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Let $f: R \rightarrow R$ be defined as $f(x)=\frac{x^{2}-x+4}{x^{2}+x+4}$. Then, range of the function $f(x)$ is
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The correct answer is:
$\left[\frac{3}{5}, \frac{5}{3}\right]$
Let $y=\frac{x^{2}-x+4}{x^{2}+x+4}$
$\Rightarrow \quad x^{2} y+x y+4 y=x^{2}-x+4$
$\Rightarrow(y-1) x^{2}+(y+1) x+4 y-4=0$
For $x$ to be real, discriminant of the above quadratic equation should be greater than equal to 0
$\\ \Rightarrow \quad(y+1)^{2}-4(y-1)(4 y-4) \geq 0$
$\Rightarrow \quad(y+1)^{2}-16(y-1)^{2} \geq$
$\Rightarrow(y+1+4 y-4)(y+1-4 y+4) \geq$
$(5 y-3)(5-3 y) \geq 0$
$\therefore \quad y \in\left[\frac{3}{5}, \frac{5}{3}\right]$
$\Rightarrow \quad x^{2} y+x y+4 y=x^{2}-x+4$
$\Rightarrow(y-1) x^{2}+(y+1) x+4 y-4=0$
For $x$ to be real, discriminant of the above quadratic equation should be greater than equal to 0
$\\ \Rightarrow \quad(y+1)^{2}-4(y-1)(4 y-4) \geq 0$
$\Rightarrow \quad(y+1)^{2}-16(y-1)^{2} \geq$
$\Rightarrow(y+1+4 y-4)(y+1-4 y+4) \geq$
$(5 y-3)(5-3 y) \geq 0$
$\therefore \quad y \in\left[\frac{3}{5}, \frac{5}{3}\right]$
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