OCR C4 2015 June — Question 10 14 marks

Exam BoardOCR
ModuleC4 (Core Mathematics 4)
Year2015
SessionJune
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPartial Fractions
TypeImproper fraction with linear factors – division then partial fractions
DifficultyStandard +0.3 This is a structured multi-part question that guides students through standard C4 techniques: partial fractions (routine), polynomial division (standard), parametric-to-cartesian conversion (algebraic manipulation), and integration using partial fractions. While it requires multiple steps and careful algebra, each part uses well-practiced methods with no novel insight required. The scaffolding makes it slightly easier than average.
Spec1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02y Partial fractions: decompose rational functions1.03g Parametric equations: of curves and conversion to cartesian1.08f Area between two curves: using integration

10
  1. Express \(\frac { x + 8 } { x ( x + 2 ) }\) in partial fractions.
  2. By first using division, express \(\frac { 7 x ^ { 2 } + 16 x + 16 } { x ( x + 2 ) }\) in the form \(P + \frac { Q } { x } + \frac { R } { x + 2 }\). A curve has parametric equations \(x = \frac { 2 t } { 1 - t } , y = 3 t + \frac { 4 } { t }\).
  3. Show that the cartesian equation of the curve is \(y = \frac { 7 x ^ { 2 } + 16 x + 16 } { x ( x + 2 ) }\).
  4. Find the area of the region bounded by the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 2\). Give your answer in the form \(L + M \ln 2 + N \ln 3\).

Question 10(i):
AnswerMarks Guidance
AnswerMarks Guidance
\(\frac{A}{x} + \frac{B}{x+2}\)B1 Award if only implied by answer
\(x + 8 = A(x+2) + Bx\) soiM1 Allow one sign error; clearing fractions successfully
\(A = 4\) and \(B = -3\)A1 [3] If M0, B1 for each value www
Question 10(ii):
AnswerMarks Guidance
AnswerMarks Guidance
Quotient \((P)\) is \(7\)B1
\(2x + 16\) seenB1 If B0, B1 for \(Q=8\) and B1 for \(R=-6\) www; e.g. as remainder or in division chunking
\(7 + \frac{8}{x} - \frac{6}{x+2}\)B1 [3] Or allow \(P=7\), \(Q=8\), \(R=-6\)
Question 10(iii):
AnswerMarks Guidance
AnswerMarks Guidance
\(t = f(x)\)M1* From \(x = \frac{2t}{1-t}\); M0 for \(t = g(y)\)
\(t = \frac{x}{x+2}\)A1 Or B2 if unsupported
\(y = 3\times\frac{x}{x+2} + \frac{4}{\frac{x}{x+2}}\)M1dep* At least one correct, constructive, intermediate step shown; if M0M0, SC2 for substitution of \(x=\frac{2t}{1-t}\) in RHS of given equation and completion with at least two correct, constructive intermediate steps to \(y = 3t + \frac{4}{t}\) www
e.g. \(\frac{3x^2+(8+4x)(x+2)}{x(x+2)}\) and completion to
\(y = \frac{7x^2+16x+16}{x(x+2)}\) www AGA1 [4]
Question 10(iv):
AnswerMarks Guidance
AnswerMark Guidance
\(\int \text{their} \left(P + \frac{Q}{x} + \frac{R}{x+2}\right) dx\)M1\* Where \(P\), \(Q\) and \(R\) are constants obtained in (ii); allow omission of \(dx\)
\(F[x] = 7x + 8\ln x - 6\ln(x+2)\)A1FT Allow recovery from omission of brackets in subsequent working. If M0, SC1 for \(Px + Q\ln x + R\ln(x+2)\) where constants are unspecified or arbitrary
\(F[2] - F[1]\)M1dep\*
\(7 - 4\ln 2 + 6\ln 3\)A1
[4]
## Question 10(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{A}{x} + \frac{B}{x+2}$ | B1 | Award if only implied by answer |
| $x + 8 = A(x+2) + Bx$ soi | M1 | Allow one sign error; clearing fractions successfully |
| $A = 4$ and $B = -3$ | A1 **[3]** | If M0, B1 for each value www |

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## Question 10(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Quotient $(P)$ is $7$ | B1 | |
| $2x + 16$ seen | B1 | If B0, B1 for $Q=8$ and B1 for $R=-6$ www; e.g. as remainder or in division chunking |
| $7 + \frac{8}{x} - \frac{6}{x+2}$ | B1 **[3]** | Or allow $P=7$, $Q=8$, $R=-6$ |

---

## Question 10(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $t = f(x)$ | M1* | From $x = \frac{2t}{1-t}$; M0 for $t = g(y)$ |
| $t = \frac{x}{x+2}$ | A1 | Or B2 if unsupported |
| $y = 3\times\frac{x}{x+2} + \frac{4}{\frac{x}{x+2}}$ | M1dep* | At least one correct, constructive, intermediate step shown; if M0M0, SC2 for substitution of $x=\frac{2t}{1-t}$ in RHS of given equation and completion with at least two correct, constructive intermediate steps to $y = 3t + \frac{4}{t}$ www |
| e.g. $\frac{3x^2+(8+4x)(x+2)}{x(x+2)}$ and completion to | | |
| $y = \frac{7x^2+16x+16}{x(x+2)}$ www **AG** | A1 **[4]** | |

## Question 10(iv):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\int \text{their} \left(P + \frac{Q}{x} + \frac{R}{x+2}\right) dx$ | **M1\*** | Where $P$, $Q$ and $R$ are constants obtained in (ii); allow omission of $dx$ |
| $F[x] = 7x + 8\ln x - 6\ln(x+2)$ | **A1FT** | Allow recovery from omission of brackets in subsequent working. If **M0**, **SC1** for $Px + Q\ln x + R\ln(x+2)$ where constants are unspecified or arbitrary |
| $F[2] - F[1]$ | **M1dep\*** | |
| $7 - 4\ln 2 + 6\ln 3$ | **A1** | |
| | **[4]** | |
10 (i) Express $\frac { x + 8 } { x ( x + 2 ) }$ in partial fractions.\\
(ii) By first using division, express $\frac { 7 x ^ { 2 } + 16 x + 16 } { x ( x + 2 ) }$ in the form $P + \frac { Q } { x } + \frac { R } { x + 2 }$.

A curve has parametric equations $x = \frac { 2 t } { 1 - t } , y = 3 t + \frac { 4 } { t }$.\\
(iii) Show that the cartesian equation of the curve is $y = \frac { 7 x ^ { 2 } + 16 x + 16 } { x ( x + 2 ) }$.\\
(iv) Find the area of the region bounded by the curve, the $x$-axis and the lines $x = 1$ and $x = 2$. Give your answer in the form $L + M \ln 2 + N \ln 3$.

\hfill \mbox{\textit{OCR C4 2015 Q10 [14]}}