Additional Mathematics – Hong Kong Certificate of Education (HKCE) – Page 7

July 5, 2022October 24, 2022

202207051339 Solution to 1974-CE-AMATH-I-XX

(a) In the figure below, $ABCDEF$ is a regular hexagon. Which one of the $6$ vectors $\overrightarrow{AD}$ , $\overrightarrow{DA}$ , $\overrightarrow{FC}$ , $\overrightarrow{CF}$ , $\overrightarrow{EB}$ , $\overrightarrow{BE}$ is equal to $\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{DC}$ ?

(b) $\mathbf{u}$ and $\mathbf{v}$ are unit vectors making an angle $60^\circ$ with each other as shown in the figure below. $AB$ is a vector making an angle $30^\circ$ with $\mathbf{u}$ and $|\overrightarrow{AB}|=2$ . Express $\overrightarrow{AB}$ in terms of $\mathbf{u}$ and $\mathbf{v}$ .

(c) In the figure below, $\angle B=90^\circ$ and $m(\overrightarrow{AB})=10$ . Calculate $\overrightarrow{AB}\cdot\overrightarrow{AC}$ .

(a)

$\begin{aligned} \mathbf{AD} & = -\mathbf{DA} \\ |\mathbf{AD}| & = |\mathbf{DA}| \\ \mathbf{FC} & = -\mathbf{CF} \\ |\mathbf{FC}| & = |\mathbf{CF}| \\ \mathbf{EB} & = -\mathbf{BE} \\ |\mathbf{EB}| & = |\mathbf{BE}| \\ \end{aligned}$

Ans. $\mathbf{FC}$ $\textrm{\scriptsize{OR}}$ $\overrightarrow{FC}$ by inspection.

Working.

$\begin{aligned} &\quad \mathbf{AB} + \mathbf{BC} + \mathbf{DC} \\ & = (\mathbf{AB} + \mathbf{BC}) + \mathbf{DC} \\ & = \mathbf{AC} + \mathbf{DC} \\ \dots & \textrm{ as }\mathbf{AC}=\mathbf{FD} \enspace\dots \\ & = \mathbf{FD} + \mathbf{DC} \\ & = \mathbf{FC} \\ \end{aligned}$

(b)

Write, in Cartesian components of unit vectors $\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$ , the following:

$\begin{aligned} \mathbf{AB} & = |\mathbf{AB}|\cos 30^\circ\,\hat{\mathbf{i}} + |\mathbf{AB}|\sin 30^\circ\,\hat{\mathbf{j}} \\ & = 2\cos 30^\circ\,\hat{\mathbf{i}} + 2\sin 30^\circ\,\hat{\mathbf{j}} \\ & = \sqrt{3}\,\hat{\mathbf{i}} + 1\,\hat{\mathbf{j}} \\ \end{aligned}$

and similarly,

$\begin{aligned} \mathbf{u} & = |\mathbf{u}|\,\hat{\mathbf{i}} \\ \mathbf{v} & = |\mathbf{v}|\cos 60^\circ\,\hat{\mathbf{i}} + |\mathbf{v}|\sin 60^\circ\,\hat{\mathbf{j}}\\ & \\ \because\enspace & \mathbf{u}, \mathbf{v}\textrm{ are unit vectors} \\ \therefore\enspace & |\mathbf{u}|=|\mathbf{v}|=1 \\ & \\ \mathbf{u} & = \hat{\mathbf{i}} \\ \mathbf{v} & = \frac{1}{2}\,\hat{\mathbf{i}} + \frac{\sqrt{3}}{2} \hat{\mathbf{j}} \\ \end{aligned}$

$\begin{aligned} \begin{bmatrix} \mathbf{u} \\ \mathbf{v} \end{bmatrix} & = \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix} \begin{bmatrix} \hat{\mathbf{i}} \\ \hat{\mathbf{j}} \end{bmatrix} \\ \begin{bmatrix} \hat{\mathbf{i}} \\ \hat{\mathbf{j}} \end{bmatrix} & = \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}^{-1}\begin{bmatrix} \mathbf{u} \\ \mathbf{v} \end{bmatrix} \\ \end{aligned}$

