| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Session | Specimen |
| 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 application of Leibnitz's theorem for the nth derivative of a product, which is a specialized technique beyond standard A-level. While the algebraic manipulation is substantial and requires careful bookkeeping of terms, the approach is direct: apply the formula systematically and simplify. It's harder than typical calculus questions but doesn't require deep insight once you know the theorem. |
| Spec | 1.07q Product and quotient rules: differentiation4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = x^3e^{kx}\), set \(u = x^3\); \(\frac{du}{dx} = 3x^2\), \(\frac{d^2u}{dx^2} = 6x\), \(\frac{d^3u}{dx^3} = 6\), \(\frac{d^4u}{dx^4} = 0\) | M1 | Differentiate \(u = x^3\) three times |
| \(v = e^{kx}\); \(\frac{d^n v}{dx^n} = k^n e^{kx}\), \(\frac{d^{n-1}v}{dx^{n-1}} = k^{n-1}e^{kx}\), \(\frac{d^{n-2}v}{dx^{n-2}} = k^{n-2}e^{kx}\) (and...) | M1 | Use \(v = e^{kx}\) and establish form of derivatives, at least three shown |
| \(\frac{d^n y}{dx^n} = x^3 k^n e^{kx} + n3x^2 k^{n-1}e^{kx} + \frac{n(n-1)}{2}6xk^{n-2}e^{kx} + \frac{n(n-1)(n-2)}{3!}6k^{n-3}e^{kx}\); remaining terms disappear | M1 | Uses correct formula with 2 and 3! (or 6), terms shown to disappear after fourth term |
| \(\frac{d^n y}{dx^n} = k^{n-3}e^{kx}\left(k^3x^3 + 3nk^2x^2 + 3n(n-1)kx + n(n-1)(n-2)\right)\) | A1* | Correct solution leading to given answer, no errors seen |
## Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = x^3e^{kx}$, set $u = x^3$; $\frac{du}{dx} = 3x^2$, $\frac{d^2u}{dx^2} = 6x$, $\frac{d^3u}{dx^3} = 6$, $\frac{d^4u}{dx^4} = 0$ | M1 | Differentiate $u = x^3$ three times |
| $v = e^{kx}$; $\frac{d^n v}{dx^n} = k^n e^{kx}$, $\frac{d^{n-1}v}{dx^{n-1}} = k^{n-1}e^{kx}$, $\frac{d^{n-2}v}{dx^{n-2}} = k^{n-2}e^{kx}$ (and...) | M1 | Use $v = e^{kx}$ and establish form of derivatives, at least three shown |
| $\frac{d^n y}{dx^n} = x^3 k^n e^{kx} + n3x^2 k^{n-1}e^{kx} + \frac{n(n-1)}{2}6xk^{n-2}e^{kx} + \frac{n(n-1)(n-2)}{3!}6k^{n-3}e^{kx}$; remaining terms disappear | M1 | Uses correct formula with 2 and 3! (or 6), terms shown to disappear after fourth term |
| $\frac{d^n y}{dx^n} = k^{n-3}e^{kx}\left(k^3x^3 + 3nk^2x^2 + 3n(n-1)kx + n(n-1)(n-2)\right)$ | A1* | Correct solution leading to given answer, no errors seen |
---\begin{enumerate}
\item Given $k$ is a constant and that
\end{enumerate}
$$y = x ^ { 3 } \mathrm { e } ^ { k x }$$
use Leibnitz theorem to show that
$$\frac { \mathrm { d } ^ { n } y } { \mathrm {~d} x ^ { n } } = k ^ { n - 3 } \mathrm { e } ^ { k x } \left( k ^ { 3 } x ^ { 3 } + 3 n k ^ { 2 } x ^ { 2 } + 3 n ( n - 1 ) k x + n ( n - 1 ) ( n - 2 ) \right)$$
\hfill \mbox{\textit{Edexcel FP1 Q2 [4]}}