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