| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Find eigenvalues/vectors of matrix combination |
| Difficulty | Standard +0.3 This is a standard Further Maths question on eigenvalues requiring: (1) a straightforward proof that if Ae=λe then A²e=λ²e (routine algebraic manipulation), (2) finding eigenvalues of a 3×3 matrix via characteristic equation, and (3) applying the result that if λ is an eigenvalue of B, then λ⁴+2λ²+3 is an eigenvalue of B⁴+2B²+3I. All steps are textbook procedures with no novel insight required, though the matrix combination adds slight complexity beyond basic eigenvalue problems. |
| Spec | 4.03a Matrix language: terminology and notation4.03b Matrix operations: addition, multiplication, scalar4.03c Matrix multiplication: properties (associative, not commutative)4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04d Angles: between planes and between line and plane |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = -1\) | B1 | |
| \(y = px + 4 - p + (p-3)(x+1)^{-1} \Rightarrow y = px + 4 - p\) | M1A1 | Part mark: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(p = 4 \Rightarrow x\)-axis is a tangent | B1 | |
| Correct location of turning points and asymptotes | B1 | |
| Each branch correct | B1B1 | Part mark: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| \(p=1 \Rightarrow y = x+3-2(x+1)^{-1} \Rightarrow y' = 1+2(x+1)^{-2}\ (\geq 1)\) | M1A1 | |
| Intersections on \(x\)-axis at \(\left(-2\pm\sqrt{3},\ 0\right)\) | B1 | |
| Each branch correct | B1B1 | Part mark: 5, Total: [12] |
| Answer | Marks | Guidance |
|---|---|---|
| \(z = \cos\dfrac{2k\pi}{5} + i\sin\dfrac{2k\pi}{5},\quad k=0,\pm1,\pm2\) | B1B1 | Part mark: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(x^2 - 2\cos\dfrac{2\pi}{5}x + 1\) | M1A1 | Part mark: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(x^2 - 2\cos\dfrac{2\pi}{5}+1\right)\left(x^2 - 2\cos\dfrac{4\pi}{5}+1\right)(x-1)\) | M1A1 | Part mark: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(x^3 = \dfrac{1}{2} \pm i\dfrac{\sqrt{3}}{2} = \cos\dfrac{\pi}{3} \pm i\sin\dfrac{\pi}{3}\) | M1A1, A1 | |
| or \(\cos\dfrac{7\pi}{3} \pm i\sin\dfrac{7\pi}{3}\) or \(\cos\dfrac{13\pi}{3} \pm i\sin\dfrac{13\pi}{3}\) | A1 | |
| \(x = \cos\dfrac{\pi}{9} \pm i\sin\dfrac{\pi}{9},\ \cos\dfrac{7\pi}{9} \pm i\sin\dfrac{7\pi}{9},\ \cos\dfrac{13\pi}{9} \pm i\sin\dfrac{13\pi}{9}\) | M1A1 | Part mark: 6 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(x^2 - 2\cos\dfrac{\pi}{9}x+1\right)\left(x^2 - 2\cos\dfrac{7\pi}{9}x+1\right)\left(x^2 - 2\cos\dfrac{13\pi}{9}x+1\right)\) (ACF) | M1A1 | Part mark: 2, Total: [14] |
| Answer | Marks | Guidance |
|---|---|---|
| \(v = y^3 \Rightarrow v' = 3y^2\dfrac{dy}{dx} \Rightarrow v'' = 6y\left(\dfrac{dy}{dx}\right)^2 + 3y^2\dfrac{d^2y}{dx^2}\) | B1B1 | |
