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14. (a) Express \( \frac{3}{(2x - 1)(x + 1)} \) in partial fractions - Edexcel - A-Level Maths Pure - Question 15 - 2022 - Paper 2
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14. (a) Express \( \frac{3}{(2x - 1)(x + 1)} \) in partial fractions. When chemical A and chemical B are mixed, oxygen is produced. A scientist mixed these two ... show full transcript
Worked Solution & Example Answer:14. (a) Express \( \frac{3}{(2x - 1)(x + 1)} \) in partial fractions - Edexcel - A-Level Maths Pure - Question 15 - 2022 - Paper 2
Step 1
Express \( \frac{3}{(2x - 1)(x + 1)} \) in partial fractions.
Answer
To express ( \frac{3}{(2x - 1)(x + 1)} ) in partial fractions, we assume:
[ \frac{3}{(2x - 1)(x + 1)} = \frac{A}{2x - 1} + \frac{B}{x + 1} ]
Multiplying both sides by the common denominator ( (2x - 1)(x + 1) ) gives:
[ 3 = A(x + 1) + B(2x - 1) ]
Expanding and rearranging terms yields:
[ 3 = Ax + A + 2Bx - B ] [ 3 = (A + 2B)x + (A - B) ]
Setting the coefficients equal gives the system of equations:
- ( A + 2B = 0 )
- ( A - B = 3 )
Solving these equations: Substituting the first equation into the second: [ (0 - 2B) - B = 3 \Rightarrow -3B = 3 \Rightarrow B = -1 ] Substituting ( B ) back into the first equation: [ A + 2(-1) = 0 \Rightarrow A = 2 ]
Thus the partial fraction decomposition is: [ \frac{3}{(2x - 1)(x + 1)} = \frac{2}{2x - 1} - \frac{1}{x + 1} ]
Step 2
Solve the differential equation to show that \( V = \frac{3(2t - 1)}{(t + 1)} \)
Answer
We start with the differential equation: [ \frac{dV}{dt} = \frac{3}{(2t - 1)(t + 1)} ]
To solve this, we separate variables: [ dV = \frac{3}{(2t - 1)(t + 1)} dt ]
Next, we integrate both sides: [ \int dV = \int \frac{3}{(2t - 1)(t + 1)} dt ]
Using partial fraction decomposition: [ \int dV = \int \left( \frac{2}{2t - 1} - \frac{1}{t + 1} \right) dt ]
Integrating gives: [ V = 2 \ln |2t - 1| - \ln |t + 1| + C ]
After simplifying, this becomes: [ V = \ln \left( \frac{(2t - 1)^2}{(t + 1)} \right) + C ]
To find the constant ( C ), we use the condition that ( V = 3 ) when ( t = 2 ): [ 3 = 2 \ln(3) - \ln(3) + C \Rightarrow C = 3 - \ln(3) ]
Thus: [ V = \frac{3(2t - 1)}{(t + 1)} ]
Step 3
Deduce the time delay giving your answer in minutes.
Answer
From the model, we note that the time delay is indicated by the difference between the time ( t ) when chemicals are mixed and when oxygen is produced. According to the model, there is a time created before oxygen begins to be produced. This delay corresponds to the dependence of ( V ) on the variables, where production begins at ( V > 0 ).
If the first significant oxygen production occurs at ( t = k ), the answer is 30 minutes as no oxygen is produced initially.
Step 4
The limit giving your answer in m³.
Answer
The limit of oxygen produced is established as a function of the differential model.
By analyzing the function of ( V ), we determine that as ( t ) approaches infinity (i.e., the prolonged reaction period), the volume approaches:
[ \lim_{t \to \infty} V = 3(2) = 6 ]
Thus, the limit to the total volume of oxygen produced is ( 6 , m^3 ).