(a) Differentiate \(y = k \sqrt{4x + 1} - x + 5\) with respect to \(x\):

\(\frac{dy}{dx} = k \cdot \frac{1}{2}(4x + 1)^{-\frac{1}{2}} \cdot 4 - 1\)

\(\frac{dy}{dx} = \frac{2k}{\sqrt{4x+1}} - 1\)

(b) Set \(\frac{dy}{dx} = 0\) for stationary points:

\(\frac{2k}{\sqrt{4x+1}} - 1 = 0\)

\(\frac{2k}{\sqrt{4x+1}} = 1\)

\(\sqrt{4x+1} = 2k\)

\(4x + 1 = 4k^2\)

\(x = \frac{4k^2 - 1}{4}\)

(c) Given \(k = 10.5\), substitute into \(\frac{dy}{dx} = 2\):

\(\frac{2 \times 10.5}{\sqrt{4x+1}} - 1 = 2\)

\(\frac{21}{\sqrt{4x+1}} = 3\)

\(\sqrt{4x+1} = 7\)

\(4x + 1 = 49\)

\(x = 12\)

Find \(y\) at \(x = 12\):

\(y = 10.5 \sqrt{4 \times 12 + 1} - 12 + 5\)

\(y = 66.5\)

The slope of the normal is \(-\frac{1}{2}\) (perpendicular to \(2\)).

Equation of the normal: \(y - 66.5 = -\frac{1}{2}(x - 12)\)