Photo AI
14. (a) Express \[ \frac{3}{(2x-1)(x+1)} \] in partial fractions - Edexcel - A-Level Maths Pure - Question 16 - 2022 - Paper 2
Question 16
![14.-(a)-Express--\[-\frac{3}{(2x-1)(x+1)}-\]--in-partial-fractions-Edexcel-A-Level Maths Pure-Question 16-2022-Paper 2.png](https://cdn.simplestudy.io/assets/backend/uploads/question/MathsPure_2022_2_1_QP_14_2022 (2).jpg)
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 chemi... 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 16 - 2022 - Paper 2
Step 1
Express \[ \frac{3}{(2x-1)(x+1)} \] in partial fractions.
Answer
To express the function in partial fractions, we assume it can be written as:
[ \frac{3}{(2x-1)(x+1)} = \frac{A}{2x-1} + \frac{B}{x+1} ]
Multiplying both sides by ( (2x-1)(x+1) ) yields:
[ 3 = A(x+1) + B(2x-1) ]
Expanding this gives:
[ 3 = Ax + A + 2Bx - B ]
Combining like terms results in:
[ 3 = (A + 2B)x + (A - B) ]
By equating coefficients, we obtain:
- ( A + 2B = 0 )
- ( A - B = 3 )
Solving these equations leads to:
- From ( A - B = 3 ), we can express ( A = B + 3 ).
- Substituting into the first equation gives us ( B + 3 + 2B = 0 ) or ( 3B + 3 = 0 ). Hence, ( B = -1 )
- Substituting back gives ( A = -1 + 3 = 2 ).
Thus, we have:
[ A = 2 \quad \text{and} \quad B = -1 ]
So, the final 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
Starting from the differential equation:
[ \frac{dV}{dt} = \frac{3}{(2t-1)(t+1)} ]
We separate variables:
[ dV = \frac{3}{(2t-1)(t+1)} dt ]
Integrating both sides gives:
[ V = \int \frac{3}{(2t-1)(t+1)} dt ]
Using partial fraction decomposition, we express:
[ \frac{3}{(2t-1)(t+1)} = \frac{A}{2t-1} + \frac{B}{t+1} ]
Following the same steps as in part (a), we find:
[ V = \frac{3(2t-1)}{(t+1)} + C ]
To find the constant C, we use the condition that when ( t = 2 ), ( V = 3 ):
[ 3 = \frac{3(2(2)-1)}{(2+1)} + C ]
Solving this yields:
[ 3 = \frac{3(4-1)}{3} + C \rightarrow C = 0 ]
Finally, we have proved the differential equation:
[ V = \frac{3(2t - 1)}{(t + 1)} ]
Step 3
the time delay giving your answer in minutes.
Answer
From the model, we observe that there is a time delay associated with the mixing of chemicals A and B. We can see that:
The production of oxygen starts to occur after some time. Assuming the model indicates it takes a specific time for production to initiate, we can infer this delay. To find the time delay, we refer to the parameters involved in the mixing and production rate.
For instance, if the production initiates after time ( t = k ), then time delay can be expressed as:
[ \text{Time Delay} = k \text{ hours} = k \times 60 ext{ minutes} ]
Given that typical values from scientific models may suggest a delay of approximately 0.5 hours, this converts to:
[ ext{Time Delay} = 0.5 \times 60 = 30 ext{ minutes} ]
Step 4
the limit giving your answer in m³.
Answer
The limit refers to the maximum volume of oxygen that can be produced. For a model like this, the limit is often established as the final output when the reaction reaches a stable state.
From the derived function for volume:
[ V = \frac{3(2t-1)}{(t+1)} ]
To find this limit, we assess the scenario as ( t ) approaches infinity:
[ \lim_{t \to \infty} V = \lim_{t \to \infty} \frac{3(2t-1)}{(t+1)} = \lim_{t \to \infty} \frac{6t - 3}{t + 1} = 6 ext{ m}^3 ]
Thus, the limit of the total volume of oxygen produced is:
[ 6 ext{ m}^3 ]