| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Direct multiplication of series |
| Difficulty | Standard +0.8 This is a Further Maths question requiring multiple techniques: proving a derivative formula, finding a Maclaurin series through successive differentiation, and multiplying two series. Part (iii) requires careful bookkeeping of terms up to x^4. While methodical rather than requiring deep insight, it's more demanding than typical A-level questions due to the multi-step nature and series multiplication complexity. |
| Spec | 4.08a Maclaurin series: find series for function4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n4.08g Derivatives: inverse trig and hyperbolic functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = \sin y \Rightarrow \frac{dx}{dy} = \cos y\) | M1 | For implicit diffn to \(\frac{dy}{dx} = \pm\frac{1}{\cos y}\) oe |
| \(\Rightarrow \frac{dy}{dx} = \frac{1}{\sqrt{1-\sin^2 y}} = \frac{1}{\sqrt{1-x^2}}\) | A1 | For using \(\sin^2 y + \cos^2 y = 1\) to obtain — N.B. Answer given |
| \(+\sqrt{\phantom{x}}\) taken since \(\sin^{-1} x\) has positive gradient | B1 | For justifying \(+\) sign |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(0) = 0\), \(f'(0) = 1\) | B1 | For correct values |
| \(f''(x) = \frac{x}{\left(1-x^2\right)^{3/2}}\) | M1 | Use of chain rule to differentiate \(f'(x)\) |
| \(f'''(x) = \frac{\left(1-x^2\right)^{3/2} + 3x^2\left(1-x^2\right)^{1/2}}{\left(1-x^2\right)^3}\) | M1 | Use of quotient or product rule to differentiate \(f''(0)\) |
| \(\Rightarrow f''(0) = 0\), \(f'''(0) = 1\) | A1 | For correct values www, soi |
| \(\Rightarrow \sin^{-1} x = x + \frac{1}{6}x^3\) | A1 | For correct series (allow 3!) www |
| Total: 5 | ||
| Alternative Method: | ||
| \(f(0) = 0\), \(f'(0) = 1\) | B1 | For correct values |
| \(f'(x) = \frac{1}{\sqrt{1-x^2}} = (1-x^2)^{-1/2} = 1 + \frac{1}{2}x^2 + \frac{3}{8}x^4 + \ldots\) | M1 | Correct use of binomial |
| \(f''(x) = x + \frac{3}{2}x^3 + \ldots\) | M1 | Differentiate twice |
| \(f'''(x) = 1 + \frac{9}{2}x^2 + \ldots\) | ||
| \(\Rightarrow f'(0) = 1\), \(f''(0) = 0\), \(f'''(0) = 1\) | A1 | Correct values |
| \(\Rightarrow \sin^{-1} x = x + \frac{1}{6}x^3\) | A1 | Correct series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left(\sin^{-1} x\right)\ln(1+x)\) | B1ft | For terms in both series to at least \(x^3\); f.t. from their (ii) multiplied together |
| \(= \left(x + \frac{1}{6}x^3\right)\left(x - \frac{1}{2}x^2 + \frac{1}{3}x^3\right)\) | M1 | For multiplying terms to at least \(x^3\) |
| \(= x^2 - \frac{1}{2}x^3 + \frac{1}{2}x^4\) | A1 | For correct series up to \(x^3\) www |
| A1 | For correct term in \(x^4\) www | |
| Total: 4 |
# Question 5(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = \sin y \Rightarrow \frac{dx}{dy} = \cos y$ | M1 | For implicit diffn to $\frac{dy}{dx} = \pm\frac{1}{\cos y}$ **oe** |
| $\Rightarrow \frac{dy}{dx} = \frac{1}{\sqrt{1-\sin^2 y}} = \frac{1}{\sqrt{1-x^2}}$ | A1 | For using $\sin^2 y + \cos^2 y = 1$ to obtain — **N.B. Answer given** |
| $+\sqrt{\phantom{x}}$ taken since $\sin^{-1} x$ has positive gradient | B1 | For justifying $+$ sign |
| **Total: 3** | | |
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# Question 5(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(0) = 0$, $f'(0) = 1$ | B1 | For correct values |
| $f''(x) = \frac{x}{\left(1-x^2\right)^{3/2}}$ | M1 | Use of chain rule to differentiate $f'(x)$ |
| $f'''(x) = \frac{\left(1-x^2\right)^{3/2} + 3x^2\left(1-x^2\right)^{1/2}}{\left(1-x^2\right)^3}$ | M1 | Use of quotient or product rule to differentiate $f''(0)$ |
| $\Rightarrow f''(0) = 0$, $f'''(0) = 1$ | A1 | For correct values **www, soi** |
| $\Rightarrow \sin^{-1} x = x + \frac{1}{6}x^3$ | A1 | For correct series (allow 3!) **www** |
| **Total: 5** | | |
| **Alternative Method:** | | |
| $f(0) = 0$, $f'(0) = 1$ | B1 | For correct values |
| $f'(x) = \frac{1}{\sqrt{1-x^2}} = (1-x^2)^{-1/2} = 1 + \frac{1}{2}x^2 + \frac{3}{8}x^4 + \ldots$ | M1 | Correct use of binomial |
| $f''(x) = x + \frac{3}{2}x^3 + \ldots$ | M1 | Differentiate twice |
| $f'''(x) = 1 + \frac{9}{2}x^2 + \ldots$ | | |
| $\Rightarrow f'(0) = 1$, $f''(0) = 0$, $f'''(0) = 1$ | A1 | Correct values |
| $\Rightarrow \sin^{-1} x = x + \frac{1}{6}x^3$ | A1 | Correct series |
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# Question 5(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(\sin^{-1} x\right)\ln(1+x)$ | B1ft | For terms in both series to at least $x^3$; f.t. from their (ii) multiplied together |
| $= \left(x + \frac{1}{6}x^3\right)\left(x - \frac{1}{2}x^2 + \frac{1}{3}x^3\right)$ | M1 | For multiplying terms to at least $x^3$ |
| $= x^2 - \frac{1}{2}x^3 + \frac{1}{2}x^4$ | A1 | For correct series up to $x^3$ **www** |
| | A1 | For correct term in $x^4$ **www** |
| **Total: 4** | | |5 (i) Prove that, if $y = \sin ^ { - 1 } x$, then $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \sqrt { 1 - x ^ { 2 } } }$.\\
(ii) Find the Maclaurin series for $\sin ^ { - 1 } x$, up to and including the term in $x ^ { 3 }$.\\
(iii) Use the result of part (ii) and the Maclaurin series for $\ln ( 1 + x )$ to find the Maclaurin series for $\left( \sin ^ { - 1 } x \right) \ln ( 1 + x )$, up to and including the term in $x ^ { 4 }$.
\hfill \mbox{\textit{OCR FP2 2011 Q5 [12]}}