Additional Mathematics – Hong Kong Certificate of Education (HKCE) – Page 4

November 21, 2022November 21, 2022

202211211714 Solution to 1976-CE-AMATH-I-XX

Express in polar form the following complex numbers:

$z=-\sqrt{3}-\mathrm{i}$ and $w=1-\mathrm{i}$ .

Hence or otherwise express $\displaystyle{\frac{z^3}{w^4}}$ in the form of $a+\mathrm{i}b$ .

Roughwork.

$\begin{aligned} \mathrm{mod}(z) & = |z| \\ & = |-\sqrt{3}-\mathrm{i}|\\ & = \sqrt{(-\sqrt{3})^2+(-1)^2} \\ & = 2 \\ \mathrm{arg}(z) & = \arctan\bigg(\frac{-1}{-\sqrt{3}}\bigg) \\ & = \frac{4\pi}{3} \\ \mathrm{mod}(w) & = |w| \\ & = |1-\mathrm{i}| \\ & = \sqrt{(1)^2+(-1)^2} \\ & = \sqrt{2} \\ \mathrm{arg}(w) & = \arctan\bigg(\frac{-1}{1}\bigg) \\ & = \frac{3\pi}{4} \\ \end{aligned}$

$\begin{cases} z=2e^{\mathrm{i}\frac{4\pi}{3}} \\ w=\sqrt{2}e^{\mathrm{i}\frac{3\pi}{4}} \\ \end{cases}$

$\begin{aligned} \frac{z^3}{w^4} & = z^3w^{-4} \\ & = \big( 2e^{\mathrm{i}\frac{4\pi}{3}}\big)^3\big(\sqrt{2}e^{\mathrm{i}\frac{3\pi}{4}}\big)^{-4} \\ & = \dots \\ \end{aligned}$

This problem is not to be attempted.

November 21, 2022November 21, 2022

202211211635 Solution to 1976-CE-AMATH-I-XX

Let the complex number $z_1=-1+\mathrm{i}\sqrt{3}$ .

i. Represent $z_1$ and its conjugate $\overline{z_1}$ on the Argand diagram.
ii. Calculate the modulus and the argument of $z_1$ .
iii. If $z_2=(z_1)^2$ , express $z_2$ in the form of $a+\mathrm{i}b$ .
iv. Show that both $z_1$ and $\displaystyle{\frac{1}{2}z_2}$ are roots of the equation $z^3-8=0$ .

Roughwork.

$\begin{aligned} \textrm{modulus }\mathrm{mod}(z_1) & = |z_1| \\ & = |-1+\mathrm{i}\sqrt{3}| \\ & = \sqrt{(-1)^2+(\sqrt{3})^2} \\ & = 2 \\ \textrm{argument }\mathrm{arg}(z_1) & = \arctan\bigg(\frac{\sqrt{3}}{-1}\bigg) \\ & = \frac{2\pi}{3}\\ \end{aligned}$

Thus, $z_1=2e^{\mathrm{i}\frac{2\pi}{3}}$ .

As is given,

$\begin{aligned} z_2 & = (z_1)^2 \\ & = \big( 2e^{\mathrm{i}\frac{2\pi}{3}}\big)^2 \\ & = 4e^{\mathrm{i}\frac{4\pi}{3}} \\ \mathrm{mod}(z_2) & = 4 \\ \mathrm{arg}(z_2) & = \frac{4\pi}{3} \\ \end{aligned}$

or,

$\begin{aligned} z_2 & = (z_1)^2 \\ & = (-1+\mathrm{i}\sqrt{3})^2 \\ & = (-1)^2 +2(-1)(\mathrm{i}\sqrt{3}) + (\mathrm{i}\sqrt{3})^2 \\ & = 1 -\mathrm{i}(2\sqrt{3}) -3 \\ & = -2-\mathrm{i}(2\sqrt{3}) \\ \end{aligned}$

This problem is not to be attempted.

November 4, 2022November 4, 2022

202211041452 Solution to 1972-CE-AMATH-I-10

A point $P$ moves such that its distance from the point $(1,5)$ is equal to its distance from the line $y=x$ . Write down and simplify the equation of the locus of $P$ .

Roughwork.

Let the locus of $P$ be $t$ -parametrized as $(x_P(t),y_P(t))=(t,f(t))$ , and let the line $L:y=x$ be $s$ -parametrized as $(x_L(s),y_L(s))=(s,s)$ .

