| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Inverse functions (inverse trig/hyperbolic) |
| Difficulty | Standard +0.8 This question requires differentiation of inverse hyperbolic functions using the chain rule, finding second derivatives, then computing a Maclaurin series requiring evaluation at x=0 (including inverse tanh evaluation). While systematic, it combines multiple A-level techniques with the less familiar inverse hyperbolic functions, placing it moderately above average difficulty. |
| Spec | 4.07e Inverse hyperbolic: definitions, domains, ranges4.07f Inverse hyperbolic: logarithmic forms4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\frac{d}{dx}\left(\frac{3-x}{6+x}\right)=\frac{-(6+x)-(3-x)}{(6+x)^2}\) | M1 | Attempts quotient/product rule to obtain form \(\frac{A(6+x)-B(3-x)}{(6+x)^2}\), \(B>0\) |
| \(=\frac{-9}{(6+x)^2}\) | A1 | Correct expression in any form |
| \(f'(x)=\frac{1}{1-\left(\frac{3-x}{6+x}\right)^2}\times\frac{-9}{(6+x)^2}\) | dM1 | Complete chain rule method; dependent on first M mark |
| \(=\frac{(6+x)^2}{36+12x+x^2-9+6x-x^2}\times\frac{-9}{(6+x)^2}=\frac{-9}{18x+27}=\frac{-1}{2x+3}\) | A1* | Reaches printed answer with sufficient working, at least one intermediate line |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\frac{d}{dx}\left(\frac{3-x}{6+x}\right)=\frac{-9}{(6+x)^2}\) | M1, A1 | See main scheme |
| \(\tanh y=\frac{3-x}{6+x}\Rightarrow \text{sech}^2 y\frac{dy}{dx}=\frac{-9}{(6+x)^2}\), then \(\frac{dy}{dx}=\frac{1}{1-\tanh^2 y}\times\frac{-9}{(6+x)^2}\) | dM1 | Must proceed from \(\text{sech}^2 y\) or \(1-\tanh^2 y\), substitute \(\tanh y\); dependent on first M |
| \(=\frac{-1}{2x+3}\) | A1* | See main scheme |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(f(x)=\frac{1}{2}\ln\left(\frac{1+\frac{3-x}{6+x}}{1-\frac{3-x}{6+x}}\right)=\frac{1}{2}\ln\left(\frac{9}{3+2x}\right)\) | M1, A1 | Uses logarithmic form of artanh; correct simplified expression |
| \(f'(x)=\frac{1}{2}\times\frac{3+2x}{9}\times\frac{-18}{(3+2x)^2}\) | dM1 | Complete method using chain rule; dependent on first M |
| \(=\frac{-1}{2x+3}\) | A1* | Correct |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\tanh y=\frac{3-x}{6+x}\Rightarrow e^{2y}=\frac{9}{2x+3}\) | M1, A1 | Takes tanh of both sides, expresses correctly in exponentials, makes \(e^{2y}\) subject |
| \(y=\frac{1}{2}\ln\left(\frac{9}{2x+3}\right)\Rightarrow f'(x)=\frac{1}{2}\times\frac{2x+3}{9}\times\frac{-18}{(2x+3)^2}\) | dM1 | Rearranges to form \(k\ln\left(\frac{a}{bx+c}\right)\), uses chain rule; dependent on first M |
| \(=\frac{-1}{2x+3}\) | A1* | Correct |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(f''(x)=\frac{2}{(2x+3)^2}\) | B1 | Correct second derivative in any form, e.g. \(\frac{2}{4x^2+12x+9}\) or \(2(2x+3)^{-2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(f(0)=\tanh^{-1}\!\left(\tfrac{1}{2}\right)=\tfrac{1}{2}\ln 3,\quad f'(0)=-\tfrac{1}{3},\quad f''(0)=\tfrac{2}{9}\) | M1 | Attempts at least two of \(f(0)\), \(f'(0)\), \(f''(0)\) |
| \(f(x)=f(0)+xf'(0)+\frac{x^2}{2}f''(0)\) | M1 | Correct application of Maclaurin series for all three values (all non-zero); not dependent on first M |
| \(=\ln\sqrt{3}-\frac{1}{3}x+\frac{1}{9}x^2\) | A1 | Correct expansion; ignore extra terms in \(x^3\) and above |
