CAIE Further Paper 1 2023 November — Question 2 6 marks

Exam BoardCAIE
ModuleFurther Paper 1 (Further Paper 1)
Year2023
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof by induction
TypeProve derivative formula
DifficultyChallenging +1.2 This is a standard induction proof on differentiation requiring the product rule and algebraic manipulation. While it involves Further Maths content (nth derivatives), the structure is routine: verify base case n=1, assume for n=k, differentiate to prove n=k+1, then expand and collect terms. The algebra is moderately involved but follows a predictable pattern for this type of question.
Spec1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07q Product and quotient rules: differentiation4.01a Mathematical induction: construct proofs
2 Prove by mathematical induction that, for all positive integers \(n\), $$\frac { d ^ { n } } { d x ^ { n } } \left( x ^ { 2 } e ^ { x } \right) = \left( x ^ { 2 } + 2 n x + n ( n - 1 ) \right) e ^ { x }$$
Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
\(\frac{d}{dx}(x^2e^x) = x^2e^x + 2xe^x = (x^2+2x)e^x\) so true when \(n=1\)M1 A1 Differentiates once using the product rule
Assume that \(\frac{d^k}{dx^k}(x^2e^x) = (x^2+2kx+k(k-1))e^x\) for some value of \(k\)B1 States inductive hypothesis
\(\frac{d^{k+1}}{dx^{k+1}}(x^2e^x) = (x^2+2kx+k(k-1))e^x + e^x(2x+2k)\)M1 Differentiates \(k\)th derivative
\((x^2+2(k+1)x+k(k+1))e^x\)A1
So true when \(n=k+1\). By induction, true for all positive integers \(n\).A1 States conclusion 
## Question 2:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{d}{dx}(x^2e^x) = x^2e^x + 2xe^x = (x^2+2x)e^x$ so true when $n=1$ | M1 A1 | Differentiates once using the product rule |
| Assume that $\frac{d^k}{dx^k}(x^2e^x) = (x^2+2kx+k(k-1))e^x$ for some value of $k$ | B1 | States inductive hypothesis |
| $\frac{d^{k+1}}{dx^{k+1}}(x^2e^x) = (x^2+2kx+k(k-1))e^x + e^x(2x+2k)$ | M1 | Differentiates $k$th derivative |
| $(x^2+2(k+1)x+k(k+1))e^x$ | A1 | |
| So true when $n=k+1$. By induction, true for all positive integers $n$. | A1 | States conclusion |

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2 Prove by mathematical induction that, for all positive integers $n$,

$$\frac { d ^ { n } } { d x ^ { n } } \left( x ^ { 2 } e ^ { x } \right) = \left( x ^ { 2 } + 2 n x + n ( n - 1 ) \right) e ^ { x }$$

\hfill \mbox{\textit{CAIE Further Paper 1 2023 Q2 [6]}}
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Q1 9 Q2 6 Q3 8 Q4 10 Q5 13 Q6 13 Q7 16