| Exam Board | WJEC |
|---|---|
| Module | Further Unit 4 (Further Unit 4) |
| Year | 2023 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Express hyperbolic in exponential form |
| Difficulty | Challenging +1.2 Part (a) requires recognizing that sinh x = (e^x - e^{-x})/2, leading to straightforward exponential integration. Part (b) is a standard partial fractions problem with a quadratic factor, requiring decomposition and arctan integration. Both are textbook techniques for Further Maths, but the 14-mark allocation and combination of hyperbolic functions with partial fractions places this above average difficulty. |
| Spec | 1.08j Integration using partial fractions4.05c Partial fractions: extended to quadratic denominators4.07d Differentiate/integrate: hyperbolic functions |
| Answer | Marks | Guidance |
|---|---|---|
| a) | METHOD 1: | |
| Using integration by parts: | ||
| \(I = \left[\frac{1}{2}e^{2x}\sin x\right]_{-2}^0 - \int_{-2}^0\frac{1}{2}e^{2x}\cos x\,dx\) | M1, A1 | Attempt by parts |
| \(I = \left[\frac{1}{2}e^{2x}\sin x - \frac{1}{4}e^{2x}\cos x\right]_{-2}^0 + \int_{-2}^0\frac{1}{4}e^{2x}\sinh x\,dx\) | A1 | |
| \(\frac{3}{4}I = \left[\frac{1}{2}e^{2x}\sinh x - \frac{1}{4}e^{2x}\cosh x\right]_{-2}^0\) | A1 | Fully correct including limits |
| \(I = \frac{-0.199599\ldots}{3/4} = -0.266\ldots\) | A1 | |
| METHOD 2: | ||
| Using integration by parts: | ||
| \(I = [e^{2x}\cosh x]_{-2}^0 - \int_{-2}^0 2e^{2x}\cosh x\,dx\) | (M1), (A1) | Attempt by parts |
| \(I = [e^{2x}\cosh x - 2e^{2x}\sinh x]_{-2}^0 + \int_{-2}^0 4e^{2x}\sinh x\,dx\) | (A1) | |
| \(-3I = [e^{2x}\cosh x - 2e^{2x}\sinh x]_{-2}^0\) | (A1) | Fully correct including limits |
| \(I = \frac{0.798236\ldots}{-3} = -0.266\ldots\) | (A1) | |
| METHOD 3: | ||
| Using substitution: | ||
| \(I = \int_{-2}^0 e^{2x}\left(\frac{e^x - e^{-x}}{2}\right)dx\) | (M1) | Substitution of \(\frac{e^x - e^{-x}}{2}\) |
| \(I = \frac{1}{2}\int_{-2}^0 (e^{3x} - e^x)dx\) | (A1) | |
| \(I = \frac{1}{2}\left[\frac{1}{3}e^{3x} - e^x\right]_{-2}^0\) | (A1) | Both terms |
| \(I = \frac{1}{2}\left[\left(\frac{1}{3} - 1\right) - \left(\frac{1}{3}e^{-6} - e^{-2}\right)\right]\) | (A1) | Correct use of limits |
| A1 | ||
| \(= -0.266\ldots\) | ||
| (5) | No marks awarded for answers only. | |
| b) | \(\frac{5}{(x-1)(x^2+9)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+9}\) | M1 |
| \(5 = A(x^2+9) + (Bx+C)(x-1)\) | A1 | |
| When \(x = 1\), \(5 = 10A\), \(\therefore A = \frac{1}{2}\) | ||
| When \(x = 0\), \(5 = 9A - C\), \(\therefore C = -\frac{1}{2}\) | ||
| Co-efficient of \(x^2\): \(0 = A + B\), \(\therefore B = -\frac{1}{2}\) | A2 | A1 any 2 coefficients |
| \(\int_{1.5}^3\frac{5}{(x-1)(x^2+9)}dx\) | ||
| \(\int_{1.5}^3\left(\frac{1/2}{x-1} + \frac{-1/2 x - 1/2}{x^2+9}\right)dx\) | M1 | FT their \(A, B, C\) provided at least two non-zero values |
