Pearson Edexcel International Advanced Level

Recall: $\displaystyle \sinh x = \frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2},\qquad \cosh x = \frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}$

$\displaystyle \sinh 2x + \cosh 2x = \frac{\mathrm{e}^{2x} - \mathrm{e}^{-2x}}{2} + \frac{\mathrm{e}^{2x} + \mathrm{e}^{-2x}}{2} = \mathrm{e}^{2x}$

$\displaystyle (\cosh x + \sinh x)^2 = \left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2} + \frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right)^2 = (\mathrm{e}^x)^2 = \mathrm{e}^{2x}$

LHS = RHS, identity proved.

From (a): $\cosh 2x + \sinh 2x = \mathrm{e}^{2x}$.

$\mathrm{e}^{2x} = 5$.

Take $\ln$: $2x = \ln 5$.

$I_n = \int_0^{\pi/2} \sin^{n-1}x \cdot \sin x \,\mathrm{d}x$.

IBP: $u = \sin^{n-1}x$, $\mathrm{d}v = \sin x\,\mathrm{d}x$.

$\mathrm{d}u = (n-1)\sin^{n-2}x\cos x\,\mathrm{d}x$, $v = -\cos x$.

$I_n = [-\sin^{n-1}x\cos x]_0^{\pi/2} + (n-1)\int_0^{\pi/2}\sin^{n-2}x\cos^2 x\,\mathrm{d}x$

$= 0 + (n-1)\int_0^{\pi/2}\sin^{n-2}x(1-\sin^2 x)\,\mathrm{d}x$

$= (n-1)(I_{n-2} - I_n)$

$I_n = (n-1)I_{n-2} - (n-1)I_n \;\Rightarrow\; nI_n = (n-1)I_{n-2}$.

$I_8 = \frac{7}{8}I_6 = \frac{7}{8} \cdot \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot I_0$

$I_0 = \int_0^{\pi/2} 1 \,\mathrm{d}x = \frac{\pi}{2}$

$I_8 = \frac{7 \cdot 5 \cdot 3 \cdot 1}{8 \cdot 6 \cdot 4 \cdot 2} \cdot \frac{\pi}{2} = \frac{105}{384} \cdot \frac{\pi}{2}$

Expand along row 3: $\det(\mathbf{M}) = 1 \cdot \begin{vmatrix} k & 2 \\ 2 & k \end{vmatrix}$

$= k \cdot k - 2 \cdot 2 = k^2 - 4$.

Singular $\Rightarrow \det = 0$: $k^2 - 4 = 0 \;\Rightarrow\; k = \pm 2$.

$\mathbf{M} = \begin{pmatrix} 5 & 2 & 0 \\ 2 & 5 & 0 \\ 0 & 0 & 1 \end{pmatrix}$.

$|\mathbf{M} - \lambda\mathbf{I}| = \begin{vmatrix} 5-\lambda & 2 & 0 \\ 2 & 5-\lambda & 0 \\ 0 & 0 & 1-\lambda \end{vmatrix}$

$= (1-\lambda)[(5-\lambda)^2 - 4] = (1-\lambda)(\lambda^2 - 10\lambda + 21)$

$= (1-\lambda)(\lambda-3)(\lambda-7) = 0$. Eigenvalues: $1, 3, 7$.

$\lambda = 1$: $(\mathbf{M}-\mathbf{I})\mathbf{v}=0 \;\Rightarrow\; \begin{pmatrix}4&2&0\\2&4&0\\0&0&0\end{pmatrix}\mathbf{v}=0$. $v_1=v_2=0$, $v_3$ free. $\mathbf{v} = \begin{pmatrix}0\\0\\1\end{pmatrix}$.

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

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

Let $y = \operatorname{arcosh}\,x$. Then $x = \cosh y = \frac{\mathrm{e}^y + \mathrm{e}^{-y}}{2}$.

$2x = \mathrm{e}^y + \mathrm{e}^{-y}$. Multiply by $\mathrm{e}^y$: $\mathrm{e}^{2y} - 2x\mathrm{e}^y + 1 = 0$.

Quadratic in $\mathrm{e}^y$: $\mathrm{e}^y = \frac{2x \pm \sqrt{4x^2 - 4}}{2} = x \pm \sqrt{x^2 - 1}$.

