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The curve C has equation $y = 4x^2 + \frac{5 - x}{x}$ - Edexcel - A-Level Maths Pure - Question 9 - 2005 - Paper 2
Question 9
The curve C has equation $y = 4x^2 + \frac{5 - x}{x}$. The point P on C has x-coordinate 1. (a) Show that the value of \( \frac{dy}{dx} \) at P is 3. (b) Find an e... show full transcript
Worked Solution & Example Answer:The curve C has equation $y = 4x^2 + \frac{5 - x}{x}$ - Edexcel - A-Level Maths Pure - Question 9 - 2005 - Paper 2
Step 1
Show that the value of \( \frac{dy}{dx} \) at P is 3.
Answer
To find ( \frac{dy}{dx} ), we first differentiate the curve's equation:
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Differentiate the equation: [ y = 4x^2 + \frac{5 - x}{x} = 4x^2 + \frac{5}{x} - 1 ]
Differentiating term by term, we have: [ \frac{dy}{dx} = 8x - \frac{5}{x^2} ]
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Evaluate at ( x = 1 ): [ \frac{dy}{dx} \text{ at } x = 1 = 8(1) - \frac{5}{1^2} = 8 - 5 = 3 ]
Thus, it is shown that ( \frac{dy}{dx} ) at P is 3.
Step 2
Find an equation of the tangent to C at P.
Answer
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Use point-slope form for the tangent: The point P is at ( x = 1 ), and from part (a), we have ( \frac{dy}{dx} = 3 ). Therefore, the slope of the tangent line is 3.
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Find the y-coordinate of P: [ y = 4(1)^2 + \frac{5 - 1}{1} = 4 + 4 = 8 ]
So, P = (1, 8).
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Tangent equation: Using the point-slope form of the line: [ y - y_1 = m(x - x_1) ] Substituting in our values: [ y - 8 = 3(x - 1) ] Simplifying gives: [ y = 3x + 5 ]
Step 3