The gradient of the curve C is given by dy/dx = (3x - 1)² - Edexcel - A-Level Maths Pure - Question 10 - 2005 - Paper 2

Starting from the gradient equation:

$$\frac{dy}{dx} = (3x - 1)^{2},$$

we integrate to find y:

$$y = \int (3x - 1)^{2}dx.$$

To integrate, we expand the square:

$$(3x - 1)^{2} = 9x^{2} - 6x + 1.$$

Integrating term-by-term gives:

$$y = \int (9x^{2} - 6x + 1)dx = \frac{9}{3}x^{3} - \frac{6}{2}x^{2} + x + C = 3x^{3} - 3x^{2} + x + C.$$

Now, substituting the point (1, 4) to find C:

$$4 = 3(1)^{3} - 3(1)^{2} + 1 + C,$$

which simplifies to:

$$4 = 3 - 3 + 1 + C \ C = 0.$$

Thus, the equation of the curve C is:

$$y = 3x^{3} - 3x^{2} + x.$$

The gradient of the line y = 1 - 2x is -2. For the tangent to be parallel to this line, we must have:

$$\frac{dy}{dx} = -2.$$

Setting the gradient of the curve equal to -2:

$$(3x - 1)^{2} = -2.$$

Since the left-hand side (a square) is always non-negative, it cannot equal -2. Hence, there is no point on C where the tangent is parallel to the line y = 1 - 2x.