Question 11 EITHER - A-Level Maths

CAIE FP1 2011 November — Question 11 EITHER

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2011
Session	November
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Integration with Partial Fractions
Type	Basic partial fractions then integrate
Difficulty	Standard +0.8 This is a multi-part Further Maths question requiring mean value calculation, arc length derivation and computation, and surface of revolution. While each technique is standard for FP1, the combination of multiple integration applications and the algebraic manipulation of surds across several parts makes this moderately challenging, though still within typical Further Maths scope.
Spec	1.08d Evaluate definite integrals: between limits 4.08d Volumes of revolution: about x and y axes 4.08f Integrate using partial fractions

The curve $C$ has equation $y = \frac { 1 } { 3 } x ^ { \frac { 1 } { 2 } } ( 3 - x )$, for $0 \leqslant x \leqslant 3$. Find the mean value of $y$ with respect to $x$ over the interval $0 \leqslant x \leqslant 3$. Show that $$\frac { \mathrm { d } s } { \mathrm {~d} x } = \frac { 1 } { 2 } \left( x ^ { - \frac { 1 } { 2 } } + x ^ { \frac { 1 } { 2 } } \right)$$ where $s$ denotes arc length, and find the arc length of $C$. Find the area of the surface generated when $C$ is rotated through $2 \pi$ radians about the $x$-axis.

Show mark scheme Show mark scheme source

Question 11 (EITHER):

Mean Value, Arc Length, Surface Area

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
Mean value $= \frac{\int_0^3 y\,dx}{3-0} = \left[\frac{2}{3}x^{\frac{3}{2}}-\frac{2}{15}x^{\frac{5}{2}}\right]_0^3 \div 3$	M1M1, A1	Uses formula and integrates $y$ wrt $x$
$= \sqrt{3}\left[\frac{2}{3}\times3-\frac{2}{15}\times9\right]\div3 = \frac{4\sqrt{3}}{15}\ (=0.462)$	A1	4 marks
$y' = \frac{1}{2\sqrt{x}}-\frac{1}{2}\sqrt{x} \Rightarrow \frac{ds}{dx}=\sqrt{1+\frac{1}{4}\left(\frac{1}{x}-2+x\right)}$	B1	Uses $\sqrt{1+(y')^2}$
$\Rightarrow \frac{ds}{dx} = \sqrt{\frac{1}{4}\left(\frac{1}{x}+2+x\right)} = \left(\frac{1}{2\sqrt{x}}+\frac{1}{2}\sqrt{x}\right)$	B1 (AG)
$s = \frac{1}{2}\int_0^3\left(x^{-\frac{1}{2}}+x^{\frac{1}{2}}\right)dx = \frac{1}{2}\left[2\sqrt{x}+\frac{2}{3}x^{\frac{3}{2}}\right]_0^3 = 2\sqrt{3}\ (=3.46)$	M1A1, M1A1	6 marks
$S = 2\pi\int_0^3\left(x^{\frac{1}{2}}-\frac{1}{3}x^{\frac{3}{2}}\right)\left(\frac{1}{2}x^{-\frac{1}{2}}+\frac{1}{2}x^{\frac{1}{2}}\right)dx$	M1	Surface area formula
$= \pi\int_0^3\left(1+\frac{2}{3}x-\frac{1}{3}x^2\right)dx = \left[x+\frac{1}{3}x^2-\frac{1}{9}x^3\right]_0^3 = 3\pi$	M1A1, A1	4 marks, total [14]

Question 11 (OR):

Eigenvalues and Eigenvectors

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$\det(\mathbf{A}-\lambda\mathbf{I})=0 \Rightarrow \lambda^3-6\lambda^2+11\lambda-6=0$	M1A1	Forms characteristic equation
$\Rightarrow \lambda=1,2,3$	A1A1	Solves
$\mathbf{e}_1=-2\mathbf{j}+\mathbf{k},\quad \mathbf{e}_2=\mathbf{i}+\mathbf{j},\quad \mathbf{e}_3=2\mathbf{i}+2\mathbf{j}+\mathbf{k}$	M1A1, A1	7 marks, eigenvectors
$\mathbf{r}=s\mathbf{e}+t\mathbf{f}$; $\mathbf{A}(s\mathbf{e}+t\mathbf{f})=s\mathbf{A}\mathbf{e}+t\mathbf{A}\mathbf{f}=(s\lambda)\mathbf{e}+(t\mu)\mathbf{f}$	B1, M1A1	States equation of plane, 3 marks
$\mathbf{e}_1\times\mathbf{e}_2=-\mathbf{i}+\mathbf{j}+2\mathbf{k}\Rightarrow x-y-2z=0$	M1A1
$\mathbf{e}_1\times\mathbf{e}_3=2\mathbf{i}-\mathbf{j}-2\mathbf{k}\Rightarrow 2x-y-2z=0$	A1
$\mathbf{e}_2\times\mathbf{e}_3=\mathbf{i}-\mathbf{j}\Rightarrow x-y=0$	A1	4 marks, total [14]

