CAIE FP1 2015 November — Question 3 6 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2015
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof by induction
TypeProve derivative formula
DifficultyStandard +0.8 This is a Further Maths induction proof involving differentiation of exponential functions. While the algebraic manipulation is straightforward (product rule and collecting terms), it requires careful bookkeeping of powers of 'a' and coefficients across multiple steps. The inductive step demands more sophistication than typical A-level induction proofs on summations, placing it moderately above average difficulty.
Spec1.07j Differentiate exponentials: e^(kx) and a^(kx)4.01a Mathematical induction: construct proofs

Given that \(a\) is a constant, prove by mathematical induction that, for every positive integer \(n\), $$\frac{\mathrm{d}^n}{\mathrm{d}x^n}(xe^{ax}) = na^{n-1}e^{ax} + a^n xe^{ax}.$$ [6]

Question 3:
AnswerMarks
3n =1 in formula gives a0eax +axeax =eax +axeax
d ( )
xeax =eax×1+x.aeax =eax +axeax ⇒H is true oe
dx 1
dk ( )
xeax =kak−1eax +akxeax
Assume Hk is true, i.e. .
dxk
dk+1 ( )
xeax =kakeax +akeax +ak+1xeax
dxk+1
=(k+1)akeax +ak+1xeax
⇒H is true, hence by PMI Hn is true for all positive integers n.
AnswerMarks
k+1B 1
B1
B1
M1
A1
A1
[6]

Total

6
Question 3:
3 | n =1 in formula gives a0eax +axeax =eax +axeax
d ( )
xeax =eax×1+x.aeax =eax +axeax ⇒H is true oe
dx 1
dk ( )
xeax =kak−1eax +akxeax
Assume Hk is true, i.e. .
dxk
dk+1 ( )
xeax =kakeax +akeax +ak+1xeax
dxk+1
=(k+1)akeax +ak+1xeax
⇒H is true, hence by PMI Hn is true for all positive integers n.
k+1 | B 1
B1
B1
M1
A1
A1
[6]
Total
6
Given that $a$ is a constant, prove by mathematical induction that, for every positive integer $n$,
$$\frac{\mathrm{d}^n}{\mathrm{d}x^n}(xe^{ax}) = na^{n-1}e^{ax} + a^n xe^{ax}.$$ [6]

\hfill \mbox{\textit{CAIE FP1 2015 Q3 [6]}}