| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Year | 2021 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Maclaurin series for inverse trigonometric functions |
| Difficulty | Standard +0.8 This requires computing derivatives of arcsin x (using implicit differentiation or the chain rule with (1-x²)^(-1/2)), evaluating at x=0, constructing the Maclaurin series, then applying it to estimate π. While the individual techniques are A-level standard, combining them for an inverse trig function and connecting to π estimation requires solid understanding and careful execution—moderately harder than typical Further Maths questions but not exceptionally challenging. |
| Spec | 4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f'(x) = A(1-x^2)^{-\frac{1}{2}}\), \(f''(x) = Bx(1-x^2)^{-\frac{3}{2}}\) and \(f'''(x) = C(1-x^2)^{-\frac{3}{2}} + Dx^2(1-x^2)^{-\frac{5}{2}}\) | M1 | Finds correct form of first three derivatives, may be unsimplified |
| \(f'(x) = (1-x^2)^{-\frac{1}{2}}\), \(f''(x) = x(1-x^2)^{-\frac{3}{2}}\), \(f'''(x) = (1-x^2)^{-\frac{3}{2}} + 3x^2(1-x^2)^{-\frac{5}{2}}\) | A1 | Correct first three derivatives, may be unsimplified |
| Finds \(f(0)=0\), \(f'(0)=1\), \(f''(0)=0\), \(f'''(0)=1\) and applies \(f(x) = f(0) + f'(0)x + f''(0)\frac{x^2}{2} + f'''(0)\frac{x^3}{6}\) | M1 | Needs to go up to \(x^3\) |
| \(f(x) = x + \frac{x^3}{6}\) | A1 | cso; ignore any higher terms whether correct or not |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\arcsin\!\left(\frac{1}{2}\right) = \frac{1}{2} + \frac{\left(\frac{1}{2}\right)^3}{6} = \frac{\pi}{6} \Rightarrow \pi = \ldots\) | M1 | Substitutes \(x = \frac{1}{2}\) into both sides and rearranges to find \(\pi = \ldots\) |
| \(\pi = \frac{25}{8}\) | A1ft | Infers \(\pi = \frac{25}{8}\); follow through on their \(6f\!\left(\frac{1}{2}\right)\) |
# Question 3:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(x) = A(1-x^2)^{-\frac{1}{2}}$, $f''(x) = Bx(1-x^2)^{-\frac{3}{2}}$ and $f'''(x) = C(1-x^2)^{-\frac{3}{2}} + Dx^2(1-x^2)^{-\frac{5}{2}}$ | M1 | Finds correct form of first three derivatives, may be unsimplified |
| $f'(x) = (1-x^2)^{-\frac{1}{2}}$, $f''(x) = x(1-x^2)^{-\frac{3}{2}}$, $f'''(x) = (1-x^2)^{-\frac{3}{2}} + 3x^2(1-x^2)^{-\frac{5}{2}}$ | A1 | Correct first three derivatives, may be unsimplified |
| Finds $f(0)=0$, $f'(0)=1$, $f''(0)=0$, $f'''(0)=1$ and applies $f(x) = f(0) + f'(0)x + f''(0)\frac{x^2}{2} + f'''(0)\frac{x^3}{6}$ | M1 | Needs to go up to $x^3$ |
| $f(x) = x + \frac{x^3}{6}$ | A1 | cso; ignore any higher terms whether correct or not |
**Special case:** If $f''(0) \neq 0$ then maximum score M1 A0 M1 A0
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\arcsin\!\left(\frac{1}{2}\right) = \frac{1}{2} + \frac{\left(\frac{1}{2}\right)^3}{6} = \frac{\pi}{6} \Rightarrow \pi = \ldots$ | M1 | Substitutes $x = \frac{1}{2}$ into both sides and rearranges to find $\pi = \ldots$ |
| $\pi = \frac{25}{8}$ | A1ft | Infers $\pi = \frac{25}{8}$; follow through on their $6f\!\left(\frac{1}{2}\right)$ |
---3.
$$f ( x ) = \arcsin x \quad - 1 \leqslant x \leqslant 1$$
\begin{enumerate}[label=(\alph*)]
\item Determine the first two non-zero terms, in ascending powers of $x$, of the Maclaurin series for $\mathrm { f } ( x )$, giving each coefficient in its simplest form.
\item Substitute $x = \frac { 1 } { 2 }$ into the answer to part (a) and hence find an approximate value for $\pi$ Give your answer in the form $\frac { p } { q }$ where $p$ and $q$ are integers to be determined.
\end{enumerate}
\hfill \mbox{\textit{Edexcel CP2 2021 Q3 [6]}}