question_answer

The domain of the function \[f(x)=\sqrt{\frac{(x+1)(x-3)}{(x-2)}}\]is

A)  \[(-1,2)\cup [3,\infty )\]

B)  \[[-1,2)\cup [3,\infty )\]

C)  \[]-\infty ,-1]\cup [3,\infty [\] 

D)  None of these

Correct Answer: B

Solution :

 \[f(x)=\sqrt{\frac{(x+1)(x-3)}{(x-2)}}=\frac{\sqrt{(x+1)(x-3)(x-2)}}{(x-2)}\] \[f(x)\]will be define when\[x-2\ne 0\Rightarrow x\ne 2\] and     \[(x+1)(x-3)(x-2)\ge 0\] \[\therefore \]Domain\[=[-1,2]\cup [3,\infty ]-\{2\}[-1,2]\cup [3,\infty ]\]