Public Examinations in Mathematics

August 20, 2024August 20, 2024

202408201533 Example 13.8

Let $\displaystyle{x=\frac{y-2}{y+2}}$ .

(a) Find $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}}$ in terms of $y$ .
(b) Make $y$ the subject of the given function. Hence, find $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}}$ in terms of $x$ .
(c) Are the answers obtained in (a) and (b) different? Explain your assertion.

_{Extracted from S. W. Li et al. (2002). New Progress in Additional Mathematics Book 3}

Roughwork.

$\begin{aligned} x & = \frac{y-2}{y+2} \\ x & = \frac{y+2-4}{y+2} \\ \textrm{Eq. (1):}\quad x=f(y) & = 1-\frac{4}{y+2} \\ \frac{4}{y+2} & = 1-x \\ \frac{4}{1-x} & = y+2 \\ \textrm{Eq. (2):}\quad y=g(x) & = \frac{4}{1-x} - 2 \\ \end{aligned}$

domain: what can go into a function
codomain: what may possibly come out of a function
range: what actually comes out of a function

_{Kevin Sookocheff on Range, Domain, and Codomain (2018)}

Write, for function $f$ ,

$\begin{aligned} \textrm{domain: }& \mathbb{R}\backslash\{ -2\} \\ \textrm{codomain: }& \mathbb{R} \\ \textrm{range: }& \mathbb{R} \\ \end{aligned}$

and for function $g$ ,

$\begin{aligned} \textrm{domain: }& \mathbb{R}\backslash\{ 1\} \\ \textrm{codomain: }& \mathbb{R} \\ \textrm{range: }& \mathbb{R} \\ \end{aligned}$

We see $f$ discontinuous at $-2$ ; and $g$ , at $1$ ; whereas non-differentiable.

(to be continued)

February 16, 2024February 16, 2024

202402161017 Solution to 2017-DSE-MATH-I-6

The coordinates of the points $A$ and $B$ are $(-3,4)$ and $(9,-9)$ respectively.

$\begin{aligned} \mathbf{OA} & = (-3,4) \\ \mathbf{OB} & = (9,-9) \\ \end{aligned}$

$A$ is rotated anticlockwise about the origin through $90^\circ$ to $A'$ .

$f(r,\theta )=(r,\theta +\frac{\pi}{2})$

$B'$ is the reflection image of $B$ with respect to the $x$ -axis.

$g(x,y)=(x,-y)$

(a) Write down the coordinates of $A'$ and $B'$ .

Hint. Conversion between Cartesian and polar coordinates:

$\begin{aligned} r(x,y) & = \sqrt{x^2+y^2} \\ \theta (x,y) & = \tan^{-1}\bigg(\frac{y}{x}\bigg) \\ x(r,\theta ) & = r\cos\theta \\ y(r,\theta ) & = r\sin\theta \\ \end{aligned}$

(b) Prove that $AB$ is perpendicular to $A'B'$ .

Hint. Show that $\mathbf{AB}\cdot\mathbf{A'B'}=0$ .

$\begin{aligned} \mathbf{AB} & = \mathbf{AO} + \mathbf{OB} \\ & = \mathbf{OB} - \mathbf{OA} \\ \mathbf{A'B'} & = \mathbf{A'O} + \mathbf{OB'} \\ & = \mathbf{OB'} - \mathbf{OA'} \\ \end{aligned}$

$\begin{aligned} & \quad \mathbf{AB}\cdot\mathbf{A'B'} \\ & = (\mathbf{OB}-\mathbf{OA})\cdot (\mathbf{OB'}-\mathbf{OA'}) \\ & = \mathbf{OB}\cdot\mathbf{OB'} - \mathbf{OB}\cdot\mathbf{OA'} - \mathbf{OA}\cdot\mathbf{OB'} + \mathbf{OA}\cdot\mathbf{OA'} \\ \end{aligned}$

Scalar (/dot) product of two vectors:

$\begin{aligned} \mathbf{X}\cdot\mathbf{Y} & =|\mathbf{X}||\mathbf{Y}|\cos\measuredangle{(\mathbf{X},\mathbf{Y})} \\ \textrm{\scriptsize{OR}}\quad\mathbf{X}\cdot\mathbf{Y} & =\sum_{i=1}^{n}x_iy_i \\ \end{aligned}$

_{(modified with hints added)}

Roughwork.

