CAIE Further Paper 1 2024 November — Question 2 6 marks

Exam BoardCAIE
ModuleFurther Paper 1 (Further Paper 1)
Year2024
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof by induction
TypeProve derivative formula
DifficultyChallenging +1.2 This is a structured induction proof requiring differentiation of arctan and the product/quotient rule. While it involves Further Maths content and requires careful algebraic manipulation across multiple steps, the framework is standard: verify base case, assume for n=k, prove for n=k+1 using the product rule. The polynomial degree claim adds mild complexity but follows naturally from the differentiation. More routine than typical Further Maths proof questions requiring novel insight.
Spec4.01a Mathematical induction: construct proofs4.08g Derivatives: inverse trig and hyperbolic functions

Prove by mathematical induction that, for all positive integers \(n\), $$\frac{\mathrm{d}^n}{\mathrm{d}x^n}\left(\tan^{-1}x\right) = P_n(x)\left(1+x^2\right)^{-n},$$ where \(P_n(x)\) is a polynomial of degree \(n-1\). [6]

Question 2:
AnswerMarks
2d ( tan−1x ) 1
= so true when n=1.
AnswerMarks Guidance
dx 1+x2B1 Differentiates once.
dk
Assume that ( tan−1x ) =P (x)( 1+x2)−k , where degP (x)=k−1.
AnswerMarks Guidance
dxk k kB1 States inductive hypothesis. Must have
degP (x)=k−1.
k
dk+1( tan−1x )
'(x)( 1+x2)−k (x)( 1+x2)−k−1
=P −2kxP
AnswerMarks Guidance
dxk+1 k kM1 A1 Differentiates kth derivative using the product
rule.
= ( P '(x)( 1+x2) −2kxP (x) )( 1+x2)−k−1 so degP (x)=k
AnswerMarks Guidance
k k k+1A1 (x)( 1+x2)−k−1
Writes in the form P .
k+1
AnswerMarks Guidance
So true when n=k+1. By induction, true for all positive integers n.A1 Attempts to show degree of P (x) is at most k
k+1
(condone not showing coefficient of xk is non-
zero) and states conclusion.
6
AnswerMarks Guidance
QuestionAnswer Marks
Question 2:
2 | d ( tan−1x ) 1
= so true when n=1.
dx 1+x2 | B1 | Differentiates once.
dk
Assume that ( tan−1x ) =P (x)( 1+x2)−k , where degP (x)=k−1.
dxk k k | B1 | States inductive hypothesis. Must have
degP (x)=k−1.
k
dk+1( tan−1x )
'(x)( 1+x2)−k (x)( 1+x2)−k−1
=P −2kxP
dxk+1 k k | M1 A1 | Differentiates kth derivative using the product
rule.
= ( P '(x)( 1+x2) −2kxP (x) )( 1+x2)−k−1 so degP (x)=k
k k k+1 | A1 | (x)( 1+x2)−k−1
Writes in the form P .
k+1
So true when n=k+1. By induction, true for all positive integers n. | A1 | Attempts to show degree of P (x) is at most k
k+1
(condone not showing coefficient of xk is non-
zero) and states conclusion.
6
Question | Answer | Marks | Guidance
Prove by mathematical induction that, for all positive integers $n$,
$$\frac{\mathrm{d}^n}{\mathrm{d}x^n}\left(\tan^{-1}x\right) = P_n(x)\left(1+x^2\right)^{-n},$$
where $P_n(x)$ is a polynomial of degree $n-1$. [6]

\hfill \mbox{\textit{CAIE Further Paper 1 2024 Q2 [6]}}