| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Product of separate expansions |
| Difficulty | Standard +0.3 This is a structured multi-part question that guides students through standard binomial expansion techniques, multiplying series, partial fractions, and integration. While lengthy (7 parts), each individual step is routine C4 material with clear scaffolding. The most challenging aspect is the algebraic manipulation in part (iii), but the question explicitly guides the approach. No novel problem-solving or insight required beyond applying learned techniques. |
| Spec | 1.02y Partial fractions: decompose rational functions1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((1+t)^{-1} = 1 - t + t^2 - ...\) | B2 | B1 for first two terms |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((2+t)^{-2} = 2^{-2}\left(1+\frac{t}{2}\right)^{-2}\) | B1 | |
| \(= \frac{1}{4}\left(1 + \frac{-2}{1}\cdot\frac{t}{2} + \frac{-2\cdot-3}{1\cdot2}\cdot\left(\frac{t}{2}\right)^2 - ...\right)\) | M1, A1 | |
| \(= \frac{1}{4} - \frac{1}{4}t + \frac{3}{16}t^2 ...\) | ||
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(5000(1+t)^{-1}(2+t)^{-2} = 5000(1-t+t^2...)\left(\frac{1}{4}-\frac{1}{4}t+\frac{3}{16}t^2...\right)\) | B1 | |
| \(= \frac{5000}{4}(1-t+t^2...)\left(1-t+\frac{3}{4}t^2...\right)\) | ||
| \(= 1250(1-t-t+t^2+t^2+\frac{3}{4}t^2...) = 1250\left(1-2t+\frac{11}{4}t^2...\right)\) | E1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Displacement \(= \int_0^{\frac{1}{2}} 1250\left(1-2t+\frac{11}{4}t^2\right)dt\) | M1 | |
| \(= \left[1250\left(t - t^2 + \frac{11}{12}t^3\right)\right]_0^{\frac{1}{2}}\) | A1 | |
| \(= 1250\times\frac{35}{96} = 455.72...\) | A1 | |
| Displacement is 456 metres, to 3sf | ||
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{(1+t)(2+t)^2} = \frac{A}{1+t} + \frac{B}{2+t} + \frac{C}{(2+t)^2}\) | M1 | Partial fractions |
| \(\Rightarrow 1 = A(2+t)^2 + B(2+t)(1+t) + C(1+t)\) | A1 | |
| Any method \(\Rightarrow A=1,\ B=-1,\ C=-1\) | ||
| \(\Rightarrow \frac{1}{(1+t)(2+t)^2} = \frac{1}{1+t} - \frac{1}{2+t} - \frac{1}{(2+t)^2}\) | A2 | 3 values, A1 for 2 |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Displacement \(= \int_0^{\frac{1}{2}}\frac{5000}{(1+t)(2+t)^2}dt = \int_0^{\frac{1}{2}}5000\left(\frac{1}{1+t} - \frac{1}{(2+t)^2} - \frac{1}{2+t}\right)dt\) | M1 | Integral in PFs |
| \(= 5000\left[\ln(1+t) + \frac{1}{2+t} - \ln(2+t)\right]_0^{\frac{1}{2}}\) | M1, A1 | Integrating to lns |
| \(= 5000\left(\ln\frac{3}{2} - \frac{1}{10} - \ln\frac{5}{4}\right) = 5000\left(\ln\frac{6}{5} - \frac{1}{10}\right)\) | A1 | |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Series in (i) has region of validity only \(-1 < t < 1\) | B1 | |
| Total: 1 |
## Question 8(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(1+t)^{-1} = 1 - t + t^2 - ...$ | B2 | B1 for first two terms |
| **Total: 2** | | |
---
## Question 8(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(2+t)^{-2} = 2^{-2}\left(1+\frac{t}{2}\right)^{-2}$ | B1 | |
| $= \frac{1}{4}\left(1 + \frac{-2}{1}\cdot\frac{t}{2} + \frac{-2\cdot-3}{1\cdot2}\cdot\left(\frac{t}{2}\right)^2 - ...\right)$ | M1, A1 | |
| $= \frac{1}{4} - \frac{1}{4}t + \frac{3}{16}t^2 ...$ | | |
| **Total: 3** | | |
---
