Question 4 - A-Level Maths

CAIE Further Paper 1 2022 November — Question 4 7 marks

Exam Board	CAIE
Module	Further Paper 1 (Further Paper 1)
Year	2022
Session	November
Marks	7
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 with a clear pattern and the key property f''(x)=f(x) simplifies the algebra significantly. While it requires careful manipulation of higher derivatives using the product rule, the steps are methodical and the relationship between consecutive cases is straightforward once the pattern is recognized. More challenging than routine induction proofs due to the derivative notation, but less demanding than problems requiring novel insight or complex algebraic manipulation.
Spec	1.07g Differentiation from first principles: for small positive integer powers of x 1.07h Differentiation from first principles: for sin(x) and cos(x)4.01a Mathematical induction: construct proofs

4 The function f is such that $\mathrm { f } ^ { \prime \prime } ( x ) = \mathrm { f } ( x )$.
Prove by mathematical induction that, for every positive integer $n$, $$\frac { d ^ { 2 n - 1 } } { d x ^ { 2 n - 1 } } ( x f ( x ) ) = x f ^ { \prime } ( x ) + ( 2 n - 1 ) f ( x )$$

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Question 4:

Answer	Marks	Guidance
$\frac{d}{dx}(xf(x)) = xf'(x) + f(x) = xf'(x) + (2(1)-1)f(x)$	B1	Checks base case using product rule.
Assume true for $n = k$, so $\frac{d^{2k-1}}{dx^{2k-1}}(xf(x)) = xf'(x) + (2k-1)f(x)$	B1	States inductive hypothesis.
Then $\frac{d^{2k}}{dx^{2k}}(xf(x)) = xf(x) + 2kf'(x)$	M1 A1	Differentiates once.
$\frac{d^{2k+1}}{dx^{2k+1}}(xf(x)) = xf'(x) + f(x) + 2kf(x) = xf'(x) + (2k+1)f(x)$	M1 A1	Differentiates again.
So, it is also true for $n = k+1$. Hence, by induction, true for all positive integers.	A1	States conclusion.

Total: 7

## Question 4:

$\frac{d}{dx}(xf(x)) = xf'(x) + f(x) = xf'(x) + (2(1)-1)f(x)$ | B1 | Checks base case using product rule.

Assume true for $n = k$, so $\frac{d^{2k-1}}{dx^{2k-1}}(xf(x)) = xf'(x) + (2k-1)f(x)$ | B1 | States inductive hypothesis.

Then $\frac{d^{2k}}{dx^{2k}}(xf(x)) = xf(x) + 2kf'(x)$ | M1 A1 | Differentiates once.

$\frac{d^{2k+1}}{dx^{2k+1}}(xf(x)) = xf'(x) + f(x) + 2kf(x) = xf'(x) + (2k+1)f(x)$ | M1 A1 | Differentiates again.

So, it is also true for $n = k+1$. Hence, by induction, true for all positive integers. | A1 | States conclusion.

**Total: 7**

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4 The function f is such that $\mathrm { f } ^ { \prime \prime } ( x ) = \mathrm { f } ( x )$.\\
Prove by mathematical induction that, for every positive integer $n$,

$$\frac { d ^ { 2 n - 1 } } { d x ^ { 2 n - 1 } } ( x f ( x ) ) = x f ^ { \prime } ( x ) + ( 2 n - 1 ) f ( x )$$

\hfill \mbox{\textit{CAIE Further Paper 1 2022 Q4 [7]}}

This paper (7 questions)

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Q1 7 Q2 9 Q3 10 Q4 7 Q5 12 Q6 15 Q7 15

Answer	Marks	Guidance
\(\frac{d}{dx}(xf(x)) = xf'(x) + f(x) = xf'(x) + (2(1)-1)f(x)\)	B1	Checks base case using product rule.
Assume true for \(n = k\), so \(\frac{d^{2k-1}}{dx^{2k-1}}(xf(x)) = xf'(x) + (2k-1)f(x)\)	B1	States inductive hypothesis.
Then \(\frac{d^{2k}}{dx^{2k}}(xf(x)) = xf(x) + 2kf'(x)\)	M1 A1	Differentiates once.
\(\frac{d^{2k+1}}{dx^{2k+1}}(xf(x)) = xf'(x) + f(x) + 2kf(x) = xf'(x) + (2k+1)f(x)\)	M1 A1	Differentiates again.
So, it is also true for \(n = k+1\). Hence, by induction, true for all positive integers.	A1	States conclusion.