| Exam Board | WJEC |
|---|---|
| Module | Further Unit 4 (Further Unit 4) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Find stationary points of hyperbolic curves |
| Difficulty | Standard +0.8 This is a Further Maths hyperbolic functions question requiring differentiation using chain rule, solving a transcendental equation (sinh x = 0), second derivative test, and range analysis. While the techniques are standard, working with hyperbolic functions and their identities (cosh²x - sinh²x = 1) requires solid understanding beyond core A-level, and the multi-part nature with 8 marks total indicates moderate complexity for Further Maths. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07d Differentiate/integrate: hyperbolic functions |
| Answer | Marks | Guidance |
|---|---|---|
| Differentiating \(f'(x) = 3\cosh^2 x \sinh x - 3\sinh x\) | M1 A1 | If identities used, must be a valid attempt at differentiation |
| At a stationary point, \(f'(x) = 0\), \(\therefore 3\cosh^2 x \sinh x - 3\sinh x = 0\) | m1 | |
| \(3\sinh x(\cosh^2 x - 1) = 0\) | A1 | Award for any solution of hyperbolic equation; Must be seen to discard equations with no solutions and show all remaining equations lead to \(x = 0\) |
| THEN: \(3\sinh x = 0 \therefore x = 0\) or \(\cosh^2 x - 1 = 0 \therefore \cosh x = 1\) or \(\cosh x = -1\) (no solutions) | A1 | |
| \(\therefore\) The only stationary point is at \(x = 0\) | A1 | |
| OR: As \(\cosh^2 x - 1 = \sinh^2 x\), \(3\sinh x(\cosh^2 x - 1) = 3\sinh^3 x = 0\) | (A1) | Correct use of identity |
| \(\therefore \sinh^3 x = 0 \therefore \sinh x = 0 \therefore x = 0\) | A1 | |
| \(\therefore\) The only stationary point is at \(x = 0\) | (A1) |
| Answer | Marks | Guidance |
|---|---|---|
| METHOD 1: Find gradient or value of \(f\) either side of \(x = 0\) | M1 | Accept graphical method |
| e.g. \(f'(-1) = -4.869... < 0\) and \(f'(1) = 4.869... > 0\) | A1 | cao |
| e.g. \(\sinh x < 0\) and therefore \(\sinh^3 x < 0\) for \(x < 0\) and \(\sinh x > 0\) and therefore \(\sinh^3 x > 0\) for \(x > 0\) | A1 | cao |
| Therefore, the stationary point at \(x = 0\) is a minimum | A1 | cao |
| METHOD 2: Differentiating and substituting \(x = 0\) | (M1) | |
| \(f''(x) = 3\cosh^3 x + 6\cosh x \sinh^2 x - 3\cosh x\) oe | A1 | cao |
| \(f''(0) = 3 + 0 - 3 = 0\) | A1 | cao |
| Finding the gradient either side of \(x = 0\) AND Stating the stationary point at \(x = 0\) is a minimum | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| When \(x = 0\), \(f(x) = -2\) | B1 | Allow 'Range \(f(x) \geq -2\)' |
| Therefore, largest range is \([-2, \infty)\) | B1 |
**Part a)**
Differentiating $f'(x) = 3\cosh^2 x \sinh x - 3\sinh x$ | M1 A1 | If identities used, must be a valid attempt at differentiation
At a stationary point, $f'(x) = 0$, $\therefore 3\cosh^2 x \sinh x - 3\sinh x = 0$ | m1 |
$3\sinh x(\cosh^2 x - 1) = 0$ | A1 | Award for any solution of hyperbolic equation; Must be seen to discard equations with no solutions and show all remaining equations lead to $x = 0$
THEN: $3\sinh x = 0 \therefore x = 0$ or $\cosh^2 x - 1 = 0 \therefore \cosh x = 1$ or $\cosh x = -1$ (no solutions) | A1 |
$\therefore$ The only stationary point is at $x = 0$ | A1 |
OR: As $\cosh^2 x - 1 = \sinh^2 x$, $3\sinh x(\cosh^2 x - 1) = 3\sinh^3 x = 0$ | (A1) | Correct use of identity
$\therefore \sinh^3 x = 0 \therefore \sinh x = 0 \therefore x = 0$ | A1 |
$\therefore$ The only stationary point is at $x = 0$ | (A1) |
**Part b)**
METHOD 1: Find gradient or value of $f$ either side of $x = 0$ | M1 | Accept graphical method
e.g. $f'(-1) = -4.869... < 0$ and $f'(1) = 4.869... > 0$ | A1 | cao
e.g. $\sinh x < 0$ and therefore $\sinh^3 x < 0$ for $x < 0$ and $\sinh x > 0$ and therefore $\sinh^3 x > 0$ for $x > 0$ | A1 | cao
Therefore, the stationary point at $x = 0$ is a minimum | A1 | cao
METHOD 2: Differentiating and substituting $x = 0$ | (M1) |
$f''(x) = 3\cosh^3 x + 6\cosh x \sinh^2 x - 3\cosh x$ oe | A1 | cao
$f''(0) = 3 + 0 - 3 = 0$ | A1 | cao
Finding the gradient either side of $x = 0$ AND Stating the stationary point at $x = 0$ is a minimum | A1 | cao
**Part c)**
When $x = 0$, $f(x) = -2$ | B1 | Allow 'Range $f(x) \geq -2$'
Therefore, largest range is $[-2, \infty)$ | B1 |
---A function $f$ has domain $(-\infty,\infty)$ and is defined by $f(x) = \cosh^3 x - 3\cosh x$.
\begin{enumerate}[label=(\alph*)]
\item Show that the graph of $y = f(x)$ has only one stationary point. [5]
\item Find the nature of this stationary point. [2]
\item State the largest possible range of $f(x)$. [1]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 4 2022 Q1 [8]}}