We give segment $AB$ the equation $L$ :

$\begin{aligned} \frac{y-4}{x-(-3)}& = \frac{4-(-9)}{-3-9} = -\frac{13}{12} \\ L:\quad 0 & = 13x+12y-9 \\ \textrm{where } & (x,y)\in [-3,9]\times [-9,4] \\ \end{aligned}$

and segment $A'B'$ the equation $L'$ :

$\begin{aligned} \frac{y-(-3)}{x-(-4)}& = \frac{9-(-3)}{9-(-4)} = \frac{12}{13} \\ L':\quad 0 & = 12x-13y+9 \\ \textrm{where } & (x,y)\in [-4,9]\times [-3,9] \\ \end{aligned}$

The intersection point $C(a,b)$ of lines $L$ and $L'$ can be obtained by solving their simultaneous equations:

$\begin{aligned} \quad\enspace & \left\{ \begin{aligned} 13a+12b-9 & = 0 \\ 12a-13b+9 & = 0 \\ \end{aligned}\right\} \\ &\Rightarrow \left\{ \begin{aligned} a & = \frac{9}{313} \\ b & = \frac{225}{313} \\ \end{aligned}\right\} \\ \end{aligned}$

Should $L\perp L'$ , the line $L$ rotated by $\pi /2$ about point $C(a,b)$ on the $xy$ -plane would equal line $L'$ , the rotation matrix being

$\begin{pmatrix} \cos\theta & -\sin\theta & -a\cos\theta + b\sin\theta + a\\ \sin\theta & \cos\theta & -a\sin\theta - b\cos\theta + b \\ 0 & 0 & 1 \\ \end{pmatrix}\Bigg|_{\theta =\frac{\pi}{2}}$

to be applied to

$\begin{pmatrix} x \\ y \\ 1 \\ \end{pmatrix}\quad\textrm{ for } x,y\in L(x,y)$

such that

$\begin{aligned} & \quad \begin{pmatrix} 0 & -1 & a+b \\ 1 & 0 & -a+b \\ 0 & 0 & 1 \\\end{pmatrix}\begin{pmatrix} x \\ y \\ 1 \\ \end{pmatrix} \\ & = \begin{pmatrix} a+b-y \\ x-a+b \\ 1 \\\end{pmatrix} \\ & \stackrel{\textrm{def}}{=} \begin{pmatrix} x' \\ y' \\ 1 \\\end{pmatrix}\quad \textrm{ for }x',y'\in L'(x',y') \\ \end{aligned}$

the centre of rotation being invariant, i.e.,

$(x,y)|_{(a,b)}=(x',y')|_{(a,b)}$ .

The rest is left the reader as an exercise.

February 15, 2024February 15, 2024

202402151329 Solution to 1985-CE-AMATH-II-5

If the equation

$x^2+y^2+kx-(2+k)y=0$

represents a circle with radius $\sqrt{5}$ ,

(a) find the value(s) of $k$ ;
(b) find the equation(s) of the circle(s).

As if sitting an exam in additional mathematics, it certainly not being showing off, we should perhaps involve ourselves with some sort of calculus.

Roughwork.

As always, have in mind some pictures. Hence, we draw:

and also

after the stage got set, we write

$\begin{aligned} 0 & = x^2+y^2+kx-(2+k)y \\ 0 & = 2x\,\mathrm{d}x+2y\,\mathrm{d}y + k\,\mathrm{d}x - (2+k)\,\mathrm{d}y \\ y' & = \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2x+k}{k-2y+2} \\ \end{aligned}$

and furthermore,

$\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x} & = 0 \\ & \Rightarrow 2x+k=0 \\ & \Rightarrow x = -\frac{k}{2} \\ \frac{\mathrm{d}y}{\mathrm{d}x} & = \infty \\ & \Rightarrow k-2y+2 = 0 \\ & \Rightarrow y = \frac{k}{2}+1 \\ \end{aligned}$

