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3.5.2 Differentiation (A-level only)

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Question 82

The vertical profile of a sculpted roller coaster rail is modeled by the function h(x)=(x−4)(2x+5)2h(x) = (x - 4)(2x + 5)^2h(x)=(x−4)(2x+5)2 for x≥−3x \ge -3x≥−3, where hhh is the height in decimetres and xxx is the horizontal distance from a sensor.

The rail touches the baseline at point PPP and crosses the baseline at point QQQ.

a.

State the coordinates of the point PPP.

[2]
b.

Determine h′(x)h'(x)h′(x).

[4]
c.

Hence show that the equation of the tangent to the rail at the point where x=116x = \frac{11}{6}x=611​ can be expressed in the form y=ky = ky=k, where kkk is a constant to be found.

[4]
d.

A modification shifts the track horizontally so the equation becomes y=h(x+b)y = h(x + b)y=h(x+b), where bbb is a constant. The modified track now passes through the sensor's origin O(0,0)O(0,0)O(0,0).

State the possible values of bbb.

[2]

3.5.2 Differentiation (A-level only) Questions

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