## Question 8(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $5000(1+t)^{-1}(2+t)^{-2} = 5000(1-t+t^2...)\left(\frac{1}{4}-\frac{1}{4}t+\frac{3}{16}t^2...\right)$ | B1 | |
| $= \frac{5000}{4}(1-t+t^2...)\left(1-t+\frac{3}{4}t^2...\right)$ | | |
| $= 1250(1-t-t+t^2+t^2+\frac{3}{4}t^2...) = 1250\left(1-2t+\frac{11}{4}t^2...\right)$ | E1 | |
| **Total: 2** | | |
---
## Question 8(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Displacement $= \int_0^{\frac{1}{2}} 1250\left(1-2t+\frac{11}{4}t^2\right)dt$ | M1 | |
| $= \left[1250\left(t - t^2 + \frac{11}{12}t^3\right)\right]_0^{\frac{1}{2}}$ | A1 | |
| $= 1250\times\frac{35}{96} = 455.72...$ | A1 | |
| Displacement is 456 metres, to 3sf | | |
| **Total: 3** | | |
---
## Question 8(v):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{(1+t)(2+t)^2} = \frac{A}{1+t} + \frac{B}{2+t} + \frac{C}{(2+t)^2}$ | M1 | Partial fractions |
| $\Rightarrow 1 = A(2+t)^2 + B(2+t)(1+t) + C(1+t)$ | A1 | |
| Any method $\Rightarrow A=1,\ B=-1,\ C=-1$ | | |
| $\Rightarrow \frac{1}{(1+t)(2+t)^2} = \frac{1}{1+t} - \frac{1}{2+t} - \frac{1}{(2+t)^2}$ | A2 | 3 values, A1 for 2 |
| **Total: 4** | | |
---
## Question 8(vi):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Displacement $= \int_0^{\frac{1}{2}}\frac{5000}{(1+t)(2+t)^2}dt = \int_0^{\frac{1}{2}}5000\left(\frac{1}{1+t} - \frac{1}{(2+t)^2} - \frac{1}{2+t}\right)dt$ | M1 | Integral in PFs |
| $= 5000\left[\ln(1+t) + \frac{1}{2+t} - \ln(2+t)\right]_0^{\frac{1}{2}}$ | M1, A1 | Integrating to lns |
| $= 5000\left(\ln\frac{3}{2} - \frac{1}{10} - \ln\frac{5}{4}\right) = 5000\left(\ln\frac{6}{5} - \frac{1}{10}\right)$ | A1 | |
| **Total: 4** | | |
---
## Question 8(vii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Series in (i) has region of validity only $-1 < t < 1$ | B1 | |
| **Total: 1** | | |
---8 Scientists predict the velocity ( $v$ kilometres per minute) for the new "outer explorer" spacecraft over the first minute of its entry to the atmosphere of the planet Titan to be modelled by the equation:
$$v = \frac { 5000 } { ( 1 + t ) ( 2 + t ) ^ { 2 } } , 0 \leq t \leq 1 \text { where } t \text { represents time in minutes. }$$
(i) Use a binomial expansion to expand $( 1 + t ) ^ { - 1 }$ up to and including the term in $t ^ { 2 }$.\\
(ii) Use a binomial expansion to expand $( 2 + t ) ^ { - 2 }$ up to and including the term in $t ^ { 2 }$.\\
(iii) Hence, or otherwise, show that $v \approx 1250 \left( 1 - 2 t + \frac { 11 t ^ { 2 } } { 4 } \right)$.\\
(iv) The displacement of the spacecraft can be found by calculating the area under the velocity time graph. Use the approximation found in part (iii) to estimate the displacement of the spacecraft over the first half minute.\\
(v) Write $\frac { 1 } { ( 1 + t ) ( 2 + t ) ^ { 2 } }$ in partial fractions.\\
(vi) The displacement of the spacecraft in the first $T$ minutes is given by $\int _ { 0 } ^ { T } v \mathrm {~d} t$
Calculate the exact value of the displacement of the spacecraft over the first half minute given by the model.\\
(vii) On further investigation the scientists believe the original model may be valid for up to three minutes. Explain why the approximation in (iii) will be no longer be valid for this time interval.
\hfill \mbox{\textit{OCR MEI C4 Q8 [19]}}