| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2020 |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Maclaurin series for inverse hyperbolics |
| Difficulty | Challenging +1.8 This is a substantial Further Maths question requiring: (a) algebraic proof from exponential definitions, (b) implicit differentiation with chain rule, and (c) Maclaurin series construction using the differential equation. While each part uses standard techniques, the multi-step nature, the need to connect parts strategically, and the topic being Further Maths content places this well above average difficulty but not at the extreme end. |
| Spec | 4.07e Inverse hyperbolic: definitions, domains, ranges4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = \frac{e^u - e^{-u}}{e^u + e^{-u}}\) | M1 | Write in exponential form |
| \(e^{2u}(1-x) = 1+x\) | M1 | Rearrange |
| \(u = \tanh^{-1}x = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)\) | A1 | AG CAO |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| EITHER Solution 1: | Use result of (a) | |
| \(y = \frac{1}{2}\ln\left(\frac{1-\frac{1-x}{2+x}}{\frac{1-x}{2+x}}\right)\) | (M1) | |
| \(= \frac{1}{2}\ln\left(\frac{3}{2x+1}\right)\) | A1 | |
| Differentiate: \(y' = \frac{1}{2} \times \frac{2x+1}{3} \times \frac{-6}{(2x+1)^2}\) | M1 | Any equivalent work, e.g. splitting the log into 2 terms |
| \(= \frac{-1}{2x+1}\) giving \((2x+1)\frac{dy}{dx} + 1 = 0\) | A1 | AG |
| OR Solution 2: Differentiate first: \(\text{sech}^2 y\frac{dy}{dx} = \frac{-3}{(2+x)^2}\) | (M1A1) | |
| Use \(\text{sech}^2 y = 1 - \tanh^2 y\) | M1 | |
| to give \((2x+1)\frac{dy}{dx} + 1 = 0\) | A1 | AG |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Differentiate again: \(y'' = \frac{2}{(2x+1)^2}\) | B1 | Or differentiate result in (b): \((2x+1)y'' + 2y' = 0\) |
| \(y(0) = -1;\ y''(0) = 2\) | M1 | Values of derivatives at 0 |
| \(y'(0) = -1;\ y''(0) = 2\) | M1 | Use Maclaurin's series |
| \(y = \frac{1}{2}\ln 3 - x + 2 \times \frac{x^2}{2}\) | M1 | In terms of ln |
| \(= \frac{1}{2}\ln 3 - x + x^2\) | A1 | CAO |
| Total: 5 |
# Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = \frac{e^u - e^{-u}}{e^u + e^{-u}}$ | M1 | Write in exponential form |
| $e^{2u}(1-x) = 1+x$ | M1 | Rearrange |
| $u = \tanh^{-1}x = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$ | A1 | AG CAO |
| **Total: 3** | | |
# Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **EITHER** Solution 1: | | Use result of (a) |
| $y = \frac{1}{2}\ln\left(\frac{1-\frac{1-x}{2+x}}{\frac{1-x}{2+x}}\right)$ | (M1) | |
| $= \frac{1}{2}\ln\left(\frac{3}{2x+1}\right)$ | A1 | |
| Differentiate: $y' = \frac{1}{2} \times \frac{2x+1}{3} \times \frac{-6}{(2x+1)^2}$ | M1 | Any equivalent work, e.g. splitting the log into 2 terms |
| $= \frac{-1}{2x+1}$ giving $(2x+1)\frac{dy}{dx} + 1 = 0$ | A1 | AG |
| **OR** Solution 2: Differentiate first: $\text{sech}^2 y\frac{dy}{dx} = \frac{-3}{(2+x)^2}$ | (M1A1) | |
| Use $\text{sech}^2 y = 1 - \tanh^2 y$ | M1 | |
| to give $(2x+1)\frac{dy}{dx} + 1 = 0$ | A1 | AG |
| **Total: 4** | | |
# Question 7(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Differentiate again: $y'' = \frac{2}{(2x+1)^2}$ | B1 | Or differentiate result in (b): $(2x+1)y'' + 2y' = 0$ |
| $y(0) = -1;\ y''(0) = 2$ | M1 | Values of derivatives at 0 |
| $y'(0) = -1;\ y''(0) = 2$ | M1 | Use Maclaurin's series |
| $y = \frac{1}{2}\ln 3 - x + 2 \times \frac{x^2}{2}$ | M1 | In terms of ln |
| $= \frac{1}{2}\ln 3 - x + x^2$ | A1 | CAO |
| **Total: 5** | | |7 (a) Starting from the definition of tanh in terms of exponentials, prove that $\tanh ^ { - 1 } x = \frac { 1 } { 2 } \ln \left( \frac { 1 + x } { 1 - x } \right)$. [3]\\
(b) Given that $y = \tanh ^ { - 1 } \left( \frac { 1 - x } { 2 + x } \right)$, show that $( 2 x + 1 ) \frac { \mathrm { d } y } { \mathrm {~d} x } + 1 = 0$.\\
(c) Hence find the first three terms in the Maclaurin's series for $\tanh ^ { - 1 } \left( \frac { 1 - x } { 2 + x } \right)$ in the form
$$a \ln 3 + b x + c x ^ { 2 } ,$$
where $a , b$ and $c$ are constants to be determined.\\
\hfill \mbox{\textit{CAIE Further Paper 2 2020 Q7 [12]}}