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If $x$ is real, then the range of $\frac{x^2+2 x+1}{x^2+2 x+7}$ is
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Verified Answer
The correct answer is:
$[0,1)$
$$
\begin{array}{lc}
\text {} & \text { Let } \frac{x^2+2 x+1}{x^2+2 x+7}=y, \\
\because & y \neq 1 \\
\Rightarrow & (y-1) x^2+2(y-1) x+(7 y-1)=0 \\
\because & x \in R \\
\text { so, } & D \geq 0 \\
\Rightarrow & 4(y-1)^2-4(y-1)(7 y-1) \geq 0 \\
\Rightarrow & (y-1)[y-1-7 y+1] \geq 0 \\
\Rightarrow & y(y-1) \leq 0 \\
\Rightarrow & y \in[0,1]
\end{array}
$$
From Eqs. (i) and (ii), we are getting
$$
y \in[0,1)
$$
\begin{array}{lc}
\text {} & \text { Let } \frac{x^2+2 x+1}{x^2+2 x+7}=y, \\
\because & y \neq 1 \\
\Rightarrow & (y-1) x^2+2(y-1) x+(7 y-1)=0 \\
\because & x \in R \\
\text { so, } & D \geq 0 \\
\Rightarrow & 4(y-1)^2-4(y-1)(7 y-1) \geq 0 \\
\Rightarrow & (y-1)[y-1-7 y+1] \geq 0 \\
\Rightarrow & y(y-1) \leq 0 \\
\Rightarrow & y \in[0,1]
\end{array}
$$
From Eqs. (i) and (ii), we are getting
$$
y \in[0,1)
$$
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