Lemma. (Inversion of 2-by-2 matrices)

For $2\times 2$ matrices, inversion can be done as follows:

$\begin{aligned} \mathbf{A}^{-1} & = \begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1} \\ & = \frac{1}{\mathrm{det}\mathbf{A}}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \\ & = \frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \\ \end{aligned}$

_{Wikipedia on Invertible matrix}

Then,

$\begin{bmatrix} 1 & 0 \\ \frac{1}{2} & \frac{\sqrt{3}}{2}\end{bmatrix}^{-1} = \begin{bmatrix} 1 & 0 \\ -\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{3}} \end{bmatrix}$

so that

$\begin{bmatrix} \hat{\mathbf{i}} \\ \hat{\mathbf{j}} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{3}} \end{bmatrix}\begin{bmatrix}\mathbf{u} \\ \mathbf{v}\end{bmatrix}$

Thus

$\begin{aligned} \mathbf{AB} & =\begin{pmatrix} \sqrt{3} & 1\end{pmatrix}\begin{bmatrix}\hat{\mathbf{i}} \\ \hat{\mathbf{j}} \end{bmatrix} \\ & = \begin{pmatrix} \sqrt{3} & 1\end{pmatrix} \begin{bmatrix} 1 & 0 \\ -\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{3}} \end{bmatrix} \begin{bmatrix}\mathbf{u} \\ \mathbf{v}\end{bmatrix} \\ & = \begin{bmatrix} \frac{2}{\sqrt{3}} & \frac{2}{\sqrt{3}} \end{bmatrix}\begin{bmatrix}\mathbf{u} \\ \mathbf{v}\end{bmatrix} \\ &=\frac{2}{\sqrt{3}}\,\mathbf{u}+\frac{2}{\sqrt{3}}\,\mathbf{v} \\ \end{aligned}$

There might be some mistake if conceptually.

(c)

I don’t know how. Skip it.

March 7, 2022October 24, 2022

202203071350 Solution to 1968-CE-AMATH-I-XX

Find the indefinite integrals:

i. $\displaystyle{\int\frac{\mathrm{d}x}{x^4}}$

ii. $\displaystyle{\int\frac{\mathrm{d}x}{(2x-3)^4}}$

Solution.

i.

$\begin{aligned} &\quad \int\frac{\mathrm{d}x}{x^4} \\ & = \int x^{-4}\,\mathrm{d}x \\ & = \frac{x^{((-4)+1)}}{((-4)+1)}+C\textrm{ for some arbitrary constant }C \\ & = -\frac{1}{3x^3}+C \\ \end{aligned}$

ii.

$\begin{aligned} &\quad \int \frac{\mathrm{d}x}{(2x-3)^4} \\ & \cdots\enspace\cdots\enspace\cdots \\ & \textrm{let }y=2x-3\\ & \textrm{then }\frac{\mathrm{d}y}{\mathrm{d}x}=2 \\ & \textrm{and }\mathrm{d}x=\frac{\mathrm{d}y}{2} \\ & \cdots\enspace\cdots\enspace\cdots \\ & = \int\frac{\frac{\mathrm{d}y}{2}}{y^4} \\ & = \frac{1}{2}\int\frac{\mathrm{d}y}{y^4} \\ & \stackrel{\textrm{(i)}}{=} \frac{1}{2}\bigg( -\frac{1}{3x^3}+C\bigg) \\ & = -\frac{1}{6x^3}+C'\textrm{ for some arbitrary constant }C' \\ \end{aligned}$

March 6, 2022October 24, 2022

202203061946 Solution to 1967-CE-AMATH-I-XX

Find the indefinite integral $\int x\sqrt{x^2-2}\,\mathrm{d}x$ .

Solution.

