Question

Let \( Y = Y(X) \) be a curve lying in the first quadrant such that the area enclosed by the line \( Y - y = Y'(x) (X - x) \) and the coordinate axes, where \( (x, y) \) is any point on the curve, is always\[\frac{-y^2}{2Y'(x)} + 1, \quad Y'(x) \neq 0.\]If \( Y(1) = 1 \), then \( 12Y(2) \) equals ______.

Answer 1

Correct Answer: 20

The line \( Y - y = Y'(x)(X - x) \) represents the tangent to \( Y = Y(X) \) at \( (x, y) \).

The area \( A = -\frac{Y(x)^2}{2Y'(x)} + 1 \) relates \( y \) and \( Y'(x) \) on the curve.

Differentiate with respect to \( x \) and solve for \( Y(x) \) using the initial condition \( Y(1) = 1 \).

\[ 1 = \frac{2}{3} + c \]

\[ c = \frac{1}{3} \]

\[ Y = \frac{2}{3} \times \frac{1}{X} + \frac{1}{3}X^2 \]

\[ 12Y(2) = \frac{5}{3} \times 12 = 20 \]