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1.11 H: Integration

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Question 105

The mass, m m\,m grams, of a certain chemical during a reaction is modeled by a differential equation involving time ttt, where 0≤t<π4\displaystyle 0 \le t < \frac{\pi}{4}0≤t<4π​.

a.

Find the derivative with respect to m m\,m of

1(1+2ln⁡m)2 \frac{1}{(1 + 2\ln m)^2} (1+2lnm)21​
[2]
b.

Hence find the general solution to the differential equation

4sec⁡(2t)dmdt=m(1+2ln⁡m)3tan⁡(2t) 4\sec(2t) \frac{\text{d}m}{\text{d}t} = m(1 + 2\ln m)^3 \tan(2t) 4sec(2t)dtdm​=m(1+2lnm)3tan(2t)

for m>e−1/2m > e^{-1/2}m>e−1/2.

[4]
c.

Show that the particular solution of this differential equation for which the initial mass is 1 g (so m=1m = 1m=1 when t=0t = 0t=0) is given by

m=eAsec⁡t−12 m = e^{A\sec t - \frac{1}{2}} m=eAsect−21​

where A A\,A is a constant to be found.

[4]
Markscheme

1.11 H: Integration Questions

  1. A Level
  2. /Maths
  3. /1.11 H: Integration

480 exam-style questions on AQA A Level Maths 1.11 H: Integration, covering 1.11.1 Fundamental Theorem of Calculus, 1.11.2 Integrating standard functions, 1.11.3 Definite integrals and areas, 1.11.4 Integration as the limit of a sum (A-level only), 1.11.5 Integration by substitution and by parts (A-level only), 1.11.6 Integration using partial fractions (A-level only), 1.11.7 Differential equations with separable variables (A-level only), and 1.11.8 Interpreting solutions of differential equations (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.

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