$\begin{aligned} &\quad \int x\sqrt{x^2-2}\,\mathrm{d}x \\ &\enspace \cdots \enspace \cdots \enspace\cdots \\ \qquad & \textrm{let }y=x^2-2 \\ \qquad & \textrm{then }\mathrm{d}y=2x\,\mathrm{d}x \\ &\enspace \cdots \enspace\cdots \enspace\cdots \\ & = \int (x)(\sqrt{y})\bigg(\frac{\mathrm{d}y}{2x}\bigg) \\ & = \frac{1}{2}\int y^{\frac{1}{2}}\,\mathrm{d}y \\ & = \frac{y^{\frac{3}{2}}}{3} + C\textrm{ for some arbitrary constant }C\\ & = \frac{\sqrt{(x^2-2)^3}}{3} + C \\ \end{aligned}$

March 6, 2022October 24, 2022

202203061208 Solution to 1973-CE-AMATH-I-XX

A circle passes through the points $(-1,-1)$ , $(3,3)$ , and $(7,-1)$ . Find the equation of this circle.

Solution.

Let the centre of this circle be located at point $(a,b)$ and its radius be $R$ long. The equation of this circle is thus in the form

$(x-a)^2+(y-b)^2=R^2$ ,

where $a$ , $b$ , and $R$ are three unknowns to be obtained in the following three equations:

$\begin{aligned} (-1-a)^2+(-1-b)^2 & = R^2 \\ (3-a)^2+(3-b)^2 & = R^2 \\ (7-a)^2+(-1-b)^2 & = R^2 \\ \end{aligned}$

$\begin{aligned} (a^2+2a+1) + (b^2+2b+1) & = R^2 \\ (a^2-6a+9) + (b^2-6b+9) & = R^2 \\ (a^2-14a+49) + (b^2+2b+1) & = R^2 \\ \end{aligned}$

$\begin{aligned} a & = 3 \\ b & = -1 \\ R & = 4 \\ \end{aligned}$

$\therefore$ The equation of this circle is

$(x-3)^2+(y+1)^2=16$ .

February 7, 2022October 24, 2022

202202071146 Solution to 1969-CE-AMATH-I-XX

Find $\displaystyle{\int\frac{2x+5}{(x^2+5x+1)^3}\,\mathrm{d}x}$ .

Solution.

Set-up.

$\begin{aligned} \textrm{Let }u & =x^2+5x+1 \\ \frac{\mathrm{d}u}{\mathrm{d}x} & = 2x+5 \\ \mathrm{d}x & = \frac{\mathrm{d}u}{2x+5} \\ \end{aligned}$

Then,

$\begin{aligned} &\quad \int\frac{2x+5}{(x^2+5x+1)^3}\,\mathrm{d}x \\ & = \int \frac{2x+5}{u^3}\bigg( \frac{\mathrm{d}u}{2x+5}\bigg) \\ & = \int \frac{1}{u^3} \,\mathrm{d}u \\ & = -\frac{1}{2u^2}+C\textrm{ for some constant }C \\ & = -\frac{1}{2(x^2+5x+1)^2}+C\\ \end{aligned}$

February 6, 2022October 24, 2022

202202070744 Solution to 1997-CE-AMATH-II-2

Find $\displaystyle{\int x\sqrt{x-1}\,\mathrm{d}x}$ .

[Hint: Let $u=x-1$ .]

Solution.

$\begin{aligned} \textrm{Let } & u = x-1 \\ \textrm{then } & x = u+1 \\ \textrm{and }&\mathrm{d}x = \mathrm{d}u\\ \end{aligned}$

$\begin{aligned} &\quad \int x\sqrt{x-1}\,\mathrm{d}x\\ & = \int (u+1)(u^{\frac{1}{2}})\,\mathrm{d}u \\ & = \int (u^{\frac{3}{2}}+u^{\frac{1}{2}})\,\mathrm{d}u \\ & = \frac{2u^{\frac{5}{2}}}{5} + \frac{2u^{\frac{3}{2}}}{3} + C\textrm{ for some constant }C \\ & = \frac{2(x-1)^{\frac{5}{2}}}{5} + \frac{2(x-1)^{\frac{3}{2}}}{3} + C \\ \end{aligned}$

December 2, 2021October 24, 2022

202112021444 Solution to 1979-CE-AMATH-I-1

Let $\displaystyle{y=x+\frac{1}{x^2}}$ . Find $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}}$ from first principles.

