| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Polynomial division before integration |
| Difficulty | Moderate -0.3 This is a standard C4 integration question requiring algebraic manipulation followed by reverse chain rule. Part (a) involves equating coefficients to find constants (routine algebra), and part (b) requires recognizing that the numerator contains the derivative of the denominator, then integrating and applying a boundary condition. While it requires multiple steps, these are all standard techniques practiced extensively at this level, making it slightly easier than average. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4x^3 - 2x^2 + 16x - 3 = Ax(2x^2-x+2) + B(4x-1)\) | M1 | Multiply through |
| Comparing coefficients or substitution to find \(A=2\) | A1 | |
| \(B = 3\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \int \left(2x + \frac{3(4x-1)}{2x^2-x+2}\right)dx\) | M1 | Attempt to integrate using part (a) |
| \(y = x^2 + \frac{3}{2}\ln(2x^2-x+2) + c\) | A1 A1 | A1 for \(x^2\), A1 for \(\frac{3}{2}\ln(2x^2-x+2)\) |
| Substituting \((-1, 2)\): \(2 = 1 + \frac{3}{2}\ln(5) + c\), so \(c = 1 - \frac{3}{2}\ln 5\) | M1 | Use of point |
| \(y = x^2 + \frac{3}{2}\ln(2x^2-x+2) + 1 - \frac{3}{2}\ln 5\) | A1 | Accept equivalent forms |
# Question 2:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4x^3 - 2x^2 + 16x - 3 = Ax(2x^2-x+2) + B(4x-1)$ | M1 | Multiply through |
| Comparing coefficients or substitution to find $A=2$ | A1 | |
| $B = 3$ | A1 | |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \int \left(2x + \frac{3(4x-1)}{2x^2-x+2}\right)dx$ | M1 | Attempt to integrate using part (a) |
| $y = x^2 + \frac{3}{2}\ln(2x^2-x+2) + c$ | A1 A1 | A1 for $x^2$, A1 for $\frac{3}{2}\ln(2x^2-x+2)$ |
| Substituting $(-1, 2)$: $2 = 1 + \frac{3}{2}\ln(5) + c$, so $c = 1 - \frac{3}{2}\ln 5$ | M1 | Use of point |
| $y = x^2 + \frac{3}{2}\ln(2x^2-x+2) + 1 - \frac{3}{2}\ln 5$ | A1 | Accept equivalent forms |
---2
\begin{enumerate}[label=(\alph*)]
\item Given that $\frac { 4 x ^ { 3 } - 2 x ^ { 2 } + 16 x - 3 } { 2 x ^ { 2 } - x + 2 }$ can be expressed as $A x + \frac { B ( 4 x - 1 ) } { 2 x ^ { 2 } - x + 2 }$, find the values of the constants $A$ and $B$.
\item The gradient of a curve is given by
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 x ^ { 3 } - 2 x ^ { 2 } + 16 x - 3 } { 2 x ^ { 2 } - x + 2 }$$
The point $( - 1,2 )$ lies on the curve. Find the equation of the curve.\\[0pt]
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2014 Q2 [7]}}