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If $f: R \rightarrow A$ defined by $f(x)=\frac{1}{x^2+2 x+2}$, $\forall x \in R$ is surjective, then $A=$
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Verified Answer
The correct answer is:
$(0,1]$
Since, the quadratic expression
$$
\begin{aligned}
& x^2+2 x+2=(x+1)^2+1 \in[1, \infty), \forall x \in R \\
\Rightarrow \quad & \frac{1}{(x+1)^2+1} \in(0,1]
\end{aligned}
$$
For $f(x)=\frac{1}{x^2+2 x+2}, \forall x \in R$ is surjective, then set
$$
A=(0,1]
$$
$$
\begin{aligned}
& x^2+2 x+2=(x+1)^2+1 \in[1, \infty), \forall x \in R \\
\Rightarrow \quad & \frac{1}{(x+1)^2+1} \in(0,1]
\end{aligned}
$$
For $f(x)=\frac{1}{x^2+2 x+2}, \forall x \in R$ is surjective, then set
$$
A=(0,1]
$$
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