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1.6 Differentiation

1.6 Differentiation

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Question 30

The temperature, TTT degrees Celsius, of a cooling laser component ttt minutes after activation is modeled by the function T=f(t)T = f(t)T=f(t) for t>0t > 0t>0.

It is given that:

  • the point P(4,10)P(4, 10)P(4,10) lies on the curve T=f(t)T = f(t)T=f(t)
  • the rate of change of temperature is f′(t)=12t+kt2f'(t) = 12\sqrt{t} + \frac{k}{t^2}f′(t)=12t​+t2k​, where kkk is a constant
  • the rate of change of temperature, f′(t)f'(t)f′(t), has a stationary point when t=4t = 4t=4
a.

Find the exact value of kkk.

[3]
b.

Determine an expression for f(t)f(t)f(t), giving your answer in its simplest form.

[5]
Markscheme

1.6 Differentiation Questions

  1. A Level
  2. /Maths
  3. /1.6 Differentiation

115 exam-style questions on CCEA A Level Maths 1.6 Differentiation, covering 1.6.1 Differentiation, 1.6.2 Differentiation, 1.6.3 Differentiation, 1.6.4 Differentiation, 1.6.5 Differentiation, 1.6.6 Differentiation, and 1.6.7 Differentiation. Each one has a worked solution and a mark scheme showing where the marks go.

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