| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Use series to approximate integral |
| Difficulty | Standard +0.8 This is a standard Further Maths question combining Maclaurin series derivation with integration. Part (a) requires differentiating arcsin(x) repeatedly and evaluating at x=0 (routine but calculation-heavy). Part (b) recognizes that 1/√(1-u²) is the derivative of arcsin(u) and applies the series to approximate the integral. While methodical, it requires careful algebraic manipulation and understanding of the connection between differentiation and integration of series—typical of Further Maths but not requiring exceptional insight. |
| Spec | 1.08d Evaluate definite integrals: between limits4.08a Maclaurin series: find series for function4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(f'(x) = (1-x^2)^{-\frac{1}{2}}\) | B1 | Correct first derivative |
| \(f''(x) = x(1-x^2)^{-\frac{3}{2}}\) | B1 | Correct second derivative |
| \(f'''(x) = 3x^2(1-x^2)^{-\frac{5}{2}} + (1-x^2)^{-\frac{3}{2}}\) | M1 | Differentiates their \(f''(x)\) using product rule |
| \(f(0)=0\), \(f'(0)=1\), \(f''(0)=0\), \(f'''(0)=1\) | M1 | Evaluates their derivatives at \(x=0\). Must have attempted all three derivatives |
| \(\sin^{-1}x = x + \frac{1}{6}x^3\) | A1 | CWO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\sin y = x \Rightarrow f'(x) = \sec y\) | (B1) | Finds first derivative |
| \(f''(x) = \tan y(\sec y)^2\) | (B1) | Finds second derivative |
| \(f'''(x) = (2\tan^2 y\sec^2 y + \sec^4 y)\sec y = 3\sec^5 y - 2\sec^3 y\) | (M1) | Differentiates \(\tan y(\sec y)^2\) using product and chain rule |
| \(f(0)=0\), \(f'(0)=1\), \(f''(0)=0\), \(f'''(0)=1\) | (M1) | Evaluates their derivatives at \(x=0\). Must have attempted all three derivatives |
| \(\sin^{-1}x = x + \frac{1}{6}x^3\) | (A1) | CWO |
| Answer | Marks |
|---|---|
| \(\frac{151}{750}\) | B1 |
## Question 1(a):
**Main Method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $f'(x) = (1-x^2)^{-\frac{1}{2}}$ | B1 | Correct first derivative |
| $f''(x) = x(1-x^2)^{-\frac{3}{2}}$ | B1 | Correct second derivative |
| $f'''(x) = 3x^2(1-x^2)^{-\frac{5}{2}} + (1-x^2)^{-\frac{3}{2}}$ | M1 | Differentiates their $f''(x)$ using product rule |
| $f(0)=0$, $f'(0)=1$, $f''(0)=0$, $f'''(0)=1$ | M1 | Evaluates their derivatives at $x=0$. Must have attempted all three derivatives |
| $\sin^{-1}x = x + \frac{1}{6}x^3$ | A1 | CWO |
**Alternative Method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sin y = x \Rightarrow f'(x) = \sec y$ | (B1) | Finds first derivative |
| $f''(x) = \tan y(\sec y)^2$ | (B1) | Finds second derivative |
| $f'''(x) = (2\tan^2 y\sec^2 y + \sec^4 y)\sec y = 3\sec^5 y - 2\sec^3 y$ | (M1) | Differentiates $\tan y(\sec y)^2$ using product and chain rule |
| $f(0)=0$, $f'(0)=1$, $f''(0)=0$, $f'''(0)=1$ | (M1) | Evaluates their derivatives at $x=0$. Must have attempted all three derivatives |
| $\sin^{-1}x = x + \frac{1}{6}x^3$ | (A1) | CWO |
**Total: 5 marks**
## Question 1(b):
| $\frac{151}{750}$ | B1 | |
---1
\begin{enumerate}[label=(\alph*)]
\item Find the Maclaurin series for $\sin ^ { - 1 } x$ up to and including the term in $x ^ { 3 }$.
\item Deduce an approximation to $\int _ { 0 } ^ { \frac { 1 } { 5 } } \frac { 1 } { \sqrt { 1 - u ^ { 2 } } } \mathrm {~d} u$, giving your answer as a fraction.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2023 Q1 [6]}}