Additional Mathematics – Hong Kong Certificate of Education (HKCE) – Page 6

October 19, 2022October 19, 2022

202210191554 Solution to 1969-CE-AMATH-II-XX

Find the $x$ -coordinate of the point on the curve $y=ax^2+bx+c$ where the tangent is horizontal.

Roughwork.

The gradient $y'(x)$ of the curve $y(x)$ is

$y'(x)=2ax+b$ .

When the tangent is horizontal,

$\begin{aligned} y'(x) & =0 \\ 2ax+b & =0 \\ x & = -\frac{b}{2a} \\ y & = a\bigg( \frac{-b}{2a}\bigg)^2+b\bigg( \frac{-b}{2a}\bigg)+c \\ & = \dots \\ \end{aligned}$

October 19, 2022October 19, 2022

202210191506 Solution to 1970-CE-AMATH-II-XX

Calculate the gradient of the curve $x^2+xy+y^2=1$ at the point $(1,-1)$ .

Roughwork.

Let the function $f$ of variables $x$ and $y$ be

$f(x,y):=x^2+xy+y^2-1=0$

Then,

$\begin{aligned} \Delta f=2x\Delta x+x\Delta y+y\Delta x+2y\Delta y & =0 \\ 2x + x\frac{\Delta y}{\Delta x}+y+2y\frac{\Delta y}{\Delta x} & = 0 \\ (x+2y)\frac{\Delta y}{\Delta x} & = -2x-y \\ \frac{\Delta y}{\Delta x} & = - \frac{2x+y}{x+2y} \\ \end{aligned}$

The gradient $\nabla$ at point $(x=1,y=-1)$ is

$\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}\bigg|_{(1,-1)}=-\frac{2(1)+(-1)}{(1)+2(-1)}=1}$ .

October 19, 2022October 19, 2022

202210191436 Solution to 1969-CE-AMATH-II-XX

Calculate the gradient of the curve $x=3y+y^2$ at the point $(0,-3)$ .

Roughwork.

Differentiating $x(y)$ wrt to $y$ and taking reciprocal,

$\begin{aligned} x(y) & = 3y+y^2 \\ \frac{\mathrm{d}x}{\mathrm{d}y} & = 3 + 2y \\ \frac{\mathrm{d}y}{\mathrm{d}x} & = \frac{1}{\frac{\mathrm{d}x}{\mathrm{d}y}} \\ & = \frac{1}{3+2y} \\ \end{aligned}$

the gradient $\nabla y(x)$ at point $(x,y) = (0,-3)$ is

$\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}\bigg|_{(0,-3)} = \frac{1}{3+2(-3)} = -\frac{1}{3}}$ .

October 19, 2022October 19, 2022

202210191139 Solution to 1973-CE-AMATH-II-XX

Find the area bounded by the curve $y=4x-x^2$ and the straight line $x-y+2=0$ .

Roughwork.

Rewriting,

$\begin{cases} C:y=-x^2+4x \\ L:y=x+2 \\ \end{cases}$

Solving,

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

$\begin{aligned} x=1 \Rightarrow y=(1)+2=3 \\ x=2 \Rightarrow y=(2)+2=4 \\ \end{aligned}$

Now that the points of intersection are $(1,3)$ and $(2,4)$ , plot a graph below.

By either definite integral,

$\textrm{Area} & = \begin{cases} \displaystyle{\int_{1}^{2}\big(y_C(x)-y_L(x)\big)\,\mathrm{d}x }\\ \displaystyle{\int_{3}^{4}\big(x_L(y)-x_C(y)\big)\,\mathrm{d}y }\\ \end{cases}$

The upper integral is easier solvable than the lower one. Exercise.

October 19, 2022October 19, 2022

202210191050 Solution to 1977-CE-AMATH-II-XX

Prove that

$\displaystyle{\cos^4\theta = \frac{1}{8}(\cos 4\theta + 4\cos 2\theta + 3)}$ .

Hence solve the equation

$\cos 4\theta + 4\cos 2\theta +1=0$

for $0^\circ\leqslant\theta\leqslant 360^\circ$ .

Roughwork.

