MCQ 1511 Mark
The domain of definition of $\text{f(x)}=\sqrt{\frac{\text{x}+3}{(2-\text{x})(\text{x}-5)}}$ is:
- ✓$(-\infty,-3]\cup(2,5)$
- B$(-\infty,-3]\cup(2,5)$
- C$(-\infty,-3]\cup[2,5]$
- DNone of these.
Answer
View full question & answer→Correct option: A.
$(-\infty,-3]\cup(2,5)$
$\text{f(x)}=\sqrt{\frac{\text{x}+3}{(2-\text{x})(\text{x}-5)}}$
For $f(x)$ to be defined,
$(2-\text{x})(\text{x}-5)\neq0$
$\Rightarrow\text{x}\neq2,5\ ...(\text{i})$
Also, $\frac{(\text{x}+3)}{(2-\text{x})(\text{x}-5)}\geq0$
$\Rightarrow\frac{(\text{x}+3)(2-\text{x})(\text{x}-5)}{(2-\text{x})^2(\text{x}-5)^2}\geq0$
$\Rightarrow(\text{x}+3)(\text{x}-2)(\text{x}-5)\leq0$
$\Rightarrow\text{x}\in\big(-\infty,-3\big]\cap(2,5)\ ...(\text{ii})$
From $(i)$ and $(ii)$
$\text{x}\in\big(-\infty,-3\big]\cup(2,5)$
For $f(x)$ to be defined,
$(2-\text{x})(\text{x}-5)\neq0$
$\Rightarrow\text{x}\neq2,5\ ...(\text{i})$
Also, $\frac{(\text{x}+3)}{(2-\text{x})(\text{x}-5)}\geq0$
$\Rightarrow\frac{(\text{x}+3)(2-\text{x})(\text{x}-5)}{(2-\text{x})^2(\text{x}-5)^2}\geq0$
$\Rightarrow(\text{x}+3)(\text{x}-2)(\text{x}-5)\leq0$
$\Rightarrow\text{x}\in\big(-\infty,-3\big]\cap(2,5)\ ...(\text{ii})$
From $(i)$ and $(ii)$
$\text{x}\in\big(-\infty,-3\big]\cup(2,5)$
