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The range of the real valued function $f(x)=\large\frac{x^2+x+2}{x^2+x+1}$ is

$\begin{array}{1 1} (1,\infty ) \\ (1, \large\frac{11}{7} ] \\ (1, \large\frac{7}{3} ] \\ (1,\large\frac{7}{5} ] \end{array}$

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Ans-(C)
Let $y=f(x)=\large\frac{x^2+x+2}{x^2+x+1}$
$\Rightarrow\:y(x^2+x+1)=x^2+x+2$
$\Rightarrow\:x^2(y-1)-x(y-1)+(y-2)=0$
Since it is real valued function, x is real.
$\Rightarrow\:$ Discriminent of the equation is $\geq 0$
$\Rightarrow\:(y-1)^2-4(y-1)(y-2)\geq 0$
$\Rightarrow 3y^2-10y+7\leq 0$
$\Rightarrow y\in[1,\frac{7}{3}]$ but $y\neq 1$ for any x
$\Rightarrow Range=(1,\frac{7}{3}]$
answered May 7, 2013 by rvidyagovindarajan_1
edited May 17, 2014 by rohanmaheshwari0831_1
 

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