{{{x}_{1}}}}-{{\sqrt{7}}^{-{{x}_{1}}}}}{{{\sqrt{7}}^{{{x}_{1}}}}+{{\sqrt{7}}^{-{{x}_{1}}}}}=\frac{{{\sqrt{7}}^{{{x}_{2}}}}-{{\sqrt{7}}^{-{{x}_{2}}}}}{{{\sqrt{7}}^{{{x}_{2}}}}+{{\sqrt{7}}^{-{{x}_{2}}}}}\]

\[\Rightarrow \]   \[({{\sqrt{7}}^{{{x}_{1}}}}-{{\sqrt{7}}^{-{{x}_{1}}}})({{\sqrt{7}}^{{{x}_{2}}}}+{{\sqrt{7}}^{-{{x}_{2}}}})\]

\[=({{\sqrt{7}}^{{{x}_{1}}}}+{{\sqrt{7}}^{-{{x}_{1}}}})({{\sqrt{7}}^{{{x}_{2}}}}-{{\sqrt{7}}^{-{{x}_{2}}}})\]  \[\Rightarrow \]   \[{{\sqrt{7}}^{{{x}_{1}}+{{x}_{2}}}}+{{\sqrt{7}}^{{{x}_{1}}-{{x}_{2}}}}-{{\sqrt{7}}^{{{x}_{2}}-{{x}_{1}}}}-{{\sqrt{7}}^{{{x}_{1}}-{{x}_{2}}}}\]

\[={{\sqrt{7}}^{{{x}_{1}}+{{x}_{2}}}}-{{\sqrt{7}}^{{{x}_{1}}-{{x}_{2}}}}+{{\sqrt{7}}^{{{x}_{2}}-{{x}_{1}}}}-{{\sqrt{7}}^{{{x}_{1}}-{{x}_{2}}}}\]

\[\Rightarrow \]   \[2\cdot {{\sqrt{7}}^{{{x}_{1}}-{{x}_{2}}}}=2\cdot {{\sqrt{7}}^{{{x}_{2}}-{{x}_{1}}}}\]

\[\Rightarrow \]   \[{{\sqrt{7}}^{{{x}_{1}}-{{x}_{2}}}}={{\sqrt{7}}^{{{x}_{2}}-{{x}_{1}}}}\]

\[\Rightarrow \]   \[{{x}_{1}}-{{x}_{2}}={{x}_{2}}-{{x}_{1}}\]

\[[\because \,\,{{a}^{b}}={{a}^{c}}\,\,\,\Rightarrow \,\,\,b=c]\]

\[\Rightarrow \]   \[2{{x}_{1}}=2{{x}_{2}}\]

\[\Rightarrow \]   \[{{x}_{1}}={{x}_{2}}\]

\[\Rightarrow \] f is one-one

For f to be onto

Let \[y\in (-\,1,\,\,1)\] be any arbitrary element. Then,

\[y=f(x)=\frac{{{\sqrt{7}}^{x}}-{{\sqrt{7}}^{-x}}}{{{\sqrt{7}}^{x}}+{{\sqrt{7}}^{-x}}}\]

\[\Rightarrow \]   \[({{\sqrt{7}}^{x}}+{{\sqrt{7}}^{-x}})y={{\sqrt{7}}^{x}}-{{\sqrt{7}}^{-x}}\]

\[\Rightarrow \]   \[\left( \frac{{{\sqrt{7}}^{2x}}+1}{{{\sqrt{7}}^{x}}} \right)\,\,y=\frac{{{\sqrt{7}}^{2x}}-1}{{{\sqrt{7}}^{x}}}\]

\[\Rightarrow \]   \[{{\sqrt{7}}^{2x}}y-{{\sqrt{7}}^{2x}}=-1-y\]

\[\Rightarrow \]   \[{{\sqrt{7}}^{2x}}(y-1)=-1-y\]

\[\Rightarrow \]   \[{{\sqrt{7}}^{2x}}=\frac{1+y}{1-y}\] \[\Rightarrow \] \[{{7}^{x}}=\frac{1+y}{1-y}\]

Taking log on both sides, we get

\[x{{\log }_{10}}7={{\log }_{10}}\left( \frac{1+y}{1-y} \right)\]

\[\Rightarrow \]   \[x=\frac{1}{{{\log }_{10}}7}{{\log }_{10}}\left( \frac{1+y}{1-y} \right)\in R,\] for

\[y\in (-\,1,\,\,1)\] ? (i)

Thus, for each y e \[y\in (-\,1,\,\,1)\] there exist

\[x=\frac{1}{{{\log }_{10}}7}{{\log }_{10}}\left( \frac{1+y}{1-y} \right)\in R,\] such that f(x) = y.

So, f is onto and hence invertible.

Also, inverse of f is given by \[{{f}^{-1}}:(-\,1,\,\,1)\to R,\]defined as

\[{{f}^{-1}}(y)=\frac{1}{{{\log }_{10}}7}{{\log }_{10}}\left( \frac{1+y}{1-y} \right).\]