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1.8 Integration

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Question 29

The cross-section of a precision-engineered lens is modeled by a curve y=f(x)y = f(x)y=f(x), for x>0x > 0x>0. The rate of change of the gradient of the profile is given by

f′′(x)=154x7−6x f''(x) = \frac{15}{4\sqrt{x^7}} - 6x f′′(x)=4x7​15​−6x

A point P(1,1.5)P(1, 1.5)P(1,1.5) lies on the boundary of the lens profile.

Given that the gradient of the curve f′(x)=0.75f'(x) = 0.75f′(x)=0.75 at point PPP,

a.

find the equation of the normal at PPP, writing your answer in the form y=mx+cy = mx + cy=mx+c, where mmm and ccc are constants,

[3]
b.

find f(x)f(x)f(x).

[5]
Markscheme

1.8 Integration Questions

  1. A Level
  2. /Maths
  3. /1.8 Integration

160 exam-style questions on WJEC A Level Maths 1.8 Integration, covering 1.8.1 Integration, 1.8.2 Integration, and 1.8.3 Integration. Each one has a worked solution and a mark scheme showing where the marks go.

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