Pearson Edexcel International Advanced Level

(a)   [2 marks]
$\sinh x = \frac{1}{2}(\mathrm{e}^x - \mathrm{e}^{-x})$, $\cosh x = \frac{1}{2}(\mathrm{e}^x + \mathrm{e}^{-x})$ B1
$(\cosh x + \sinh x)^2 = (\mathrm{e}^x)^2 = \mathrm{e}^{2x}$ M1
$\sinh 2x + \cosh 2x = \frac{1}{2}(\mathrm{e}^{2x} - \mathrm{e}^{-2x}) + \frac{1}{2}(\mathrm{e}^{2x} + \mathrm{e}^{-2x}) = \mathrm{e}^{2x}$
Hence $\sinh 2x + \cosh 2x \equiv (\cosh x + \sinh x)^2$ A1

(b)   [4 marks]
From (a): $\cosh 2x + \sinh 2x = \mathrm{e}^{2x}$ M1
$\mathrm{e}^{2x} = 5$ M1
$2x = \ln 5$ M1
$x = \frac{1}{2}\ln 5$ A1

(a)   [4 marks]
$I_n = \int_0^{\pi/2} \sin^{n-1}x \cdot \sin x \,\mathrm{d}x$ M1
Parts: $u = \sin^{n-1}x$, $\mathrm{d}v = \sin x\,\mathrm{d}x$
$\mathrm{d}u = (n-1)\sin^{n-2}x\cos x\,\mathrm{d}x$, $v = -\cos x$ M1
$I_n = [-\sin^{n-1}x\cos x]_0^{\pi/2} + (n-1)\int_0^{\pi/2}\sin^{n-2}x\cos^2 x\,\mathrm{d}x$
    $= 0 + (n-1)\int_0^{\pi/2}\sin^{n-2}x(1-\sin^2 x)\,\mathrm{d}x$ M1
    $= (n-1)(I_{n-2} - I_n)$
$I_n = (n-1)I_{n-2} - (n-1)I_n \;\Rightarrow\; nI_n = (n-1)I_{n-2}$ A1

(b)   [3 marks]
$I_8 = \frac{7}{8}I_6 = \frac{7}{8}\cdot\frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}\cdot I_0$ M1
$I_0 = \int_0^{\pi/2}1\,\mathrm{d}x = \frac{\pi}{2}$ B1
$I_8 = \frac{7\cdot5\cdot3\cdot1}{8\cdot6\cdot4\cdot2} \cdot \frac{\pi}{2} = \frac{35\pi}{256}$ A1

(a)   [3 marks]
$\det(\mathbf{M}) = k\begin{vmatrix}k & 0\\0 & 1\end{vmatrix} - 2\begin{vmatrix}2 & 0\\0 & 1\end{vmatrix} + 0$ M1
$= k(k\cdot 1 - 0) - 2(2\cdot 1 - 0)$ M1
$= k^2 - 4$ A1

(b)   [2 marks]
Singular when $\det(\mathbf{M}) = 0$: $k^2 - 4 = 0$ M1
$k = \pm 2$ A1

(c)   [5 marks]
For $k = 5$: $\mathbf{M} = \begin{pmatrix}5&2&0\\2&5&0\\0&0&1\end{pmatrix}$
Characteristic: $|\mathbf{M} - \lambda\mathbf{I}| = \begin{vmatrix}5-\lambda&2&0\\2&5-\lambda&0\\0&0&1-\lambda\end{vmatrix}$ M1
$= (1-\lambda)[(5-\lambda)^2 - 4] = (1-\lambda)(\lambda^2 - 10\lambda + 21)$ M1
$= (1-\lambda)(\lambda-3)(\lambda-7) = 0$
Eigenvalues: $\lambda = 1, 3, 7$ A1

$\lambda = 1$: $(\mathbf{M}-\mathbf{I})\mathbf{v}=0 \Rightarrow v_3 = 1$, $4v_1+2v_2=0$, $2v_1+4v_2=0 \Rightarrow v_1=v_2=0$
Eigenvector: $\begin{pmatrix}0\\0\\1\end{pmatrix}$ A1
$\lambda = 3$: $(\mathbf{M}-3\mathbf{I})\mathbf{v}=0 \Rightarrow \begin{pmatrix}2&2&0\\2&2&0\\0&0&-2\end{pmatrix}\mathbf{v}=0$
$v_1+v_2=0$, $v_3=0$. Eigenvector: $\begin{pmatrix}1\\-1\\0\end{pmatrix}$ A1
$\lambda = 7$: $(\mathbf{M}-7\mathbf{I})\mathbf{v}=0 \Rightarrow \begin{pmatrix}-2&2&0\\2&-2&0\\0&0&-6\end{pmatrix}\mathbf{v}=0$
$v_1=v_2$, $v_3=0$. Eigenvector: $\begin{pmatrix}1\\1\\0\end{pmatrix}$ A1 (final)

