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If $f(x)=x^2-2 x+4$, then the set of values of $x$ satisfying $f(x-1)=f(x+1)$ is
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Verified Answer
The correct answer is:
$\{1\}$
Given, $f(x)=x^2-2 x+4$
Now, $f(x-1)=f(x+1)$
$$
\begin{aligned}
& \Rightarrow \quad(x-1)^2-2(x-1)+4=(x+1)^2-2(x+1)+4 \\
& \Rightarrow \quad x^2-2 x+1-2 x+2+4 \\
& =x^2+2 x+1-2 x-2+4 \\
& \Rightarrow x^2-4 x+7=x^2+3 \\
& \Rightarrow \quad \quad-4 x=-4 \\
& \Rightarrow \quad x=1 \\
& \therefore \quad \quad x=\{1\}
\end{aligned}
$$
Now, $f(x-1)=f(x+1)$
$$
\begin{aligned}
& \Rightarrow \quad(x-1)^2-2(x-1)+4=(x+1)^2-2(x+1)+4 \\
& \Rightarrow \quad x^2-2 x+1-2 x+2+4 \\
& =x^2+2 x+1-2 x-2+4 \\
& \Rightarrow x^2-4 x+7=x^2+3 \\
& \Rightarrow \quad \quad-4 x=-4 \\
& \Rightarrow \quad x=1 \\
& \therefore \quad \quad x=\{1\}
\end{aligned}
$$
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