Question 11 EITHER - A-Level Maths

CAIE FP1 2013 November — Question 11 EITHER

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2013
Session	November
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Reduction Formulae
Type	Compound expressions with binomial expansion
Difficulty	Challenging +1.2 This is a standard reduction formula question requiring integration by parts to derive the recurrence relation, then straightforward substitution to find specific values. The derivation follows a well-practiced technique (integrating by parts with u=(1+x²)^n), and the evaluation steps are mechanical arithmetic. While it requires multiple steps and careful algebra, it's a textbook example of the reduction formula method with no novel insight needed. Slightly above average difficulty due to the algebraic manipulation and the negative index extension, but well within the standard Further Maths repertoire.
Spec	1.08d Evaluate definite integrals: between limits 4.08a Maclaurin series: find series for function 8.06a Reduction formulae: establish, use, and evaluate recursively

Let $I _ { n } = \int _ { 0 } ^ { 1 } \left( 1 + x ^ { 2 } \right) ^ { n } \mathrm {~d} x$. Show that, for all integers $n$, $$( 2 n + 1 ) I _ { n } = 2 n I _ { n - 1 } + 2 ^ { n }$$ Evaluate $I _ { 0 }$ and hence find $I _ { 3 }$. Given that $I _ { - 1 } = \frac { 1 } { 4 } \pi$, find $I _ { - 3 }$.

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Question 11 (EITHER – reduction formula):

Part (proving reduction formula):

Answer	Marks	Guidance
$I_n = \left[x(1+x^2)^n\right]_0^1 - 2\int_0^1 nx^2(1+x^2)^{n-1}dx$	M1A1
$= 2^n - 2n\int_0^1(1+x^2-1)(1+x^2)^{n-1}dx$	M1A1
$\Rightarrow I_n = 2^n - 2nI_n + 2nI_{n-1}$
$\Rightarrow (2n+1)I_n = 2nI_{n-1} + 2^n$ (AG)	A1	(5)

Part (finding $I_3$):

Answer	Marks	Guidance
$I_0 = \int_0^1 1\,dx = 1$	B1
$3I_1 = 2\times1\times1 + 2 \Rightarrow I_1 = \frac{4}{3}$	M1
$5I_2 = \frac{16}{3} + 4 \Rightarrow I_2 = \frac{28}{15}$	A1
$7I_3 = \frac{56}{5} + 8 \Rightarrow I_3 = \frac{96}{35}$ (2.74)	A1	(4)

Part (negative $n$):

Answer	Marks	Guidance
$2nI_{n-1} = (2n+1)I_n - 2^n$	M1
$\Rightarrow 2I_{-2} = I_{-1} + \frac{1}{2}$ and $4I_{-3} = 3I_{-2} + \frac{1}{4}$	A1A1
$\Rightarrow I_{-3} = \frac{3}{8}I_{-1} + \frac{1}{4} = \frac{3}{32}\pi + \frac{1}{4}$	M1A1	(5)

Question 11 (OR – eigenvalues/matrices):

Part (eigenvalue proof):

Answer	Marks	Guidance
$\mathbf{Ae} = \lambda\mathbf{e}$ and $\mathbf{Be} = \mu\mathbf{e}$	B1
Adding: $\mathbf{Ae} + \mathbf{Be} = \lambda\mathbf{e} + \mu\mathbf{e} \Rightarrow (\mathbf{A}+\mathbf{B})\mathbf{e} = (\lambda+\mu)\mathbf{e}$	M1
$\Rightarrow \lambda + \mu$ is an eigenvalue (since $\mathbf{e} \neq \mathbf{0}$)	A1	(3)

Part (eigenvalues of $\mathbf{A}$):

Answer	Marks	Guidance
$\mathbf{A}\begin{pmatrix}1\\1\\1\end{pmatrix} = \begin{pmatrix}-1\\-1\\-1\end{pmatrix} \Rightarrow \lambda_1 = -1$	M1A1
$\mathbf{A}\begin{pmatrix}1\\-1\\1\end{pmatrix} = \begin{pmatrix}1\\-1\\1\end{pmatrix} \Rightarrow \lambda^2 = 1$	A1
$\mathbf{A}\begin{pmatrix}2\\0\\1\end{pmatrix} = \begin{pmatrix}6\\0\\3\end{pmatrix} \Rightarrow \lambda^3 = 3$	A1	(4)

Part (eigenvalues of $\mathbf{M}$):

Answer	Marks	Guidance
Eigenvalues of $\mathbf{M}$ are $-8, -2, 2$	B1	(1)

Part (diagonalisation):

