Show that the function f: \[R\to R\] defined by |
\[f(x)=\frac{3x-1}{2},\] \[x\in R\] is one-one and onto functions. Also, find the inverse of the function f. |
OR |
Examine which of the following is a binary operation and check whether the operation is commutative and associative? |
(i) On \[{{Z}^{+}},\] define \[a*b={{2}^{ab}}.\] |
(ii) On Q, define \[a*b=\frac{ab}{2}.\] |
Answer:
Given, \[f(x)=\frac{3x-1}{2},\] \[x\in R\] For one-one Let \[{{x}_{1}},\,\,{{x}_{2}}\in R\] such that \[f({{x}_{1}})=f({{x}_{2}})\] \[\Rightarrow \] \[\frac{3{{x}_{1}}-1}{2}=\frac{3{{x}_{2}}-1}{2}\] \[\Rightarrow \] \[3{{x}_{1}}-1=3{{x}_{2}}-1\] \[\Rightarrow \] \[3{{x}_{1}}=3{{x}_{2}}\] \[\Rightarrow \] \[{{x}_{1}}={{x}_{2}}\] \[\therefore \] f is one-one. For onto Let \[y\in R,\,\,\text{then}\,\,f(x)=y\] \[\Rightarrow \] \[\frac{3x-1}{2}=y\] \[\Rightarrow \] \[3x-1=2y\] \[\Rightarrow \] \[3x=2y+1\] \[\Rightarrow \] \[x=\frac{2y+1}{3}\in R\] ?(i) Thus, for each \[y\in R,\] there exists \[x=\frac{2y+1}{3}\in R\] such that \[f\left( \frac{2y+1}{3} \right)=y\] Hence, f is onto. Since, f is one-one and onto. Therefore \[{{f}^{-1}}\] exists. Hence proved. Now, from Eq. (i), we get \[x=\frac{2y+1}{3}\] \[\Rightarrow \] \[{{f}^{-1}}(y)=\frac{2y+1}{3}\] \[[\because \,\,y=f(x)\Rightarrow x={{f}^{-1}}(y)]\] \[\therefore \] \[{{f}^{-1}}(x)=\frac{2x+1}{3}\] OR (i) On \[{{Z}^{+}},\] operation ?*? is defined by \[a*b={{2}^{ab}}.\] Here, \[a\,\,b\in {{Z}^{+}};\,\,\forall \,\,a,\,\,b\in {{Z}^{+}}.\] therefore, \[a*b={{2}^{ab}}\in {{Z}^{+}}.\] Thus, the operation \['*'\] is a binary operation, We know that, \[ab=ba;\forall a,\,\,b\in {{Z}^{+}}\] \[\therefore \] \[{{2}^{ab}}={{2}^{ab}};\] \[\forall a,\] \[b\in {{Z}^{+}}\] \[\Rightarrow \] \[a*b=b*a;\] \[\forall a,\] \[b\in {{Z}^{+}}\] \[\therefore \] Thus, the operation \['*'\] is commutative. Now consider, \[1,\,\,2,\,\,3\in {{Z}^{+}}\] Then, \[(1*2)*3={{2}^{1\,\,\times \,\,2}}*3=4*3={{2}^{4\,\,\times \,\,3}}={{2}^{12}}\] and \[1*(2*3)=1*{{2}^{2\,\,\times \,\,3}}=1*{{2}^{6}}\] \[=1*64={{2}^{1\,\,\times \,\,64}}={{2}^{64}}\] \[\because \] \[(1*2)*3\ne 1*(2*3),\] where \[1,\,\,2,\,\,3\in {{Z}^{+}}\] \[\therefore \] The operation \['*'\] is not associative (ii) On Q, operation \[*\] is defined by \[a*b=\frac{ab}{2}.\] Here, \[a\,\,b\in Q;\forall a,\,\,b\in Q,\] therefore \[a*b=\frac{ab}{2}\in Q.\] Thus, the operation \['*'\] is a binary operation. We know that, \[ab=ba;\,\,\forall a,\,\,b\in Q\] \[\therefore \] \[\frac{ab}{2}=\frac{ba}{2};\] \[\forall \,\,a,\] \[b\in Q\] \[\Rightarrow \] \[a*b=b*a;\] \[\forall a,\] \[b\in Q\] \[\therefore \] The operation \['*'\] is commutative Now, let \[a,\,\,b,\,\,c\in Q\] be any arbitrary elements. Then, \[(a*b)*c=\left( \frac{ab}{2} \right)*c=\frac{\left( \frac{ab}{2} \right)c}{2}=\frac{abc}{4}\] and \[a*(b*c)=a*\left( \frac{bc}{2} \right)=\frac{\frac{a(bc)}{2}}{2}=\frac{abc}{4}\] \[\because \] \[(a*b)*c=a*(b*c)\] \[\therefore \] The operation \['*'\] is associative.
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