| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2014 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration with Partial Fractions |
| Type | Improper algebraic form then partial fractions |
| Difficulty | Standard +0.3 This is a structured partial fractions question with explicit guidance. Part (a) involves straightforward algebraic manipulation to find constants by comparing coefficients. Part (b) applies the result to integrate, requiring recognition that the numerator splits into a polynomial plus a proper fraction, leading to standard integrals (polynomial, ln, and arctan terms). While it requires multiple techniques, the question scaffolds the approach and uses routine methods without requiring novel insight. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.08d Evaluate definite integrals: between limits1.08j Integration using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(4x^3 + 2x^2 + 17x + 8 \equiv (Ax+B)(x^2+4) + Cx + D\) | — | — |
| Compare \(x^3\) terms: \(A = 4\) | B1 | States \(A=4\). May be implied by writing rhs as \((4x+B)(x^2+4)+Cx+D\) |
| Compare \(x^2\) terms: \(B = 2\) | B1 | States \(B=2\). May be implied by writing rhs as \((Ax+2)(x^2+4)+Cx+D\) |
| Compare \(x\) term or constant: \(4A+C=17\) or \(4B+D=8 \Rightarrow C=1,\ D=0\) | M1, A1 | M1: Compares \(x\) or constant terms to find \(C\) or \(D\). A1: Both values correct \(C=1,\ D=0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Quotient \(4x+2\) obtained | B1, B1 | B1, B1 for sight of 4 and 2 in quotient \(4x+2\) |
| Proceeding to get linear remainder | M1 | — |
| Remainder \(= x\) (correct) | A1 | Accept \(x\). If division unclear, accept answers in correct place in part (b) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int_1^4 \frac{4x^3+2x^2+17x+8}{x^2+4}\,dx = \int_1^4 \left(4x+2+\frac{x}{x^2+4}\right)dx\) | M1 | For using part (a) answers to rewrite integral in form \(\int \left('A'x+'B'+\frac{'C'x+'D'}{x^2+4}\right)dx\) with numerical values |
| \(= \left[2x^2+2x+\frac{1}{2}\ln(x^2+4)\right]_1^4\) | M1, M1A1 | M1: Correct integration of \('A'x+'B'\) part. Accept \(\frac{A}{2}x^2+Bx\) or \(\frac{(Ax+B)^2}{2}\). M1: Correct method for integrating \(\frac{'C'x}{x^2+4}\); accept constant \(\times \ln(x^2+4)\). A1: Correct integral |
| \(= \left[2(16)+2(4)+\frac{1}{2}\ln 20\right] - \left[2(1)+2(1)+\frac{1}{2}\ln 5\right]\) | — | — |
| \(= 36 + \frac{1}{2}\ln\!\left(\frac{20}{5}\right)\) | dM1 | Putting in limits, subtracting and correctly collecting ln terms using subtraction law. Dependent on previous M |
| \(= 36 + \ln(2)\) | A1 | CSO and CAO. \(\frac{1}{2}\ln 20 - \frac{1}{2}\ln 5 = \ln 2\) |
## Question 3:
**Part (a):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $4x^3 + 2x^2 + 17x + 8 \equiv (Ax+B)(x^2+4) + Cx + D$ | — | — |
| Compare $x^3$ terms: $A = 4$ | B1 | States $A=4$. May be implied by writing rhs as $(4x+B)(x^2+4)+Cx+D$ |
| Compare $x^2$ terms: $B = 2$ | B1 | States $B=2$. May be implied by writing rhs as $(Ax+2)(x^2+4)+Cx+D$ |
| Compare $x$ term or constant: $4A+C=17$ or $4B+D=8 \Rightarrow C=1,\ D=0$ | M1, A1 | M1: Compares $x$ or constant terms to find $C$ or $D$. A1: Both values correct $C=1,\ D=0$ |
*Alternative via division:*
| Answer/Working | Marks | Guidance |
|---|---|---|
| Quotient $4x+2$ obtained | B1, B1 | B1, B1 for sight of 4 and 2 in quotient $4x+2$ |
| Proceeding to get linear remainder | M1 | — |
| Remainder $= x$ (correct) | A1 | Accept $x$. If division unclear, accept answers in correct place in part (b) |
**Part (b):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int_1^4 \frac{4x^3+2x^2+17x+8}{x^2+4}\,dx = \int_1^4 \left(4x+2+\frac{x}{x^2+4}\right)dx$ | M1 | For using part (a) answers to rewrite integral in form $\int \left('A'x+'B'+\frac{'C'x+'D'}{x^2+4}\right)dx$ with numerical values |
| $= \left[2x^2+2x+\frac{1}{2}\ln(x^2+4)\right]_1^4$ | M1, M1A1 | M1: Correct integration of $'A'x+'B'$ part. Accept $\frac{A}{2}x^2+Bx$ or $\frac{(Ax+B)^2}{2}$. M1: Correct method for integrating $\frac{'C'x}{x^2+4}$; accept constant $\times \ln(x^2+4)$. A1: Correct integral |
| $= \left[2(16)+2(4)+\frac{1}{2}\ln 20\right] - \left[2(1)+2(1)+\frac{1}{2}\ln 5\right]$ | — | — |
| $= 36 + \frac{1}{2}\ln\!\left(\frac{20}{5}\right)$ | dM1 | Putting in limits, subtracting and correctly collecting ln terms using subtraction law. Dependent on previous M |
| $= 36 + \ln(2)$ | A1 | CSO and CAO. $\frac{1}{2}\ln 20 - \frac{1}{2}\ln 5 = \ln 2$ |
---3. Given that
$$4 x ^ { 3 } + 2 x ^ { 2 } + 17 x + 8 \equiv ( A x + B ) \left( x ^ { 2 } + 4 \right) + C x + D$$
\begin{enumerate}[label=(\alph*)]
\item find the values of the constants $A , B , C$ and $D$.
\item Hence find
$$\int _ { 1 } ^ { 4 } \frac { 4 x ^ { 3 } + 2 x ^ { 2 } + 17 x + 8 } { x ^ { 2 } + 4 } d x$$
giving your answer in the form $p + \ln q$, where $p$ and $q$ are integers.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2014 Q3 [10]}}