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Range of $y=\large\frac{x+2}{2x^2+3x+6}$ if $x$ is real is :

$\begin{array}{1 1}(A)\;\big[-\large\frac{1}{13},\frac{1}{3}\big]&(B)\;\big[-\large\frac{1}{3},\frac{-1}{13}\big]\\(C)\;\big[-\large\frac{1}{3},\frac{1}{3}\big]&(D)\;\big[-\large\frac{1}{13},\frac{1}{13}\big]\end{array} $

1 Answer

$y=\large\frac{x+2}{2x^2+3x+6}$
$\Rightarrow 2yx^2+3yx+6y=x+2$
$\Rightarrow 2yx^2+(3y-1)x+6y-2$
$x$ is real
$\therefore D \geq 0$
$\Rightarrow (3y-1)^2-4(2y)(6y-2)\geq 0$
$\Rightarrow (3y-1)-4(2y)(6y-2)\geq 0$
$\Rightarrow (3y-1)(13y+1)\leq 0$
$\Rightarrow y\in \big[-\large\frac{1}{13},\frac{1}{3}\big]$
Hence (A) is the correct answer.
answered Apr 14, 2014 by sreemathi.v
 
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