MCQ
Let $f(x) = x, \text{g(x)}=\frac{1}{\text{x}}$ and $h(x) = f(x)\ g(x).$ Then, $h(x) = 1$
- A$\text{x}\in\text{R}$
- B$\text{x}\in\text{Q}$
- C$\text{x}\in\text{R}-\text{Q}$
- ✓$\text{x}\in\text{R},\text{ x}\neq0$
✓
Answer
Correct option: D.
$\text{x}\in\text{R},\text{ x}\neq0$
Given,
$\text{f(x)}=\text{x},\text{ g(x)}=\frac{1}{\text{x}}$ and $\text{h(x)}=\text{f(x)}\ \text{g(x)}$
Now,
$\text{h(x)}=\text{x}\times\frac{1}{\text{x}}=1$
We observe that the domain of $f$ is $R$ and the domain of $g$ is $R - \{0\}$
$\therefore\ \text{Domain of h}=\text{Domain of f }\cap\text{ Domain of g}\\\ \ \ =\text{R }\cap\big[\text{R}-\{0\}\big]=\text{R}-\{0\}$
$\Rightarrow\text{x}\in\text{R},\text{ x}\neq0$
$\text{f(x)}=\text{x},\text{ g(x)}=\frac{1}{\text{x}}$ and $\text{h(x)}=\text{f(x)}\ \text{g(x)}$
Now,
$\text{h(x)}=\text{x}\times\frac{1}{\text{x}}=1$
We observe that the domain of $f$ is $R$ and the domain of $g$ is $R - \{0\}$
$\therefore\ \text{Domain of h}=\text{Domain of f }\cap\text{ Domain of g}\\\ \ \ =\text{R }\cap\big[\text{R}-\{0\}\big]=\text{R}-\{0\}$
$\Rightarrow\text{x}\in\text{R},\text{ x}\neq0$
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