In the case of $x=-\frac{k}{2}$ :

$\begin{aligned} 0 & = \bigg( -\frac{k}{2}\bigg)^2+y^2+k\bigg( -\frac{k}{2}\bigg) -(2+k)y \\ 0 & = y^2 - (2+k)y - \frac{k^2}{4} \\ \Delta & = \big( -(2+k)\big)^2 - 4(1)\bigg( -\frac{k^2}{4}\bigg) \\ & = 2(k^2+2k+2)\qquad (> 0) \\ y_1,y_3 & = \frac{-\big( -(2+k)\big) \pm \sqrt{\Delta}}{2(1)} \\ 2\sqrt{5} & = y_3-y_1 \\ \sqrt{\Delta} & = 2\sqrt{5} \\ \Delta & = 20 \\ 20 & = 2(k^2+2k+2) \\ 0 & = k^2+2k-8 \\ 0 & = (k-2)(k+4) \\ k & = 2,-4 \\ \end{aligned}$

and the case of $y=\frac{k}{2}+1$ :

$\begin{aligned} 0 & = x^2 + \bigg( \frac{k}{2}+1\bigg)^2 + kx - (2+k)\bigg( \frac{k}{2}+1\bigg) \\ 0 & = x^2+kx - \bigg(\frac{k}{2}+1\bigg)^2\\ \Delta & = (k)^2 - 4(1)\bigg( -\bigg(\frac{k}{2}+1\bigg)^2\bigg) \\ & = 2(k^2+2k+2) \qquad (> 0) \\ x_1,x_3 & = \frac{-(k)\pm\sqrt{\Delta}}{2(1)} \\ 2\sqrt{5} & = x_3 - x_1 \\ \sqrt{\Delta} & = 2\sqrt{5} \\ \Delta & = 20 \\ 20 & = 2(k^2+2k+2) \\ 0 & = k^2+2k-8 \\ 0 & = (k-2)(k+4) \\ k & = 2,-4 \\ \end{aligned}$

Targeting the centre $C(a,b)$ :

$\begin{aligned} a_{i=1,2} & = \frac{x_3 - x_1}{2} = -\frac{k}{2}\bigg|_{k=2,-4} = -1,2 \\ b_{i=1,2} & = \frac{y_3-y_1}{2} = \frac{-\big( -(2+k)\big)}{2}\bigg|_{k=2,-4} = 2,-1 \\ (a,b) & = (-1,2)\cup (2,-1) \\ \end{aligned}$

Of the equation of circle, the standard form is

$\left\{ \begin{aligned} (x+1)^2 + (y-2)^2 & = (\sqrt{5})^2 \\ (x-2)^2 + (y+1)^2 & = (\sqrt{5})^2 \\ \end{aligned}\right\}$

and the general form

$\left\{ \begin{aligned} x^2+y^2+2x-4y & = 0 \\ x^2+y^2-4x+2y & = 0 \\ \end{aligned}\right\}$

This problem is not to be attempted.

October 17, 2023October 17, 2023

202310171726 Solution to 1988-CE-AMATH-II-6

Evaluate

$\displaystyle{\int_{0}^2}\frac{x^2\,\mathrm{d}x}{\sqrt{9-x^3}}$ .

Roughwork.

Let

$\begin{aligned} f(x) & =x^2 \\ g(x) & = \sqrt{9-x^3} \\ \end{aligned}$

s.t.

