| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 1 (Further Pure Core 1) |
| Year | 2018 |
| Session | September |
| Marks | 6 |
| Topic | Taylor series |
| Type | Direct substitution into standard series |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question requiring direct substitution into the standard Maclaurin series for ln(1+x), finding radius of convergence using |x|<1, and recognizing the given series matches the expansion at a specific value. While it's Further Maths content (inherently harder), the question is entirely procedural with no novel insight required—just careful algebraic manipulation and pattern recognition. |
| Spec | 4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots\); \(\Rightarrow \ln(1+3x^2) = (3x^2) - \frac{(3x^2)^2}{2} + \frac{(3x^2)^3}{3} - \frac{(3x^2)^4}{4} + \ldots\) | M1 | Substitute \(3x^2\) into standard formula |
| \(= 3x^2 - \frac{9x^4}{2} + 9x^6 - \frac{81x^8}{4} + \ldots\) | A1 [2] | or \(\frac{27x^6}{3}\) |
| (ii) Valid for \(3x^2 \leq 1 \Rightarrow | x | \leq \frac{1}{\sqrt{3}}\) oe |
| (iii) Substitute \(x = \frac{1}{2}\); \(\Rightarrow \ln(1+3x^2) = \ln(1 + \frac{3}{4}) = 3(\frac{1}{2})^2 - 9(\frac{1}{2})^2 + \frac{27(\frac{1}{2})^6}{3} - \frac{81(\frac{1}{2})^8}{4} + \ldots\) | M1 [3] | 3.1a |
| \(\Rightarrow \ln(\frac{7}{4}) = 3(\frac{1}{2})^2 - \frac{9(\frac{1}{2})^4}{2} + \frac{3^3(\frac{1}{2})^6}{3} - \frac{3^4(\frac{1}{2})^8}{4} + \ldots\) | M1 [3] | 1.1 |
| \(\Rightarrow \frac{3(\frac{1}{2})^2 + \frac{9(\frac{1}{2})^4}{3} + \frac{3^4(\frac{1}{2})^8}{4} + \ldots = 1 - \frac{4}{3}\ln(\frac{7}{4})\) | A1 [3] | 3.2a |
**(i)** $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots$; $\Rightarrow \ln(1+3x^2) = (3x^2) - \frac{(3x^2)^2}{2} + \frac{(3x^2)^3}{3} - \frac{(3x^2)^4}{4} + \ldots$ | M1 | Substitute $3x^2$ into standard formula
$= 3x^2 - \frac{9x^4}{2} + 9x^6 - \frac{81x^8}{4} + \ldots$ | A1 [2] | or $\frac{27x^6}{3}$
**(ii)** Valid for $3x^2 \leq 1 \Rightarrow |x| \leq \frac{1}{\sqrt{3}}$ oe | B1 [1] | 2.2a
**(iii)** Substitute $x = \frac{1}{2}$; $\Rightarrow \ln(1+3x^2) = \ln(1 + \frac{3}{4}) = 3(\frac{1}{2})^2 - 9(\frac{1}{2})^2 + \frac{27(\frac{1}{2})^6}{3} - \frac{81(\frac{1}{2})^8}{4} + \ldots$ | M1 [3] | 3.1a | Deducing the required value for $x$
$\Rightarrow \ln(\frac{7}{4}) = 3(\frac{1}{2})^2 - \frac{9(\frac{1}{2})^4}{2} + \frac{3^3(\frac{1}{2})^6}{3} - \frac{3^4(\frac{1}{2})^8}{4} + \ldots$ | M1 [3] | 1.1 | Rearranging either series (with either $\frac{1}{2}$ or $x$) to show the relationship with the other
$\Rightarrow \frac{3(\frac{1}{2})^2 + \frac{9(\frac{1}{2})^4}{3} + \frac{3^4(\frac{1}{2})^8}{4} + \ldots = 1 - \frac{4}{3}\ln(\frac{7}{4})$ | A1 [3] | 3.2a
---10 (i) Using the Maclaurin series for $\ln ( 1 + x )$, find the first four terms in the series expansion for $\ln \left( 1 + 3 x ^ { 2 } \right)$.\\
(ii) Find the range of $x$ for which the expansion is valid.\\
(iii) Find the exact value of the series
$$\frac { 3 ^ { 1 } } { 2 \times 2 ^ { 2 } } - \frac { 3 ^ { 2 } } { 3 \times 2 ^ { 4 } } + \frac { 3 ^ { 3 } } { 4 \times 2 ^ { 6 } } - \frac { 3 ^ { 4 } } { 5 \times 2 ^ { 8 } } + \ldots .$$
\hfill \mbox{\textit{OCR Further Pure Core 1 2018 Q10 [6]}}