Given that 6x + 3x^{rac{5}{2}} can be written in the form 6x^{p} + 3x^{q}, a) write down the value of p and the value of q - Edexcel - A-Level Maths Pure - Question 9 - 2011 - Paper 1

Starting with ( \frac{dy}{dx} = \frac{6x + 3x^{\frac{5}{2}}}{\sqrt{x}} ), we can break this down:

[ \frac{dy}{dx} = \frac{6x}{\sqrt{x}} + \frac{3x^{\frac{5}{2}}}{\sqrt{x}} = 6x^{\frac{1}{2}} + 3x^{2} = 6\sqrt{x} + 3x^{2} ]

Now, we integrate ( dy = (6\sqrt{x} + 3x^{2})dx ):

[ y = \int (6\sqrt{x} + 3x^{2}) , dx = 6 \cdot \frac{2}{3} x^{\frac{3}{2}} + 3 \cdot \frac{1}{3} x^{3} + C = 4x^{\frac{3}{2}} + x^{3} + C ]

Next, we apply the initial condition ( y = 90 ) when ( x = 4 ):

[ 90 = 4(4^{\frac{3}{2}}) + (4^{3}) + C ]

Calculating each term:

• ( 4^{\frac{3}{2}} = 4 \cdot 2 = 8 ) so ( 4 \cdot 8 = 32 )
• ( 4^{3} = 64 )

Putting it all together: [ 90 = 32 + 64 + C \Rightarrow C = 90 - 96 = -6 ]

Thus, we have: [ y = 4x^{\frac{3}{2}} + x^{3} - 6 ]