Question

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

• A. \(y = 3x - \frac{1}{x}\)
• B. \(y = x + \frac{1}{x}\)
• C. \(y = 2x + \frac{1}{x} - 1\)
• D. \(y = x + \frac{2}{x} - 1\)

Hint:

When the slope (dy/dx) is given, always integrate to find the original curve. Use initial conditions to find the constant.

Answer 1

The Correct Option is B

We are given the slope of the tangent: \[ \frac{dy}{dx} = 1 - \frac{1}{x^2} \] To find the curve, integrate: \[ \int \frac{dy}{dx} \, dx = \int \left(1 - \frac{1}{x^2}\right) dx = x + \frac{1}{x} + C \] So the general solution is: \[ y = x + \frac{1}{x} + C \] Now use the point \((1, 2)\) to find \(C\): \[ 2 = 1 + 1 + C \Rightarrow C = 0 \] Hence, the required equation is: \[ y = x + \frac{1}{x} \] \[ \boxed{y = x + \frac{1}{x}} \]