A2 June 2024 Paper 2 Q2
2.
\[\mathrm{f}(x) = \tanh^{-1}\left(\frac{3 - x}{6 + x}\right) \qquad |x| \lt \frac{3}{2}\]| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}\left(\dfrac{3 - x}{6 + x}\right)}{\mathrm{d}x} = \dfrac{-(6 + x) - (3 - x)}{(6 + x)^2}\) | M1 A1 | 3.1a 1.1b |
| \(\mathrm{f}(x) = \tanh^{-1}\left(\dfrac{3 - x}{6 + x}\right) \Rightarrow \mathrm{f}^{\prime}(x) = \dfrac{1}{1 - \left(\dfrac{3 - x}{6 + x}\right)^2} \times \dfrac{-9}{(6 + x)^2}\) | dM1 | 3.1a |
| \(= \dfrac{(6 + x)^2}{36 + 12x + x^2 - 9 + 6x - x^2} \times \dfrac{-9}{(6 + x)^2} = \dfrac{-9}{18x + 27} = \dfrac{-1}{2x + 3}\,*\) | A1* | 2.1 |
| (4) |
Notes
Alternative 1 for part (a)
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}\left(\dfrac{3 - x}{6 + x}\right)}{\mathrm{d}x} = \dfrac{-(6 + x) - (3 - x)}{(6 + x)^2}\) | M1 A1 | 3.1a 1.1b |
| \(y = \tanh^{-1}\left(\dfrac{3 - x}{6 + x}\right) \Rightarrow \tanh y = \dfrac{3 - x}{6 + x} \Rightarrow \operatorname{sech}^2 y\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-9}{(6 + x)^2}\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{1 - \tanh^2 y} \times \dfrac{-9}{(6 + x)^2} = \dfrac{1}{1 - \left(\dfrac{3 - x}{6 + x}\right)^2} \times \dfrac{-9}{(6 + x)^2}\) | dM1 | 3.1a |
| \(= \dfrac{(6 + x)^2}{36 + 12x + x^2 - 9 + 6x - x^2} \times \dfrac{-9}{(6 + x)^2} = \dfrac{-9}{18x + 27} = \dfrac{-1}{2x + 3}\,*\) | A1* | 2.1 |
| (4) |
Alternative 2 for part (a)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{f}(x) = \tanh^{-1}\left(\dfrac{3 - x}{6 + x}\right) = \dfrac{1}{2}\ln\left(\dfrac{1 + \dfrac{3 - x}{6 + x}}{1 - \dfrac{3 - x}{6 + x}}\right) = \dfrac{1}{2}\ln\left(\dfrac{9}{3 + 2x}\right)\) | M1 A1 | 3.1a 1.1b |
| \(\mathrm{f}(x) = \tanh^{-1}\left(\dfrac{3 - x}{6 + x}\right) \Rightarrow \mathrm{f}^{\prime}(x) = \dfrac{1}{2} \times \dfrac{3 + 2x}{9} \times \dfrac{-18}{(3 + 2x)^2}\) | dM1 | 3.1a |
| \(= \dfrac{-1}{2x + 3}\,*\) | A1* | 1.1b |
| (4) |
Alternative 3 for part (a)
| Scheme | Marks | AO |
|---|---|---|
| \(\tanh y = \mathrm{f}(x) = \left(\dfrac{3 - x}{6 + x}\right) \Rightarrow \dfrac{\mathrm{e}^y - \mathrm{e}^{-y}}{\mathrm{e}^y + \mathrm{e}^{-y}} = \dfrac{3 - x}{6 + x}\) or \(\dfrac{\mathrm{e}^{2y} - 1}{\mathrm{e}^{2y} + 1} = \dfrac{3 - x}{6 + x}\) \(\mathrm{e}^{2y} = \dfrac{9}{2x + 3}\) | M1 A1 | 3.1a 1.1b |
| \(y = \dfrac{1}{2}\ln\left(\dfrac{9}{2x + 3}\right) \Rightarrow \mathrm{f}^{\prime}(x) = \dfrac{1}{2} \times \dfrac{2x + 3}{9} \times \dfrac{-18}{(2x + 3)^2}\) | dM1 | 3.1a |
| \(= \dfrac{-1}{2x + 3}\,*\) | A1 | 1.1b |
| (4) |
Notes
(a) Do not allow misreads of \(\tan x\) for \(\tanh x\) in this question; this would be a maximum of M1A1 for part (a)
M1: Attempts to use the quotient (or product) rule on \(\dfrac{3 - x}{6 + x}\) to obtain an expression of the form
\[\frac{A(6 + x) - B(3 - x)}{(6 + x)^2},\ B \gt 0 \qquad \textbf{or} \qquad C(3 - x)(6 + x)^{-2} + D(6 + x)^{-1}\]Alternatively, candidates may also write \(\dfrac{3 - x}{6 + x}\) as \(-1 + \dfrac{9}{6 + x}\) and then differentiate to find an expression of the form \(\dfrac{E}{(6 + x)^2}\) which may be seen embedded in their working.