| \(= \int_{1.5}^3\left(\frac{1}{2(x-1)} - \frac{x}{2(x^2+9)} - \frac{1}{2(x^2+9)}\right)dx\) | A1 | Three terms |
| $= \left[\frac{1}{2}\ln\ | x-1\ | - \frac{1}{4}\ln |
a) | **METHOD 1:** | | |
| Using integration by parts: | | |
| $I = \left[\frac{1}{2}e^{2x}\sin x\right]_{-2}^0 - \int_{-2}^0\frac{1}{2}e^{2x}\cos x\,dx$ | M1, A1 | Attempt by parts |
| | | |
| $I = \left[\frac{1}{2}e^{2x}\sin x - \frac{1}{4}e^{2x}\cos x\right]_{-2}^0 + \int_{-2}^0\frac{1}{4}e^{2x}\sinh x\,dx$ | A1 | |
| | | |
| $\frac{3}{4}I = \left[\frac{1}{2}e^{2x}\sinh x - \frac{1}{4}e^{2x}\cosh x\right]_{-2}^0$ | A1 | Fully correct including limits |
| | | |
| $I = \frac{-0.199599\ldots}{3/4} = -0.266\ldots$ | A1 | |
| | | |
| **METHOD 2:** | | |
| Using integration by parts: | | |
| $I = [e^{2x}\cosh x]_{-2}^0 - \int_{-2}^0 2e^{2x}\cosh x\,dx$ | (M1), (A1) | Attempt by parts |
| | | |
| $I = [e^{2x}\cosh x - 2e^{2x}\sinh x]_{-2}^0 + \int_{-2}^0 4e^{2x}\sinh x\,dx$ | (A1) | |
| | | |
| $-3I = [e^{2x}\cosh x - 2e^{2x}\sinh x]_{-2}^0$ | (A1) | Fully correct including limits |
| | | |
| $I = \frac{0.798236\ldots}{-3} = -0.266\ldots$ | (A1) | |
| | | |
| **METHOD 3:** | | |
| Using substitution: | | |
| $I = \int_{-2}^0 e^{2x}\left(\frac{e^x - e^{-x}}{2}\right)dx$ | (M1) | Substitution of $\frac{e^x - e^{-x}}{2}$ |
| | | |
| $I = \frac{1}{2}\int_{-2}^0 (e^{3x} - e^x)dx$ | (A1) | |
| | | |
| $I = \frac{1}{2}\left[\frac{1}{3}e^{3x} - e^x\right]_{-2}^0$ | (A1) | Both terms |
| | | |
| $I = \frac{1}{2}\left[\left(\frac{1}{3} - 1\right) - \left(\frac{1}{3}e^{-6} - e^{-2}\right)\right]$ | (A1) | Correct use of limits |
| | A1 | |
| $= -0.266\ldots$ | | |
| | (5) | No marks awarded for answers only. |
b) | $\frac{5}{(x-1)(x^2+9)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+9}$ | M1 | |
| | | |
| $5 = A(x^2+9) + (Bx+C)(x-1)$ | A1 | |
| | | |
| When $x = 1$, $5 = 10A$, $\therefore A = \frac{1}{2}$ | | |
| | | |
| When $x = 0$, $5 = 9A - C$, $\therefore C = -\frac{1}{2}$ | | |
| | | |
| Co-efficient of $x^2$: $0 = A + B$, $\therefore B = -\frac{1}{2}$ | A2 | A1 any 2 coefficients |
| | | |
| $\int_{1.5}^3\frac{5}{(x-1)(x^2+9)}dx$ | | |
| | | |
| $\int_{1.5}^3\left(\frac{1/2}{x-1} + \frac{-1/2 x - 1/2}{x^2+9}\right)dx$ | M1 | FT their $A, B, C$ provided at least two non-zero values |
| | | |
| $= \int_{1.5}^3\left(\frac{1}{2(x-1)} - \frac{x}{2(x^2+9)} - \frac{1}{2(x^2+9)}\right)dx$ | A1 | Three terms |
| | | |
| $= \left[\frac{1}{2}\ln\|x-1\| - \frac{1}{4}\ln|x^2+9| - \frac{1}{6}\tan^{-1}\frac{x}{3}\right]_{1.5}^Evaluate the integrals
\begin{enumerate}[label=(\alph*)]
\item $\int_{-2}^{0} e^{2x} \sinh x \, \mathrm{d}x$, [5]
\item $\int_{\frac{1}{2}}^{3} \frac{5}{(x-1)(x^2+9)} \, \mathrm{d}x$. [9]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 4 2023 Q11 [14]}}