Answer	Marks	Guidance
$\mathbf{R} = \begin{pmatrix}1 & 1 & 2\\1 & -1 & 0\\1 & 1 & 1\end{pmatrix}$	B1
$\mathbf{S} = \mathbf{R}^{-1}$	M1
Determinant $= 2$, $\mathbf{S} = \frac{1}{2}\begin{pmatrix}-1 & 1 & 2\\-1 & -1 & 2\\2 & 0 & -2\end{pmatrix}$	M1A1
$\mathbf{D} = \begin{pmatrix}-32768 & 0 & 0\\0 & -32 & 0\\0 & 0 & 32\end{pmatrix}$	M1A1	(6)

# Question 11 (EITHER – reduction formula):

## Part (proving reduction formula):
| $I_n = \left[x(1+x^2)^n\right]_0^1 - 2\int_0^1 nx^2(1+x^2)^{n-1}dx$ | M1A1 | |
|---|---|---|
| $= 2^n - 2n\int_0^1(1+x^2-1)(1+x^2)^{n-1}dx$ | M1A1 | |
| $\Rightarrow I_n = 2^n - 2nI_n + 2nI_{n-1}$ | | |
| $\Rightarrow (2n+1)I_n = 2nI_{n-1} + 2^n$ (AG) | A1 | **(5)** |

## Part (finding $I_3$):
| $I_0 = \int_0^1 1\,dx = 1$ | B1 | |
|---|---|---|
| $3I_1 = 2\times1\times1 + 2 \Rightarrow I_1 = \frac{4}{3}$ | M1 | |
| $5I_2 = \frac{16}{3} + 4 \Rightarrow I_2 = \frac{28}{15}$ | A1 | |
| $7I_3 = \frac{56}{5} + 8 \Rightarrow I_3 = \frac{96}{35}$ (2.74) | A1 | **(4)** |

## Part (negative $n$):
| $2nI_{n-1} = (2n+1)I_n - 2^n$ | M1 | |
|---|---|---|
| $\Rightarrow 2I_{-2} = I_{-1} + \frac{1}{2}$ and $4I_{-3} = 3I_{-2} + \frac{1}{4}$ | A1A1 | |
| $\Rightarrow I_{-3} = \frac{3}{8}I_{-1} + \frac{1}{4} = \frac{3}{32}\pi + \frac{1}{4}$ | M1A1 | **(5)** |

---

# Question 11 (OR – eigenvalues/matrices):

## Part (eigenvalue proof):
| $\mathbf{Ae} = \lambda\mathbf{e}$ and $\mathbf{Be} = \mu\mathbf{e}$ | B1 | |
|---|---|---|
| Adding: $\mathbf{Ae} + \mathbf{Be} = \lambda\mathbf{e} + \mu\mathbf{e} \Rightarrow (\mathbf{A}+\mathbf{B})\mathbf{e} = (\lambda+\mu)\mathbf{e}$ | M1 | |
| $\Rightarrow \lambda + \mu$ is an eigenvalue (since $\mathbf{e} \neq \mathbf{0}$) | A1 | **(3)** |

## Part (eigenvalues of $\mathbf{A}$):
| $\mathbf{A}\begin{pmatrix}1\\1\\1\end{pmatrix} = \begin{pmatrix}-1\\-1\\-1\end{pmatrix} \Rightarrow \lambda_1 = -1$ | M1A1 | |
|---|---|---|
| $\mathbf{A}\begin{pmatrix}1\\-1\\1\end{pmatrix} = \begin{pmatrix}1\\-1\\1\end{pmatrix} \Rightarrow \lambda^2 = 1$ | A1 | |
| $\mathbf{A}\begin{pmatrix}2\\0\\1\end{pmatrix} = \begin{pmatrix}6\\0\\3\end{pmatrix} \Rightarrow \lambda^3 = 3$ | A1 | **(4)** |

## Part (eigenvalues of $\mathbf{M}$):
| Eigenvalues of $\mathbf{M}$ are $-8, -2, 2$ | B1 | **(1)** |
|---|---|---|

## Part (diagonalisation):
| $\mathbf{R} = \begin{pmatrix}1 & 1 & 2\\1 & -1 & 0\\1 & 1 & 1\end{pmatrix}$ | B1 | |
|---|---|---|
| $\mathbf{S} = \mathbf{R}^{-1}$ | M1 | |
| Determinant $= 2$, $\mathbf{S} = \frac{1}{2}\begin{pmatrix}-1 & 1 & 2\\-1 & -1 & 2\\2 & 0 & -2\end{pmatrix}$ | M1A1 | |
| $\mathbf{D} = \begin{pmatrix}-32768 & 0 & 0\\0 & -32 & 0\\0 & 0 & 32\end{pmatrix}$ | M1A1 | **(6)** |

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Let $I _ { n } = \int _ { 0 } ^ { 1 } \left( 1 + x ^ { 2 } \right) ^ { n } \mathrm {~d} x$. Show that, for all integers $n$,

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

Evaluate $I _ { 0 }$ and hence find $I _ { 3 }$.

