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7. (a) Show that \( \frac{(3 - \sqrt{x})^3}{\sqrt{x}} \) can be written as \( 9x^{\frac{1}{2}} - 6 + x^{\frac{1}{3}} \) - Edexcel - A-Level Maths Pure - Question 9 - 2005 - Paper 1
Question 9
7. (a) Show that \( \frac{(3 - \sqrt{x})^3}{\sqrt{x}} \) can be written as \( 9x^{\frac{1}{2}} - 6 + x^{\frac{1}{3}} \). Given that \( \frac{dy}{dx} = \frac{(3 - ... show full transcript
Worked Solution & Example Answer:7. (a) Show that \( \frac{(3 - \sqrt{x})^3}{\sqrt{x}} \) can be written as \( 9x^{\frac{1}{2}} - 6 + x^{\frac{1}{3}} \) - Edexcel - A-Level Maths Pure - Question 9 - 2005 - Paper 1
Step 1
Show that \( \frac{(3 - \sqrt{x})^3}{\sqrt{x}} \) can be written as \( 9x^{\frac{1}{2}} - 6 + x^{\frac{1}{3}} \)
Answer
To show that ( \frac{(3 - \sqrt{x})^3}{\sqrt{x}} ) can be expressed in the desired form, we start by expanding ( (3 - \sqrt{x})^3 ):
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Use the binomial expansion:
[(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 ]
Here, let ( a = 3 ) and ( b = \sqrt{x} ):
[ (3 - \sqrt{x})^3 = 27 - 27\sqrt{x} + 9x - x^{\frac{3}{2}} ] -
Substitute this back into the original expression:
[ \frac{(3 - \sqrt{x})^3}{\sqrt{x}} = \frac{27 - 27\sqrt{x} + 9x - x^{\frac{3}{2}}}{\sqrt{x}} ] -
Split the fraction:
[ \frac{27}{\sqrt{x}} - 27 + 9\sqrt{x} - \frac{x^{\frac{3}{2}}}{\sqrt{x}} = 9x^{\frac{1}{2}} - 6 + x^{\frac{1}{3}} ]
Thus, we have proven the expression as required.
Step 2
find \( y \) in terms of \( x \)
Answer
To find ( y ) in terms of ( x ), we first integrate ( \frac{dy}{dx} = \frac{(3 - \sqrt{x})^3}{\sqrt{x}} ):
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We substitute and find the integral:
[ \int \frac{(3 - \sqrt{x})^3}{\sqrt{x}} dx = \int (9x^{\frac{1}{2}} - 6 + x^{\frac{1}{3}}) dx ]
The integral can be solved term by term:- For ( 9x^{\frac{1}{2}} ): [ \int 9x^{\frac{1}{2}} dx = 6x^{\frac{3}{2}} + C ]
- For ( -6 ): [ \int -6 dx = -6x + C ]
- For ( x^{\frac{1}{3}} ): [ \int x^{\frac{1}{3}} dx = \frac{3}{4}x^{\frac{4}{3}} + C ]
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Adding them together, we get:
[ y = 6x^{\frac{3}{2}} - 6x + \frac{3}{4}x^{\frac{4}{3}} + C ]
Using the initial condition ( y = \frac{2}{3} ) when ( x = 1 ):
[ \frac{2}{3} = 6(1)^{\frac{3}{2}} - 6(1) + \frac{3}{4}(1)^{\frac{4}{3}} + C ]
This leads to:
[ \frac{2}{3} = 6 - 6 + \frac{3}{4} + C ]
Solving gives ( C = -\frac{12}{3} = -12 ). -
Therefore, the expression for ( y ) in terms of ( x ) is:
[ y = 6x^{\frac{3}{2}} - 6x + \frac{3}{4}x^{\frac{4}{3}} - 12 ]