CAIE FP1 2011 November — Question 3 7 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2011
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof by induction
TypeProve derivative formula
DifficultyChallenging +1.2 This is a standard induction proof on a derivative formula requiring the product rule and trigonometric identities. While it involves multiple techniques (induction structure, repeated differentiation, trig addition formulas), it follows a predictable pattern once the inductive step is set up. The factor of 2^(n/2) and phase shift of nπ/4 emerge naturally from the calculation, making this a moderately challenging but routine Further Maths question.
Spec1.07q Product and quotient rules: differentiation4.01a Mathematical induction: construct proofs

3 Prove by mathematical induction that, for all positive integers \(n\), $$\frac { \mathrm { d } ^ { n } } { \mathrm {~d} x ^ { n } } \left( \mathrm { e } ^ { x } \sin x \right) = 2 ^ { \frac { 1 } { 2 } n } \mathrm { e } ^ { x } \sin \left( x + \frac { 1 } { 4 } n \pi \right)$$

Question 3:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(H_n\): \(\frac{d^n}{dx^n}(e^x \sin x) = 2^{\frac{n}{2}} e^x \sin\left(x + \frac{n\pi}{4}\right)\) Proves base case
\(\frac{d}{dx}(e^x \sin x) = \sin x \cdot e^x + e^x \cos x\)M1
\(= \sqrt{2}e^x\left(\frac{\sin x}{\sqrt{2}} + \frac{\cos x}{\sqrt{2}}\right) = 2^{\frac{1}{2}}e^x\sin\left(x + \frac{\pi}{4}\right)\)
\(\Rightarrow H_1\) is trueA1
Assume \(H_k\) is trueB1 States inductive hypothesis
\(\frac{d^{k+1}}{dx^{k+1}}(e^x \sin x) = 2^{\frac{k}{2}}\left\{e^x\sin\left(x+\frac{k\pi}{4}\right) + e^x\cos\left(x+\frac{k\pi}{4}\right)\right\}\)M1 Proves inductive step
\(= 2^{\frac{k+1}{2}}e^x\left\{\frac{1}{\sqrt{2}}\sin\left(x+\frac{k\pi}{4}\right) + \frac{1}{\sqrt{2}}\cos\left(x+\frac{k\pi}{4}\right)\right\}\)A1
\(= 2^{\frac{k+1}{2}}e^x\left\{\sin\left(x+\frac{k\pi}{4}+\frac{\pi}{4}\right)\right\}\)
\(= 2^{\frac{k+1}{2}}e^x\sin\left(x+\frac{(k+1)\pi}{4}\right)\)A1
\(\therefore H_k \Rightarrow H_{k+1}\); Hence true for all positive integers by PMIA1 States conclusion; Total: 7 marks
# Question 3:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_n$: $\frac{d^n}{dx^n}(e^x \sin x) = 2^{\frac{n}{2}} e^x \sin\left(x + \frac{n\pi}{4}\right)$ | | Proves base case |
| $\frac{d}{dx}(e^x \sin x) = \sin x \cdot e^x + e^x \cos x$ | M1 | |
| $= \sqrt{2}e^x\left(\frac{\sin x}{\sqrt{2}} + \frac{\cos x}{\sqrt{2}}\right) = 2^{\frac{1}{2}}e^x\sin\left(x + \frac{\pi}{4}\right)$ | | |
| $\Rightarrow H_1$ is true | A1 | |
| Assume $H_k$ is true | B1 | States inductive hypothesis |
| $\frac{d^{k+1}}{dx^{k+1}}(e^x \sin x) = 2^{\frac{k}{2}}\left\{e^x\sin\left(x+\frac{k\pi}{4}\right) + e^x\cos\left(x+\frac{k\pi}{4}\right)\right\}$ | M1 | Proves inductive step |
| $= 2^{\frac{k+1}{2}}e^x\left\{\frac{1}{\sqrt{2}}\sin\left(x+\frac{k\pi}{4}\right) + \frac{1}{\sqrt{2}}\cos\left(x+\frac{k\pi}{4}\right)\right\}$ | A1 | |
| $= 2^{\frac{k+1}{2}}e^x\left\{\sin\left(x+\frac{k\pi}{4}+\frac{\pi}{4}\right)\right\}$ | | |
| $= 2^{\frac{k+1}{2}}e^x\sin\left(x+\frac{(k+1)\pi}{4}\right)$ | A1 | |
| $\therefore H_k \Rightarrow H_{k+1}$; Hence true for all positive integers by PMI | A1 | States conclusion; Total: 7 marks |

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3 Prove by mathematical induction that, for all positive integers $n$,

$$\frac { \mathrm { d } ^ { n } } { \mathrm {~d} x ^ { n } } \left( \mathrm { e } ^ { x } \sin x \right) = 2 ^ { \frac { 1 } { 2 } n } \mathrm { e } ^ { x } \sin \left( x + \frac { 1 } { 4 } n \pi \right)$$

\hfill \mbox{\textit{CAIE FP1 2011 Q3 [7]}}