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\[=[-1,2]\cup [3,\infty ]-\{2\}[-1,2]\cup [3,\infty ]\]You need to login to perform this action.
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