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(A) The slope of the tangent to the curve at any point $(x, y)$ is given by $\frac{dy}{dx} = 1 - \frac{1}{x^2}$.Integrating both sides with respect to

Vedclass · Accepted Answer

$y = x + \frac{1}{x}$

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(A) The slope of the tangent to the curve at any point $(x, y)$ is given by $\frac{dy}{dx} = 1 - \frac{1}{x^2}$.Integrating both sides with respect to $x$:$\int dy = \int \left( 1 - \frac{1}{x^2} \right) dx$$y = x - (-\frac{1}{x}) + C$$y = x + \frac{1}{x} + C$Since the curve passes through the point $(1, 2)$,we substitute $x = 1$ and $y = 2$ to find the constant $C$:$2 = 1 + \frac{1}{1} + C$$2 = 1 + 1 + C$$2 = 2 + C$$C = 0$Therefore,the equation of the curve is $y = x + \frac{1}{x}$.

If a curve passes through $(1, 2)$ and has the slope of its tangent $1 - \frac{1}{x^2}$ at any point $(x, y)$,then the equation of that curve is:

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