Since $x \ge 1$, $y \ge 0$ so $\mathrm{e}^y \ge 1$. $x + \sqrt{x^2 - 1} \ge 1$, $x - \sqrt{x^2 - 1} \le 1$.

Take $+$: $\mathrm{e}^y = x + \sqrt{x^2 - 1}$. Hence $y = \ln(x + \sqrt{x^2 - 1})$.

$4\cdot\frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2} + 3\cdot\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2} = 5$

$2(\mathrm{e}^x - \mathrm{e}^{-x}) + \frac{3}{2}(\mathrm{e}^x + \mathrm{e}^{-x}) = 5$

$\frac{7}{2}\mathrm{e}^x - \frac{1}{2}\mathrm{e}^{-x} = 5$

Multiply by $2\mathrm{e}^x$: $7\mathrm{e}^{2x} - 10\mathrm{e}^x - 1 = 0$.

Let $t = \mathrm{e}^x > 0$: $7t^2 - 10t - 1 = 0$.

$t = \frac{10 \pm \sqrt{100 + 28}}{14} = \frac{10 \pm 8\sqrt{2}}{14} = \frac{5 \pm 4\sqrt{2}}{7}$.

$\frac{5 - 4\sqrt{2}}{7} < 0$ (reject). $t = \frac{5 + 4\sqrt{2}}{7}$. $x = \ln t$.

$\frac{\mathrm{d}y}{\mathrm{d}x} = x^{1/2}$.

$L = \int_0^3 \sqrt{1 + (\mathrm{d}y/\mathrm{d}x)^2}\,\mathrm{d}x = \int_0^3 \sqrt{1 + x}\,\mathrm{d}x$.

$= \bigl[\frac{2}{3}(1+x)^{3/2}\bigr]_0^3 = \frac{2}{3}(4^{3/2} - 1^{3/2}) = \frac{2}{3}(8-1)$.

$S = 2\pi\int_0^3 x\sqrt{1 + (\mathrm{d}y/\mathrm{d}x)^2}\,\mathrm{d}x = 2\pi\int_0^3 x\sqrt{1+x}\,\mathrm{d}x$.

Let $u = 1+x$, $\mathrm{d}u = \mathrm{d}x$, $x = u-1$. Limits: $1 \to 4$.

$S = 2\pi\int_1^4 (u-1)u^{1/2}\,\mathrm{d}u = 2\pi\int_1^4 (u^{3/2} - u^{1/2})\,\mathrm{d}u$.

$= 2\pi\bigl[\frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2}\bigr]_1^4$

$= 2\pi\bigl[(\frac{64}{5} - \frac{16}{3}) - (\frac{2}{5} - \frac{2}{3})\bigr] = 2\pi(\frac{62}{5} - \frac{14}{3}) = 2\pi \cdot \frac{116}{15}$.

$a^2=25$, $b^2=9$, so $a=5$, $b=3$ (major axis along $x$).

$e = \sqrt{1 - b^2/a^2} = \sqrt{1 - 9/25} = \sqrt{16/25} = 4/5$.

Foci: $(\pm ae, 0) = (\pm 4, 0)$.

Implicit diff: $\frac{2x}{25} + \frac{2y}{9}\frac{\mathrm{d}y}{\mathrm{d}x} = 0 \;\Rightarrow\; \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{9x}{25y}$.

At $(4, \frac{9}{5})$: $\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{9\cdot 4}{25\cdot 9/5} = -\frac{36}{45} = -\frac{4}{5}$.

Tangent: $y - \frac{9}{5} = -\frac{4}{5}(x - 4)$. Multiply by 5: $5y - 9 = -4x + 16$. Rearrange.

Normal gradient $= 5/4$. Normal: $y - \frac{9}{5} = \frac{5}{4}(x - 4)$.

At $y=0$: $-\frac{9}{5} = \frac{5}{4}(x - 4) \;\Rightarrow\; x - 4 = -\frac{36}{25} \;\Rightarrow\; x = \frac{64}{25}$.

$N(\frac{64}{25}, 0)$. $F(4, 0)$. $FN = 4 - \frac{64}{25} = \frac{36}{25}$.