$\begin{aligned} \textrm{RHS} & = \frac{1}{8}(\cos 4\theta + 4\cos 2\theta + 3) \\ & = \frac{1}{8}\big(\cos 2(2\theta ) + 4\cos 2\theta + 3\big) \\ & = \frac{1}{8}\big( (2\cos^22\theta -1) + 4\cos 2\theta + 3\big) \\ & = \frac{1}{8}(2\cos^22\theta +4\cos 2\theta +2) \\ \dots\,\because\enspace & \cos2\theta = 2\cos^2\theta -1\,\dots \\ & = \frac{1}{8}\big( 2(2\cos^2\theta -1)^2+4(2\cos^2\theta -1)+2\big) \\ & = \dots \\ & = \cos^4\theta \\ & = \textrm{LHS} \\ \end{aligned}$

The missing steps are left the reader.

October 18, 2022October 18, 2022

202210181552 Solution to 1971-CE-AMATH-I-XX

Show that the straight line $x+2y+4=0$ touches the curve $y^2=4x$ .

Roughwork.

If there exists some point $(a,b)$ of intersection of the straight line and the curve, we have

$\begin{cases} a+2b+4 = 0 \\ b^2 = 4a \\ \end{cases}$

Substituting $-2b-4$ for $a$ in the second equation, we have

$\begin{aligned} & b^2 = 4(-2b-4) \\ \Leftrightarrow\quad & b^2 + 8b +16 = 0 \\ \Leftrightarrow\quad & (b+4)^2 = 0 \\ \Leftrightarrow\quad & b = -4\quad\textrm{(rep.)} \\ \Rightarrow\quad & a = -2(-4)-4= 4 \\ \end{aligned}$

$\begin{cases} a = 4 \\ b = -4 \\ \end{cases}$

Such point as $(4,-4)$ exists.

$\because$ The intersection of $\begin{cases} L: x+2y+4=0 \\ C: y^2 = 4x \\ \end{cases}$ is non-empty.

$\therefore$ The straight line touches the curve.

October 14, 2022October 20, 2022

202210141629 Solution to 1976-CE-AMATH-II-XX

By using

$\displaystyle{\int_{a}^{b}f(x)\,\mathrm{d}x = \int_{a}^{c}f(x)\,\mathrm{d}x + \int_{c}^{b}f(x)\,\mathrm{d}x}$

or otherwise, evaluate

$\displaystyle{\int_{0}^{\pi}\cos^{7}x\,\mathrm{d}x}$ .

Warm-up.

$\begin{aligned} f(x) & =\cos^7x \\ f(-x) & = \cos^7(-x) \\ & = \cos^7(x) \\ & = f(x) \\ \end{aligned}$

$\therefore$ $f(x)$ is an even function such that

$\displaystyle{\int_{-a}^{a}f(x)\,\mathrm{d}x = 2\int_{0}^{a}f(x)\,\mathrm{d}x=2\int_{-a}^{0}f(x)\,\mathrm{d}x}$ .

Assume $f(x+T)=f(x)$ is a periodic function for some $T\in (0,2\pi ]$ .

$\begin{aligned} & x\overset{f} \longrightarrow \cos^7x \\ \sim\enspace & x\overset{g}\longrightarrow \cos x\overset{h}\longrightarrow \cos^7x \\ \end{aligned}$

where $g(x)=\cos x$ , $h(x)=x^7$ , and $f=h\circ g$ .

$g:[0,\pi ]\to [-1,1]$ by $x\mapsto \cos x$ is one-to-one and onto.

$h:[-1,1]\to [-1,1]$ by $x\mapsto x^7$ is injective and surjective as well.

Hence $f$ , the composition $h\circ g$ of bijections $g$ and $h$ , is also bijective. Besides $f:[0,\pi ]\to [-1,1]$ is a continuously differentiable function that $f(0)=1$ , $f(\frac{\pi}{2}) =0$ , and $f(\pi )=-1$ .

How do you evaluate the integral below:

$\displaystyle{F(\theta )=\int_{0}^{\theta}f(x)\,\mathrm{d}x}$

where $F$ is an odd function such that

$\begin{aligned} F(0) & = F(2\pi ) = 0 \\ F(\theta ) & = -F(-\theta ) \\ \end{aligned}$

The curve of function $f$ is flat at point(s) whose slope $f'$ is zero, i.e.,

$\begin{aligned} f'(x) & = 0 \\ \frac{\mathrm{d}}{\mathrm{d}x}(\cos^7x) & = 0 \\ -7\cos^6x\sin x & = 0 \\ x & = 0,\frac{\pi}{2}, \pi \\ \end{aligned}$

