MCQ
The domain of the function $\text{f(x)}=\sqrt{\frac{(\text{x}+1)(\text{x}-3)}{\text{x}-2}}$ is:
- ✓$\big[-1,2\big)\cap\big[3,\infty\big)$
- B$(-1,2)\cap[3,\infty)$
- C$[-1,2]\cap[3,\infty]$
- DNone of these.
✓
Answer
Correct option: A.
$\big[-1,2\big)\cap\big[3,\infty\big)$
$\text{f(x)}=\sqrt{\frac{(\text{x}+1)(\text{x}-3)}{\text{x}-2}}$
For f(x) to be defined,
$(\text{x}-2)\neq0$
$\Rightarrow\text{x}\neq2\ ...(\text{i})$
Also,
$\frac{(\text{x}+1)(\text{x}-3)}{\text{x}-2}\geq0$
$\Rightarrow\frac{(\text{x}+1)(\text{x}-3)(\text{x}-2)}{(\text{x}-2)^2}\geq0$
$\Rightarrow(\text{x}+1)(\text{x}-3)(\text{x}-2)\geq0$
$\Rightarrow\text{x}\in\big[-1,2\big)\cup\big[3,\infty\big)\ ...(\text{ii})$
From (i) and (ii),
$\text{x}\in\big[-1,2\big)\cap\big[3,\infty\big)$
For f(x) to be defined,
$(\text{x}-2)\neq0$
$\Rightarrow\text{x}\neq2\ ...(\text{i})$
Also,
$\frac{(\text{x}+1)(\text{x}-3)}{\text{x}-2}\geq0$
$\Rightarrow\frac{(\text{x}+1)(\text{x}-3)(\text{x}-2)}{(\text{x}-2)^2}\geq0$
$\Rightarrow(\text{x}+1)(\text{x}-3)(\text{x}-2)\geq0$
$\Rightarrow\text{x}\in\big[-1,2\big)\cup\big[3,\infty\big)\ ...(\text{ii})$
From (i) and (ii),
$\text{x}\in\big[-1,2\big)\cap\big[3,\infty\big)$
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