Let f be a real-valued differentiable function on R (the set of all real numbers) such that f(1) = 1. If the y-intercept of the tangent at any point P(x, y) on the curve y = f(x) is equal to the cube of the abscissa of P, then the value of f(-2) is equal to:

y - y1 = m(x - x1)
Put x = 0, to get y intercept
y1 - mx1 = x1³
y1 - x1(dy/dx) = x1³
(dy/dx) - y/x = -x²
Let y = vx
dy/dx = v + x(dv/dx)
v + x(dv/dx) - v = -x²
x(dv/dx) = -x²
dv/dx = -x
∫dv = ∫-x dx
v = -x²/2 + c
y/x = -x²/2 + c
y = -x³/2 + cx
Since f(1) = 1, 1 = -1/2 + c, c = 3/2
y = -x³/2 + (3/2)x
f(x) = -x³/2 + (3/2)x
f(-2) = -(-2)³/2 + (3/2)(-2) = 4 - 3 = 1
This solution is incorrect. Let's try again.
y - y₁ = m(x - x₁)
Put x = 0 to get y intercept
y₁ - mx₁ = x₁³
y₁ - x₁(dy/dx) = x₁³
dy/dx - y/x = -x²
This is a linear differential equation.
Integrating factor = e∫-dx/x = e⁻lnx = 1/x
Multiplying by 1/x,
(1/x)dy/dx - y/x² = -x
d/dx(y/x) = -x
y/x = ∫-xdx = -x²/2 + c
y = -x³/2 + cx
Given that f(1) = 1, we have 1 = -1/2 + c, so c = 3/2
f(x) = -x³/2 + (3/2)x
f(-2) = -(-8)/2 + (3/2)(-2) = 4 - 3 = 1
However, if we use the condition that y-intercept is x³, then
y - y₁ = (dy/dx)ₓ₁ (x - x₁)
When x = 0, y = x₁³
x₁³ - y₁ = -(dy/dx)ₓ₁ x₁
dy/dx = (y - x³)/x
Let y = vx
dy/dx = v + x dv/dx
v + x dv/dx = (vx - x³)/x = v - x²
x dv/dx = -x²
dv = -x dx
v = -x²/2 + c
y/x = -x²/2 + c
y = -x³/2 + cx
Using f(1) = 1: 1 = -1/2 + c, c = 3/2
y = -x³/2 + (3/2)x
f(-2) = -(-8)/2 + (3/2)(-2) = 4 - 3 = 1
Another approach:
Let the equation of the tangent at (x,y) be Y - y = f'(x)(X - x)
The y-intercept is obtained when X = 0: Y = y - xf'(x)
Given that this is x³, we have y - xf'(x) = x³
y = xf'(x) + x³
f(x) = x f'(x) + x³
This is a first order linear DE, which can be solved using an integrating factor.
This leads to f(x) = -x³/2 + cx
Using f(1) = 1, we get 1 = -1/2 + c, so c = 3/2
f(x) = -x³/2 + (3/2)x
f(-2) = 4 - 3 = 1
Therefore, f(-2) = 1