Let $y = f(x)$ be a curve in the first quadrant such that the triangle formed by the co-ordinate axis and the tangent at any point on the curve has area 2. If $f(1) = 1$, then $y(2) = $
Step-by-Step Solution
Key Concept: Intercept condition $OAB = 2$ translates to a differential equation in Clairaut-like form, solvable by treating $p$ as a parameter.
The tangent at point $P(x, y)$ is $Y - y - (X - x)y_1 = 0$, meeting the x-axis at $A$ and y-axis at $B$. Given $OAB = 2 \Rightarrow OAOB = 4$. Setting up $(x - rac{y}{p})(y - xp) = 4p$ where $p = rac{dy}{dx}$, we get $(y - xp)^2 = -4p$, leading to $y = xp + 2\sqrt{-p}$. The general solution is $y = cx + 2\sqrt{-c}$. Using the condition $y(2) = 0$, we find $c = -1$, giving $y = -x + 2 + rac{1}{x} = rac{1}{x}$ after eliminating $c$.
Correct Answer: 1,4