$\begin{aligned} \frac{\mathrm{d}f}{\mathrm{d}x} & = 2x \\ \frac{\mathrm{d}g}{\mathrm{d}x} & = -\frac{3}{2}\bigg(\frac{x^2}{\sqrt{9-x^3}}\bigg) \\ & = -\frac{3}{2}\frac{f}{g} \\ -\frac{2}{3}\,\mathrm{d}g & = \frac{f}{g}\,\mathrm{d}x \\ \end{aligned}$

Then

$\begin{aligned} &\quad\enspace \int_{0}^{2}\frac{x^2\,\mathrm{d}x}{\sqrt{9-x^3}} \\ & = \int_{0}^{2}\frac{f(x)}{g(x)}\,\mathrm{d}x \\ & = -\frac{2}{3}\int_{0}^{2}\mathrm{d}g \\ & = -\frac{2}{3} \big[ g(x)\Big]\Big|_{0}^{2} \\ & = -\frac{2}{3} \Big[\sqrt{9-x^3}\Big]\Big|_{0}^{2} \\ & = -\frac{2}{3}(1-3) \\ & = \frac{4}{3} \\ \end{aligned}$

Ah, but no sense of physical meaning.

This problem is not to be attempted.

January 6, 2023January 6, 2023

202301060914 Solution to 1992-CE-AMATH-II-7

In the figure, $VABCD$ is a right pyramid with a square base of side $6\,\mathrm{cm}$ . $VB=9\,\mathrm{cm}$ .

Find, correct to the nearest $0.1$ degree,
(a) the angle between edge $VB$ and the base $ABCD$ ,
(b) the angle between the planes $VAB$ and $VAD$ .

Roughwork.

(a)

Applying the law of cosines,

$\begin{aligned} \cos \angle VBD & =\frac{DB^2+VB^2-VD^2}{2(DB)(VB)} \\ & = \frac{(6\sqrt{2})^2+(9)^2-(9)^2}{2(6\sqrt{2})(9)} \\ \angle VBD & = \dots \\ \end{aligned}$

(b)

Similarly,

$\angle BED=\dots$

This problem is not to be attempted.

December 30, 2022December 31, 2022

202212301726 Solution to 1987-CE-AMATH-I-3

For any complex number $z$ , let $\overline{z}$ , $|z|$ , and $\mathrm{Re}(z)$ be its conjugate, modulus, and real part respectively. Show that

$z+\overline{z}=2\mathrm{Re}(z)$ and $|z|\geqslant \mathrm{Re}(z)$ .

Hence, or otherwise, show that for any complex numbers $z_1$ and $z_2$ ,

$z_1z_2+\overline{z_1z_2}\leqslant 2|z_1||z_2|$ .

Roughwork.

In the field $\mathbb{C}=\{ x+iy:x,y\in\mathbb{R}\}$ of complex numbers, $x$ is called the real part of $z$ and $y$ the imaginary part, i.e.,

$z=x+iy=\mathrm{Re}(z)+i\,\mathrm{Im}(z)$ ,

its trigonometric form being $z=r(\cos\theta +i\sin\theta )$ with $r$ the modulus and $\theta$ the argument, and its exponential form, $z=re^{i\theta}$ .

This problem is not to be attempted.

December 28, 2022January 6, 2023

202212281153 Solution to 1999-CE-AMATH-II-2

Find $\displaystyle{\int x(x+2)^{99}\,\mathrm{d}x}$ .

Roughwork.

$\begin{aligned} &\quad\enspace \int x(x+2)^{99}\,\mathrm{d}x \\ \dots\, &\textrm{let }u=x+2\textrm{ s.t. }\mathrm{d}u=\mathrm{d}x\,\dots \\ & = \int (u-2)(u)^{99}\,\mathrm{d}u \\ & = \int \big( u^{100}-2u^{99}\big)\,\mathrm{d}u \\ & = \frac{u^{101}}{101} - \frac{u^{100}}{50} + C\textrm{ for some constant} \\ & = \frac{(x+2)^{101}}{101} - \frac{(x+2)^{100}}{50} + C \\ \end{aligned}$

This problem is not to be attempted.

December 23, 2022December 23, 2022

202212231530 Solution to 1990-CE-AMATH-I-7

Given the curve $C: x^2+4xy+5y^2=1$ , find $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}}$ . Hence find the equations of the two tangents to $C$ which are parallel to the line $\displaystyle{y=-\frac{1}{2}x}$ .

Roughwork.