Solution.

From first principles,

$\begin{aligned} y:=f(x) & =x+\frac{1}{x^2} \\ f(x+\Delta x) & =(x+\Delta x) + \frac{1}{(x+\Delta x)^2} \\ f(x+\Delta x)-f(x) & = \Delta x + \frac{1}{(x+\Delta x)^2} - \frac{1}{x^2} \\ & = \Delta x + \frac{x^2-(x+\Delta x)^2}{x^2(x+\Delta x)^2} \\ & = \Delta x - \frac{2x\Delta x+(\Delta x)^2}{x^2(x+\Delta x)^2} \\ \frac{f(x+\Delta x)-f(x)}{\Delta x} & = 1 - \frac{2x + \Delta x}{x^2(x+\Delta x)^2} \\ \lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} & = 1 - \frac{2x + 0}{x^2(x+0)^2} \\ \frac{\mathrm{d}y}{\mathrm{d}x} & = 1 - \frac{2}{x^3} \\ \end{aligned}$

Playground.

To make fun of calculus, do let

$\begin{aligned} y & =x+\frac{1}{x^2} \\ x^2y & =x^3+1 \\ 0 & = x^3-x^2y+1 := f(x,y) \\ \end{aligned}$

where $f(x,y)$ is a degree $3$ polynomial in two variables $x$ and $y$ .

The set of solutions to $f(x,y)=0$ is

$\begin{cases} \enspace x \neq 0 \\ \enspace y = \frac{x^3+1}{x^2} \end{cases}$

Computing $\partial_xf$ and $\partial_yf$ as follow:

$\begin{aligned} \partial_xf & = \frac{\partial}{\partial x}f(x,y) \\ & = 3x^2-2xy \\ \end{aligned}$

and

$\begin{aligned} \partial_yf & = \frac{\partial}{\partial y}f(x,y) \\ & = -x^2 \\ \end{aligned}$ ,

we oversee the relation

$\mathrm{d}f = (\partial_xf)(\Delta x)+(\partial_yf)(\Delta y)$ .

Definition. (Singularity and Smoothness)

A point $p=(a,b)$ on a curve $\mathcal{C}=\{ (x,y)\in\mathbb{C}^2:f(x,y)=0\}$ is said to be singular if

$\begin{aligned} \frac{\partial f}{\partial x}(a,b) & = 0 \\ \frac{\partial f}{\partial y}(a,b) & = 0 \\ \end{aligned}$ .

A point that is not singular is called smooth. If there is at least one singular point on $\mathcal{C}$ , then curve $\mathcal{C}$ is called a singular curve. If there are no singular points on $\mathcal{C}$ , the curve $\mathcal{C}$ is called a smooth curve.

_{C.f. Definition 1.9.1, T. Carrity, et al., Algebraic Geometry: A Problem Solving Approach}

That said, in our scenario

$f(x,y)=x^3-x^2y+1$ ,

the set of candidates for singularity

$\{ (a,b)\in\mathbb{C}^2\enspace \mathrm{s.t.}\enspace f(a,b)=0\textrm{ and }\partial_xf(a,b)=\partial_yf(a,b)=0\}$ .

is empty.

$\therefore$ The curve $\mathcal{C}=\{ (x,y)\in\mathbb{C}^2:x^3-x^2y+1=0\}$ is smooth, thus everywhere differentiable.

Wait, the function in question should be the curve

$F(x)=\displaystyle{x+\frac{1}{x^2}}$ ;

still, for better or worse, $F(x)$ is $\textrm{\scriptsize{NOT}}$ a differentiable function because the derivative does $\textrm{\scriptsize{NOT}}$ exist at $x=0$ , for that point is a discontinuity.

November 24, 2021October 24, 2022

202111241152 Solution to 1971-CE-AMATH-3

Prove by mathematical induction that $n(n^2+2)$ is divisible by $3$ for all positive integral values of $n$ .

