| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Prove formula by induction |
| Difficulty | Challenging +1.8 This is a Further Maths question requiring systematic application of Leibnitz's theorem for the 5th derivative of a product, involving careful bookkeeping of binomial coefficients, derivatives of x³ and sin(kx), and algebraic manipulation to match the given form. While methodical rather than requiring deep insight, the extended calculation with multiple terms and the need to collect and simplify coefficients correctly makes it significantly harder than standard A-level calculus. |
| Spec | 4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u = x^3 \Rightarrow \frac{du}{dx} = 3x^2,\ \frac{d^2u}{dx^2} = 6x,\ \frac{d^3u}{dx^3} = 6\) | M1 | Differentiates \(u = x^3\) three times. Need to see \(x^3 \to \ldots x^2 \to \ldots x \to k\) |
| \(v = \sin kx \Rightarrow \frac{dv}{dx} = k\cos kx,\ \frac{d^2v}{dx^2} = -k^2\sin kx,\ \frac{d^3v}{dx^3} = -k^3\cos kx,\ \frac{d^4v}{dx^4} = k^4\sin kx,\ \frac{d^5v}{dx^5} = k^5\cos kx\) | M1 | Uses \(v = \sin kx\) to establish derivatives. Need alternating \(k^n\sin kx\) and \(k^n\cos kx\) with increasing powers of \(k\) for at least 3 derivatives. |
| \(\frac{d^5y}{dx^5} = x^3 k^5\cos kx + 5 \times 3x^2 \times k^4\sin kx + \frac{5\times4}{2}\times 6x \times(-k^3\cos kx) + \frac{5\times4\times3}{3!}\times 6\times(-k^2\sin kx)\) | M1 | Uses correct formula with 2 and 3! (or 6) with terms shown to disappear after fourth term. Correct binomial coefficients paired with correct derivatives. |
| \(= (k^2x^2 - 60)k^3x\cos kx + 15(k^2x^2 - 4)k^2\sin kx\) | A1 | Correct expression with correct values of \(A\), \(B\), \(C\). NB \(A = -60, B = 15, C = -4\). Apply isw if necessary. |
# Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u = x^3 \Rightarrow \frac{du}{dx} = 3x^2,\ \frac{d^2u}{dx^2} = 6x,\ \frac{d^3u}{dx^3} = 6$ | M1 | Differentiates $u = x^3$ three times. Need to see $x^3 \to \ldots x^2 \to \ldots x \to k$ |
| $v = \sin kx \Rightarrow \frac{dv}{dx} = k\cos kx,\ \frac{d^2v}{dx^2} = -k^2\sin kx,\ \frac{d^3v}{dx^3} = -k^3\cos kx,\ \frac{d^4v}{dx^4} = k^4\sin kx,\ \frac{d^5v}{dx^5} = k^5\cos kx$ | M1 | Uses $v = \sin kx$ to establish derivatives. Need alternating $k^n\sin kx$ and $k^n\cos kx$ with increasing powers of $k$ for at least 3 derivatives. |
| $\frac{d^5y}{dx^5} = x^3 k^5\cos kx + 5 \times 3x^2 \times k^4\sin kx + \frac{5\times4}{2}\times 6x \times(-k^3\cos kx) + \frac{5\times4\times3}{3!}\times 6\times(-k^2\sin kx)$ | M1 | Uses correct formula with 2 and 3! (or 6) with terms shown to disappear after fourth term. Correct binomial coefficients paired with correct derivatives. |
| $= (k^2x^2 - 60)k^3x\cos kx + 15(k^2x^2 - 4)k^2\sin kx$ | A1 | Correct expression with correct values of $A$, $B$, $C$. NB $A = -60, B = 15, C = -4$. Apply isw if necessary. |
**Note: If there is no use of Leibnitz's theorem, this scores no marks.**
---\begin{enumerate}
\item Given that $k$ is a real non-zero constant and that
\end{enumerate}
$$y = x ^ { 3 } \sin k x$$
use Leibnitz's theorem to show that
$$\frac { \mathrm { d } ^ { 5 } y } { \mathrm {~d} x ^ { 5 } } = \left( k ^ { 2 } x ^ { 2 } + A \right) k ^ { 3 } x \cos k x + B \left( k ^ { 2 } x ^ { 2 } + C \right) k ^ { 2 } \sin k x$$
where $A$, $B$ and $C$ are integers to be determined.
\hfill \mbox{\textit{Edexcel FP1 2019 Q2 [4]}}