question_answer

Consider the function \[f:{{R}^{+}}\to [4,\,\,\infty )\] defined by \[f(x)={{x}^{2}}+4,\] where \[{{R}^{+}}\] is the set of all non-negative real numbers. Show that f is invertible. Also, find the inverse of f.

OR

Show that the relation S in the set

\[A=\{x\in Z:0\,\,\le x\le 12\}\] given by

\[S=\{(a,\,\,b):a,\,\,b\in Z,\,\,|a-b|\]is divisible by 4} is an equivalence relation. Find the set of all elements related to 4.

Answer:

We have a mapping \[f:{{R}^{+}}\to [4,\,\,\infty )\]given by\[f\,(x)={{x}^{2}}+4\].

To prove f is invertible.

For f to be one-one

Let \[{{x}_{1}},\]\[{{x}_{2}}\in {{R}^{+}}\] be any arbitrary elements, such that

\[f\,({{x}_{1}})=f\,({{x}_{2}})\]

\[\Rightarrow \]   \[x_{1}^{2}+4=x_{2}^{2}+4\Rightarrow x_{1}^{2}-x_{2}^{2}=0\]

\[\Rightarrow \]\[({{x}_{1}}-{{x}_{2}})\,\,({{x}_{1}}+{{x}_{2}})=0\]

\[\Rightarrow \]\[{{x}_{1}}-{{x}_{2}}=0\]             \[[\because {{x}_{1}}+{{x}_{2}}\ne 0\,\,as\,\,{{x}_{1}},\,\,{{x}_{2}}\in {{R}^{+}}]\]

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

So, f is one-one.

For f to be onto

Let \[y\,\,\in [4,\,\,\infty )\]be any arbitrary element and let y\[y=f\,(x)\]

Then,                \[y={{x}^{2}}+4\]

\[{{x}^{2}}=y-4\]

\[{{x}^{2}}\pm \sqrt{y-4}\]

\[\therefore \]\[x\in {{R}^{+}},\]therefore \[x\ne -\sqrt{y-4}\]

Now, \[x=\sqrt{y-4}\in {{R}^{+}}\]

\[[\because 4\le y<\infty \Rightarrow 0\le y-4<\infty \Rightarrow 0\le \sqrt{y-4}<\infty ]\]

Thus, for each\[y\,\,\in [4,\,\,\infty )\] there exist \[x=\sqrt{y-4}\in {{R}^{+}}\]such that \[f\,(x)=y\]

So, f is onto.

\[\Rightarrow \]   f is invertible and inverse of f is given by

\[{{f}^{-\,1}}:[4,\,\,\infty ]\to {{R}^{+}},\]defined as

\[{{f}^{-\,1}}(y)=\sqrt{y-4}\]              Hence proved.

Or

The given relation is\[S=\{(a,\,\,b):|a-b|\]is divisible by 4, where \[a,\,\,b\in A\}\]

and       \[A=\{x:x\in Z\,\,and\,\,\le x\le 12\}\]

Now, A can be written as

\[A=\{0,\,\,1,\,\,2,\,\,3,\,\,....,\,\,12\}\]

Reflexive For any \[x\in A,\] we have \[|x-x|\,\,=0\]which is divisible by 4.

\[\Rightarrow \]               \[(x,\,\,x)\in S,\,\,\forall \,\,x\in A\]

So, S is reflexive.

Symmetric For any \[(x,\,\,y)\in S,\] we have \[|x-y|\]is divisible by 4.           [using definition of given relation]

\[\Rightarrow \]   \[|x-y|\,\,=4\lambda \]for some \[\lambda \in N\]

\[\Rightarrow \]   \[|y-x|\,\,=4\lambda \]for some \[\lambda \in N\]

\[\Rightarrow \]   \[(y,\,\,x)\in S\]

\[\therefore \]We have shown that\[(x,\,\,y)\in S\].

\[\Rightarrow \]               \[(y,\,\,x)\in S,\,\,\forall \,\,x,\,\,y\in A\]

So, S is symmetric.

Transitive For any \[(x,\,\,y)\in S,\]and \[(y,\,\,z)\in S,\]we have \[|x-y|\]is divisible by 4 and \[|y-z|\]is divisible by 4.

[using definition of given relation]

\[\Rightarrow \]   \[|x\,-y|\,\,=4\lambda \]and \[|y\,\,-z|\,\,=4\mu \]for some \[\lambda ,\]\[\mu \in N\]

\[\Rightarrow \]   \[(x-y)=\pm \,\,4\lambda \]and \[(y-z)=\pm \,\,4\mu \]

Now, \[(x-z)=(x-y)+(y-z)=(\pm \,\,4\lambda )+(\pm \,\,4\mu )\]

\[=\pm \,\,4\,\,(\lambda +\mu )\].

\[\Rightarrow \]   \[|x-z|\]is divisible by 4.

\[\Rightarrow \]               \[(x,\,\,z)\in S\]

\[\therefore \]We have shown that\[(x,\,\,y)\in S,\]and\[(y,\,\,z)\in S\]

\[\Rightarrow \]   \[(x,\,\,z)\in S,\,\,\forall \,\,x,\,\,y,\,\,x\in A\]

Since, S is reflexive, symmetric and transitive, so it is an equivalence relation.

Now, set of all elements related to 4,

\[[4]=\{x\in A:4Rx\}\]

\[=\{x\in A:|4-x|\,\,is\,\,divisible\,\,by\,\,4\}\]

\[=\{0,\,\,4,\,\,8,\,\,12\}\]