| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Inverse functions (inverse trig/hyperbolic) |
| Difficulty | Challenging +1.2 This is a structured Further Maths question combining inverse hyperbolic functions with Taylor series. Part (a) guides students through implicit differentiation using hyperbolic identities, while part (b) requires systematic application of Maclaurin series formula. The question is methodical rather than insightful—students follow standard procedures (differentiate, evaluate at x=0) with moderate algebraic manipulation. Slightly above average due to Further Maths content and multi-step nature, but the scaffolding and routine techniques keep it accessible. |
| Spec | 4.07e Inverse hyperbolic: definitions, domains, ranges4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{sech}\, y = \frac{1}{\cosh y} = x + \tfrac{1}{2} \Rightarrow \cosh y = \left(x+\tfrac{1}{2}\right)^{-1}\) | B1 | Relates to \(\cosh y\). |
| \(\sinh y \frac{dy}{dx}\) | B1 | Differentiates LHS. |
| \(-\left(x+\tfrac{1}{2}\right)^{-2}\) | B1 | Differentiates RHS, AG. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sinh\frac{d^2y}{dx^2} + \cosh y\left(\frac{dy}{dx}\right)^2 = 2\left(x+\tfrac{1}{2}\right)^{-3}\) | M1 A1 | M1 A1 for LHS. B1 for RHS. |
| (B1 for RHS) | B1 | |
| \(y(0) = \text{sech}^{-1}\!\left(\tfrac{1}{2}\right) = \cosh^{-1}(2) = \ln\!\left(2+\sqrt{3}\right)\) | M1 A1 | Relates to \(\cosh^{-1}\) and uses logarithmic form. |
| \(y'(0) = -\dfrac{4}{\sqrt{3}}, \quad y''(0) = \dfrac{16}{3\sqrt{3}}\) | M1 | Evaluates derivatives at \(x=0\). |
| \(y = \ln\!\left(2+\sqrt{3}\right) - \dfrac{4}{\sqrt{3}}x + \dfrac{8}{3\sqrt{3}}x^2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = -\dfrac{1}{\left(x+\frac{1}{2}\right)^2\sqrt{\left(x+\frac{1}{2}\right)^{-2}-1}} = -\dfrac{1}{\left(x+\frac{1}{2}\right)\sqrt{\frac{3}{4}-x-x^2}}\) | B1 | |
| \(\frac{d^2y}{dx^2} = \tfrac{1}{2}\left(x+\tfrac{1}{2}\right)^{-1}\!\left(\tfrac{3}{4}-x-x^2\right)^{-\frac{3}{2}}(-1-2x) + \left(\tfrac{3}{4}-x-x^2\right)^{-\frac{1}{2}}\!\left(x+\tfrac{1}{2}\right)^{-2}\) | M1 A1 | |
| \(y(0) = \text{sech}^{-1}\!\left(\tfrac{1}{2}\right) = \cosh^{-1}(2) = \ln\!\left(2+\sqrt{3}\right)\) | M1 A1 | Relates to \(\cosh^{-1}\) and uses logarithmic form. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y'(0) = -\frac{4}{\sqrt{3}}\), \(y''(0) = \frac{16}{3\sqrt{3}}\) | M1 | Evaluates derivatives at \(x = 0\) |
| \(y = \ln(2+\sqrt{3}) - \frac{4}{\sqrt{3}}x + \frac{8}{3\sqrt{3}}x^2\) | A1 | |
| 7 (total) |
## Question 7(a):
$\text{sech}\, y = \frac{1}{\cosh y} = x + \tfrac{1}{2} \Rightarrow \cosh y = \left(x+\tfrac{1}{2}\right)^{-1}$ | **B1** | Relates to $\cosh y$.
$\sinh y \frac{dy}{dx}$ | **B1** | Differentiates LHS.
$-\left(x+\tfrac{1}{2}\right)^{-2}$ | **B1** | Differentiates RHS, AG.
---
## Question 7(b):
$\sinh\frac{d^2y}{dx^2} + \cosh y\left(\frac{dy}{dx}\right)^2 = 2\left(x+\tfrac{1}{2}\right)^{-3}$ | **M1 A1** | M1 A1 for LHS. B1 for RHS.
(B1 for RHS) | **B1** |
$y(0) = \text{sech}^{-1}\!\left(\tfrac{1}{2}\right) = \cosh^{-1}(2) = \ln\!\left(2+\sqrt{3}\right)$ | **M1 A1** | Relates to $\cosh^{-1}$ and uses logarithmic form.
$y'(0) = -\dfrac{4}{\sqrt{3}}, \quad y''(0) = \dfrac{16}{3\sqrt{3}}$ | **M1** | Evaluates derivatives at $x=0$.
$y = \ln\!\left(2+\sqrt{3}\right) - \dfrac{4}{\sqrt{3}}x + \dfrac{8}{3\sqrt{3}}x^2$ | **A1** |
**Alternative method for 7(b):**
$\frac{dy}{dx} = -\dfrac{1}{\left(x+\frac{1}{2}\right)^2\sqrt{\left(x+\frac{1}{2}\right)^{-2}-1}} = -\dfrac{1}{\left(x+\frac{1}{2}\right)\sqrt{\frac{3}{4}-x-x^2}}$ | **B1** |
$\frac{d^2y}{dx^2} = \tfrac{1}{2}\left(x+\tfrac{1}{2}\right)^{-1}\!\left(\tfrac{3}{4}-x-x^2\right)^{-\frac{3}{2}}(-1-2x) + \left(\tfrac{3}{4}-x-x^2\right)^{-\frac{1}{2}}\!\left(x+\tfrac{1}{2}\right)^{-2}$ | **M1 A1** |
$y(0) = \text{sech}^{-1}\!\left(\tfrac{1}{2}\right) = \cosh^{-1}(2) = \ln\!\left(2+\sqrt{3}\right)$ | **M1 A1** | Relates to $\cosh^{-1}$ and uses logarithmic form.
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y'(0) = -\frac{4}{\sqrt{3}}$, $y''(0) = \frac{16}{3\sqrt{3}}$ | **M1** | Evaluates derivatives at $x = 0$ |
| $y = \ln(2+\sqrt{3}) - \frac{4}{\sqrt{3}}x + \frac{8}{3\sqrt{3}}x^2$ | **A1** | |
| | **7** (total) | |
---7
\begin{enumerate}[label=(\alph*)]
\item It is given that $\mathrm { y } = \operatorname { sech } ^ { - 1 } \left( \mathrm { x } + \frac { 1 } { 2 } \right)$.\\
Express cosh $y$ in terms of $x$ and hence show that $\sinh y \frac { d y } { d x } = - \frac { 1 } { \left( x + \frac { 1 } { 2 } \right) ^ { 2 } }$.
\item Find the first three terms in the Maclaurin's series for $\operatorname { sech } ^ { - 1 } \left( x + \frac { 1 } { 2 } \right)$ in the form
$$\ln a + b x + c x ^ { 2 }$$
where $a$, $b$ and $c$ are constants to be determined.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2021 Q7 [10]}}