October 2020 Paper 1 Q15

Notes

M1: Attempts to differentiate \(\tan y\) implicitly. Eg. \(\tan y \to \sec^2 y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(\cot y \to -\operatorname{cosec}^2 y\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
You may well see an attempt \(\tan y = \dfrac{9}{x^2} \Rightarrow \sec^2 y\dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\)
When a candidate writes \(x^2\tan y = 9 \Rightarrow x = 3\tan^{-\frac{1}{2}} y\) the mark is scored for \(\tan^{-\frac{1}{2}} y \to \ldots\tan^{-\frac{3}{2}} y\sec^2 y\)

A1: Correct differentiation \(2x\tan y + x^2\sec^2 y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\)
Allow also \(\sec^2 y\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{18}{x^3}\) or \(2x = -9\operatorname{cosec}^2 y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) amongst others

M1: Full method to get \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(x\) using \(\sec^2 y = 1 + \tan^2 y = 1 + \mathrm{f}(x)\)

A1*: Proceeds correctly to the given answer of \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-18x}{x^4 + 81}\)

Alternative part (a) using arctan

M1: Sets \(y = \arctan\dfrac{9}{x^2} \to \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{1 + \left(\frac{9}{x^2}\right)^2} \times \ldots\) where … could be 1

A2: \(y = \arctan\dfrac{9}{x^2} \to \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{1 + \left(\frac{9}{x^2}\right)^2} \times -\dfrac{18}{x^3}\)

A1*: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{1 + \frac{81}{x^4}} \times -\dfrac{18}{x^3} = \dfrac{-18x}{x^4 + 81}\) showing correct intermediate step and no errors.
(corrected from the printed mark scheme: the final fraction is printed as \(\dfrac{-18x}{x^4 + 1}\))

Notes

M1: Attempts to differentiate the given expression using the product or quotient rule.
For example look for a correct attempt at \(\dfrac{vu^{\prime} - uv^{\prime}}{v^2}\) with \(u = -18x, v = x^4 + 81, u^{\prime} = \pm 18, v^{\prime} = \ldots x^3\)
If no method is seen or implied award for \(\dfrac{\pm 18 \times \left(x^4 + 81\right) \pm 18x\left(ax^3\right)}{\left(x^4 + 81\right)^2}\)
Using the product rule award for \(\pm 18\left(x^4 + 81\right)^{-1} \pm 18x\left(x^4 + 81\right)^{-2} \times cx^3\)

A1: Correct simplified \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = \dfrac{54\left(x^4 - 27\right)}{\left(x^4 + 81\right)^2}\) o.e. such as \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = \dfrac{54x^4 - 1458}{\left(x^4 + 81\right)^2}\)
Alternatively score for showing that when a correct (unsimplified) \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 0 \Rightarrow x^4 = 27 \Rightarrow x = \sqrt[4]{27}\)
Or for substituting \(x = \sqrt[4]{27}\) into an unsimplified but correct \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) and showing that it is 0

A1: Correct explanation with a minimal conclusion and correct second derivative.
See scheme.
It can be also be argued from \(x^4 \lt 27,\ x^4 = 27\) and \(x^4 \gt 27\) provided the conclusion states that the point of inflection is at \(x = \sqrt[4]{27}\)
Alternatively substitutes values of \(x\) either side of \(\sqrt[4]{27}\) and at \(\sqrt[4]{27}\), into \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\), finds all three values and makes a minimal conclusion.
A different method involves finding \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}\) and showing that \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} \neq 0\) and \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 0\) when \(x = \sqrt[4]{27}\)
FYI \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = \dfrac{216x^3\left(135 - x^4\right)}{\left(x^4 + 81\right)^3} = 0.219\) when \(x = \sqrt[4]{27}\)
(corrected from the printed mark scheme: the FYI expression is printed as \(\dfrac{23328x^3}{\left(x^4 + 81\right)^3}\), which equals \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}\) only at \(x = \sqrt[4]{27}\))