Challenging +1.2 This question requires differentiation of inverse hyperbolic functions and solving f'(x)=0, but uses standard formulas from the formula book. Part (a) is routine application of the chain rule with arsech, while part (b) involves combining two derivatives and algebraic manipulation to find exact values. It's above average due to the Further Maths content and multi-step algebraic work, but follows predictable patterns for this topic.
3. (a) Given that \(y = \operatorname { arsech } \left( \frac { x } { 2 } \right)\), where \(0 < x \leqslant 2\), show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { p } { x \sqrt { q - x ^ { 2 } } }$$
where \(p\) and \(q\) are constants to be determined.
In part (b) solutions based entirely on calculator technology are not acceptable.
$$\mathrm { f } ( x ) = \operatorname { artanh } ( x ) + \operatorname { arsech } \left( \frac { x } { 2 } \right) \quad 0 < x \leqslant 1$$
(b) Determine, in simplest form, the exact value of \(x\) for which \(\mathrm { f } ^ { \prime } ( x ) = 0\)
M1: Replaces \(\text{sech}\,y\) with \(\frac{x}{2}\) and \(\tanh y\) with \(\sqrt{1-\left(\frac{x}{2}\right)^2}\). A1: Correct equation involving \(\frac{dx}{dy}\) or \(\frac{dy}{dx}\) in any form in terms of \(x\) only
\(\frac{dy}{dx}=\frac{-2}{x\sqrt{4-x^2}}\)
A1
Correct derivative in required form or correct values for \(p\) and \(q\)
M1: Differentiates to form \(\frac{k}{x^2\sqrt{\left(\frac{2}{x}\right)^2-1}}\). A1: Correct equation in terms of \(x\) only. Score as first M mark and first A mark on EPEN
M1: Differentiates to form \(\frac{kx\left(1-\frac{x^2}{4}\right)^{-\frac{1}{2}}}{1-\left(1-\frac{x^2}{4}\right)}\). A1: Correct in terms of \(x\) only. Score as first M and first A on EPEN
Correct \(f'(x)\) following through their (a) of the form \(\frac{p}{x\sqrt{q-x^2}}\). Also allow with "made up" \(p\), \(q\) or the letters \(p\) and \(q\)
Solves quartic to obtain value for \(x^2\) and proceeds to value of \(x\). Apply usual rules for solving and check if necessary. Allow complex roots
\(x=\sqrt{\frac{2}{5}}\)
A1
Correct exact answer (allow equivalents e.g. \(\frac{\sqrt{10}}{5}\)). If any extra answers given score A0. e.g. \(x=\pm\sqrt{\frac{2}{5}}\)
Special case: A correct solution in (b) following a sign error in (a) giving \(\frac{dy}{dx}=\frac{2}{x\sqrt{4-x^2}}\) is likely to score M1M1A0A0 in (a) but allow full recovery in (b) if it leads to the correct answer.
## Question 3(a):
**Way 1:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y=\text{arsech}\!\left(\frac{x}{2}\right)\Rightarrow\text{sech}\,y=\frac{x}{2}\Rightarrow\frac{dx}{dy}=-2\text{sech}\,y\tanh y$ | M1 | Takes "sech" of both sides and differentiates to obtain $\frac{dx}{dy}=k\,\text{sech}\,y\tanh y$ or equivalent |
| $\frac{dx}{dy}=-2\!\left(\frac{x}{2}\right)\!\sqrt{1-\left(\frac{x}{2}\right)^2}$ | M1A1 | M1: Replaces $\text{sech}\,y$ with $\frac{x}{2}$ and $\tanh y$ with $\sqrt{1-\left(\frac{x}{2}\right)^2}$. A1: Correct equation involving $\frac{dx}{dy}$ or $\frac{dy}{dx}$ in any form in terms of $x$ only |
| $\frac{dy}{dx}=\frac{-2}{x\sqrt{4-x^2}}$ | A1 | Correct derivative in required form or correct values for $p$ and $q$ |
**Way 2:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y=\text{arsech}\!\left(\frac{x}{2}\right)\Rightarrow\cosh y=\frac{2}{x}\Rightarrow\sinh y\frac{dy}{dx}=-\frac{2}{x^2}$ | M1 | Takes "sech" of both sides, changes to "cosh" and differentiates to obtain $\sinh y\frac{dy}{dx}=\frac{k}{x^2}$ or equivalent |
| $\frac{dy}{dx}=-\frac{2}{x^2\sinh y}=-\frac{2}{x^2\sqrt{\left(\frac{2}{x}\right)^2-1}}$ | M1A1 | M1: Replaces $\sinh y$ with $\sqrt{\left(\frac{2}{x}\right)^2-1}$. A1: Correct equation in terms of $x$ only |
| $\frac{dy}{dx}=\frac{-2}{x\sqrt{4-x^2}}$ | A1 | Correct derivative in required form |
