Answer: \(A=5\), \(B=11\).

Differentiate first. Stationary points occur where the derivative is zero, while tangent and normal problems use the gradient at the given point.

(a) The expression under the square root is \(5x-2\).

For the curve to exist, this must be non-negative:

\(5x-2\geq0\).

So

\(x\geq\frac25\).

Therefore the curve does not exist when \(x\lt \frac25\), because then \(5x-2\) is negative and \(\sqrt{5x-2}\) is not real.

(b) Write

\(y=\frac{(5x-2)^{1/2}}{x-3}\).

Using the quotient rule,

\(\frac{dy}{dx}=\frac{(x-3)\cdot\frac52(5x-2)^{-1/2}-(5x-2)^{1/2}}{(x-3)^2}\).

Put the numerator over the common factor \((5x-2)^{1/2}\):

\(\frac{dy}{dx}=\frac{\frac52(x-3)-(5x-2)}{(x-3)^2\sqrt{5x-2}}\).

Simplify the numerator:

\(\frac52(x-3)-(5x-2)=\frac{5x-15}{2}-5x+2\).

So

\(\frac{5x-15}{2}-5x+2=\frac{5x-15-10x+4}{2}=\frac{-5x-11}{2}\).

Therefore

\(\frac{dy}{dx}=\frac{-(5x+11)}{2(x-3)^2\sqrt{5x-2}}\).

This is of the required form with \(A=5\) and \(B=11\).