| \(\dfrac{1}{3}\dfrac{d^2v}{dx^2} - 2\dfrac{dv}{dx} + 3v = 25e^{-2x}\) | M1 | |
| \(\Rightarrow \dfrac{d^2v}{dx^2} - 6\dfrac{dv}{dx} + 9v = 75e^{-2x}\) (AG) | A1 | Part mark: 4 |
| Answer | Marks |
|---|---|
| \(v = Ae^{3x} + Bxe^{3x}\) | M1, A1 |
| Answer | Marks |
|---|---|
| \(v = Ae^{3x} + Bxe^{3x} + 3e^{-2x}\) | M1, A1, A1 |
| Answer | Marks |
|---|---|
| \(x=0,\ y=2,\ v=8 \Rightarrow 8 = A+3 \Rightarrow A=5\) | B1 |
| \(v' = 15e^{3x} + 3Bxe^{3x} + Be^{3x} - 6e^{-2x}\) | M1A1 |
| Answer | Marks |
|---|---|
| \(12 = 15 + B - 6 \Rightarrow B = 3\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = \left\{5e^{3x} + 3xe^{3x} + 3e^{-2x}\right\}^{\frac{1}{3}}\) | A1 | Part mark: 10, Total: [14] |
## Question 10:
**(i) Asymptotes:**
$x = -1$ | B1 |
$y = px + 4 - p + (p-3)(x+1)^{-1} \Rightarrow y = px + 4 - p$ | M1A1 | Part mark: 3
**(ii) Value of p, sketch:**
$p = 4 \Rightarrow x$-axis is a tangent | B1 |
Correct location of turning points and asymptotes | B1 |
Each branch correct | B1B1 | Part mark: 4
**(iii) Proof and sketch:**
$p=1 \Rightarrow y = x+3-2(x+1)^{-1} \Rightarrow y' = 1+2(x+1)^{-2}\ (\geq 1)$ | M1A1 |
Intersections on $x$-axis at $\left(-2\pm\sqrt{3},\ 0\right)$ | B1 |
Each branch correct | B1B1 | Part mark: 5, Total: **[12]**
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## Question 11E:
**Fifth roots:**
$z = \cos\dfrac{2k\pi}{5} + i\sin\dfrac{2k\pi}{5},\quad k=0,\pm1,\pm2$ | B1B1 | Part mark: 2
**Simplification:**
$x^2 - 2\cos\dfrac{2\pi}{5}x + 1$ | M1A1 | Part mark: 2
**Factors:**
$\left(x^2 - 2\cos\dfrac{2\pi}{5}+1\right)\left(x^2 - 2\cos\dfrac{4\pi}{5}+1\right)(x-1)$ | M1A1 | Part mark: 2
**Solving quadratic in $x^3$:**
$x^3 = \dfrac{1}{2} \pm i\dfrac{\sqrt{3}}{2} = \cos\dfrac{\pi}{3} \pm i\sin\dfrac{\pi}{3}$ | M1A1, A1 |
or $\cos\dfrac{7\pi}{3} \pm i\sin\dfrac{7\pi}{3}$ or $\cos\dfrac{13\pi}{3} \pm i\sin\dfrac{13\pi}{3}$ | A1 |
$x = \cos\dfrac{\pi}{9} \pm i\sin\dfrac{\pi}{9},\ \cos\dfrac{7\pi}{9} \pm i\sin\dfrac{7\pi}{9},\ \cos\dfrac{13\pi}{9} \pm i\sin\dfrac{13\pi}{9}$ | M1A1 | Part mark: 6
**Final factors:**
$\left(x^2 - 2\cos\dfrac{\pi}{9}x+1\right)\left(x^2 - 2\cos\dfrac{7\pi}{9}x+1\right)\left(x^2 - 2\cos\dfrac{13\pi}{9}x+1\right)$ (ACF) | M1A1 | Part mark: 2, Total: **[14]**
---
## Question 11O:
**Substitution to obtain v–x equation:**
$v = y^3 \Rightarrow v' = 3y^2\dfrac{dy}{dx} \Rightarrow v'' = 6y\left(\dfrac{dy}{dx}\right)^2 + 3y^2\dfrac{d^2y}{dx^2}$ | B1B1 |
$\dfrac{1}{3}\dfrac{d^2v}{dx^2} - 2\dfrac{dv}{dx} + 3v = 25e^{-2x}$ | M1 |
$\Rightarrow \dfrac{d^2v}{dx^2} - 6\dfrac{dv}{dx} + 9v = 75e^{-2x}$ (AG) | A1 | Part mark: 4
**Complementary function:**
$m^2 - 6m + 9 = 0 \Rightarrow m = 3$
$v = Ae^{3x} + Bxe^{3x}$ | M1, A1 |
**Particular integral:**
$v = ke^{-2x} \Rightarrow v' = -2ke^{-2x} \Rightarrow v'' = 4ke^{-2x}$
