The given curve is \(y = 2kx^2 + x + 1\). The gradient of the curve at any point \(x\) is given by the derivative of the curve with respect to \(x\), which is \(\frac{\mathrm d y}{\mathrm d x} = 4kx + 1\). At \(x = 1\), the gradient of the curve is \(4k + 1\). We are given that this value is equal to 9, so we have:

4k + 1 = 9.

Solving for \(k\), we find that:

\(k = \frac{9 - 1}{4} = \frac{8}{4} = 2\)

So, the value of \(k\) is 2.