For the shortest line joining any point $P$ on the locus to its perpendicular projection on $L$ , its slope is:

$\textrm{Slope}_\alpha = \displaystyle{\frac{-1}{y_L'(s)} = \frac{-1}{1} = -1}$

with angle of inclination $\displaystyle{\alpha =\frac{3\pi}{4}}$ , and for the line joining any point $P$ on the locus to point $(1,5)$ , its slope is

$\textrm{Slope}_\beta =\displaystyle{\frac{y_P(t_0)-5}{x_P(t_0)-1}}$ ,

with angle of inclination $\displaystyle{\beta =\tan^{-1}\bigg(\frac{y_P-5}{x_P-1}\bigg)}$ , whereas the slope of tangent at any point $(x_P(t_0),y_P(t_0))$ on the locus is

$\textrm{Slope}_\gamma =f'(t_0)=\displaystyle{\frac{\mathrm{d}f(t)}{\mathrm{d}t}\bigg|_{t=t_0}}$ .

with angle of inclination $\gamma =\tan^{-1}(f'(t_0))$ .

Have a picture in mind

Recall

$\displaystyle{\tan (A\pm B)=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}}$ .

Hence,

$\begin{aligned} \tan (\gamma -\alpha ) & = \tan (\gamma -\beta ) \\ \frac{\tan\gamma -\tan\alpha}{1+\tan\gamma\tan\alpha} & = \frac{\tan\gamma -\tan\beta}{1+\tan\gamma\tan\beta} \\ \frac{(f'(t_0))-(-1)}{1+(f'(t_0))(-1)} & = \frac{(f'(t_0))-\bigg(\displaystyle{\frac{y(t_0)-5}{x(t_0)-1}}\bigg)}{1+(f'(t_0))\bigg(\displaystyle{\frac{y(t_0)-5}{x(t_0)-1}}\bigg)} \\ \end{aligned}$

The remaining are left the reader as an exercise.

November 4, 2022November 4, 2022

202211041443 Solution to 1978-CE-AMATH-I-1

Find the range of $x$ satisfying the inequality

$(x-2)(x+3)<2(x-2)$ .

Roughwork.

$\begin{aligned} (x-2)(x+3) & <2(x-2) \\ x^2+x-6 & < 2x-4 \\ x^2-x-2 & < 0 \\ (x-2)(x+1) & < 0 \\ -2<x & <1 \\ \end{aligned}$

$\therefore$ The range of $x$ is $x\in (-2,1)$ .

November 4, 2022November 4, 2022

202211041431 Solution to 1977-CE-AMATH-I-2

Find all the values of $x$ which satisfy

$(x+1)^2>16\textrm{ \scriptsize{OR} }2x+5>7$ .

Roughwork.

First,

$\begin{aligned} (x+1)^2 & >16 \\ x^2+2x+1 & >16 \\ x^2+2x-15 & > 0 \\ (x+5)(x-3) & > 0 \\ x<-5 & \textrm{ \scriptsize{OR} }x>3 \\ \end{aligned}$

Second,

$\begin{aligned} 2x+5 & > 7 \\ 2x & > 2 \\ x & > 1 \\ \end{aligned}$

In sum, $x<-5\textrm{ \scriptsize{OR} }x>1$ .

November 4, 2022December 14, 2022

202211041415 Solution to 1975-CE-AMATH-I-2

Find the values of $x$ satisfying both of the following inequalities:

$\begin{aligned} (1+x)(6-x)&\leqslant -8 \\ 3x-1 & \geqslant 5\\ \end{aligned}$

Roughwork.

The first inequality is

$\begin{aligned} (1+x)(6-x) & \leqslant -8 \\ 6+5x-x^2 \leqslant -8 \\ x^2-5x-14 \geqslant 0 \\ (x-7)(x+2) \geqslant 0 \\ x\leqslant -2\textrm{ \scriptsize{OR} }x\geqslant 7 \\ \end{aligned}$

whereas the second

$\begin{aligned} 3x-1 & \geqslant 5 \\ 3x & \geqslant 6 \\ x & \geqslant 2 \\ \end{aligned}$

In sum, $x\geqslant 7$ .

November 4, 2022November 4, 2022

202211041204 Solution to 2002-CE-AMATH-II-4

Find $\displaystyle{\int_{0}^{\frac{1}{2}}\frac{1}{\sqrt{1-x^2}}\,\mathrm{d}x}$ .
[Hint: Let $x=\sin\theta$ .]

