| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Differentiate inverse trigonometric functions |
| Difficulty | Standard +0.8 This is an FP2 question requiring a standard proof of an inverse trig derivative (bookwork), then applying chain rule to a composite function, and finally verifying a second derivative relationship. While it involves multiple steps and careful algebraic manipulation, these are standard Further Maths techniques without requiring novel insight. The difficulty is elevated above average due to the FP2 content and the need for precise differentiation and simplification across multiple parts. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.07l Derivative of ln(x): and related functions1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos y = x \Rightarrow -\sin y \frac{dy}{dx} = 1\) | M1 | For differentiating cos y wrt x |
| \(\Rightarrow \frac{dy}{dx} = \frac{1}{\sin y} = \frac{1}{\sqrt{1-x^2}}\) | A1 | For using \(\cos^2 y + \sin^2 y = 1\) to obtain AG |
| \(-\) sign since \(\frac{dy}{dx} < 0\) (e.g. by graph) | B1 | For justification of \(+\sqrt{}\) taken |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = \frac{-2x}{\sqrt{1-(1-x^2)^2}}\) | M1 | For differentiating \(\cos^{-1}(1-x^2)\) (as a function of a function) |
| A1 | For correct \(\frac{dy}{dx}\) (unsimplified) | |
| \(= \frac{2x}{\sqrt{2x^2-x^4}} = \frac{2}{\sqrt{2-x^2}}\) | A1 | For correct \(\frac{dy}{dx}\) (simplified) |
| \(\frac{d^2y}{dx^2} = 2 - \frac{1}{2} - 2x\left(2-x^2\right)^{-\frac{3}{2}} = \frac{2x}{(2-x^2)^{\frac{3}{2}}}\) | M1 | For differentiating \(\frac{dy}{dx}\) using chain rule correctly (or product or quotient if y' is wrong) |
| \(\Rightarrow (2-x^2)\frac{d^2y}{dx^2} = \frac{2x}{\sqrt{2-x^2}} = x\frac{dy}{dx}\) | A1 | For verification of AG |
| [5] |
### (i)
$\cos y = x \Rightarrow -\sin y \frac{dy}{dx} = 1$ | M1 | For differentiating cos y wrt x
$\Rightarrow \frac{dy}{dx} = \frac{1}{\sin y} = \frac{1}{\sqrt{1-x^2}}$ | A1 | For using $\cos^2 y + \sin^2 y = 1$ to obtain AG
$-$ sign since $\frac{dy}{dx} < 0$ (e.g. by graph) | B1 | For justification of $+\sqrt{}$ taken
| [3] |
SC1 if in fractions $\frac{14}{3}$ and $\frac{2047}{441}$
### (ii)
$\frac{dy}{dx} = \frac{-2x}{\sqrt{1-(1-x^2)^2}}$ | M1 | For differentiating $\cos^{-1}(1-x^2)$ (as a function of a function)
| A1 | For correct $\frac{dy}{dx}$ (unsimplified)
$= \frac{2x}{\sqrt{2x^2-x^4}} = \frac{2}{\sqrt{2-x^2}}$ | A1 | For correct $\frac{dy}{dx}$ (simplified)
$\frac{d^2y}{dx^2} = 2 - \frac{1}{2} - 2x\left(2-x^2\right)^{-\frac{3}{2}} = \frac{2x}{(2-x^2)^{\frac{3}{2}}}$ | M1 | For differentiating $\frac{dy}{dx}$ using chain rule correctly (or product or quotient if y' is wrong)
$\Rightarrow (2-x^2)\frac{d^2y}{dx^2} = \frac{2x}{\sqrt{2-x^2}} = x\frac{dy}{dx}$ | A1 | For verification of AG
| [5] |
---\begin{enumerate}[label=(\roman*)]
\item Prove that the derivative of $\cos^{-1} x$ is $-\frac{1}{\sqrt{1 - x^2}}$. [3]
\end{enumerate}
A curve has equation $y = \cos^{-1}(1 - x^2)$, for $0 < x < \sqrt{2}$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find and simplify $\frac{dy}{dx}$, and hence show that
$$(2 - x^2)\frac{d^2y}{dx^2} = x\frac{dy}{dx}.$$ [5]
\end{enumerate}
\hfill \mbox{\textit{OCR FP2 2012 Q6 [8]}}