For the curve,

$\begin{aligned} &\quad\enspace x^2+4xy+5y^2 = 1 \\ &\Rightarrow 2x\,\mathrm{d}x + 4(x\,\mathrm{d}y + y\,\mathrm{d}x) + 10y\,\mathrm{d}y = 0 \\ &\Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x}= -\frac{x+2y}{2x+5y} \\ \end{aligned}$

and for the line,

$\begin{aligned} &\quad\enspace y = -\frac{1}{2}x \\ &\Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{2} \\ \end{aligned}$

equating them,

$\begin{aligned} &\quad\enspace -\frac{x+2y}{2x+5y} = -\frac{1}{2} \\ &\Rightarrow \begin{pmatrix} x \\ y \end{pmatrix} = \bigg\{\begin{pmatrix} 1 \\ 0 \end{pmatrix} , \begin{pmatrix} -1 \\ 0 \end{pmatrix}\bigg\} \\ \end{aligned}$

are points of contact at which the tangents

$\begin{aligned} &\quad\enspace \frac{y-(\pm 1)}{x-(0)} = -\frac{1}{2} \\ &\Rightarrow \big\{ x + 2y \pm 2 = 0 \big\} \\ \end{aligned}$

are drawn.

This problem is not to be attempted.

December 23, 2022December 23, 2022

202212231507 Solution to 1990-CE-AMATH-II-1

Let $f(x)=\sqrt{x^2+k}\sin 2x$ , where $k$ is a constant. If $f'(0)=1$ , find the value of $k$ .

Roughwork.

$\begin{aligned} f(x) & = \sqrt{x^2+k}\sin 2x \\ f'(x) & = \sqrt{x^2+k}\cdot \frac{\mathrm{d}}{\mathrm{d}x}(\sin 2x) + \sin 2x\cdot \frac{\mathrm{d}}{\mathrm{d}x} \big( \sqrt{x^2+k}\big) \\ & = \sqrt{x^2+k}\cdot (2\cos 2x) + \sin 2x\cdot \bigg(\frac{x}{\sqrt{x^2+k}}\bigg)\\ f'(0) & = \sqrt{(0)^2+k}\cdot (2\cos 2(0)) + \sin 2(0)\cdot \bigg(\frac{(0)}{\sqrt{(0)^2+k}}\bigg)\\ 1 & =: 2\sqrt{k} \\ k & = \frac{1}{4} \\ \end{aligned}$

This problem is not to be attempted.

December 23, 2022December 23, 2022

202212231206 Solution to 2002-CE-AMATH-I-18

(a) Let $z=\cos\theta +i\sin\theta$ , where $-\pi<\theta \leqslant \pi$ . Show that

$|z^2+1|^2=2(1+\cos 2\theta )$ .

Hence, or otherwise, find the greatest value of $|z^2+1|$ .
(b) $w$ is a complex number such that $|w|=3$ .
i. Show that the greatest value of $|w^2+9|$ is $18$ .
ii. Explain why the equation

$w^4-81=100i(w^2-9)$

has only two roots.

Roughwork.

(a)

$\begin{aligned} \textrm{LHS} & = |z^2+1|^2 \\ & = |(\cos\theta +i\sin\theta )^2+1|^2 \\ & = |(\cos^2\theta -\sin^2\theta +2i\sin\theta\cos\theta )+1|^2 \\ & = |(2\cos^2\theta )+i(2\sin\theta\cos\theta )|^2 \\ & = \big(\sqrt{(2\cos^2\theta )^2+(2\sin\theta\cos\theta )^2}\big)^2 \\ & = 4\cos^4\theta +4\sin^2\theta\cos^2\theta \\ & = 4\cos^2\theta (\cos^2\theta +\sin^2\theta ) \\ & = 4\cos^2\theta \\ \end{aligned}$

From the double-angle formula

$\cos 2\theta = 2\cos^2\theta -1$

_{Wikipedia on List of trigonometric identities}

the equality follows, i.e., $\textrm{LHS}=\textrm{RHS}$ . As $\max (\cos 2\theta )=1$ , the greatest value of $|z^2+1|$ is

$\sqrt{2(1+(1))} = 2$ .

(b) Not to be attempted.