| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Inverse functions (inverse trig/hyperbolic) |
| Difficulty | Standard +0.8 This is a structured Further Maths question requiring chain rule application, implicit differentiation to establish differential equations, and systematic use of these relations to generate Maclaurin coefficients. While methodical, it demands careful algebraic manipulation across multiple derivatives and connecting differential equations to series expansion—more demanding than standard FP2 questions but follows a clear guided path. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = \sin^{-1}(2x) \Rightarrow \dfrac{dy}{dx} = \dfrac{1}{\sqrt{1-(2x)^2}} \cdot \dfrac{d(2x)}{dx} = \dfrac{2}{\sqrt{1-4x^2}}\) | B1 | Oe |
| Answer | Marks | Guidance |
|---|---|---|
| \(\dfrac{d^2y}{dx^2} = 2\times\left(-\dfrac{1}{2}\right)(1-4x^2)^{-\frac{3}{2}}(-8x) = \dfrac{8x}{(1-4x^2)^{\frac{3}{2}}} = \dfrac{4x}{(1-4x^2)} \cdot \dfrac{dy}{dx}\) | B1 | For correct 2nd derivative |
| \((1-4x^2)\dfrac{d^2y}{dx^2} = 4x\dfrac{dy}{dx}\) | M1, A1 | Using their answers to connect 1st and 2nd derivatives; ft to achieve ag |
| Answer | Marks | Guidance |
|---|---|---|
| \((1-4x^2)\dfrac{d^3y}{dx^3} - 12x\dfrac{d^2y}{dx^2} - 4\dfrac{dy}{dx} = 0\) | M1, A1 | Using result of (ii) and product rule correctly; ans www |
| Answer | Marks | Guidance |
|---|---|---|
| \(y_0, y'_0, y''_0, y'''_0 = \{0, 2, 0, 8\}\) | B1 | Soi |
| \(y = 0 + 2x + 0 + \dfrac{8x^3}{6} \Rightarrow y = 2x + \dfrac{4x^3}{3}\) | M1, A1 | Correctly substituting their 4 values into correct Maclaurin; ignore higher order terms |
# Question 5(i):
$y = \sin^{-1}(2x) \Rightarrow \dfrac{dy}{dx} = \dfrac{1}{\sqrt{1-(2x)^2}} \cdot \dfrac{d(2x)}{dx} = \dfrac{2}{\sqrt{1-4x^2}}$ | B1 | Oe
## Question 5(ii):
$\dfrac{d^2y}{dx^2} = 2\times\left(-\dfrac{1}{2}\right)(1-4x^2)^{-\frac{3}{2}}(-8x) = \dfrac{8x}{(1-4x^2)^{\frac{3}{2}}} = \dfrac{4x}{(1-4x^2)} \cdot \dfrac{dy}{dx}$ | B1 | For correct 2nd derivative
$(1-4x^2)\dfrac{d^2y}{dx^2} = 4x\dfrac{dy}{dx}$ | M1, A1 | Using their answers to connect 1st and 2nd derivatives; ft to achieve ag
## Question 5(iii):
$(1-4x^2)\dfrac{d^3y}{dx^3} - 8x\dfrac{d^2y}{dx^2} = 4\dfrac{dy}{dx} + 4x\dfrac{d^2y}{dx^2}$
$(1-4x^2)\dfrac{d^3y}{dx^3} - 12x\dfrac{d^2y}{dx^2} - 4\dfrac{dy}{dx} = 0$ | M1, A1 | Using result of (ii) and product rule correctly; ans www
## Question 5(iv):
$y_0, y'_0, y''_0, y'''_0 = \{0, 2, 0, 8\}$ | B1 | Soi
$y = 0 + 2x + 0 + \dfrac{8x^3}{6} \Rightarrow y = 2x + \dfrac{4x^3}{3}$ | M1, A1 | Correctly substituting their 4 values into correct Maclaurin; ignore higher order terms5 It is given that $y = \sin ^ { - 1 } 2 x$.\\
(i) Using the derivative of $\sin ^ { - 1 } x$ given in the List of Formulae (MF1), find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.\\
(ii) Show that $\left( 1 - 4 x ^ { 2 } \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 4 x \frac { \mathrm {~d} y } { \mathrm {~d} x }$.\\
(iii) Hence show that $\left( 1 - 4 x ^ { 2 } \right) \frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } - 12 x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 \frac { \mathrm {~d} y } { \mathrm {~d} x } = 0$.\\
(iv) Using your results from parts (i), (ii) and (iii), find the Maclaurin series for $\sin ^ { - 1 } 2 x$ up to and including the term in $x ^ { 3 }$.
\hfill \mbox{\textit{OCR FP2 2015 Q5 [9]}}