Roughwork. Show

Integrating by parts,

$\begin{aligned} \int_{0}^{\pi}\cos^7x\,\mathrm{d}x & = \int_{0}^{\pi}\cos^6x\,\mathrm{d}(\sin x) \\ & = \big[\cos^6x\sin x\big]\big|_{0}^{\pi} - \int_{0}^{\pi}\sin x\,\mathrm{d}(\cos^6x)\\ & = 0 + \int_{0}^{\pi}6\sin^2x\cos^5x\,\mathrm{d}x \\ & = \int_{0}^{\pi}6(1-\cos^2x)\cos^5x\,\mathrm{d}x \\ & = \int_{0}^{\pi}6\cos^5x\,\mathrm{d}x - \int_{0}^{\pi}6\cos^7x\,\mathrm{d}x \\ \int_{0}^{\pi}7\cos^7x\,\mathrm{d}x & = \int_{0}^{\pi}6\cos^5x\,\mathrm{d}x \\ \int_{0}^{\pi}\cos^7x\,\mathrm{d}x & = \frac{6}{7}\int_{0}^{\pi}\cos^5x\,\mathrm{d}x \\ & = \frac{6}{7}\int_{0}^{\pi}\cos^4x\,\mathrm{d}(\sin x) \\ &= \frac{6}{7}\bigg\{ \big[\cos^4x\sin x\big]\big|_{0}^{\pi} - \int_{0}^{\pi}\sin x\,\mathrm{d}(\cos^4x)\bigg\} \\ &= \frac{6}{7}\bigg\{ 0 + \int_{0}^{\pi}4\sin^2 x\cos^3x\,\mathrm{d}x\bigg\} \\ & = \frac{24}{7}\int_{0}^{\pi} (1-\cos^2x)\cos^3x\,\mathrm{d}x \\ \frac{30}{7}\int_{0}^{\pi}\cos^5x\,\mathrm{d}x & = \frac{24}{7} \int_{0}^{\pi}\cos^3x\,\mathrm{d}x \\ \int_{0}^{\pi}\cos^5x\,\mathrm{d}x & = \frac{4}{5}\int_{0}^{\pi}\cos^3x\,\mathrm{d}x \\ \end{aligned}$

$\begin{aligned} \int_{0}^{\pi}\cos^7x\,\mathrm{d}x & = \frac{6}{7}\bigg\{ \frac{4}{5}\int_{0}^{\pi}\cos^3x\,\mathrm{d}x \bigg\} \\ & = \frac{24}{35} \int_{0}^{\pi} \cos^2x\,\mathrm{d}(\sin x) \\ & = \frac{24}{35}\bigg\{ \big[ \cos^2x\sin x\big]\big|_{0}^{\pi} - \int_{0}^{\pi}\sin x\,\mathrm{d}(\cos^2x) \bigg\} \\ & = \frac{24}{35} \bigg\{ 0 + \int_{0}^{\pi}2\sin^2x\cos x\,\mathrm{d}x \bigg\} \\ & = \frac{48}{35} \int_{0}^{\pi}(1-\cos^2x)\cos x\,\mathrm{d}x \\ \bigg(\frac{24}{35}+\frac{48}{35}\bigg) \int_{0}^{\pi} \cos^3x\,\mathrm{d}x & = \frac{48}{35}\int_{0}^{\pi}\cos x\,\mathrm{d}x \\ \int_{0}^{\pi} \cos^3x\,\mathrm{d}x & = \frac{2}{3}\int_{0}^{\pi}\cos x\,\mathrm{d}x \\ \int_{0}^{\pi}\cos^7x\,\mathrm{d}x & = \frac{24}{35}\bigg\{ \frac{2}{3}\int_{0}^{\pi}\cos x\,\mathrm{d}x \bigg\} \\ & = \frac{16}{35}\big[ \sin x\big]\big|_{0}^{\pi} \\ & = 0 \\ \end{aligned}$

Solution.

$\begin{aligned} \int_{0}^{\pi}\cos^7x\,\mathrm{d}x & = \int_{0}^{\frac{\pi}{2}}\cos^7x\,\mathrm{d}x + \int_{\frac{\pi}{2}}^{\pi}\cos^7x\,\mathrm{d}x \\ & = \int_{0}^{\frac{\pi}{2}}\cos^7x\,\mathrm{d}x + \int_{0}^{\frac{\pi}{2}}\cos^7(x-\pi /2)\,\mathrm{d}(x-\pi /2) \\ & = \int_{0}^{\frac{\pi}{2}}\cos^7x\,\mathrm{d}x - \int_{0}^{\frac{\pi}{2}}\sin^7x\,\mathrm{d}x \\ & = \int_{0}^{\frac{\pi}{2}}(\cos^7x-\sin^7x)\,\mathrm{d}x \\ \end{aligned}$

The rest is left as an exercise to the reader.