Solution.

Let $P(n)$ be the statement that

$P(n):\enspace n(n^2+2)\textrm{ is divisible by }3\textrm{ for }n\in\mathbb{Z}^{+}$

First, we wish to show $P(1)$ holds true.

$\begin{aligned} &\quad (1)\big( (1)^2 + 2\big) \\ & = 1(1+2) \\ & = 3\qquad (\textrm{divisible by }3)\\ \end{aligned}$

$\therefore$ $P(1)$ is true.

Once we assume that $P(n)$ is true for some positive integral values, we wish to show $P(n+1)$ is true.

$\begin{aligned} &\quad (n+1)\big( (n+1)^2 + 2\big) \\ & = (n+1)(n^2+2n+3) \\ & = n^3 + 3n^2 + 5n +3 \\ \end{aligned}$

Let $f(x)=x^3+3x^2+5x+3$ .

Suppose we can divide $f(x)$ by $3$ , i.e.,

$f(x)=3\cdot g(x)$

for some $g(x)=Ax^3+Bx^2+Cx+D$ where $A,B,C,D\in\mathbb{R}$ .

Then, let’s see

$\begin{aligned} \frac{f(x)}{g(x)} & = 3 \\ \frac{\mathrm{d}}{\mathrm{d}x}\bigg( \frac{f(x)}{g(x)}\bigg) & = \frac{\mathrm{d}}{\mathrm{d}x}(3) \\ \frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2} & = 0 \\ g(x)f'(x)-f(x)g'(x) & = 0 \\ \end{aligned}$

Plugging in $f(x)$ , $f'(x)$ , $g(x)$ , and $g'(x)$ , we have

$\begin{aligned} &\quad (Ax^3+Bx^2+Cx+D)(3x^2+6x+5) \\ & = (x^3+3x^2+5x+3)(3Ax^2+2Bx+C) \\ \end{aligned}$

$\begin{aligned} &\quad \enspace \textrm{LHS} \\ & = (3A)x^5 + (6A+3B)x^4 + (5A+6B+3C)x^3 \\ &\qquad\quad + (5B+6C+3D)x^2 + (5C+6D)x + (5D) \\ \end{aligned}$

$\begin{aligned} &\quad \enspace \textrm{RHS} \\ & = (3A)x^5 + (9A+2B)x^4 + (15A+6B+C)x^3 \\ &\qquad\quad + (9A+10B+3C)x^2 + (6B+5C)x + (3C) \\ \end{aligned}$

We have a set of five equations below:

$\begin{aligned} 6A + 3B & = 9A + 2B \\ 5A+6B+3C & = 15A+6B+C \\ 5B+6C+3D & = 9A+10B+3C \\ 5C +6D & = 6B+5C \\ 5D & = 3C \\ \end{aligned}$

which can be summarized to

$\displaystyle{A=\frac{B}{3}=\frac{C}{5}=\frac{D}{3}}$ ,

that is equivalent to saying

$g(x)=\displaystyle{\frac{f(x)}{3}=\frac{x^3}{3}+x^2+\frac{5x}{3}+1}$ ,

viz. I have been wasting time over a circular reasoning of no use.

I should have tried to attest otherwise that $g(x)$ is of some positive integral values, even though this is $\textrm{\scriptsize{NOT}}$ proving the statement $P(n)$ by mathematical induction firsthand.

Reformulation.

To show that the following statement $Q(x)$ holds true:

$Q(x):\quad \displaystyle{\frac{x^3}{3}+x^2+\frac{5x}{3}+1}\textrm{ is an integer for any }x\in\mathbb{Z}^{+}$ .

$Q(1)$ is true because

$\displaystyle{\frac{(1)^3}{3}+(1)^2+\frac{5(1)}{3}+1}=4$

is an integer.

Assume $Q(x)$ is true for some $x\in\mathbb{Z}^{+}$ .

