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

A-Level Edexcel Maths Pure Integration: (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 7 - 2005 - Paper 1

Question 7

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

(a) Show that $ \frac{(3 - \sqrt{x})^3}{\sqrt{x}} $ can be written as $ 9x^{\frac{1}{2}} - 6x + x^{\frac{3}{2}} $. (b) Given that \( \frac{dy}{dx} = \frac{(3 - ... 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 7 - 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 show that ( \frac{(3 - \sqrt{x})^3}{\sqrt{x}} ) can be written in the specified form, we start by expanding the numerator:

Expand ( (3 - \sqrt{x})^3 ): [ (3 - \sqrt{x})^3 = 3^3 - 3 \times 3^2 \times \sqrt{x} + 3 \times 3 \times x - x^{\frac{3}{2}} = 27 - 27\sqrt{x} + 9x - x^{\frac{3}{2}}. ]
Thus, substituting this into the original fraction: [ \frac{(3 - \sqrt{x})^3}{\sqrt{x}} = \frac{27 - 27\sqrt{x} + 9x - x^{\frac{3}{2}}}{\sqrt{x}}. ]
Divide each term by ( \sqrt{x} ): [ = \frac{27}{\sqrt{x}} - 27 + 9\sqrt{x} - \sqrt{x}^2 = 9x^{\frac{1}{2}} - 6x + x^{\frac{3}{2}}. ]

This confirms the required result.

Step 2

find $ y $ in terms of $ x $

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Answer

To find ( y ) in terms of ( x ), we start with the differential equation:

[ \frac{dy}{dx} = \frac{(3 - \sqrt{x})^3}{\sqrt{x}} ]

Substitute the left side with ( y' ): [ y' = \frac{(3 - \sqrt{x})^3}{\sqrt{x}} ]
Integrate both sides: [ \int dy = \int \frac{(3 - \sqrt{x})^3}{\sqrt{x}} dx. ]
Using the earlier result, we have: [ y = \int (9x^{\frac{1}{2}} - 6x + x^{\frac{3}{2}})dx = 6x^{\frac{3}{2}} - 3x^2 + \frac{2}{5} x^{\frac{5}{2}} + c. ]
Substitute ( x = 1 ) and ( y = \frac{1}{3} ) to find ( c ): [ \frac{1}{3} = 6\left(\frac{3}{2}\right) - 3(1)^2 + \frac{2}{5}(1) + c. ]

Evaluating this: [ \frac{1}{3} = 9 - 3 + \frac{2}{5} + c \Rightarrow c = \frac{1}{3} - 6 + \frac{2}{5} \Rightarrow c = -\frac{12}{15} + \frac{6}{15} \Rightarrow c = -\frac{6}{15} = -\frac{2}{5}. ]
Thus, substitute ( c ) back to the expression for ( y ): [ y = 6x^{\frac{3}{2}} - 3x^2 + \frac{2}{5} x^{\frac{5}{2}} - \frac{2}{5}. ]

(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 7 - 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 7 - 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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