| Exam Board | WJEC |
|---|---|
| Module | Further Unit 4 (Further Unit 4) |
| Year | 2019 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration using inverse trig and hyperbolic functions |
| Type | Partial fractions then inverse trig integration |
| Difficulty | Standard +0.3 This is a structured multi-part question from Further Maths covering standard techniques: proving a derivative formula for inverse trig (bookwork), partial fractions with linear and quadratic factors (routine), integration using the partial fractions (straightforward application), and identifying a discontinuity. While it's Further Maths content, each part follows well-established methods with no novel insight required, making it slightly easier than average overall. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08j Integration using partial fractions4.08g Derivatives: inverse trig and hyperbolic functions |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = \cot y\) | M1 | |
| \(\frac{dx}{dy} = -\cosec^2 y\) | A1 | |
| \(\frac{dx}{dy} = -(\cot^2 y + 1)\) | A1 | |
| \(\frac{dx}{dy} = -(x^2 + 1)\) | A1 | |
| \(\frac{dy}{dx} = \frac{-1}{x^2 + 1}\) | A1 | Convincing |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{6x^2 - 10x - 9}{(2x + 3)(x^2 + 1)} = \frac{A}{2x + 3} + \frac{Bx + C}{x^2 + 1}\) | M1 | Use of |
| \(6x^2 - 10x - 9 = A(x^2 + 1) + (Bx + C)(2x + 3)\) When \(x = -1.5\), \(19.5 = 3.25A\) | A1 | |
| \(\therefore A = 6\) | A1 | |
| When \(x = 0\), \(-9 = A + 3C\) | A1 | |
| \(\therefore C = -5\) | A1 | |
| Co-efficients of \(x^2\): \(6 = A + 2B\) | A1 | |
| \(\therefore B = 0\) | A1 | |
| \(\frac{6x^2 - 10x - 9}{(2x + 3)(x^2 + 1)} = \frac{6}{2x + 3} - \frac{5}{x^2 + 1}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{6x^2 - 8x - 6}{(2x + 3)(x^2 + 1)}\) | ||
| \(= \frac{6}{2x + 3} - \frac{5}{x^2 + 1} + \frac{2x + 3}{(2x + 3)(x^2 + 1)}\) | M1 | FT (b) |
| \(= \frac{6}{2x + 3} - \frac{4}{x^2 + 1}\) | A1 | |
| \(\therefore \int\left(\frac{6}{2x + 3} - \frac{4}{x^2 + 1}\right) dx\) | M1 | |
| \(= 3\ln | 2x + 3 | + 4\cot^{-1}x + c\) |
| Answer | Marks |
|---|---|
| The expression is undefined when \(x = -1.5\) (and -1.5 is in the range of integration.) | E1 |
## Part a)
$x = \cot y$ | M1 |
$\frac{dx}{dy} = -\cosec^2 y$ | A1 |
$\frac{dx}{dy} = -(\cot^2 y + 1)$ | A1 |
$\frac{dx}{dy} = -(x^2 + 1)$ | A1 |
$\frac{dy}{dx} = \frac{-1}{x^2 + 1}$ | A1 | Convincing
## Part b)
$\frac{6x^2 - 10x - 9}{(2x + 3)(x^2 + 1)} = \frac{A}{2x + 3} + \frac{Bx + C}{x^2 + 1}$ | M1 | Use of
$6x^2 - 10x - 9 = A(x^2 + 1) + (Bx + C)(2x + 3)$ When $x = -1.5$, $19.5 = 3.25A$ | A1 |
$\therefore A = 6$ | A1 |
When $x = 0$, $-9 = A + 3C$ | A1 |
$\therefore C = -5$ | A1 |
Co-efficients of $x^2$: $6 = A + 2B$ | A1 |
$\therefore B = 0$ | A1 |
$\frac{6x^2 - 10x - 9}{(2x + 3)(x^2 + 1)} = \frac{6}{2x + 3} - \frac{5}{x^2 + 1}$ | |
## Part c)
$\frac{6x^2 - 8x - 6}{(2x + 3)(x^2 + 1)}$ | |
$= \frac{6}{2x + 3} - \frac{5}{x^2 + 1} + \frac{2x + 3}{(2x + 3)(x^2 + 1)}$ | M1 | FT (b)
$= \frac{6}{2x + 3} - \frac{4}{x^2 + 1}$ | A1 |
$\therefore \int\left(\frac{6}{2x + 3} - \frac{4}{x^2 + 1}\right) dx$ | M1 |
$= 3\ln|2x + 3| + 4\cot^{-1}x + c$ | A1A1 | Accept $-4\tan^{-1}x$; Penalise -1 for no constant term
## Part d)
The expression is undefined when $x = -1.5$ (and -1.5 is in the range of integration.) | E1 |\begin{enumerate}[label=(\alph*)]
\item Given that $y = \cot^{-1} x$, show that $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-1}{x^2 + 1}$. [5]
\item Express $\frac{6x^2 - 10x - 9}{(2x + 3)(x^2 + 1)}$ in terms of partial fractions. [5]
\item Hence find $\int \frac{6x^2 - 8x - 6}{(2x + 3)(x^2 + 1)} \mathrm{d}x$. [5]
\item Explain why $\int_{-2}^{5} \frac{6x^2 - 8x - 6}{(2x + 3)(x^2 + 1)} \mathrm{d}x$ cannot be evaluated. [1]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 4 2019 Q4 [16]}}