$4k + 12k + 9k = 75 \Rightarrow k = 3$
$v = Ae^{3x} + Bxe^{3x} + 3e^{-2x}$ | M1, A1, A1 |
**Using initial conditions:**
$x=0,\ y=2,\ v=8 \Rightarrow 8 = A+3 \Rightarrow A=5$ | B1 |
$v' = 15e^{3x} + 3Bxe^{3x} + Be^{3x} - 6e^{-2x}$ | M1A1 |
$x=0,\ y=2,\ y'=1 \Rightarrow v'=12$
$12 = 15 + B - 6 \Rightarrow B = 3$ | A1 |
**Explicit solution:**
$y^3 = v = 5e^{3x} + 3xe^{3x} + 3e^{-2x}$
$y = \left\{5e^{3x} + 3xe^{3x} + 3e^{-2x}\right\}^{\frac{1}{3}}$ | A1 | Part mark: 10, Total: **[14]**10 \\
22
\end{array} \right)$$
has the form
$$\mathbf { x } = \left( \begin{array} { r }
1 \\
- 2 \\
- 3 \\
- 4
\end{array} \right) + \lambda \mathbf { e } _ { 1 } + \mu \mathbf { e } _ { 2 } ,$$
where $\lambda$ and $\mu$ are real numbers and $\left\{ \mathbf { e } _ { 1 } , \mathbf { e } _ { 2 } \right\}$ is a basis for $K$.
7 The square matrix $\mathbf { A }$ has $\lambda$ as an eigenvalue with $\mathbf { e }$ as a corresponding eigenvector. Show that $\mathbf { e }$ is an eigenvector of $\mathbf { A } ^ { 2 }$ and state the corresponding eigenvalue.
Find the eigenvalues of the matrix $\mathbf { B }$, where
$$\mathbf { B } = \left( \begin{array} { l l l }
1 & 3 & 0 \\
2 & 0 & 2 \\
1 & 1 & 2
\end{array} \right)$$
Find the eigenvalues of $\mathbf { B } ^ { 4 } + 2 \mathbf { B } ^ { 2 } + 3 \mathbf { I }$, where $\mathbf { I }$ is the $3 \times 3$ identity matrix.
8 The plane $\Pi _ { 1 }$ has equation $\mathbf { r } = \left( \begin{array} { r } 2 \\ 3 \\ - 1 \end{array} \right) + s \left( \begin{array} { l } 1 \\ 0 \\ 1 \end{array} \right) + t \left( \begin{array} { r } 1 \\ - 1 \\ - 2 \end{array} \right)$. Find a cartesian equation of $\Pi _ { 1 }$.
The plane $\Pi _ { 2 }$ has equation $2 x - y + z = 10$. Find the acute angle between $\Pi _ { 1 }$ and $\Pi _ { 2 }$.
Find an equation of the line of intersection of $\Pi _ { 1 }$ and $\Pi _ { 2 }$, giving your answer in the form $\mathbf { r } = \mathbf { a } + \lambda \mathbf { b }$.
9 The curve $C$ has parametric equations
$$x = t ^ { 2 } , \quad y = t - \frac { 1 } { 3 } t ^ { 3 } , \quad \text { for } 0 \leqslant t \leqslant 1 .$$
Find the surface area generated when $C$ is rotated through $2 \pi$ radians about the $x$-axis.
Find the coordinates of the centroid of the region bounded by $C$, the $x$-axis and the line $x = 1$.
10 The curve $C$ has equation
$$y = \frac { p x ^ { 2 } + 4 x + 1 } { x + 1 } ,$$
where $p$ is a positive constant and $p \neq 3$.\\
(i) Obtain the equations of the asymptotes of $C$.\\
(ii) Find the value of $p$ for which the $x$-axis is a tangent to $C$, and sketch $C$ in this case.\\
(iii) For the case $p = 1$, show that $C$ has no turning points, and sketch $C$, giving the exact coordinates of the points of intersection of $C$ with the $x$-axis.
\hfill \mbox{\textit{CAIE FP1 2013 Q10 [12]}}