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5. (a) Write \( \frac{2\sqrt{x}+3}{x} \) in the form \( 2p + 3x^r \) where \( p \) and \( q \) are constants - Edexcel - A-Level Maths Pure - Question 6 - 2008 - Paper 2
Question 6
5. (a) Write \( \frac{2\sqrt{x}+3}{x} \) in the form \( 2p + 3x^r \) where \( p \) and \( q \) are constants. Given that \( y = 5x - 7 + \frac{2\sqrt{x}+3}{x} \),... show full transcript
Worked Solution & Example Answer:5. (a) Write \( \frac{2\sqrt{x}+3}{x} \) in the form \( 2p + 3x^r \) where \( p \) and \( q \) are constants - Edexcel - A-Level Maths Pure - Question 6 - 2008 - Paper 2
Step 1
(a) Write \( \frac{2\sqrt{x}+3}{x} \) in the form \( 2p + 3x^r \)
Answer
To rewrite the expression ( \frac{2\sqrt{x}+3}{x} ), start by breaking it down:
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Separate the terms:
[ \frac{2\sqrt{x}}{x} + \frac{3}{x} ]
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Simplify each term:
- The first term becomes ( \frac{2\sqrt{x}}{x} = 2\frac{1}{\sqrt{x}} = \frac{2}{x^{1/2}} )
- The second term is just ( \frac{3}{x} = 3x^{-1} )
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Combine the simplified terms:
[ \frac{2}{x^{1/2}} + 3x^{-1} ]
The expression is now in the form ( 2p + 3x^r ), where ( p = \frac{1}{2} ) and ( q = -1 ).
Step 2
(b) find \( \frac{dy}{dx} \), simplifying the coefficient of each term.
Answer
Given ( y = 5x - 7 + \frac{2\sqrt{x}+3}{x} ), differentiate it term by term:
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Differentiate ( 5x ) which gives:
[ \frac{dy}{dx} = 5 ]
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The term ( -7 ) differentiates to 0:
[ \frac{dy}{dx} = 5 + 0 ]
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Now, differentiate ( \frac{2\sqrt{x}+3}{x} ). Use the quotient rule:
Let ( u = 2\sqrt{x}+3 ) and ( v = x ).
[ \frac{du}{dx} = \frac{2}{2\sqrt{x}} \cdot \frac{1}{2}\sqrt{x} = \frac{1}{\sqrt{x}} ] and ( \frac{dv}{dx} = 1 )
Using the quotient rule:\n [ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} ]
Applying it:
[ \frac{dy}{dx} = 5 + \frac{x \cdot \frac{1}{\sqrt{x}} - (2\sqrt{x}+3) \cdot 1}{x^2} ]
which simplifies to:
[ \frac{dy}{dx} = 5 + \frac{\sqrt{x} - 2\sqrt{x} - 3}{x^2} = 5 + \frac{-\sqrt{x} - 3}{x^2} ]
Thus,\n [ \frac{dy}{dx} = 5 - \frac{\sqrt{x} + 3}{x^2} ]