Question 2 - A-Level Maths

CAIE FP1 2019 November — Question 2 6 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2019
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 standard induction proof for derivatives with a clear pattern. While it requires familiarity with higher-order derivatives and factorial notation, the structure is straightforward: verify base case (n=1), assume for n=k, differentiate to prove n=k+1. The algebraic manipulation is routine for Further Maths students, and this type of question appears regularly in FP1 syllabi.
Spec	1.06d Natural logarithm: ln(x) function and properties 4.01a Mathematical induction: construct proofs

2 It is given that $y = \ln ( a x + 1 )$, where $a$ is a positive constant. Prove by mathematical induction that, for every positive integer $n$, $$\frac { \mathrm { d } ^ { n } y } { \mathrm {~d} x ^ { n } } = ( - 1 ) ^ { n - 1 } \frac { ( n - 1 ) ! a ^ { n } } { ( a x + 1 ) ^ { n } }$$

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Question 2:

Answer	Marks	Guidance
Answer	Marks	Guidance
$\frac{dy}{dx} = \frac{a}{ax+1} = (-1)^0 \frac{0! \, a^1}{(ax+1)^1}$, so true for $n=1$	M1 A1	Proves base case
Assume $\frac{d^k y}{dx^k} = (-1)^{k-1} \frac{(k-1)! \, a^k}{(ax+1)^k}$ for some positive integer $k$	B1	States inductive hypothesis
Then $\frac{d^{k+1}y}{dx^{k+1}} = -ka(-1)^{k-1} \frac{(k-1)! \, a^k}{(ax+1)^{k+1}} = (-1)^k \frac{k! \, a^{k+1}}{(ax+1)^{k+1}}$, so true for $n = k+1$	M1 A1	Differentiates $k^{\text{th}}$ derivative
By induction, true for every positive integer $n$	A1	States conclusion
6

## Question 2:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{a}{ax+1} = (-1)^0 \frac{0! \, a^1}{(ax+1)^1}$, so true for $n=1$ | **M1 A1** | Proves base case |
| Assume $\frac{d^k y}{dx^k} = (-1)^{k-1} \frac{(k-1)! \, a^k}{(ax+1)^k}$ for some positive integer $k$ | **B1** | States inductive hypothesis |
| Then $\frac{d^{k+1}y}{dx^{k+1}} = -ka(-1)^{k-1} \frac{(k-1)! \, a^k}{(ax+1)^{k+1}} = (-1)^k \frac{k! \, a^{k+1}}{(ax+1)^{k+1}}$, so true for $n = k+1$ | **M1 A1** | Differentiates $k^{\text{th}}$ derivative |
| By induction, true for every positive integer $n$ | **A1** | States conclusion |
| | **6** | |

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2 It is given that $y = \ln ( a x + 1 )$, where $a$ is a positive constant. Prove by mathematical induction that, for every positive integer $n$,

$$\frac { \mathrm { d } ^ { n } y } { \mathrm {~d} x ^ { n } } = ( - 1 ) ^ { n - 1 } \frac { ( n - 1 ) ! a ^ { n } } { ( a x + 1 ) ^ { n } }$$

\hfill \mbox{\textit{CAIE FP1 2019 Q2 [6]}}

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