# Question 2(a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\frac{d}{dx}\left(\frac{3-x}{6+x}\right)=\frac{-(6+x)-(3-x)}{(6+x)^2}$ | M1 | Attempts quotient/product rule to obtain form $\frac{A(6+x)-B(3-x)}{(6+x)^2}$, $B>0$ |
| $=\frac{-9}{(6+x)^2}$ | A1 | Correct expression in any form |
| $f'(x)=\frac{1}{1-\left(\frac{3-x}{6+x}\right)^2}\times\frac{-9}{(6+x)^2}$ | dM1 | Complete chain rule method; dependent on first M mark |
| $=\frac{(6+x)^2}{36+12x+x^2-9+6x-x^2}\times\frac{-9}{(6+x)^2}=\frac{-9}{18x+27}=\frac{-1}{2x+3}$ | A1* | Reaches printed answer with sufficient working, at least one intermediate line |
**Alternative 1:**
| Working | Marks | Guidance |
|---------|-------|----------|
| $\frac{d}{dx}\left(\frac{3-x}{6+x}\right)=\frac{-9}{(6+x)^2}$ | M1, A1 | See main scheme |
| $\tanh y=\frac{3-x}{6+x}\Rightarrow \text{sech}^2 y\frac{dy}{dx}=\frac{-9}{(6+x)^2}$, then $\frac{dy}{dx}=\frac{1}{1-\tanh^2 y}\times\frac{-9}{(6+x)^2}$ | dM1 | Must proceed from $\text{sech}^2 y$ or $1-\tanh^2 y$, substitute $\tanh y$; dependent on first M |
| $=\frac{-1}{2x+3}$ | A1* | See main scheme |
**Alternative 2:**
| Working | Marks | Guidance |
|---------|-------|----------|
| $f(x)=\frac{1}{2}\ln\left(\frac{1+\frac{3-x}{6+x}}{1-\frac{3-x}{6+x}}\right)=\frac{1}{2}\ln\left(\frac{9}{3+2x}\right)$ | M1, A1 | Uses logarithmic form of artanh; correct simplified expression |
| $f'(x)=\frac{1}{2}\times\frac{3+2x}{9}\times\frac{-18}{(3+2x)^2}$ | dM1 | Complete method using chain rule; dependent on first M |
| $=\frac{-1}{2x+3}$ | A1* | Correct |
**Alternative 3:**
| Working | Marks | Guidance |
|---------|-------|----------|
| $\tanh y=\frac{3-x}{6+x}\Rightarrow e^{2y}=\frac{9}{2x+3}$ | M1, A1 | Takes tanh of both sides, expresses correctly in exponentials, makes $e^{2y}$ subject |
| $y=\frac{1}{2}\ln\left(\frac{9}{2x+3}\right)\Rightarrow f'(x)=\frac{1}{2}\times\frac{2x+3}{9}\times\frac{-18}{(2x+3)^2}$ | dM1 | Rearranges to form $k\ln\left(\frac{a}{bx+c}\right)$, uses chain rule; dependent on first M |
| $=\frac{-1}{2x+3}$ | A1* | Correct |
---
# Question 2(b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $f''(x)=\frac{2}{(2x+3)^2}$ | B1 | Correct second derivative in any form, e.g. $\frac{2}{4x^2+12x+9}$ or $2(2x+3)^{-2}$ |
---
# Question 2(c):
| Working | Marks | Guidance |
|---------|-------|----------|
| $f(0)=\tanh^{-1}\!\left(\tfrac{1}{2}\right)=\tfrac{1}{2}\ln 3,\quad f'(0)=-\tfrac{1}{3},\quad f''(0)=\tfrac{2}{9}$ | M1 | Attempts at least two of $f(0)$, $f'(0)$, $f''(0)$ |
| $f(x)=f(0)+xf'(0)+\frac{x^2}{2}f''(0)$ | M1 | Correct application of Maclaurin series for all three values (all non-zero); not dependent on first M |
| $=\ln\sqrt{3}-\frac{1}{3}x+\frac{1}{9}x^2$ | A1 | Correct expansion; ignore extra terms in $x^3$ and above |
---2.
$$f ( x ) = \tanh ^ { - 1 } \left( \frac { 3 - x } { 6 + x } \right) \quad | x | < \frac { 3 } { 2 }$$
\begin{enumerate}[label=(\alph*)]
\item Show that
$$f ^ { \prime } ( x ) = - \frac { 1 } { 2 x + 3 }$$
\item Hence determine $\mathrm { f } ^ { \prime \prime } ( x )$
\item Hence show that the Maclaurin series for $\mathrm { f } ( x )$, up to and including the term in $x ^ { 2 }$, is
$$\ln p + q x + r x ^ { 2 }$$
where $p , q$ and $r$ are constants to be determined.
\end{enumerate}
\hfill \mbox{\textit{Edexcel CP2 2024 Q2 [8]}}