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If $x$ is real, then the value of $\frac{x^2-3 x+4}{x^2+3 x+4}$ lies in the interval
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Verified Answer
The correct answer is:
$\left[\frac{1}{7}, 7\right]$
$\begin{aligned} & \text { Let } y=\frac{x^2-3 x+4}{x^2+3 x+4} \\ & \Rightarrow x^2(y-1)+x(3 y+3)+(4 y-4)=0\end{aligned}$
Since, $x$ is real.
$\begin{array}{lc}
\therefore & D \geq 0 \\
\Rightarrow & B^2-4 A C \geq 0 \\
\Rightarrow & (3 y+3)^2-4(y-1)(4 y-4) \geq 0 \\
\Rightarrow & 9 y^2+9+18 y-16\left(y^2+1-2 y\right) \geq 0 \\
\Rightarrow & 7 y^2-50 y+7 \leq 0 \\
\Rightarrow & (7 y-1)(y-7) \leq 0 \\
\Rightarrow & \frac{1}{7} \leq y \leq 7
\end{array}$
Since, $x$ is real.
$\begin{array}{lc}
\therefore & D \geq 0 \\
\Rightarrow & B^2-4 A C \geq 0 \\
\Rightarrow & (3 y+3)^2-4(y-1)(4 y-4) \geq 0 \\
\Rightarrow & 9 y^2+9+18 y-16\left(y^2+1-2 y\right) \geq 0 \\
\Rightarrow & 7 y^2-50 y+7 \leq 0 \\
\Rightarrow & (7 y-1)(y-7) \leq 0 \\
\Rightarrow & \frac{1}{7} \leq y \leq 7
\end{array}$
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