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Relations and Functions — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsRelations and Functions1 Mark
MCQ
Let $\text{f(x)}=\frac{1}{1-\text{x}}.$ Then$, \{fo(fof)\}(x):$
  • ✓
    $x$ for all $\text{x}\in\text{R}$
  • B
    $x$ for all $\text{x}\in\text{R}-\{1\}$
  • C
    $x$ for all $\text{x}\in\text{R}-\{0,1\}$
  • D
    None of these.
✓

Answer

Correct option: A.
$x$ for all $\text{x}\in\text{R}$
Domain of $f: 1-\text{x}\neq0$
$\Rightarrow\ \text{x}\neq1$
Domain of $f = R - \{1\}$
Range of $f:\ \text{y}=\frac{1}{1-\text{x}}$
$\Rightarrow\ 1-\text{x}=\frac{1}{\text{y}}$
$\Rightarrow\ \text{x}=1-\frac{1}{\text{y}}$
$\Rightarrow\ \text{y}\neq0$
Range of $f = R - \{0\}$
So$, f : R - \{1\} \rightarrow R - \{0\}$ and $f : R - \{1\} \rightarrow R - \{0\}$
Range of $f$ is not a subset of the domain of $f.$
Domain $\text{(fof)} = \{x : \text{x}\in$ domain of $f$ and $\text{f(x)}\in$ domain of $f\}$
Domain $\text{(fof)} =\big\{\text{x}:\text{x}\in\text{R}-\{1\}$ and $\frac{1}{1-\text{x}}\in\text{R}-\{1\}\big\}$
Domain $\text{(fof)} =\big\{\text{x}:\text{x}\neq1$ and $\frac{1}{1-\text{x}}\neq1\big\}$
Domain $\text{(fof)} =\{\text{x}:\text{x}\neq1$ and $1-\text{x}\neq1\}$
Domain $\text{(fof)} =\{\text{x}:\text{x}\neq1$ and $\text{x}\neq0\}$
Domain $\text{(fof)} = R - \{0, 1\}$
$\text{(fof)(x) = f(f(x))}$
$=\text{f}\Big(\frac{1}{1-\text{x}}\Big)=\frac{1}{1-\frac{1}{1-\text{x}}}=\frac{1-\text{x}}{1-\text{x}-1}$
$=\frac{1-\text{x}}{-\text{x}}=\frac{\text{x}-1}{\text{x}}$
For range of fof, $\text{x}\neq0$
Now$, \text{fof} : R \rightarrow \{0, 1\} \rightarrow R - \{0\}$ and $f : R - {1} \rightarrow R - \{0\}$
Range of fof is not a subset of domain of $f.$
Domain $\text{(fo(fof))} =\{\text{x}:\text{x}\in$ domain of $\text{fof}$ and $(fof)(x) \in$ domain of f$\}$
Domain $\text{(fo(fof))} =\Big\{\text{x}:\text{x}\in\text{R}-\{0,1\}$ and $\frac{\text{x}-1}{\text{x}}\in\text{R}-\{1\}\Big\}$
Domain $\text{(fo(fof))} =\Big\{\text{x}:\text{x}\neq0,1$ and $\frac{\text{x}-1}{\text{x}}\neq1\Big\}$
Domain $\text{(fo(fof))} =\{\text{x}:\text{x}\neq0,1$ and $\text{x}-1\neq\text{x}\}$
Domain $\text{(fo(fof))} =\{\text{x}:\text{x}\neq0,1$ and $\text{x}\in\text{R}\}$
Domain $\text{(fo(fof))} = R - \{0, 1\}$
Domain $\text{(fo(fof)) = f((fof)(x))}$
$=\text{f}\Big(\frac{\text{x}-1}{\text{x}}\Big)$
$=\frac{1}{1-\frac{\text{x}-1}{\text{x}}}$
$=\frac{\text{x}}{\text{x}-\text{x}+1}$
$=\text{x}$
So$, \text{(fo(fof))(x) = x},$ where $\text{x}\neq0,1$

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