Roughwork.

$\begin{aligned} &\quad\, \int_{0}^{\frac{1}{2}}\frac{1}{\sqrt{1-x^2}}\,\mathrm{d}x \\ \dots\enspace & \textrm{let }x=\sin\theta\textrm{ s.t. }\mathrm{d}x=\cos\theta\,\mathrm{d}\theta \enspace\dots \\ & = \int_{x=0}^{x=\frac{1}{2}}\frac{\cos\theta}{\sqrt{1-(\sin\theta )^2}}\,\mathrm{d}\theta \\ & = \int_{x=0}^{x=\frac{1}{2}}\frac{\cos\theta}{\cos\theta}\,\mathrm{d}\theta \\ & = \int_{x=0}^{x=\frac{1}{2}}\mathrm{d}\theta \\ & = \big[\theta \big]\big|_{x=0}^{x=\frac{1}{2}} \\ & = \big[\sin^{-1}x\big]\big|_{0}^{\frac{1}{2}} \\ & = \frac{\pi}{6}\\ \end{aligned}$

November 4, 2022November 4, 2022

202211041147 Solution to 2003-CE-AMATH-II-1

Find $\displaystyle{\int \cos^2\theta\,\mathrm{d}\theta}$ .

Roughwork.

We need the following

Lemma. (Double-angle formulae)

$\begin{aligned} \cos 2\theta & = \cos^2\theta -\sin^2\theta \\ & = \cos^2\theta - (1-\cos^2\theta )\\ & = 2\cos^2\theta -1 \\ \cos^2\theta & = \frac{1+\cos 2\theta}{2} \\ \end{aligned}$

Hence,

$\begin{aligned} &\quad\, \int\cos^2\theta\,\mathrm{d}\theta \\ & = \int \bigg( \frac{1+\cos 2\theta}{2}\bigg) \,\mathrm{d}\theta \\ & = \frac{1}{2}\int \mathrm{d}\theta + \frac{1}{2}\int \cos 2\theta\,\mathrm{d}\theta \\ & = \frac{\theta}{2} + \frac{1}{4}\sin 2\theta + C\textrm{ for some constant} \\ \end{aligned}$

November 4, 2022November 4, 2022

202211041133 Solution to 2004-CE-AMATH-II-1

Find

(a) $\displaystyle{\int\cos (3x+1)\,\mathrm{d}x}$ ,
(b) $\displaystyle{\int (2-x)^{2004}\,\mathrm{d}x}$ .

Roughwork.

(a)

$\begin{aligned} &\quad\, \int\cos (3x+1)\,\mathrm{d}x \\ \dots\enspace & \textrm{let }u=3x+1\textrm{ s.t. } \mathrm{d}x=\frac{\mathrm{d}u}{3}\enspace\dots \\ & = \int (\cos u)\bigg(\frac{\mathrm{d}u}{3}\bigg) \\ & = \frac{\sin u}{3}+C\textrm{ for some constant} \\ & = \frac{\sin (3x+1)}{3} + C \\ \end{aligned}$

(b)

$\begin{aligned} &\quad\, \int (2-x)^{2004}\,\mathrm{d}x \\ \dots\enspace &\textrm{let }u=2-x\textrm{ s.t. }\mathrm{d}x=-\mathrm{d}u \\ & = -\int u^{2004}\,\mathrm{d}u \\ & = -\frac{u^{2005}}{2005} + C\textrm{ for some constant} \\ & = -\frac{(2-x)^{2005}}{2005} + C \\ \end{aligned}$

November 4, 2022November 4, 2022

202211041123 Solution to 2004-CE-AMATH-II-3

The slope at any point $(x,y)$ of a curve $C$ is given by

$\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}=3x^2+1}$ .

If the $x$ -intercept of $C$ is $1$ , find the equation of $C$ .

Roughwork.

$\begin{aligned} y & = \int\frac{\mathrm{d}y}{\mathrm{d}x}\,\mathrm{d}x \\ & = \int (3x^2+1)\,\mathrm{d}x \\ y(x) & = x^3 + x + C\textrm{ for some constant} \\ \end{aligned}$

Substituting $(1,0)$ for $(x,y(x))$ :

$\begin{aligned} 0 & = (1)^3+(1) + C \\ C & = -2 \\ \end{aligned}$

In conclusion, the equation of $C$ is

$y(x) = x^3 + x -2$ .