| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2016 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Partial fractions for differential equations |
| Difficulty | Standard +0.8 This is a two-part question combining partial fractions with separable differential equations. Part (i) requires handling a repeated linear factor, which is more challenging than simple distinct factors. Part (ii) requires separating variables, integrating the partial fractions (including ln|x+1| terms and 1/(x+1)² integration), applying initial conditions, and manipulating to reach the exact exponential form. The multi-step nature, integration of multiple terms, and algebraic manipulation to reach the final form elevate this above average difficulty. |
| Spec | 1.02y Partial fractions: decompose rational functions1.06g Equations with exponentials: solve a^x = b1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{A}{(x+4)} + \frac{B}{(x+1)} + \frac{C}{(x+1)^2}\) | B1 | May be awarded later |
| \([16+5x-2x^2] = A(x+1)^2 + B(x+1)(x+4) + C(x+4)\) | M1 | Allow sign errors only |
| \(A = -4\) | A1 | NB \(36 = -9A\) |
| \(C = 3\) | A1 | \(9 = 3C\) |
| \(B = 2\) isw | A1 | \(-2 = A+B,\ 5 = 2A+5B+C,\ 16 = A+4B+4C\); NB \(\frac{-4}{(x+4)} + \frac{2}{(x+1)} + \frac{3}{(x+1)^2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int\frac{dy}{y} = \int\frac{16+5x-2x^2}{(x+1)^2(x+4)}dx\) | B1 | Separation of variables; allow omission of integral signs; allow omission of \(dy\) or \(dx\) but not both |
| \(\frac{3}{(x+1)^2} + \frac{2}{(x+1)} - \frac{4}{(x+4)}\) seen in RHS, may be embedded | M1* | FT their partial fractions if two or three terms; ignore LHS; may be implied by correct integration of two of their terms |
| \(\frac{-3}{x+1} + 2\ln(x+1) - 4\ln(x+4) + c\) | A1FT | FT their non-zero 3, 2 and 4; allow recovery from \(x+1^2\) in denominator; if brackets in log terms omitted, allow A1 if recovery seen in substitution; allow omission of \(+c\) here |
| \(\ln\!\left(\frac{1}{256}\right) = -3 + 2\ln 1 - 4\ln 4 + c\) | M1*dep | Substitution of \(x=0\) and \(y=\frac{1}{256}\); allow if error in manipulation following integration; \(+c\) must be included and LHS must be correctly obtained |
| \(c = 3\) cao | A1 | or \(A = e^{-3}\) from \(y = Ae^{\frac{-3}{x+1}}\frac{(x+1)^2}{(x+4)^4}\) |
| \(\ln y = \frac{-3}{2+1} + 2\ln(2+1) - 4\ln(2+4) + 3\) | M1*dep | Substitution of \(x=2\); dependent on award of previous M1M1 and numerical value found for \(c\); allow M1 if substitution follows incorrect manipulation e.g. to find expression for \(y\) |
| \(y = \frac{e^2}{144}\) oe | A1 |
## Question 10(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{A}{(x+4)} + \frac{B}{(x+1)} + \frac{C}{(x+1)^2}$ | B1 | May be awarded later |
| $[16+5x-2x^2] = A(x+1)^2 + B(x+1)(x+4) + C(x+4)$ | M1 | Allow sign errors only |
| $A = -4$ | A1 | **NB** $36 = -9A$ |
| $C = 3$ | A1 | $9 = 3C$ |
| $B = 2$ isw | A1 | $-2 = A+B,\ 5 = 2A+5B+C,\ 16 = A+4B+4C$; **NB** $\frac{-4}{(x+4)} + \frac{2}{(x+1)} + \frac{3}{(x+1)^2}$ |
---
## Question 10(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int\frac{dy}{y} = \int\frac{16+5x-2x^2}{(x+1)^2(x+4)}dx$ | B1 | Separation of variables; allow omission of integral signs; allow omission of $dy$ or $dx$ but not both |
| $\frac{3}{(x+1)^2} + \frac{2}{(x+1)} - \frac{4}{(x+4)}$ seen in RHS, may be embedded | M1* | **FT** their partial fractions if two or three terms; ignore LHS; may be implied by correct integration of two of their terms |
| $\frac{-3}{x+1} + 2\ln(x+1) - 4\ln(x+4) + c$ | A1FT | **FT** their non-zero 3, 2 and 4; allow recovery from $x+1^2$ in denominator; if brackets in log terms omitted, allow **A1** if recovery seen in substitution; allow omission of $+c$ here |
| $\ln\!\left(\frac{1}{256}\right) = -3 + 2\ln 1 - 4\ln 4 + c$ | M1*dep | Substitution of $x=0$ and $y=\frac{1}{256}$; allow if error in manipulation following integration; $+c$ must be included and LHS must be correctly obtained |
| $c = 3$ cao | A1 | or $A = e^{-3}$ from $y = Ae^{\frac{-3}{x+1}}\frac{(x+1)^2}{(x+4)^4}$ |
| $\ln y = \frac{-3}{2+1} + 2\ln(2+1) - 4\ln(2+4) + 3$ | M1*dep | Substitution of $x=2$; dependent on award of previous **M1M1** and numerical value found for $c$; allow **M1** if substitution follows incorrect manipulation e.g. to find expression for $y$ |
| $y = \frac{e^2}{144}$ oe | A1 | |
The image you've shared appears to be only the **contact/back page** of an OCR mark scheme document — it contains only OCR's address, contact details, and legal information. There is **no mark scheme content** (no questions, answers, mark allocations, or guidance notes) visible on this page.
To get the mark scheme content extracted, please share the **interior pages** of the mark scheme document that contain the actual question answers and mark allocations.10 (i) Express $\frac { 16 + 5 x - 2 x ^ { 2 } } { ( x + 1 ) ^ { 2 } ( x + 4 ) }$ in partial fractions.\\
(ii) It is given that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { \left( 16 + 5 x - 2 x ^ { 2 } \right) y } { ( x + 1 ) ^ { 2 } ( x + 4 ) }$$
and that $y = \frac { 1 } { 256 }$ when $x = 0$. Find the exact value of $y$ when $x = 2$. Give your answer in the form $A \mathrm { e } ^ { n }$.
\hfill \mbox{\textit{OCR C4 2016 Q10 [12]}}