To find the value of \(k\), we set the curve equation equal to the line equation since the line is tangent to the curve:

\(2x^2 + kx + k - 1 = 2x + 3\)

Rearrange to form a quadratic equation:

\(2x^2 + (k-2)x + (k-4) = 0\)

For the line to be tangent to the curve, the discriminant of the quadratic must be zero:

\(b^2 - 4ac = 0\)

Here, \(a = 2\), \(b = k-2\), \(c = k-4\). Substitute these into the discriminant formula:

\((k-2)^2 - 4 imes 2 imes (k-4) = 0\)

Simplify:

\((k-2)^2 = 8(k-4)\)

Expanding both sides:

\(k^2 - 4k + 4 = 8k - 32\)

Rearrange to form a quadratic equation:

\(k^2 - 12k + 36 = 0\)

Factorize:

\((k-6)^2 = 0\)

Thus, \(k = 6\).