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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 7 - 2008 - Paper 2
Question 7
5. (a) Write \(\frac{2\sqrt{x}+3}{x} \) in the form \( 2p + 3x^r \) 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 \( 2p + 3x^r \) 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 \( 2p + 3x^r \)
Answer
To rewrite (\frac{2\sqrt{x}+3}{x} ), we can separate the terms:
[ \frac{2\sqrt{x}+3}{x} = \frac{2\sqrt{x}}{x} + \frac{3}{x} ]
Now, replace ( \sqrt{x} ) with ( x^{1/2} ):
[ \frac{2x^{1/2}}{x} + \frac{3}{x} = 2x^{-1/2} + 3x^{-1} ]
Thus, we have:
[ p = -\frac{1}{2}, q = -1 ]
This gives us the rewritten form:
[ \frac{2\sqrt{x}+3}{x} = 2(-\frac{1}{2}) + 3x^{-1} ]
Step 2
Given that \( y = 5x - 7 + \frac{2\sqrt{x}+3}{x} \), find \( \frac{dy}{dx} \)
Answer
To differentiate ( y = 5x - 7 + \frac{2\sqrt{x}+3}{x} ), we first find ( \frac{dy}{dx} ):
[ \frac{dy}{dx} = 5 + \frac{d}{dx}\left( \frac{2\sqrt{x}+3}{x} \right) ]
Using the quotient rule, where ( u = 2\sqrt{x}+3 ) and ( v = x ):
[ \frac{dy}{dx} = 5 + \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} ]
Calculating ( \frac{du}{dx} ) gives us:
[ \frac{d}{dx}(2\sqrt{x}) = 2\cdot\frac{1}{2}x^{-1/2} = x^{-1/2} \text{ and } \frac{d}{dx}(3) = 0 \Rightarrow \frac{du}{dx} = x^{-1/2} ]
And ( \frac{dv}{dx} = 1 ):
Substituting these into the quotient rule:
[ \frac{dy}{dx} = 5 + \frac{x( x^{-1/2}) - (2\sqrt{x}+3)(1)}{x^2} ]
Simplifying gives us:
[ \frac{dy}{dx} = 5 + \frac{x^{1/2} - (2\sqrt{x}+3)}{x^2} ]
This can be rewritten as:
[ \frac{dy}{dx} = 5 - \frac{2\sqrt{x}+3 - x^{1/2}}{x^2} ]
Finally, further simplification yields the final answer.