question_answer

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.