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Given the equation: y = x^3 - k rac{1}{2} x^2, where k is a constant - Edexcel - A-Level Maths Pure - Question 6 - 2010 - Paper 3

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Given-the-equation:--y-=-x^3---k--rac{1}{2}-x^2,---where-k-is-a-constant-Edexcel-A-Level Maths Pure-Question 6-2010-Paper 3.png

Given the equation: y = x^3 - k rac{1}{2} x^2, where k is a constant. (a) Find \( \frac{dy}{dx} \). (b) Given that y is decreasing at \( x = 4 \), find the set... show full transcript

Worked Solution & Example Answer:Given the equation: y = x^3 - k rac{1}{2} x^2, where k is a constant - Edexcel - A-Level Maths Pure - Question 6 - 2010 - Paper 3

Step 1

Find \( \frac{dy}{dx} \)

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Answer

To find ( \frac{dy}{dx} ), we differentiate the function with respect to x.

[ \frac{dy}{dx} = \frac{d}{dx}(x^3) - \frac{d}{dx}(k \cdot \frac{1}{2} x^2) ]

Using the power rule for differentiation:

[ \frac{dy}{dx} = 3x^2 - kx ]

Step 2

Given that y is decreasing at x = 4, find the set of possible values of k.

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Answer

For y to be decreasing at ( x = 4 ), ( \frac{dy}{dx} ) must be less than or equal to 0.

Substituting ( x = 4 ) into the derivative:

[ \frac{dy}{dx} = 3(4)^2 - k(4) ]

This leads to:

[ \frac{dy}{dx} = 48 - 4k \leq 0 ]

Rearranging gives:

[ 48 \leq 4k \quad \Rightarrow \quad k \geq 12 ]

Thus, the set of possible values for k is ( k \geq 12 ).

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