(a)   [3 marks]
Let $y = \operatorname{arcosh}\,x$ so that $x = \cosh y = \frac{\mathrm{e}^y + \mathrm{e}^{-y}}{2}$ M1
$2x = \mathrm{e}^y + \mathrm{e}^{-y} \;\Rightarrow\; \mathrm{e}^{2y} - 2x\mathrm{e}^y + 1 = 0$ M1
$\mathrm{e}^y = x \pm \sqrt{x^2 - 1}$. Since $y \ge 0$ for $x \ge 1$, take $+$: $\mathrm{e}^y = x + \sqrt{x^2 - 1}$
$\operatorname{arcosh}\,x = y = \ln(x + \sqrt{x^2 - 1})$ A1

(b)   [5 marks]
$\sinh x = \frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}$, $\cosh x = \frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}$ M1
$4\cdot\frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2} + 3\cdot\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2} = 5$
$2\mathrm{e}^x - 2\mathrm{e}^{-x} + \frac{3}{2}\mathrm{e}^x + \frac{3}{2}\mathrm{e}^{-x} = 5$ M1
$\frac{7}{2}\mathrm{e}^x - \frac{1}{2}\mathrm{e}^{-x} = 5$
Multiply by $2\mathrm{e}^x$: $7\mathrm{e}^{2x} - 10\mathrm{e}^x - 1 = 0$ M1
Let $t = \mathrm{e}^x > 0$: $7t^2 - 10t - 1 = 0$ M1
$t = \frac{10 + \sqrt{100 + 28}}{14} = \frac{10 + \sqrt{128}}{14} = \frac{10 + 8\sqrt{2}}{14} = \frac{5 + 4\sqrt{2}}{7}$
$x = \ln\!\bigl(\frac{5 + 4\sqrt{2}}{7}\bigr)$ A1

(a)   [5 marks]
$y = \frac{2}{3}x^{3/2}$, $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = x^{1/2}$ B1
Arc length $L = \displaystyle \int_0^3\!\sqrt{1 + \bigl(\frac{\mathrm{d}y}{\mathrm{d}x}\bigr)^2}\,\mathrm{d}x$ M1
$= \displaystyle \int_0^3\!\sqrt{1 + x}\,\mathrm{d}x$ M1
$= \displaystyle \Bigl[\frac{2}{3}(1+x)^{3/2}\Bigr]_0^3$ M1
$= \frac{2}{3}(8 - 1) = \frac{14}{3}$ A1

(b)   [4 marks]
Surface area $S = 2\pi\!\displaystyle \int_0^3\! x\sqrt{1 + \bigl(\frac{\mathrm{d}y}{\mathrm{d}x}\bigr)^2}\,\mathrm{d}x$ (rotation about $y$-axis) M1
$= 2\pi\!\displaystyle \int_0^3\! x\sqrt{1 + x}\,\mathrm{d}x$ M1
Substitute $u = 1 + x$, $\mathrm{d}u = \mathrm{d}x$, $x = u-1$, limits: $1 \to 4$
$S = 2\pi\!\displaystyle \int_1^4 (u-1)u^{1/2}\,\mathrm{d}u = 2\pi\!\displaystyle \int_1^4 (u^{3/2} - u^{1/2})\,\mathrm{d}u$
$= 2\pi\Bigl[\frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2}\Bigr]_1^4$ M1
$= 2\pi\bigl[(\frac{64}{5} - \frac{16}{3}) - (\frac{2}{5} - \frac{2}{3})\bigr] = 2\pi\bigl(\frac{62}{5} - \frac{14}{3}\bigr) = 2\pi\cdot\frac{116}{15} = \frac{232\pi}{15}$ A1

(a)   [3 marks]
$a^2 = 25$, $b^2 = 9$, so $a = 5$, $b = 3$ B1
$e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}$ M1
Foci: $(\pm ae,\,0) = (\pm 4,\,0)$ A1

(b)   [3 marks]
$\frac{2x}{25} + \frac{2y}{9}\frac{\mathrm{d}y}{\mathrm{d}x} = 0 \;\Rightarrow\; \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{9x}{25y}$ M1
At $(4,\,\frac{9}{5})$: $\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{9\cdot 4}{25\cdot 9/5} = -\frac{4}{5}$ M1
Tangent: $y - \frac{9}{5} = -\frac{4}{5}(x - 4) \;\Rightarrow\; 4x + 5y = 25$ A1

(c)   [3 marks]
Normal gradient $= \frac{5}{4}$ M1
Normal: $y - \frac{9}{5} = \frac{5}{4}(x - 4)$
At $y = 0$: $-\frac{9}{5} = \frac{5}{4}(x - 4)$ M1
$x - 4 = -\frac{36}{25} \;\Rightarrow\; x = \frac{64}{25}$. So $N(\frac{64}{25},\,0)$
$F = (4,\,0)$. $FN = 4 - \frac{64}{25} = \frac{36}{25}$ A1

