Question 2 - A-Level Maths

CAIE Further Paper 1 2021 November — Question 2 6 marks

Exam Board	CAIE
Module	Further Paper 1 (Further Paper 1)
Year	2021
Session	November
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Proof by induction
Type	Prove derivative formula
Difficulty	Challenging +1.2 This is a structured induction proof on a derivative formula with clear pattern recognition. While it requires product rule application and algebraic manipulation at each step, the structure is standard for Further Maths induction proofs. The formula is given (not discovered), and the inductive step follows predictably from applying the product rule to the assumed form. More challenging than basic induction but routine for Further Maths students.
Spec	4.01a Mathematical induction: construct proofs

2 It is given that $\mathrm { y } = \mathrm { xe } ^ { \mathrm { ax } }$, where $a$ is a constant.
Prove by mathematical induction that, for all positive integers $n$, $$\frac { d ^ { n } y } { d x ^ { n } } = \left( a ^ { n } x + n a ^ { n - 1 } \right) e ^ { a x }$$

Show mark scheme Show mark scheme source

Question 2:

Answer	Marks	Guidance
$\frac{dy}{dx} = axe^{ax} + e^{ax} = (ax+1)e^{ax}$ so true when $n=1$	M1 A1	Differentiates once using the product rule
Assume that $\frac{d^k y}{dx^k} = (a^k x + ka^{k-1})e^{ax}$	B1	States inductive hypothesis
$\frac{d^{k+1}y}{dx^{k+1}} = a(a^k x + ka^{k-1})e^{ax} + e^{ax}(a^k) = (a^{k+1} + (k+1)a^k)e^{ax}$	M1 A1	Differentiates $k$th derivative
So true when $n = k+1$. By induction, true for all positive integers $n$.	A1	States conclusion

## Question 2:

| $\frac{dy}{dx} = axe^{ax} + e^{ax} = (ax+1)e^{ax}$ so true when $n=1$ | M1 A1 | Differentiates once using the product rule |
|---|---|---|
| Assume that $\frac{d^k y}{dx^k} = (a^k x + ka^{k-1})e^{ax}$ | B1 | States inductive hypothesis |
| $\frac{d^{k+1}y}{dx^{k+1}} = a(a^k x + ka^{k-1})e^{ax} + e^{ax}(a^k) = (a^{k+1} + (k+1)a^k)e^{ax}$ | M1 A1 | Differentiates $k$th derivative |
| So true when $n = k+1$. By induction, true for all positive integers $n$. | A1 | States conclusion |

---

Show LaTeX source

2 It is given that $\mathrm { y } = \mathrm { xe } ^ { \mathrm { ax } }$, where $a$ is a constant.\\
Prove by mathematical induction that, for all positive integers $n$,

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

\hfill \mbox{\textit{CAIE Further Paper 1 2021 Q2 [6]}}

This paper (7 questions)

View full paper

Q1 9 Q2 6 Q3 7 Q4 10 Q5 12 Q6 14 Q7 17

Answer	Marks	Guidance
\(\frac{dy}{dx} = axe^{ax} + e^{ax} = (ax+1)e^{ax}\) so true when \(n=1\)	M1 A1	Differentiates once using the product rule
Assume that \(\frac{d^k y}{dx^k} = (a^k x + ka^{k-1})e^{ax}\)	B1	States inductive hypothesis
\(\frac{d^{k+1}y}{dx^{k+1}} = a(a^k x + ka^{k-1})e^{ax} + e^{ax}(a^k) = (a^{k+1} + (k+1)a^k)e^{ax}\)	M1 A1	Differentiates \(k\)th derivative
So true when \(n = k+1\). By induction, true for all positive integers \(n\).	A1	States conclusion