A) \[R-\{0\}\]
B) \[R-\{0,\,\,1,\,\,3\}\]
C) \[R-\{0,\,\,-1,\,\,-3\}\]
D) \[R-\left\{ 0,\,\,-1,\,\,-3,\,\,+\frac{1}{2} \right\}\]
Correct Answer: C
Solution :
Let\[A=\left\{ x\in R:\frac{2x-1}{{{x}^{3}}+4{{x}^{2}}+3x} \right\}\] Now,\[{{x}^{3}}+4{{x}^{2}}+3x=x({{x}^{2}}+4x+3)\] \[=x(x+3)(x+1)\] \[\therefore \] \[A=R-\{0,\,\,-1,\,\,-3\}\]
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