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Given that $y = x^3 - k \sqrt{x}$, where $k$ is a constant - Edexcel - A-Level Maths Pure - Question 5 - 2010 - Paper 3

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Given that $y = x^3 - k \sqrt{x}$, where $k$ is a constant. (a) Find $ rac{dy}{dx}$. (b) Given that $y$ is decreasing at $x = 4$, find the set of possible values ... show full transcript

Worked Solution & Example Answer:Given that $y = x^3 - k \sqrt{x}$, where $k$ is a constant - Edexcel - A-Level Maths Pure - Question 5 - 2010 - Paper 3

Step 1

Find $ rac{dy}{dx}$

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Answer

To find rac{dy}{dx}, we differentiate the given function with respect to xx:

rac{dy}{dx} = 3x^2 - k \cdot \frac{1}{2} x^{-\frac{1}{2}} = 3x^2 - \frac{k}{2 \sqrt{x}}.

Step 2

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

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Answer

To find when yy is decreasing, we set the derivative rac{dy}{dx} less than or equal to zero:

  1. Substitute x=4x = 4 into the derivative: dydx∣x=4=3(4)2−k24=48−k4.\frac{dy}{dx}\bigg|_{x=4} = 3(4)^2 - \frac{k}{2 \sqrt{4}} = 48 - \frac{k}{4}.
  2. Set the inequality:
    48−k4<048 - \frac{k}{4} < 0
  3. Solve for kk:
    48<k448 < \frac{k}{4}
    k>192.k > 192.
    Thus, the set of possible values of kk is k>192k > 192.

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