October 13, 2022October 18, 2022

202210131233 Solution to 1969-CE-AMATH-I-XX

Find $y$ in terms of $x$ , if $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}=8x-1}$ and $y=-1$ when $x=-1$ .

Roughwork.

$\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x} & = 8x-1 \\ y(x) & = \int \frac{\mathrm{d}y}{\mathrm{d}x}\,\mathrm{d}x \\ & = \int (8x-1)\,\mathrm{d}x \\ & = 4x^2-x+C\qquad\textrm{for some constant }C \\ \because\enspace -1 & = y(-1) \\ -1 & = 4(-1)^2-(-1)+C \\ \therefore\enspace C & = -6 \\ \end{aligned}$

In conclusion, $y(x) = 4x^2-x-6$ .

July 7, 2022October 24, 2022

202207071711 Solution to 1976-CE-AMATH-I-XX

July 6, 2022October 24, 2022

202207061152 Solution to 1971-CE-AMATH-I-XX

If $k$ is an integer, simplify

$\displaystyle{\cos\bigg[ \frac{1}{2}(4k+1)\pi +\theta \bigg]\tan\bigg[\frac{1}{2}(4k+3)\pi -\theta \bigg]}$ .

Roughwork.

First,

$\begin{aligned} & \quad \cos\bigg[ \frac{1}{2}(4k+1)\pi +\theta \bigg]\tan\bigg[\frac{1}{2}(4k+3)\pi -\theta \bigg] \\ & = \cos\bigg( 2k\pi+\frac{\pi}{2}+\theta \bigg) \tan \bigg( 2k\pi+\frac{3\pi}{2}-\theta \bigg) \\ \dots\textrm{as } &\textrm{are}\cos (\angle ),\tan (\angle )\textrm{ shifted by full periods, \textit{i.e.}, }k\cdot 2\pi\,\dots \\ & = \cos\bigg(\frac{\pi}{2}+\theta\bigg) \tan \bigg(\frac{3\pi}{2}-\theta \bigg) \\ \end{aligned}$

Second,

recall that

$\begin{aligned} \textrm{(in radians)} & \quad\textrm{(in degrees)} \\ \pi & = 180^\circ \\ \frac{\pi}{2} & = 90^\circ \\ \frac{3\pi}{2} & = 270^\circ\textrm{ \scriptsize{OR} }-90^\circ = -\frac{\pi}{2}\\ \end{aligned}$

Hence,

$\begin{aligned} & \quad \cos\bigg(\frac{\pi}{2}+\theta\bigg) \tan \bigg(\frac{3\pi}{2}-\theta \bigg) \\ & = \cos\bigg(\frac{\pi}{2}+\theta\bigg) \tan \bigg(-\frac{\pi}{2}-\theta \bigg) \\ & = \cos\bigg(\frac{\pi}{2}+\theta\bigg) \tan \bigg[ -\bigg(\frac{\pi}{2}+\theta \bigg)\bigg] \\ \dots &\textrm{ as }\tan(-\theta )=-\tan\theta \enspace\dots \\ & = -\cos\bigg(\frac{\pi}{2}+\theta\bigg) \tan \bigg(\frac{\pi}{2}+\theta\bigg) \\ \end{aligned}$ .

Recall that

$\begin{aligned} \because \enspace & \tan\theta = \frac{\sin\theta}{\cos\theta} \\ \therefore \enspace & \cos\theta\tan\theta = \sin\theta \\ \end{aligned}$ .

Hence,

$\begin{aligned} &\quad -\cos\bigg( \frac{\pi}{2}+\theta\bigg)\tan\bigg( \frac{\pi}{2}+\theta\bigg) \\ & = -\sin\bigg( \frac{\pi}{2}+\theta\bigg) \\ \end{aligned}$

Third,

recall that

$\displaystyle{\sin \bigg(\theta\pm\frac{\pi}{2}\bigg)}=\pm\cos\theta$ .

$\begin{aligned} &\quad -\sin \bigg( \frac{\pi}{2}+\theta\bigg) \\ & = -\cos\theta \\ \end{aligned}$

Answer.

$\displaystyle{\cos\bigg[ \frac{1}{2}(4k+1)\pi +\theta \bigg]\tan\bigg[\frac{1}{2}(4k+3)\pi -\theta \bigg]}=-\cos\theta$ .