TS EAMCET · Maths · Functions
\(\left\{x \in R: \frac{2 x-1}{x^3+4 x^2+3 x} \in R\right\}\) equals
- A \(R-\{0\}\)
- B \(R-\{0,1,3\}\)
- C \(R-\{0,-1,-3\}\)
- D \(R-\left\{0,-1,-3,+\frac{1}{2}\right\}\)
Answer & Solution
Correct Answer
(C) \(R-\{0,-1,-3\}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { Let } A=\left\{x \in R: \frac{2 x-1}{x^3+4 x^2+3 x}\right\} \\ & \text { Now, } x^3+4 x^2+3 x=x\left(x^2+4 x+3\right) \\ & =x(x+3)(x+1) \\ & \therefore \quad A=R-\{0,-1,-3\} \\ & \end{aligned}\)
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