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 :
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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