Given that $I _ { - 1 } = \frac { 1 } { 4 } \pi$, find $I _ { - 3 }$.

\hfill \mbox{\textit{CAIE FP1 2013 Q11 EITHER}}

This paper (12 questions)

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Q1 6 Q2 6 Q3 7 Q4 7 Q5 8 Q6 8 Q7 9 Q8 11 Q9 11 Q10 13 Q11 EITHER Q11 OR

CAIE FP1 2013 November — Question 11 EITHER

Question 11 (EITHER – reduction formula):

Part (proving reduction formula):

Part (finding \(I_3\)):

Part (negative \(n\)):

Question 11 (OR – eigenvalues/matrices):

Part (eigenvalue proof):

Part (eigenvalues of \(\mathbf{A}\)):

Part (eigenvalues of \(\mathbf{M}\)):

Part (diagonalisation):

This paper (12 questions)

Answer	Marks	Guidance
\(I_n = \left[x(1+x^2)^n\right]_0^1 - 2\int_0^1 nx^2(1+x^2)^{n-1}dx\)	M1A1
\(= 2^n - 2n\int_0^1(1+x^2-1)(1+x^2)^{n-1}dx\)	M1A1
\(\Rightarrow I_n = 2^n - 2nI_n + 2nI_{n-1}\)
\(\Rightarrow (2n+1)I_n = 2nI_{n-1} + 2^n\) (AG)	A1	(5)

Answer	Marks	Guidance
\(I_0 = \int_0^1 1\,dx = 1\)	B1
\(3I_1 = 2\times1\times1 + 2 \Rightarrow I_1 = \frac{4}{3}\)	M1
\(5I_2 = \frac{16}{3} + 4 \Rightarrow I_2 = \frac{28}{15}\)	A1
\(7I_3 = \frac{56}{5} + 8 \Rightarrow I_3 = \frac{96}{35}\) (2.74)	A1	(4)

Answer	Marks	Guidance
\(2nI_{n-1} = (2n+1)I_n - 2^n\)	M1
\(\Rightarrow 2I_{-2} = I_{-1} + \frac{1}{2}\) and \(4I_{-3} = 3I_{-2} + \frac{1}{4}\)	A1A1
\(\Rightarrow I_{-3} = \frac{3}{8}I_{-1} + \frac{1}{4} = \frac{3}{32}\pi + \frac{1}{4}\)	M1A1	(5)

Answer	Marks	Guidance
\(\mathbf{Ae} = \lambda\mathbf{e}\) and \(\mathbf{Be} = \mu\mathbf{e}\)	B1
Adding: \(\mathbf{Ae} + \mathbf{Be} = \lambda\mathbf{e} + \mu\mathbf{e} \Rightarrow (\mathbf{A}+\mathbf{B})\mathbf{e} = (\lambda+\mu)\mathbf{e}\)	M1
\(\Rightarrow \lambda + \mu\) is an eigenvalue (since \(\mathbf{e} \neq \mathbf{0}\))	A1	(3)

Answer	Marks	Guidance
\(\mathbf{A}\begin{pmatrix}1\\1\\1\end{pmatrix} = \begin{pmatrix}-1\\-1\\-1\end{pmatrix} \Rightarrow \lambda_1 = -1\)	M1A1
\(\mathbf{A}\begin{pmatrix}1\\-1\\1\end{pmatrix} = \begin{pmatrix}1\\-1\\1\end{pmatrix} \Rightarrow \lambda^2 = 1\)	A1
\(\mathbf{A}\begin{pmatrix}2\\0\\1\end{pmatrix} = \begin{pmatrix}6\\0\\3\end{pmatrix} \Rightarrow \lambda^3 = 3\)	A1	(4)

Answer	Marks	Guidance
\(\mathbf{R} = \begin{pmatrix}1 & 1 & 2\\1 & -1 & 0\\1 & 1 & 1\end{pmatrix}\)	B1
\(\mathbf{S} = \mathbf{R}^{-1}\)	M1
Determinant \(= 2\), \(\mathbf{S} = \frac{1}{2}\begin{pmatrix}-1 & 1 & 2\\-1 & -1 & 2\\2 & 0 & -2\end{pmatrix}\)	M1A1
\(\mathbf{D} = \begin{pmatrix}-32768 & 0 & 0\\0 & -32 & 0\\0 & 0 & 32\end{pmatrix}\)	M1A1	(6)