Then,

$\begin{aligned} &\qquad \frac{(x+1)^3}{3}+(x+1)^2+\frac{5(x+1)}{3}+1 \\ & = \frac{(x^3+3x^2+3x+1)}{3} + (x^2+2x+1)+\frac{(5x+5)}{3} + 1 \\ & = \bigg( \frac{x^3}{3}+x^2+\frac{5x}{3}+1\bigg) + \bigg( \frac{3x^2+3x+1}{3}+2x+1+\frac{5}{3}+1 \bigg) \\ \end{aligned}$

The first term is an integer for $Q(x)$ is assumed to be true. We then need to check whether the second term is of integral value(s).

The second term, after simplification, is

$x^2+3x+4$ ,

obviously an integer.

$\therefore$ $Q(x+1)$ thus holds.

Because $Q(1)$ is true, by mathematical induction, $Q(x)$ is true for all positive integers $x$ .

Escape to the original statement.

$3|n^3+3n^2+5n+3\Longleftrightarrow P(n+1)\textrm{ holds true.}$

Because $P(1)$ is true, by mathematical induction, $P(n)$ is true for any positive integers $n$ .

November 24, 2021October 24, 2022

202111241104 Solution to 1969-CE-AMATH-3

Prove by mathematical induction that if $n$ is a positive integer, $f(n)\equiv 3^{2n}-1$ is divisible by $8$ .

Solution.

We wish to show that the statement is true for $n=1$ , and also true for $n+1$ .

When $n=1$ , $f(1)=3^{2(1)}-1=8$ is divisible by $8$ .

Assume $f(n)$ is true for some positive integer $n$ , i.e.,

$\{f(n):3^{2n}-1=8m\, |\, \exists \, m, n\in\mathbb{Z}^{+}\}$ .

For $f(n+1)=3^{2(n+1)}-1$ , we have

$\begin{aligned} & \quad 3^{2(n+1)}-1 \\ & = 3^{2n+2} - 1 \\ & = 3^{2n}\cdot 3^2 - 1 \\ & = 9(3^{2n})-1 \\ & = 3^{2n} -1 + 8(3^{2n}) \\ \end{aligned}$

As is assumed $3^{2n}-1$ divisible by $8$ , the statement is thus also true for $n+1$ .

We have proven by mathematical induction that for any positive integer $n\enspace (\in\mathbb{Z}^+)$ , $f(n)$ is divisible by $8$ .

Afterword.

Try to prove $\textrm{\scriptsize{NOT}}$ by mathematical induction $\textrm{\scriptsize{BUT}}$ by direct proof.

$\begin{aligned} &\quad \frac{3^{2n}-1}{8} \\ & = \frac{3^{2n}-1}{9-1} \\ & = \frac{3^{2n}-1}{3^2-1} \\ & = \dots \end{aligned}$

Let $x=3^2$ , then

$\begin{aligned} &\quad \dots \\ & = \frac{x^n-1}{x-1} \\ & = 1+x+x^2+\cdots + x^{n-2} + x^{n-1} \enspace (\in\mathbb{Z}^{+}) \\ \end{aligned}$

$\therefore$ $f(n)\equiv 3^{2n}-1$ is divisible by $8$ .

November 22, 2021October 24, 2022

202111221509 Solution to 1971-CE-AMATH-1

When the functions $x^2-bx+3$ and $2b-x$ are divided by $(x-a)$ , the remainders are 1 and 4 respectively. Find $a$ and $b$ .

Solution.

Translating mathematically, we have

$\begin{aligned} x^2-bx+3 & = (x-a)F(x)+1 \\ 2b-x & = (x-a)G(x)+4\\ \end{aligned}$

Plugging in $x=a$ , we have

$\begin{aligned} (a)^2-b(a)+3 & = \big( (a)-a\big) F(a)+1 \\ 2b-(a) & = \big( (a)-a\big) G(a)+4 \\ \end{aligned}$

After simplifying it, we have

$\begin{aligned} a^2-ab & = -2\\ -a+2b & = 4\\ \end{aligned}$

Solving two equations with two unknowns,

$\begin{cases} & a =2\\ & b =3 \\ \end{cases}$

is the solution.