They may also write \(\dfrac{3 - x}{6 + x}\) as \(\dfrac{3}{6 + x} - \dfrac{x}{6 + x}\) and then differentiate to find an expression of the form \(\dfrac{E}{(6 + x)^2}\) oe
A1: Correct expression in any form.
dM1: A complete method to find the derivative using the chain rule to obtain
\[\frac{1}{1 - \left(\dfrac{3 - x}{6 + x}\right)^2} \times \left(\text{their } \frac{-9}{(6 + x)^2}\right)\]This is dependent on the first M mark.
A1*: Reaches the printed answer with sufficient working and no errors, with at least one intermediate line.
Alternative 1:
M1: see main scheme
A1: see main scheme
dM1: A complete method to find the derivative using the chain rule:
Rearranges the equation and uses implicit differentiation. Must proceed from \(\operatorname{sech}^2 y\) or equivalent to \(1 - \tanh^2 y\) and then substitute for \(\tanh y\)
This is dependent on the first M mark.
A1*: see main scheme
Alternative 2:
M1: Uses the logarithmic form of artanh to obtain \(k\ln\left(\dfrac{1 + \dfrac{3 - x}{6 + x}}{1 - \dfrac{3 - x}{6 + x}}\right)\)
A1: Correct simplified expression
dM1: A complete method to find the derivative to obtain
\(\mathrm{f}^{\prime}(x) = k \times \left(1 \div \left(\text{their } \dfrac{9}{3 + 2x}\right)\right) \times \left(\text{their } \dfrac{9}{3 + 2x}\right)^2\)
Alternatively uses log rules to partition their expression and then differentiates each term.
This is dependent on the first M mark.
A1*: Reaches the printed answer with sufficient working and no errors.
Alternative 3:
M1: Takes tanh of both sides to obtain \(\tanh y\) in terms of \(x\) and expresses \(\tanh y\) correctly in terms of exponentials and makes \(\mathrm{e}^{2y}\) the subject.
A1: A correct expression for \(\mathrm{e}^{2y}\)
dM1: Rearranges their function to the form \(k\ln\left(\dfrac{a}{bx + c}\right)\) and uses the chain rule to find a derivative of the form \(\mathrm{f}^{\prime}(x) = \dfrac{1}{2} \times \dfrac{bx + c}{a} \times \dfrac{k}{(bx + c)^2}\) or uses log rules to partition their expression and then differentiates each term.
This is dependent on the first M mark.
A1*: Reaches the printed answer with sufficient working and no errors
| Scheme | Marks | AO |
|---|---|---|
| \(\left[\mathrm{f}^{\prime\prime}(x) =\right]\ \dfrac{2}{(2x + 3)^2}\) | B1 | 1.1b |
| (1) |
Notes
B1: Correct second derivative in any form such as e.g. \(\dfrac{2}{(4x^2 + 12x + 9)}\) or \(2(2x + 3)^{-2}\)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{f}(0) = \tanh^{-1}\left(\dfrac{1}{2}\right)\left(= \dfrac{1}{2}\ln 3\right),\ \mathrm{f}^{\prime}(0) = -\dfrac{1}{3},\ \mathrm{f}^{\prime\prime}(0) = \dfrac{2}{9}\) | M1 | 1.1b |
| \(\left[\mathrm{f}(x)\right] = \mathrm{f}(0) + x\mathrm{f}^{\prime}(0) + \dfrac{x^2}{2}\mathrm{f}^{\prime\prime}(0)\) | M1 | 1.1b |
| \(= \ln\sqrt{3} - \dfrac{1}{3}x + \dfrac{1}{9}x^2\) | A1 | 1.1b |
| (3) | ||
| (8 marks) |
Notes
M1: Attempts at least two of the values of f(0), f′(0) and f″(0)
M1: Correct application of the Maclaurin series for all of \(\mathrm{f}(0)\), \(\mathrm{f}^{\prime}(0)\) and \(\mathrm{f}^{\prime\prime}(0)\) where all are non-zero values. Substitutes their values into a correct expression. This mark is not dependent.
A1: Correct expansion.
Award if written as a single expression, or \(p\), \(q\) and \(r\) written separately.
Ignore any extra terms written as powers of \(x^3\) and above, isw when a correct answer is seen.