| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2023 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Summation bounds using hyperbolic integrals |
| Difficulty | Hard +2.3 This Further Maths question requires finding a maximum using hyperbolic differentiation (product rule with sech²x), numerical verification, then crucially interpreting a Riemann sum with rectangles to establish an inequality involving hyperbolic integrals. The summation-to-integral conversion with careful bound management and the non-standard integral of xsech²x (requiring integration by parts) makes this significantly harder than typical A-level questions, though the individual techniques are within Further Maths scope. |
| Spec | 4.07d Differentiate/integrate: hyperbolic functions4.08f Integrate using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = -2x\,\text{sech}^2 x\tanh x + \text{sech}^2 x\) \(\left[= 1-\tanh^2 x - 2x\tanh x + 2x\tanh^3 x\right]\) | M1 A1 | Differentiating using product rule and chain rule for M1 |
| \(\text{sech}^2 x \neq 0 \Rightarrow 2x\tanh x - 1 = 0\) | A1 | AG |
| \(2(0.7)\tanh(0.7)-1 = -0.154 < 0\) \(2(0.8)\tanh(0.8)-1 = 0.062 > 0\) | B1 | Shows sign change. Must write down values correct to at least 1dp for B1 |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sum_{r=2}^{n} r\,\text{sech}^2 r < \int_1^n x\,\text{sech}^2 x\,dx\) | M1 A1 | Compares sum with integral. Consistent limits for M1 |
| \(\int_1^n x\,\text{sech}^2 x\,dx = \left[x\tanh x\right]_1^n - \int_1^n \tanh x\,dx\) | M1 A1 | Integrates by parts |
| \(= \left[x\tanh x + \ln\text{sech}\, x\right]_1^n\) | A1 | |
| \(= n\tanh n + \ln\text{sech}\,n - (\tanh 1 + \ln\text{sech}\,1)\) | A1 | AG. Must have gained all previous marks |
| Total: 6 |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = -2x\,\text{sech}^2 x\tanh x + \text{sech}^2 x$ $\left[= 1-\tanh^2 x - 2x\tanh x + 2x\tanh^3 x\right]$ | M1 A1 | Differentiating using product rule and chain rule for M1 |
| $\text{sech}^2 x \neq 0 \Rightarrow 2x\tanh x - 1 = 0$ | A1 | AG |
| $2(0.7)\tanh(0.7)-1 = -0.154 < 0$ $2(0.8)\tanh(0.8)-1 = 0.062 > 0$ | B1 | Shows sign change. Must write down values correct to at least 1dp for B1 |
| **Total: 4** | | |
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sum_{r=2}^{n} r\,\text{sech}^2 r < \int_1^n x\,\text{sech}^2 x\,dx$ | M1 A1 | Compares sum with integral. Consistent limits for M1 |
| $\int_1^n x\,\text{sech}^2 x\,dx = \left[x\tanh x\right]_1^n - \int_1^n \tanh x\,dx$ | M1 A1 | Integrates by parts |
| $= \left[x\tanh x + \ln\text{sech}\, x\right]_1^n$ | A1 | |
| $= n\tanh n + \ln\text{sech}\,n - (\tanh 1 + \ln\text{sech}\,1)$ | A1 | AG. Must have gained all previous marks |
| **Total: 6** | | |5\\
\begin{tikzpicture}[scale=1.2]
% Axes
\draw[->] (-0.3,0) -- (5.5,0) node[right] {$x$};
\draw[->] (0,-0.3) -- (0,2.5) node[above] {$y$};
\node[below left] at (0,0) {$O$};
% Curve y = x sech^2(x), y-axis scaled by 4 for visibility
\draw[thick] plot[smooth, tension=0.6] coordinates {
(0,0) (0.2,0.77) (0.4,1.37) (0.6,1.71) (0.77,1.79)
(1.0,1.68) (1.5,1.08) (2.0,0.57) (2.5,0.27)
(3.0,0.12) (3.5,0.05) (4.0,0.02) (5.0,0.004)
};
% Maximum point M
\fill (0.77,1.79) circle (1.5pt);
\node[above left] at (0.77,1.79) {$M$};
\end{tikzpicture}
The diagram shows part of the curve $\mathrm { y } = \mathrm { xsech } ^ { 2 } \mathrm { x }$ and its maximum point $M$.
\begin{enumerate}[label=(\alph*)]
\item Show that, at $M$,
$$2 x \tanh x - 1 = 0$$
and verify that this equation has a root between 0.7 and 0.8 .
\item By considering a suitable set of rectangles, use the diagram to show that\\
$\sum _ { r = 2 } ^ { n } r \operatorname { sech } ^ { 2 } r < n \tanh n + \operatorname { lnsechn } - \tanh 1 - \operatorname { lnsech } 1$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2023 Q5 [10]}}