If the tangent at the point $$(x_1, y_1)$$ on the curve $$y = x^3 + 3x^2 + 5$$ passes through the origin, then $$(x_1, y_1)$$ does NOT lie on the curve

We are given the curve $$y = x^3 + 3x^2 + 5$$ and the tangent at $$(x_1, y_1)$$ passes through the origin.

Find the tangent equation.

The slope at $$(x_1, y_1)$$ is $$y' = 3x_1^2 + 6x_1$$.

The tangent line: $$y - y_1 = (3x_1^2 + 6x_1)(x - x_1)$$.

Apply the condition that the tangent passes through the origin.

Substituting $$(0, 0)$$: $$-y_1 = (3x_1^2 + 6x_1)(-x_1)$$

$$y_1 = 3x_1^3 + 6x_1^2$$

Use the fact that $$(x_1, y_1)$$ lies on the curve.

$$y_1 = x_1^3 + 3x_1^2 + 5$$

Equating: $$x_1^3 + 3x_1^2 + 5 = 3x_1^3 + 6x_1^2$$

$$2x_1^3 + 3x_1^2 - 5 = 0$$

Solve the cubic.

Testing $$x_1 = 1$$: $$2 + 3 - 5 = 0$$ ✓

Factoring: $$(x_1 - 1)(2x_1^2 + 5x_1 + 5) = 0$$

The discriminant of $$2x_1^2 + 5x_1 + 5$$ is $$25 - 40 = -15 < 0$$, so the only real solution is $$x_1 = 1$$.

$$y_1 = 1 + 3 + 5 = 9$$. So $$(x_1, y_1) = (1, 9)$$.

Check which curve (1, 9) does NOT lie on.

Option A: $$1 + \frac{81}{81} = 1 + 1 = 2$$ ✓

Option B: $$\frac{81}{9} - 1 = 9 - 1 = 8$$ ✓

Option C: $$4(1) + 5 = 9$$ ✓

Option D: $$\frac{1}{3} - 81 = -\frac{242}{3} \neq 2$$ ✗

Answer: Option D