(a)   [3 marks]
For $\mathbf{r} = (1+2t,\,2+t,\,3-t)$, substitute into $\Pi$: M1
$(1+2t) + 2(2+t) + 2(3-t) = 13$ M1
$1+2t+4+2t+6-2t = 13 \;\Rightarrow\; 2t + 11 = 13 \;\Rightarrow\; t = 1$
$P = (3,\,3,\,2)$ A1

(b)   [3 marks]
Direction $\mathbf{d} = (2,\,1,\,-1)$, normal $\mathbf{n} = (1,\,2,\,2)$ M1
$\sin\theta = \frac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|} = \frac{|2+2-2|}{\sqrt{6}\cdot 3} = \frac{2}{3\sqrt{6}}$ M1
$\theta = \arcsin\bigl(\frac{2}{3\sqrt{6}}\bigr) = 15.8^{\circ}$ (to nearest $0.1^{\circ}$) A1

(c)   [3 marks]
Distance $= \frac{|(5,\,0,\,1)\cdot(1,\,2,\,2) - 13|}{\sqrt{1^2+2^2+2^2}}$ M1
$= \frac{|5+0+2-13|}{3}$ M1
$= \frac{6}{3} = 2$ A1

(a)   [3 marks]
$a^2 = 16$, $b^2 = 9$, so $a = 4$, $b = 3$ B1
$e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9}{16}} = \frac{5}{4}$ M1
Directrices: $x = \pm\frac{a}{e} = \pm\frac{16}{5}$ A1

(b)   [4 marks]
$\frac{x^2}{16} - \frac{y^2}{9} = 1 \;\Rightarrow\; \frac{x}{8} - \frac{2y}{9}\frac{\mathrm{d}y}{\mathrm{d}x} = 0$
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{9x}{16y}$ M1
At $P(4\sec t,\,3\tan t)$: $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{9\cdot 4\sec t}{16\cdot 3\tan t} = \frac{3\sec t}{4\tan t} = \frac{3}{4\sin t}$ M1
Tangent: $y - 3\tan t = \frac{3}{4\sin t}(x - 4\sec t)$ M1
Rearranging: $\frac{x\sec t}{4} - \frac{y\tan t}{3} = 1$ A1

(c)   [3 marks]
Asymptotes: $y = \pm\frac{3}{4}x$ B1
Intersect with $y = \frac{3}{4}x$: $\frac{x\sec t}{4} - \frac{x\tan t}{4} = 1 \;\Rightarrow\; x(\sec t - \tan t) = 4$
$A:\; x = \frac{4}{\sec t - \tan t}$, $y = \frac{3x}{4}$ M1
Intersect with $y = -\frac{3}{4}x$: $x(\sec t + \tan t) = 4$
$B:\; x = \frac{4}{\sec t + \tan t}$, $y = -\frac{3x}{4}$
Area $= \frac{1}{2}|x_A y_B - x_B y_A| = \frac{16}{\sec^2 t - \tan^2 t} \cdot \frac{3}{4} \cdot \frac{1}{2} = 12$
(since $\sec^2 t - \tan^2 t = 1$) A1

(a)   [4 marks]
$|\mathbf{A} - \lambda\mathbf{I}| = \begin{vmatrix}3-\lambda & 1\\1 & 3-\lambda\end{vmatrix} = (3-\lambda)^2 - 1$ M1
$= \lambda^2 - 6\lambda + 8 = (\lambda - 2)(\lambda - 4) = 0$ M1
Eigenvalues: $\lambda = 2,\,4$ A1

$\lambda = 2$: $(\mathbf{A}-2\mathbf{I})\mathbf{v}=0 \;\Rightarrow\; \begin{pmatrix}1&1\\1&1\end{pmatrix}\mathbf{v}=0$, $v_1+v_2=0$
Eigenvector $\begin{pmatrix}1\\-1\end{pmatrix}$, normalised $\frac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\end{pmatrix}$
$\lambda = 4$: $(\mathbf{A}-4\mathbf{I})\mathbf{v}=0 \;\Rightarrow\; \begin{pmatrix}-1&1\\1&-1\end{pmatrix}\mathbf{v}=0$, $v_1=v_2$
Eigenvector $\begin{pmatrix}1\\1\end{pmatrix}$, normalised $\frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix}$ A1

(b)   [3 marks]
$\mathbf{P} = \frac{1}{\sqrt{2}}\begin{pmatrix}1 & 1\\-1 & 1\end{pmatrix}$ (or its transpose) M1
Check $\mathbf{P}^{\mathrm{T}}\mathbf{P} = \mathbf{I}$ (orthogonal) M1
$\mathbf{P}^{\mathrm{T}}\!\mathbf{A}\mathbf{P} = \begin{pmatrix}2 & 0\\0 & 4\end{pmatrix}$ A1