Question 3 - A-Level Maths

CAIE FP1 2017 Specimen — Question 3 6 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2017
Session	Specimen
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Proof by induction
Type	Prove derivative formula
Difficulty	Challenging +1.2 This is a standard induction proof on differentiation requiring the product rule and algebraic manipulation. While it involves higher derivatives and exponentials (Further Maths content), the structure is routine: verify base case n=1, assume for n=k, differentiate to prove n=k+1, then factor to match the required form. The algebra is straightforward and follows a predictable pattern for this type of question.
Spec	1.07q Product and quotient rules: differentiation 4.01a Mathematical induction: construct proofs

3 Given that $a$ is a constant, prove by mathematical induction that, for every positive integer $n$, $$\frac { \mathrm { d } ^ { n } } { \mathrm {~d} x ^ { n } } \left( x \mathrm { e } ^ { a x } \right) = n a ^ { n - 1 } \mathrm { e } ^ { a x } + a ^ { n } x \mathrm { e } ^ { a x }$$

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Question 3:

Answer	Marks	Guidance
Answer	Marks	Guidance
$n=1$ in formula gives $a^0 e^{ax} + axe^{ax} = e^{ax} + axe^{ax}$	1	B1
$\frac{d}{dx}(xe^{ax}) = e^{ax} \times 1 + x.ae^{ax} = e^{ax} + axe^{ax} \Rightarrow H_1$ is true oe	1	B1
Assume $H_k$ is true, i.e. $\frac{d^k}{dx^k}(xe^{ax}) = ka^{k-1}e^{ax} + a^k xe^{ax}$	1	B1
$\frac{d^{k+1}}{dx^{k+1}}(xe^{ax}) = ka^k e^{ax} + a^k e^{ax} + a^{k+1}xe^{ax}$	1	M1
$= (k+1)a^k e^{ax} + a^{k+1}xe^{ax}$	1	A1
$\Rightarrow H_{k+1}$ is true, hence by PMI $H_n$ is true for all positive integers $n$	1	A1
Total	6

## Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $n=1$ in formula gives $a^0 e^{ax} + axe^{ax} = e^{ax} + axe^{ax}$ | 1 | B1 |
| $\frac{d}{dx}(xe^{ax}) = e^{ax} \times 1 + x.ae^{ax} = e^{ax} + axe^{ax} \Rightarrow H_1$ is true oe | 1 | B1 |
| Assume $H_k$ is true, i.e. $\frac{d^k}{dx^k}(xe^{ax}) = ka^{k-1}e^{ax} + a^k xe^{ax}$ | 1 | B1 |
| $\frac{d^{k+1}}{dx^{k+1}}(xe^{ax}) = ka^k e^{ax} + a^k e^{ax} + a^{k+1}xe^{ax}$ | 1 | M1 |
| $= (k+1)a^k e^{ax} + a^{k+1}xe^{ax}$ | 1 | A1 |
| $\Rightarrow H_{k+1}$ is true, hence by PMI $H_n$ is true for all positive integers $n$ | 1 | A1 |
| **Total** | **6** | |

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3 Given that $a$ is a constant, prove by mathematical induction that, for every positive integer $n$,

$$\frac { \mathrm { d } ^ { n } } { \mathrm {~d} x ^ { n } } \left( x \mathrm { e } ^ { a x } \right) = n a ^ { n - 1 } \mathrm { e } ^ { a x } + a ^ { n } x \mathrm { e } ^ { a x }$$

\hfill \mbox{\textit{CAIE FP1 2017 Q3 [6]}}

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