| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2020 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find second derivative d²y/dx² |
| Difficulty | Challenging +1.2 This is a structured implicit differentiation question with clear guidance. Part (a) requires product rule and implicit differentiation (standard A-level technique), while part (b) requires finding and verifying d²y/dx² = 0 with sign change. The algebra is moderately involved but the approach is straightforward, making it slightly above average difficulty. |
| Spec | 1.07p Points of inflection: using second derivative1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(x^2\tan y = 9 \Rightarrow 2x\tan y + x^2\sec^2 y\,\frac{dy}{dx}=0\) | M1, A1 | M1: attempts to differentiate \(\tan y\) implicitly, e.g. \(\tan y \to \sec^2 y\,\frac{dy}{dx}\). Also allow attempt with \(\tan y = \frac{9}{x^2} \Rightarrow \sec^2 y\,\frac{dy}{dx}=\ldots\) |
| Full method to get \(\frac{dy}{dx}\) in terms of \(x\) using \(\sec^2 y = 1+\tan^2 y = 1+f(x)\) | M1 | |
| \(\frac{dy}{dx} = \frac{-2x\times\frac{9}{x^2}}{x^2\left(1+\frac{81}{x^4}\right)} = \frac{-18x}{x^4+81}\) | A1* | Must show correct intermediate step with no errors. Also allow arctan method: \(y=\arctan\frac{9}{x^2} \to \frac{dy}{dx}=\frac{1}{1+\left(\frac{9}{x^2}\right)^2}\times-\frac{18}{x^3}=\frac{-18x}{x^4+81}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{d^2y}{dx^2} = \frac{-18(x^4+81)-(-18x)(4x^3)}{(x^4+81)^2} = \frac{54(x^4-27)}{(x^4+81)^2}\) | M1, A1 | M1: attempts product or quotient rule with \(u=-18x,\ v=x^4+81,\ u'=\pm18,\ v'=\ldots x^3\) |
| When \(x < \sqrt[4]{27} \Rightarrow \frac{d^2y}{dx^2}<0\); when \(x=\sqrt[4]{27} \Rightarrow \frac{d^2y}{dx^2}=0\); when \(x>\sqrt[4]{27} \Rightarrow \frac{d^2y}{dx^2}>0\); giving point of inflection at \(x=\sqrt[4]{27}\) | A1 | All three conditions needed. Alternatively: substitute \(x=\sqrt[4]{27}\) into correct (unsimplified) \(\frac{d^2y}{dx^2}\) to show it equals 0, then argue \(x^4=27\). Or find \(\frac{d^3y}{dx^3}\), show \(\frac{d^3y}{dx^3}\neq 0\) and \(\frac{d^2y}{dx^2}=0\) when \(x=\sqrt[4]{27}\). FYI: \(\frac{d^3y}{dx^3}=\frac{23328x^3}{(x^4+81)^3}=0.219\) when \(x=\sqrt[4]{27}\) |
# Question 15 (Implicit Differentiation):
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $x^2\tan y = 9 \Rightarrow 2x\tan y + x^2\sec^2 y\,\frac{dy}{dx}=0$ | M1, A1 | M1: attempts to differentiate $\tan y$ implicitly, e.g. $\tan y \to \sec^2 y\,\frac{dy}{dx}$. Also allow attempt with $\tan y = \frac{9}{x^2} \Rightarrow \sec^2 y\,\frac{dy}{dx}=\ldots$ |
| Full method to get $\frac{dy}{dx}$ in terms of $x$ using $\sec^2 y = 1+\tan^2 y = 1+f(x)$ | M1 | |
| $\frac{dy}{dx} = \frac{-2x\times\frac{9}{x^2}}{x^2\left(1+\frac{81}{x^4}\right)} = \frac{-18x}{x^4+81}$ | A1* | Must show correct intermediate step with no errors. Also allow arctan method: $y=\arctan\frac{9}{x^2} \to \frac{dy}{dx}=\frac{1}{1+\left(\frac{9}{x^2}\right)^2}\times-\frac{18}{x^3}=\frac{-18x}{x^4+81}$ |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{d^2y}{dx^2} = \frac{-18(x^4+81)-(-18x)(4x^3)}{(x^4+81)^2} = \frac{54(x^4-27)}{(x^4+81)^2}$ | M1, A1 | M1: attempts product or quotient rule with $u=-18x,\ v=x^4+81,\ u'=\pm18,\ v'=\ldots x^3$ |
| When $x < \sqrt[4]{27} \Rightarrow \frac{d^2y}{dx^2}<0$; when $x=\sqrt[4]{27} \Rightarrow \frac{d^2y}{dx^2}=0$; when $x>\sqrt[4]{27} \Rightarrow \frac{d^2y}{dx^2}>0$; giving point of inflection at $x=\sqrt[4]{27}$ | A1 | All three conditions needed. Alternatively: substitute $x=\sqrt[4]{27}$ into correct (unsimplified) $\frac{d^2y}{dx^2}$ to show it equals 0, then argue $x^4=27$. Or find $\frac{d^3y}{dx^3}$, show $\frac{d^3y}{dx^3}\neq 0$ and $\frac{d^2y}{dx^2}=0$ when $x=\sqrt[4]{27}$. FYI: $\frac{d^3y}{dx^3}=\frac{23328x^3}{(x^4+81)^3}=0.219$ when $x=\sqrt[4]{27}$ |\begin{enumerate}
\item The curve $C$ has equation
\end{enumerate}
$$x ^ { 2 } \tan y = 9 \quad 0 < y < \frac { \pi } { 2 }$$
(a) Show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { - 18 x } { x ^ { 4 } + 81 }$$
(b) Prove that $C$ has a point of inflection at $x = \sqrt [ 4 ] { 27 }$
\hfill \mbox{\textit{Edexcel Paper 1 2020 Q15 [7]}}