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If a curve passes through $(1,2)$ and has the slope of its tangent $1-\frac{1}{x^2}$ at a point $(x, y)$, then the equation of that curve is
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Verified Answer
The correct answer is:
$y=x+\frac{1}{x}$
Given $\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{}$
$\begin{aligned}
& \Rightarrow \int \mathrm{dy}=\int\left(1-\frac{1}{\mathrm{x}^2}\right) \mathrm{dx} \\
& \Rightarrow \mathrm{y}=\mathrm{x}+\frac{1}{\mathrm{x}}+\mathrm{C}
\end{aligned}$
Since y $(1)=2 \Rightarrow C=0$
$y(x)=x+\frac{1}{x}$
$\begin{aligned}
& \Rightarrow \int \mathrm{dy}=\int\left(1-\frac{1}{\mathrm{x}^2}\right) \mathrm{dx} \\
& \Rightarrow \mathrm{y}=\mathrm{x}+\frac{1}{\mathrm{x}}+\mathrm{C}
\end{aligned}$
Since y $(1)=2 \Rightarrow C=0$
$y(x)=x+\frac{1}{x}$
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