Solve this following
Question:

Let $f: \mathrm{R} \rightarrow \mathrm{R}$ be defined as $f(\mathrm{x})=2 \mathrm{x}-1$ and

$g: R-\{1\} \rightarrow R$ be defined as $g(x)=\frac{x-\frac{1}{2}}{x-1}$

Then the composition function $f(\mathrm{~g}(\mathrm{x}))$ is :

 

  1. onto but not one-one

  2. both one-one and onto

  3. one-one but not onto

  4. neither one-one nor onto


Correct Option: 3,

Solution:

$f(\mathrm{~g}(\mathrm{x}))=2 \mathrm{~g}(\mathrm{x})-1=2\left(\frac{2 \mathrm{x}-1}{2(\mathrm{x}-1)}\right)-1$

$=\frac{x}{x-1}=1+\frac{1}{x-1}$

Range of $f(\mathrm{~g}(\mathrm{x})=\mathbb{R}-\{1\}$

Range of $f(\mathrm{~g}(\mathrm{x}))$ is not onto

$\& f(g(x))$ is one-one

So $f(g(x))$ is one-one but not onto.

 

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