Skip to content

Course home

1.7 Integration

1.7 Integration

EasyMediumHard
1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606162636465666768697071727374757677787980818283848586878889909192939495
Question 92

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

1.7 Integration Questions

  1. A Level
  2. /Maths
  3. /1.7 Integration

133 exam-style questions on CCEA A Level Maths 1.7 Integration, covering 1.7.1 Integration, 1.7.2 Integration, and 1.7.3 Integration. Each one has a worked solution and a mark scheme showing where the marks go.

Question bank