| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2017 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove derivative formula |
| Difficulty | Challenging +1.8 This is a challenging Further Maths induction proof requiring differentiation of products, manipulation of factorial expressions, and recognition of harmonic series patterns. Part (i) requires careful application of the product rule to higher derivatives, while part (ii) demands rigorous inductive reasoning with non-trivial algebraic manipulation. The combination of higher-order derivatives, factorial algebra, and proof structure places this well above average difficulty, though it follows a standard induction framework once the pattern is recognized. |
| Spec | 4.01a Mathematical induction: construct proofs |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{d^{n+1}}{dx^{n+1}}(x^{n+1}\ln x) = \frac{d^n}{dx^n}\left(x^{n+1}\cdot\frac{1}{x} + (n+1)x^n\ln x\right)\) | M1A1 | AG |
| \(= \frac{d^n}{dx^n}\left(x^n + (n+1)x^n\ln x\right)\) | ||
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Assume \(H_k\) is true \(\Rightarrow \frac{d^k}{dx^k}(x^k\ln x) = k!\left\{\ln x + 1 + \frac{1}{2} + \ldots + \frac{1}{k}\right\}\) | B1 | Statement of \(H_k\) seen |
| \(\frac{d^{k+1}}{dx^{k+1}}(x^{k+1}\ln x) = \frac{d^k}{dx^k}\left(x^k + [k+1]x^k\ln x\right)\) | M1 | |
| \(= k! + [k+1]k!\left\{\ln x + 1 + \frac{1}{2} + \ldots + \frac{1}{k}\right\}\) | A1 | |
| \(= (k+1)!\left\{\ln x + 1 + \frac{1}{2} + \ldots + \frac{1}{k+1}\right\} \Rightarrow H_{k+1}\) is true | A1 | |
| Check \(H_1\) is true and \(H_k\) is true \(\Rightarrow H_{k+1}\) is true; hence by PMI, \(H_n\) is true for all positive integers \(n\) | A1 | |
| Total | 5 |
## Question 3(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{d^{n+1}}{dx^{n+1}}(x^{n+1}\ln x) = \frac{d^n}{dx^n}\left(x^{n+1}\cdot\frac{1}{x} + (n+1)x^n\ln x\right)$ | M1A1 | AG |
| $= \frac{d^n}{dx^n}\left(x^n + (n+1)x^n\ln x\right)$ | | |
| **Total** | **2** | |
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## Question 3(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Assume $H_k$ is true $\Rightarrow \frac{d^k}{dx^k}(x^k\ln x) = k!\left\{\ln x + 1 + \frac{1}{2} + \ldots + \frac{1}{k}\right\}$ | B1 | Statement of $H_k$ seen |
| $\frac{d^{k+1}}{dx^{k+1}}(x^{k+1}\ln x) = \frac{d^k}{dx^k}\left(x^k + [k+1]x^k\ln x\right)$ | M1 | |
| $= k! + [k+1]k!\left\{\ln x + 1 + \frac{1}{2} + \ldots + \frac{1}{k}\right\}$ | A1 | |
| $= (k+1)!\left\{\ln x + 1 + \frac{1}{2} + \ldots + \frac{1}{k+1}\right\} \Rightarrow H_{k+1}$ is true | A1 | |
| Check $H_1$ is true **and** $H_k$ is true $\Rightarrow H_{k+1}$ is true; hence by PMI, $H_n$ is true for all positive integers $n$ | A1 | |
| **Total** | **5** | |
---3 (i) Show that $\frac { \mathrm { d } ^ { n + 1 } } { \mathrm {~d} x ^ { n + 1 } } \left( x ^ { n + 1 } \ln x \right) = \frac { \mathrm { d } ^ { n } } { \mathrm {~d} x ^ { n } } \left( x ^ { n } + ( n + 1 ) x ^ { n } \ln x \right)$.\\
(ii) Prove by mathematical induction that, for all positive integers $n$,
$$\frac { \mathrm { d } ^ { n } } { \mathrm {~d} x ^ { n } } \left( x ^ { n } \ln x \right) = n ! \left( \ln x + 1 + \frac { 1 } { 2 } + \ldots + \frac { 1 } { n } \right)$$
\hfill \mbox{\textit{CAIE FP1 2017 Q3 [7]}}