CAIE FP1 2011 November — Question 7 9 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2011
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicReduction Formulae
TypeCompound expressions with binomial expansion
DifficultyChallenging +1.2 This is a standard reduction formula question requiring product rule differentiation, integration by parts setup, and recursive calculation. While it involves multiple steps and careful algebraic manipulation, the techniques are well-practiced in Further Maths and the question provides clear guidance through 'show that' statements, making it moderately above average difficulty but not requiring novel insight.
Spec1.08i Integration by parts4.08f Integrate using partial fractions
7 Show that \(\frac { \mathrm { d } } { \mathrm { d } t } \left( t \left( 1 + t ^ { 3 } \right) ^ { n } \right) = ( 3 n + 1 ) \left( 1 + t ^ { 3 } \right) ^ { n } - 3 n \left( 1 + t ^ { 3 } \right) ^ { n - 1 }\). Let \(I _ { n } = \int _ { 0 } ^ { 1 } \left( 1 + t ^ { 3 } \right) ^ { n } \mathrm {~d} t\). Using the above result, or otherwise, show that $$( 3 n + 1 ) I _ { n } = 2 ^ { n } + 3 n I _ { n - 1 }$$ Hence evaluate \(I _ { 3 }\).
Question 7:
Differentiation/Reduction Formula
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{d}{dt}\{t(1+t^3)^n\} = 3t^3n(1+t^3)^{n-1} + (1+t^3)^n\)B1 Differentiates
\(= 3n(1+t^3-1)(1+t^3)^{n-1} + (1+t^3)^n\)M1 Rearranges
\(= (3n+1)(1+t^3)^n - 3n(1+t^3)^{n-1}\)A1 (AG) 3 marks total
\((3n+1)I_n = [t(1+t^3)^n]_0^1 + 3nI_{n-1}\)M1 Integrates wrt \(t\)
\((3n+1)I_n = 2^n + 3nI_{n-1}\)A1 (AG) 2 marks total
\(I_1 = \int_0^1(1+t^3)\,dt = [t+0.25t^4]_0^1 = 1.25\)B1M1 Evaluates \(I_1\) directly, uses reduction formula
\(7I_2 = 4 + 6\times1.25 \Rightarrow I_2 = \frac{23}{14}\)A1 Obtains \(I_2\)
\(10I_3 = 8 + 9\times\frac{23}{14} \Rightarrow I_3 = \frac{319}{140}\ (=2.28)\)A1 4 marks, total [9] 
# Question 7:

## Differentiation/Reduction Formula

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{d}{dt}\{t(1+t^3)^n\} = 3t^3n(1+t^3)^{n-1} + (1+t^3)^n$ | B1 | Differentiates |
| $= 3n(1+t^3-1)(1+t^3)^{n-1} + (1+t^3)^n$ | M1 | Rearranges |
| $= (3n+1)(1+t^3)^n - 3n(1+t^3)^{n-1}$ | A1 (AG) | 3 marks total |
| $(3n+1)I_n = [t(1+t^3)^n]_0^1 + 3nI_{n-1}$ | M1 | Integrates wrt $t$ |
| $(3n+1)I_n = 2^n + 3nI_{n-1}$ | A1 (AG) | 2 marks total |
| $I_1 = \int_0^1(1+t^3)\,dt = [t+0.25t^4]_0^1 = 1.25$ | B1M1 | Evaluates $I_1$ directly, uses reduction formula |
| $7I_2 = 4 + 6\times1.25 \Rightarrow I_2 = \frac{23}{14}$ | A1 | Obtains $I_2$ |
| $10I_3 = 8 + 9\times\frac{23}{14} \Rightarrow I_3 = \frac{319}{140}\ (=2.28)$ | A1 | 4 marks, total [9] |

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7 Show that $\frac { \mathrm { d } } { \mathrm { d } t } \left( t \left( 1 + t ^ { 3 } \right) ^ { n } \right) = ( 3 n + 1 ) \left( 1 + t ^ { 3 } \right) ^ { n } - 3 n \left( 1 + t ^ { 3 } \right) ^ { n - 1 }$.

Let $I _ { n } = \int _ { 0 } ^ { 1 } \left( 1 + t ^ { 3 } \right) ^ { n } \mathrm {~d} t$. Using the above result, or otherwise, show that

$$( 3 n + 1 ) I _ { n } = 2 ^ { n } + 3 n I _ { n - 1 }$$

Hence evaluate $I _ { 3 }$.

\hfill \mbox{\textit{CAIE FP1 2011 Q7 [9]}}
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