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5. (a) Write \( \frac{2\sqrt{x}+3}{x} \) in the form \( 2p+r+3x^r \) where \( p \) and \( q \) are constants - Edexcel - A-Level Maths Pure - Question 7 - 2008 - Paper 2
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5. (a) Write \( \frac{2\sqrt{x}+3}{x} \) in the form \( 2p+r+3x^r \) where \( p \) and \( q \) are constants. (b) 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 \( 2p+r+3x^r \) where \( p \) and \( q \) are constants - Edexcel - A-Level Maths Pure - Question 7 - 2008 - Paper 2
Step 1
Write \( \frac{2\sqrt{x}+3}{x} \) in the form \( 2p + r + 3x^r \)
Answer
To rewrite ( \frac{2\sqrt{x}+3}{x} ), we first express each term separately:
- Splitting the fraction gives:
[ \frac{2\sqrt{x}}{x} + \frac{3}{x} = 2\frac{\sqrt{x}}{x} + 3\frac{1}{x} ]
Here, ( \frac{\sqrt{x}}{x} = \frac{x^{1/2}}{x^{1}} = x^{-1/2} )
Therefore, the equation becomes:
[ 2x^{-1/2} + 3x^{-1} ] - We can express this in the desired form:
[ p = 1, \quad r = -1, \quad \text{thus, } q = -1 ]
Step 2
Find \( \frac{dy}{dx} \)
Answer
To find ( \frac{dy}{dx} ) from ( y = 5x - 7 + \frac{2\sqrt{x}+3}{x} ):
- First, we differentiate each term:
[ \frac{d}{dx}(5x) = 5 ]
[ \frac{d}{dx}(-7) = 0 ]
For the third term, using the product or quotient rule: [ \frac{d}{dx}(\frac{2\sqrt{x}+3}{x}) = \frac{(x)(\frac{1}{2} \cdot 2x^{-\frac{1}{2}}) - (2\sqrt{x}+3)(1)}{x^2} = \frac{x^{-\frac{1}{2}} - (2\sqrt{x}+3)}{x^2} ] 2. Putting it all together:
[ \frac{dy}{dx} = 5 + \left( \frac{x^{-\frac{1}{2}} - (2\sqrt{x}+3)}{x^2} \right) ]
3. Simplifying the expression for the final answer will produce the specific terms.