**Way 3:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y=\text{arsech}\!\left(\frac{x}{2}\right)\Rightarrow y=\text{arcosh}\!\left(\frac{2}{x}\right)$ | M1 | Changes to "arcosh" correctly. **Score as second M mark on EPEN** |
| $\frac{dy}{dx}=\frac{1}{\sqrt{\left(\frac{2}{x}\right)^2-1}}\times-\frac{2}{x^2}$ | M1A1 | M1: Differentiates to form $\frac{k}{x^2\sqrt{\left(\frac{2}{x}\right)^2-1}}$. A1: Correct equation in terms of $x$ only. **Score as first M mark and first A mark on EPEN** |
| $\frac{dy}{dx}=\frac{-2}{x\sqrt{4-x^2}}$ | A1 | Correct derivative in required form |
**Way 4:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{sech}^2 y\frac{dy}{dx}=-x\left(1-\frac{x^2}{4}\right)^{-\frac{1}{2}}$ | M1 | Differentiates to $\text{sech}^2 y\frac{dy}{dx}=kx\left(1-\frac{x^2}{4}\right)^{-\frac{1}{2}}$ or equivalent |
| $\frac{dy}{dx}=-\frac{4}{x}\left(1-\frac{x^2}{4}\right)^{-\frac{1}{2}}$ after replacing $\text{sech}^2 y$ with $\left(\frac{2}{x}\right)^2$ | M1A1 | M1: Replaces $\text{sech}^2 y$ with $\left(\frac{2}{x}\right)^2$. A1: Correct in terms of $x$ only |
| $\frac{dy}{dx}=\frac{-2}{x\sqrt{4-x^2}}$ | A1 | Correct derivative in required form |
**Way 5:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y=\text{arsech}\!\left(\frac{x}{2}\right)\Rightarrow y=\text{artanh}\!\left(\sqrt{1-\left(\frac{x}{2}\right)^2}\right)$ | M1 | Changes to "artanh" correctly. **Score as second M mark on EPEN** |
| $\frac{dy}{dx}=\frac{\frac{1}{2}\left(1-\frac{x^2}{4}\right)^{-\frac{1}{2}}}{1-\left(1-\frac{x^2}{4}\right)}\times-\frac{x}{2}$ | M1A1 | M1: Differentiates to form $\frac{kx\left(1-\frac{x^2}{4}\right)^{-\frac{1}{2}}}{1-\left(1-\frac{x^2}{4}\right)}$. A1: Correct in terms of $x$ only. **Score as first M and first A on EPEN** |
| $\frac{dy}{dx}=\frac{-2}{x\sqrt{4-x^2}}$ | A1 | Correct derivative in required form |
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## Question 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x)=\tanh^{-1}(x)+\text{sech}^{-1}\!\left(\frac{x}{2}\right)\Rightarrow f'(x)=\frac{1}{1-x^2}-\frac{2}{x\sqrt{4-x^2}}$ | B1ft | Correct $f'(x)$ following through their (a) of the form $\frac{p}{x\sqrt{q-x^2}}$. Also allow with "made up" $p$, $q$ or the letters $p$ and $q$ |
| $\frac{1}{1-x^2}-\frac{2}{x\sqrt{4-x^2}}=0\Rightarrow 2(1-x^2)=x\sqrt{4-x^2}\Rightarrow 4(1-x^2)^2=x^2(4-x^2)$ | M1 | Sets $\frac{dy}{dx}=0$ with their (a) of the form $\frac{p}{x\sqrt{q-x^2}}$ and squares both sides to reach a quartic equation |
| $5x^4-12x^2+4=0$ | A1 | Correct quartic |
| $5x^4-12x^2+4=0\Rightarrow x^2=2,\;0.4\Rightarrow x=\ldots$ | M1 | Solves quartic to obtain value for $x^2$ and proceeds to value of $x$. Apply usual rules for solving and check if necessary. Allow complex roots |
| $x=\sqrt{\frac{2}{5}}$ | A1 | Correct exact answer (allow equivalents e.g. $\frac{\sqrt{10}}{5}$). If any extra answers given score A0. e.g. $x=\pm\sqrt{\frac{2}{5}}$ |
**Special case:** A correct solution in (b) following a sign error in (a) giving $\frac{dy}{dx}=\frac{2}{x\sqrt{4-x^2}}$ is likely to score M1M1A0A0 in (a) but allow full recovery in (b) if it leads to the correct answer.
3. (a) Given that $y = \operatorname { arsech } \left( \frac { x } { 2 } \right)$, where $0 < x \leqslant 2$, show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { p } { x \sqrt { q - x ^ { 2 } } }$$
where $p$ and $q$ are constants to be determined.
In part (b) solutions based entirely on calculator technology are not acceptable.
$$\mathrm { f } ( x ) = \operatorname { artanh } ( x ) + \operatorname { arsech } \left( \frac { x } { 2 } \right) \quad 0 < x \leqslant 1$$
(b) Determine, in simplest form, the exact value of $x$ for which $\mathrm { f } ^ { \prime } ( x ) = 0$\\
\hfill \mbox{\textit{Edexcel F3 2022 Q3 [9]}}