question_answer

If \[f:R-\{2\}\to R-\{3\}\] is defined by

\[f(x)=\frac{3x+1}{x-2},\]  where R is the set of real numbers,

show that f is invertible and hence find the value of \[{{f}^{-1}}.\]

OR

Let \[f:N\to R\] be a function defined as \[f(x)=4{{x}^{2}}+12x+15.\] Show that \[f:N\to \] range f is invertible. Find the inverse of \[{{f}^{-1}}\].

Answer:

Given, \[f\,(x)=\frac{3x+1}{x-2}\]and \[f\,(x);R-\{z\}\to R-\{z\}\]

For one-one Let\[f\,({{x}_{1}})=f\,({{x}_{2}}),\]for some \[{{x}_{1}},\]\[{{x}_{2}}\in A\]

\[\Rightarrow \]   \[\frac{3{{x}_{1}}+1}{{{x}_{1}}-2}=\frac{3{{x}_{2}}+1}{{{x}_{2}}-2}\]

\[\Rightarrow \]   \[(3{{x}_{1}}+1)\,({{x}_{2}}-2)=(3{{x}_{2}}+1)\,({{x}_{1}}-2)\]

\[\Rightarrow \]\[3{{x}_{1}}{{x}_{2}}-6{{x}_{1}}+{{x}_{2}}-2=3{{x}_{1}}{{x}_{2}}-6{{x}_{2}}+{{x}_{1}}-2\]

\[\Rightarrow \]   \[-\,6{{x}_{1}}+{{x}_{2}}=-\,6{{x}_{2}}+{{x}_{1}}\]

\[\Rightarrow \]   \[-\,6{{x}_{1}}-{{x}_{1}}=-\,6{{x}_{2}}-{{x}_{2}}\]

\[\Rightarrow \]   \[-\,7{{x}_{1}}-7{{x}_{2}}\]

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

So, \[f\,(x)\] is one-one function.

For onto

Let                    \[y=\frac{3x+1}{x-2},\]then

\[\Rightarrow \]   \[xy-2y=3x+1\]

\[\Rightarrow \]   \[xy-3x=2y+1\]

\[\Rightarrow \]   \[x\,(y-3)=2y+1\]

\[\Rightarrow \]   \[x=\frac{2y+1}{y-3}\]

Since, \[x\in R-\{2\},\,\,\forall \in R-\{3\}\].

So, range of\[f\,(x)=R-\{z\}\]

\[\therefore \]      Range = Codomain

So, \[f\,(x)\] is onto function. (1)

Also, from Eq. (i), we get

\[{{f}^{-\,1}}(y)=\frac{2y+1}{y-3}\]                  \[[\because x={{f}^{-\,1}}(y)]\]

Or         \[{{f}^{-\,1}}(x)=\frac{2x+1}{x-3}\]

OR

We have a mapping \[f:N\to R\] defined as \[f\,(x)=4{{x}^{2}}+12x+15\]

To show \[f:N\to R\] range \[(f)\] is invertible. For this it is sufficient to prove that f is one-one.

[\[\because \]range \[(f)\]= codomain \[(f)\Rightarrow f\] is onto]

For one-one

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

Then, \[4x_{1}^{2}+12{{x}_{1}}+15=4x_{2}^{2}+12{{x}_{2}}+15\]

\[\Rightarrow \]   \[4\,(x_{1}^{2}-x_{2}^{2})=12\,({{x}_{2}}-{{x}_{1}})\]

\[\Rightarrow \]   \[(4\,({{x}_{1}}+{{x}_{2}})+12)\,\,({{x}_{1}}-{{x}_{2}})=0\]

\[\Rightarrow \]   \[{{x}_{1}}-{{x}_{2}}=0\]

\[[\because 4{{x}_{1}}+4{{x}_{2}}+12\ne 0\,\,as\,\,{{x}_{1}},\,\,{{x}_{2}}\in N]\]

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

\[\Rightarrow \]   f is one-one.

Hence \[f:N\to \] range \[(f)\] is invertible.

Now, to find inverse of f ,

let         \[y=f\,(x)\Rightarrow y=4{{x}^{2}}+12x+15\]

\[\therefore \]      \[4{{x}^{2}}+12x+15-y=0\]

Now,     \[x=\frac{-\,12\,\,\pm \sqrt{144-4\,(4)\,\,(15-y)}}{8}\]

[by Sridharacharya?s formula]

\[=\frac{-\,12\,\,\pm \,\,4\sqrt{9-(15-y)}}{8}\]

\[=\frac{-\,3\pm \sqrt{y-6}}{2}\]

But       \[x\ne \frac{-\,3-\sqrt{y-6}}{2}\]         \[[\because x\in N]\]

\[\therefore \]      \[x=\frac{-\,3+\sqrt{y-6}}{2}\]

\[\Rightarrow {{f}^{-\,1}}(y)=\frac{-\,3+\sqrt{y-6}}{2}\]

So, \[{{f}^{-\,1}}:\]range \[(f)\to N\]defined as\[{{f}^{-\,1}}(x)=\frac{-\,3+\sqrt{x-6}}{2}\].