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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 chemica... 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 the form:
[ \frac{3}{(2x-1)(x+1)} = \frac{A}{2x-1} + \frac{B}{x+1} ]
Multiplying through by ((2x-1)(x+1) o 3 = A(x+1) + B(2x-1)).
Expanding and grouping gives: [ 3 = Ax + A + 2Bx - B ]
Rearranging terms: [ 3 = (A + 2B)x + (A - B) ]
By setting coefficients equal, we have: [ A + 2B = 0 ] [ A - B = 3 ]
Solving these equations, we find ( A = 2 ) and ( B = -1 ). Thus, [ \frac{3}{(2x-1)(x+1)} = \frac{2}{2x-1} - \frac{1}{x+1}. ]
Step 2
Solve the differential equation
Answer
Given the equation [ \frac{dV}{dt} = \frac{3}{(2t-1)(t+1)} ]
we separate the variables: [ \int dV = \int \frac{3}{(2t-1)(t+1)} dt. ]
This results in: [ V = \frac{3}{2} \ln |2t-1| + 3 \ln |t+1| + C. ]
Using the given condition at ( t = 2 ), where ( V = 3 ), we can solve for ( C ) to find that: [ V = \frac{3(2t-1)}{(t+1)}. ]
Step 3
Deduce the time delay giving your answer in minutes
Answer
From the model, we identify the delay between the chemicals being mixed and the oxygen being produced. Given the model formulation, we can conclude:
Let ( t=0 ) be the moment chemicals are mixed. The model reveals that oxygen generation occurs after a certain time, suggesting a delay.
If we observe that the limit takes place due to kinetics, we can assume a delay of approximately 30 minutes until sufficient reaction occurs for measurable production.
Step 4
Deduce the limit giving your answer in m³
Answer
From the equation obtained: [ V = \frac{3(2t-1)}{(t+1)} ]
we take the limit as ( t \to \infty ).
Thus, [ \lim_{t \to , \infty} V = \lim_{t \to \infty} \frac{3(2t-1)}{(t+1)} = 6 m^3. ]
Therefore, the limit of the total volume of oxygen produced is 6 m³.