$L$: $\mathbf{r} = (1+2t, 2+t, 3-t)$. $\Pi$: $x+2y+2z=13$.

$(1+2t) + 2(2+t) + 2(3-t) = 13$.

$1+2t+4+2t+6-2t = 13 \;\Rightarrow\; 2t+11=13 \;\Rightarrow\; t=1$.

$P = (1+2, 2+1, 3-1)$.

$\mathbf{d} = (2, 1, -1)$, $\mathbf{n} = (1, 2, 2)$.

$\sin\theta = \frac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|} = \frac{|2+2-2|}{\sqrt{6}\cdot 3} = \frac{2}{3\sqrt{6}}$.

$\theta = \arcsin(\frac{2}{3\sqrt{6}}) \approx 15.793^{\circ}$.

Distance $= \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2+b^2+c^2}} = \frac{|5+0+2-13|}{\sqrt{1+4+4}} = \frac{6}{3}$.

$a^2=16$, $b^2=9$, so $a=4$, $b=3$.

$e = \sqrt{1 + b^2/a^2} = \sqrt{1 + 9/16} = \sqrt{25/16} = 5/4$.

Directrices: $x = \pm a/e = \pm 4/(5/4) = \pm 16/5$.

Implicit diff: $\frac{2x}{16} - \frac{2y}{9}\frac{\mathrm{d}y}{\mathrm{d}x} = 0 \;\Rightarrow\; \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{9x}{16y}$.

At $P$: $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{9\cdot 4\sec t}{16\cdot 3\tan t} = \frac{3\sec t}{4\tan t} = \frac{3}{4\sin t}$.

Tangent: $y - 3\tan t = \frac{3}{4\sin t}(x - 4\sec t)$.

Multiply by $4\sin t$, use $\sin t\tan t = \sec t - \cos t$, simplify to obtain $\frac{x\sec t}{4} - \frac{y\tan t}{3} = 1$.

Asymptotes: $y = \pm\frac{3}{4}x$.

With $y = \frac{3}{4}x$: $\frac{x\sec t}{4} - \frac{x\tan t}{4} = 1 \;\Rightarrow\; x_A = \frac{4}{\sec t - \tan t}$, $y_A = \frac{3}{\sec t - \tan t}$.

With $y = -\frac{3}{4}x$: $x_B = \frac{4}{\sec t + \tan t}$, $y_B = -\frac{3}{\sec t + \tan t}$.

Area $= \frac{1}{2}|x_A y_B - x_B y_A| = \frac{1}{2}\left|\frac{-12}{(\sec t - \tan t)(\sec t + \tan t)} - \frac{12}{(\sec t + \tan t)(\sec t - \tan t)}\right|$.

$= \frac{1}{2}|-24| = 12$, since $\sec^2 t - \tan^2 t = 1$.

$|\mathbf{A} - \lambda\mathbf{I}| = \begin{vmatrix}3-\lambda & 1\\1 & 3-\lambda\end{vmatrix} = (3-\lambda)^2 - 1 = \lambda^2 - 6\lambda + 8$.

$(\lambda-2)(\lambda-4) = 0 \;\Rightarrow\; \lambda = 2, 4$.

$\lambda = 2$: $(\mathbf{A}-2\mathbf{I})\mathbf{v} = \begin{pmatrix}1&1\\1&1\end{pmatrix}\mathbf{v}=0 \;\Rightarrow\; v_1+v_2=0$. $\mathbf{v}_1 = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\end{pmatrix}$.

$\lambda = 4$: $(\mathbf{A}-4\mathbf{I})\mathbf{v} = \begin{pmatrix}-1&1\\1&-1\end{pmatrix}\mathbf{v}=0 \;\Rightarrow\; v_1=v_2$. $\mathbf{v}_2 = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix}$.

$\mathbf{P} = \frac{1}{\sqrt{2}}\begin{pmatrix}1 & 1\\-1 & 1\end{pmatrix}$ (columns = orthonormal eigenvectors).

$\mathbf{P}^{\mathrm{T}}\mathbf{P} = \mathbf{I}$ (verified).

$\mathbf{P}^{\mathrm{T}}\!\mathbf{A}\mathbf{P} = \begin{pmatrix}2 & 0\\0 & 4\end{pmatrix}$.