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$\bigg \{ x \in R :\large\frac{2x-1}{x^3+4x^2+3x} \in $$R \bigg \} =$

\[\begin {array} {1 1} (1)\;R -\{ 0 \} & \quad (2)\;R- \{ 0,1,3\} \\ (3)\;R-\{0,-1,-3\} & \quad (4)\;IR -\{0,-1,-3,+\frac{1}{2}\} \end {array}\]

1 Answer

$x^3+4x^2+3x= x(x^2+4x+3)$
On factorizing this we get $x =0$ and $x = -1$ and $-3.$
Hence $R- \{0.-1,-3\}$ is the correct option.
c is the correct answer.
answered Nov 3, 2013 by vijayalakshmi_ramakrishnans
edited Jan 2, 2014 by meena.p
 
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