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A curve has equation y = 2x cos 3x + (3x^2 - 4) sin 3x - AQA - A-Level Maths Pure - Question 8 - 2017 - Paper 2

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A curve has equation y = 2x cos 3x + (3x^2 - 4) sin 3x. 8 (a) Find \( \frac{dy}{dx} \), giving your answer in the form \( m x^2 + n \cos 3x \), where m and n a... show full transcript

Worked Solution & Example Answer:A curve has equation y = 2x cos 3x + (3x^2 - 4) sin 3x - AQA - A-Level Maths Pure - Question 8 - 2017 - Paper 2

Step 1

Find \( \frac{dy}{dx} \), giving your answer in the form \( mx^2 + n \cos 3x \), where m and n are integers.

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Answer

To find the derivative ( \frac{dy}{dx} ), we will use the product rule. The given function can be differentiated as follows:

  1. Differentiate the first term:

    [ \frac{d}{dx}(2x \cos 3x) = 2 \cos 3x + 2x \frac{d}{dx}(\cos 3x) = 2 \cos 3x - 6x \sin 3x ]

  2. Differentiate the second term:

    [ \frac{d}{dx}((3x^2 - 4) \sin 3x) = (3 \cdot 2x) \sin 3x + (3x^2 - 4) \frac{d}{dx}(\sin 3x) = 6x \sin 3x + 3(3x^2 - 4) \cos 3x ]

  3. Combine the derivatives from both terms, we have:

    [ \frac{dy}{dx} = (2 \cos 3x - 6x \sin 3x) + (6x \sin 3x + 9x^2 \cos 3x - 12 \cos 3x) ]

  4. Simplifying further:

    [ \frac{dy}{dx} = (9x^2 - 10) \cos 3x ]

Thus, we can write it in the required form where ( m = 9 ) and ( n = -10 ).

Step 2

Show that the x-coordinates of the points of inflection of the curve satisfy the equation \( \cot 3x = \frac{9x^2 - 10}{6x} \)

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Answer

To find the points of inflection, we need to set the second derivative ( \frac{d^2y}{dx^2} ) equal to zero.

  1. We previously found that:

    [ \frac{dy}{dx} = (9x^2 - 10) \cos 3x ]

  2. Thus, differentiating ( \frac{dy}{dx} ):

    [ \frac{d^2y}{dx^2} = \frac{d}{dx}((9x^2 - 10) \cos 3x) ]

    Using the product rule again:

    [ \frac{d^2y}{dx^2} = (18x) \cos 3x + (9x^2 - 10)(-3 \sin 3x) ]

    Setting ( \frac{d^2y}{dx^2} = 0 ) gives:

    [ 18x \cos 3x - 3(9x^2 - 10) \sin 3x = 0 ]

    1. Rearranging this gives us:

    [ 18x \cos 3x = 3(9x^2 - 10) \sin 3x ]

    Dividing both sides by 3:

    [ 6x \cos 3x = (9x^2 - 10) \sin 3x ]

    1. Dividing by ( \sin 3x ) gives us:

    [ \cot 3x = \frac{9x^2 - 10}{6x} ]

This shows that the x-coordinates of the points of inflection satisfy the given equation.

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