# Question 11 (EITHER):

## Mean Value, Arc Length, Surface Area

| Answer/Working | Marks | Guidance |
|---|---|---|
| Mean value $= \frac{\int_0^3 y\,dx}{3-0} = \left[\frac{2}{3}x^{\frac{3}{2}}-\frac{2}{15}x^{\frac{5}{2}}\right]_0^3 \div 3$ | M1M1, A1 | Uses formula and integrates $y$ wrt $x$ |
| $= \sqrt{3}\left[\frac{2}{3}\times3-\frac{2}{15}\times9\right]\div3 = \frac{4\sqrt{3}}{15}\ (=0.462)$ | A1 | 4 marks |
| $y' = \frac{1}{2\sqrt{x}}-\frac{1}{2}\sqrt{x} \Rightarrow \frac{ds}{dx}=\sqrt{1+\frac{1}{4}\left(\frac{1}{x}-2+x\right)}$ | B1 | Uses $\sqrt{1+(y')^2}$ |
| $\Rightarrow \frac{ds}{dx} = \sqrt{\frac{1}{4}\left(\frac{1}{x}+2+x\right)} = \left(\frac{1}{2\sqrt{x}}+\frac{1}{2}\sqrt{x}\right)$ | B1 (AG) | |
| $s = \frac{1}{2}\int_0^3\left(x^{-\frac{1}{2}}+x^{\frac{1}{2}}\right)dx = \frac{1}{2}\left[2\sqrt{x}+\frac{2}{3}x^{\frac{3}{2}}\right]_0^3 = 2\sqrt{3}\ (=3.46)$ | M1A1, M1A1 | 6 marks |
| $S = 2\pi\int_0^3\left(x^{\frac{1}{2}}-\frac{1}{3}x^{\frac{3}{2}}\right)\left(\frac{1}{2}x^{-\frac{1}{2}}+\frac{1}{2}x^{\frac{1}{2}}\right)dx$ | M1 | Surface area formula |
| $= \pi\int_0^3\left(1+\frac{2}{3}x-\frac{1}{3}x^2\right)dx = \left[x+\frac{1}{3}x^2-\frac{1}{9}x^3\right]_0^3 = 3\pi$ | M1A1, A1 | 4 marks, total [14] |

---

# Question 11 (OR):

## Eigenvalues and Eigenvectors

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\det(\mathbf{A}-\lambda\mathbf{I})=0 \Rightarrow \lambda^3-6\lambda^2+11\lambda-6=0$ | M1A1 | Forms characteristic equation |
| $\Rightarrow \lambda=1,2,3$ | A1A1 | Solves |
| $\mathbf{e}_1=-2\mathbf{j}+\mathbf{k},\quad \mathbf{e}_2=\mathbf{i}+\mathbf{j},\quad \mathbf{e}_3=2\mathbf{i}+2\mathbf{j}+\mathbf{k}$ | M1A1, A1 | 7 marks, eigenvectors |
| $\mathbf{r}=s\mathbf{e}+t\mathbf{f}$; $\mathbf{A}(s\mathbf{e}+t\mathbf{f})=s\mathbf{A}\mathbf{e}+t\mathbf{A}\mathbf{f}=(s\lambda)\mathbf{e}+(t\mu)\mathbf{f}$ | B1, M1A1 | States equation of plane, 3 marks |
| $\mathbf{e}_1\times\mathbf{e}_2=-\mathbf{i}+\mathbf{j}+2\mathbf{k}\Rightarrow x-y-2z=0$ | M1A1 | |
| $\mathbf{e}_1\times\mathbf{e}_3=2\mathbf{i}-\mathbf{j}-2\mathbf{k}\Rightarrow 2x-y-2z=0$ | A1 | |
| $\mathbf{e}_2\times\mathbf{e}_3=\mathbf{i}-\mathbf{j}\Rightarrow x-y=0$ | A1 | 4 marks, total [14] |

Show LaTeX source

The curve $C$ has equation $y = \frac { 1 } { 3 } x ^ { \frac { 1 } { 2 } } ( 3 - x )$, for $0 \leqslant x \leqslant 3$. Find the mean value of $y$ with respect to $x$ over the interval $0 \leqslant x \leqslant 3$.

Show that

$$\frac { \mathrm { d } s } { \mathrm {~d} x } = \frac { 1 } { 2 } \left( x ^ { - \frac { 1 } { 2 } } + x ^ { \frac { 1 } { 2 } } \right)$$

where $s$ denotes arc length, and find the arc length of $C$.

Find the area of the surface generated when $C$ is rotated through $2 \pi$ radians about the $x$-axis.

\hfill \mbox{\textit{CAIE FP1 2011 Q11 EITHER}}

This paper (12 questions)

View full paper

Q1 6 Q2 6 Q3 7 Q4 7 Q5 7 Q6 8 Q7 9 Q8 10 Q9 13 Q10 13 Q11 EITHER Q11 OR