(a) Show that \( \frac{(3-\sqrt{x})^3}{\sqrt{x}} \) can be written as \( 9x^{1/2} - 6x + x^{3/2} \) - Edexcel - A-Level Maths Pure - Question 7 - 2005 - Paper 1

Given ( \frac{dy}{dx} = \frac{(3-\sqrt{x})^3}{\sqrt{x}} ), we will integrate to find ( y ).

This implies:

$$y = \int \frac{(3-\sqrt{x})^3}{\sqrt{x}} \, dx.$$

Using the substitution, let ( u = \sqrt{x} ), then ( x = u^2 ) and ( dx = 2u , du ). The integral becomes:

$$y = \int \frac{(3-u)^3}{u} (2u \, du) = 2 \int (3-u)^3 \, du.$$

Expanding ( (3-u)^3 ):

$$(3-u)^3 = 27 - 27u + 9u^2 - u^3.$$

Thus, the integral becomes:

$$y = 2 \left( 27u - \frac{27u^2}{2} + 3u^3 - \frac{u^4}{4} + c \right).$$

Now, substituting back ( u = \sqrt{x} ):

$$y = 54\sqrt{x} - 27x + 6x^{3/2} - \frac{2x^2}{4} + c.$$

Next, we need to find the constant of integration ( c ) using the given condition ( y = \frac{1}{3} ) when ( x = 1 ):

$$\frac{1}{3} = 54(1) - 27(1) + 6(1) - \frac{2(1^2)}{4} + c.$$

This simplifies to:

$$\frac{1}{3} = 54 - 27 + 6 - \frac{1}{2} + c$$ $$\frac{1}{3} = 33.5 + c$$

Then solving for ( c ):

$$c = \frac{1}{3} - 33.5 = -\frac{100}{3}.$$

Finally, substitute ( c ) back into the equation:

$$y = 54\sqrt{x} - 27x + 6x^{3/2} - \frac{2x^2}{4} - \frac{100}{3}.$$

This gives us ( y ) in terms of ( x ).